EM2_GM1_M1_Teach_23A_881746_Updated 08.23 by Great Minds

MATHEMATICS I

A Story of Functions®

Comparing with Functions TEACH ▸ Module 1 ▸ Equations, Inequalities, and Data in One Variable

Teacher Edition: Math 1, Module 1, Cover What does this painting have to do with math? Although he is best known as a cartoonist for his comic strip Barnaby and the book Harold and the Purple Crayon, the American artist Crockett Johnson also took a keen interest in mathematics, painting more than 100 pieces relating to mathematics and mathematical physics during his life. His painting Logarithms shows two sequences of rectangles. How do the lengths of the rectangles in the top row relate to those in the bottom row? On the cover Logarithms, 1966 Crockett Johnson, American, 1906–1975 Masonite and wood Smithsonian National Museum of American History, Washington, DC, USA Crockett Johnson, Logarithms, 1966, masonite and wood, 56 cm x 66.3 cm x 3.8 cm; 22 1/16 in x 26 1/8 in x 1 1/2 in. Courtesy of Ruth Krauss in memory of Crockett Johnson, Smithsonian National Museum of American History

Teacher Edition: Math 1, Module 1, Copyright

Great Minds® is the creator of Eureka Math®, Wit & Wisdom®, Alexandria Plan™, and PhD Science®. Published by Great Minds PBC. greatminds.org © 2023 Great Minds PBC. All rights reserved. No part of this work may be reproduced or used in any form or by any means—graphic, electronic, or mechanical, including photocopying or information storage and retrieval systems—without written permission from the copyright holder. Where expressly indicated, teachers may copy pages solely for use by students in their classrooms. Printed in the USA A-Print 1 2 3 4 5 6 7 8 9 10 XXX 27 26 25 24 23 ISBN 979-8-88588-174-6

Teacher Edition: Math 1, Module 1, Title

A Story of Functions®

Comparing with Functions ▸ MATHEMATICS I TEACH Module

1 2 3 4 5 6

Equations, Inequalities, and Data in One Variable

Equations, Inequalities, and Data in Two Variables

Functions and Their Representations

Transformations of the Plane and Constructions

Linear and Exponential Functions

Modeling with Data and for Contexts

Teacher Edition: Math 1, Module 1, Module Overview Before This Module

Overview

Grade 8 Module 4

Equations, Inequalities, and Data in One Variable

Students informally apply the commutative, associative, and distributive properties and the addition and multiplication properties of equality to solve multi-step linear equations in one variable.

Grade 6 Module 6 Grade 7 Module 6 Students represent univariate data with dot plots, histograms, and box plots. They analyze the center and spread of a distribution by using the mean, median, range, mean absolute deviation, and interquartile range. They use shape, center, and spread to compare data sets.

Topic A The Structure of Expressions Students use operations and the properties of arithmetic to generate equivalent expressions and to demonstrate the equivalence of algebraic expressions. Students justify how equivalent expressions can model the same situation by interpreting coefficients, terms, and factors of the expressions in context. Expression 1: 6 + 5(n − 1) Expression 2: 1 + 5n

Figure 1

Expression 3: 2n + 2n + (n + 1)

Figure 2

Figure 3

Figure 4

Figure 5

2(4) + 2(4) + 5

2(5) + 2(5) + 3

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Topic B Solving Equations and Inequalities in One Variable Students use operations and the properties of arithmetic, equations, and inequalities to find solution sets of linear equations and inequalities in one variable. They represent the solution sets by using set notation and set-builder notation and by using graphs on the number line. Students apply the properties and operations in new situations, including rearranging formulas, solving equations with variable coefficients, and creating equations or inequalities to model contexts.

−6b + 2 − 2b < 10 {b | b > −1}

–10 –8 –6 –4 –2

Topic C Compound Statements Involving Equations and Inequalities in One Variable Students determine the solution sets of compound statements involving any combination of two equations or two inequalities joined by and or or. Students rewrite absolute value equations and absolute value inequalities as compound statements and represent the solution sets by using set-builder notation and as graphs on the number line.

0 ≤ 4x − 3 ≤ 11

4x − 3 ≥ 0 and 4x − 3 ≤ 11 3 7 x ≥ _ and x ≤ _

| _3 4

{x 4 ≤ x ≤ 2 }

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Topic D

After This Module

Univariate Data Students represent univariate data distributions with dot plots, histograms, and box plots. They use graphical displays and statistics to compare two or more data distributions based on shape, center, and spread, and they interpret the results within the context of the data. Students also analyze how outliers affect the measures of center and spread for data distributions. Maximum Speeds of Roller Coasters

Mathematics I Module 2 Students build on their understanding of solving equations and inequalities and analyzing data in one variable to explore similar topics in two variables. They represent solution sets of equations, inequalities, and systems of equations and inequalities in two variables. Additionally, they analyze bivariate data distributions.

Wooden Steel

100

110

120

130

Speed (miles per hour)

Contents Equations, Inequalities, and Data in One Variable Why. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Achievement Descriptors: Overview . . . . . . . . . . . . . . . . . . . . 10 Topic A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 The Structure of Expressions Lesson 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 A Powerful Trio • Solve problems with a combination of tools used for statistics, geometry, and algebra.

Lesson 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Looking for Patterns

Lesson 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Solution Sets of Equations and Inequalities in One Variable • Find values to assign to the variables in equations or inequalities that make the statements true. • Describe a solution set in set notation, in words, and on a graph.

Lesson 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Solving Linear Equations in One Variable • Justify each step in solving a linear equation.

Lesson 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Some Potential Dangers When Solving Equations • Explore steps in solving an equation that are not guaranteed to preserve the solution set.

• Extend patterns and write expressions to describe them.

Lesson 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

Lesson 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

• Create equations in one variable and use them to solve problems.

The Commutative, Associative, and Distributive Properties • Rewrite algebraic expressions in equivalent forms. • Show the equivalency of two algebraic expressions by using properties and operations.

Lesson 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Interpreting Linear Expressions

Writing and Solving Equations in One Variable

Lesson 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 Rearranging Formulas • Rearrange formulas to highlight a quantity of interest.

Lesson 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Solving Linear Inequalities in One Variable

• Interpret parts of linear expressions within a context.

• Solve inequalities and graph the solution sets on the number line.

Topic B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Solving Equations and Inequalities in One Variable

Topic C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 Compound Statements Involving Equations and Inequalities in One Variable

Lesson 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

Lesson 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

Printing Presses • Investigate a problem that can be solved by reasoning quantitatively or algebraically.

Solution Sets of Compound Statements • Describe the solution set of two equations or inequalities joined by and or or and graph the solution set on a number line. • Write a compound statement to describe a situation.

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Lesson 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

Lesson 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

Solving and Graphing Compound Inequalities

Describing the Center of a Distribution

• Find the solution sets of compound inequalities in one variable and graph the solution sets on the number line.

• Find the mean and median of data shown in a dot plot and estimate the mean and median of a data distribution represented by a histogram. • Identify whether the mean, the median, or both appropriately describe a typical value for a given data set.

Lesson 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 Solving Absolute Value Equations • Write absolute value equations in one variable as compound statements and solve.

Lesson 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 Solving Absolute Value Inequalities • Write absolute value inequalities in one variable as compound statements joined by and or or. • Solve absolute value inequalities and graph the solution sets on the number line.

Lesson 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 Applying Absolute Value (Optional) • Use absolute value equations and absolute value inequalities to solve real-world problems.

Topic D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 Univariate Data Lesson 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Distributions and Their Shapes • Informally describe a data distribution displayed in a dot plot.

Lesson 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366 Using Center to Compare Data Distributions • Determine the median from data distributions displayed in box plots. • Use the median to compare data distributions displayed in box plots.

Lesson 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Describing Variability in a Univariate Distribution with Standard Deviation • Calculate standard deviation to represent a typical variation from the mean of a data distribution. • Use standard deviation to compare two data distributions.

Lesson 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 Estimating Variability in Data Distributions • Estimate and compare variation in data distributions represented by histograms. • Use the interquartile range to compare the variation in data distributions represented by box plots and dot plots.

Lesson 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430 Comparing Distributions of Univariate Data • Compare two or more data sets by using shape, center, and variability. • Interpret differences in data distributions in context.

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Resources Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 Achievement Descriptors: Proficiency Indicators . . . . . . . . . . . . . . . 456 Terminology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 Math Past . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 Fluency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480 Sample Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 Works Cited. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508 Credits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510

Teacher Edition: Math 1, Module 1, Why

Why Equations, Inequalities, and Data in One Variable Why do you use set-builder notation in this module for some solution sets and set notation for others? Set-builder notation is needed to accurately represent the solution sets of inequalities without using a graph. While students may choose to represent the solution sets to equations with a finite number of solutions by using set-builder notation, using set notation may feel more accessible for students while still accurately representing the solution sets. The same reasoning applies to representing the solution sets for all real solutions and no solutions by using the symbols ℝ and { }, respectively.

Why did you include absolute value equations and inequalities in this module? Solving absolute value equations and inequalities connects to solving compound inequalities, as they can be rewritten as simple statements connected by and or or. Solving absolute value equations in one variable builds upon students’ understanding of absolute value from previous grades.

Why is lesson 16 optional? In lesson 16, students model real-world situations with absolute value equations and absolute value inequalities. Through problem solving, students learn an alternate interpretation for absolute value expressions that emphasizes the distance between any two numbers rather than the distance between a number and zero. For this reason, the lesson is optional but recommended.

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Why is univariate statistics included in this module? Because the focus of module 1 is on expressions, equations, and compound statements in one variable, it is appropriate to conclude the module with a study of data collected on a single variable. While univariate data analysis and bivariate data analysis are often taught in succession, it is not necessary. Separating the topics allows us to better incorporate statistics into a blended Mathematics I curriculum, rather than treating it as a separate entity.

Teacher Edition: Math 1, Module 1, Achievement Descriptors: Overview

Achievement Descriptors: Overview Equations, Inequalities, and Data in One Variable Achievement Descriptors (ADs) are standards-aligned descriptions that detail what students should know and be able to do based on the instruction. ADs are written by using portions of various standards to form a clear, concise description of the work covered in each module. Each module has its own set of ADs, and the number of ADs varies by module. Taken together, the sets of module-level ADs describe what students should accomplish by the end of the year. ADs and their proficiency indicators support teachers with interpreting student work on

• Exit Tickets,

inequalities in one variable as viable or nonviable options in a modeling context.

A.CED.A.3

Math1.Mod1.AD7 Rearrange formulas to highlight a quantity of interest.

A.CED.A.4

A.REI.A.1

Math1.Mod1.AD9 Explain steps required to solve linear

equations in one variable and justify solution paths.

A.REI.A.1

Math1.Mod1.AD10 Solve linear equations in one

• Topic Quizzes, and

variable.

• Module Assessments.

A.REI.B.3

Math1.Mod1.AD11 Solve linear inequalities in one

This module contains the fifteen ADs listed.

variable.

Math1.Mod1.AD1 Use units to understand and guide

A.REI.B.3

Math1.Mod1.AD12 Write equivalent expressions for N.Q.A.1

Math1.Mod1.AD2 Define appropriate quantities for

Math1.Mod1.AD6 Interpret solutions to equations and

that algebraic expressions are equivalent.

• data from other lesson-embedded formative assessments,

the purpose of descriptive modeling.

A.CED.A.3

Math1.Mod1.AD8 Explain each step in demonstrating

• informal classroom observations,

the solution path of multi-step problems.

Math1.Mod1.AD5 Represent constraints by using equations and inequalities.

contexts that can be modeled by arithmetic sequences.

F.BF.A.1.a

Math1.Mod1.AD13 Represent data with dot plots,

histograms, or box plots.

S.ID.A.1

Math1.Mod1.AD3 Interpret parts of linear expressions, such as terms, factors, and coefficients in context. A.SSE.A.1.a

Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers, and variabilities.

S.ID.A.2

Math1.Mod1.AD4 Create equations and inequalities in one variable and use them to solve problems.

Math1.Mod1.AD15 Interpret differences in summary measures in the context of the data sets.

S.ID.A.3

N.Q.A.2

A.CED.A.1

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The first page of each lesson identifies the ADs aligned with that lesson. Each AD may have up to three indicators, each aligned to a proficiency category (i.e., Partially Proficient, Proficient, Highly Proficient). While every AD has an indicator to describe Proficient performance, only select ADs have an indicator for Partially Proficient and/or Highly Proficient performance.

ADs have the following parts: • AD Code: The code indicates the grade level and the module number and then lists the ADs in no particular order. For example, the first AD for Mathematics I module 1 is coded as Math1.Mod1.AD1. • AD Language: The language is crafted from standards and concisely describes what will be assessed.

An example of one of these ADs, along with its proficiency indicators, is shown here for reference. The complete set of this module’s ADs with proficiency indicators can be found in the Achievement Descriptors: Proficiency Indicators resource.

AD Code: Grade.Mod#.AD#

• AD Indicators: The indicators describe the precise expectations of the AD for the given proficiency category. • Related Standard: This identifies the standard or parts of standards from the Common Core State Standards that the AD addresses.

AD Language

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Math1.Mod1.AD5 Represent constraints by using equations and inequalities.

Related Standard

RELATED CCSSM

A.CED.A.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.

Partially Proficient Interpret constraints of equations or inequalities in one variable.

Nina wants to have her bike repaired. The inequality 60 ≤ c ≤ 100 describes the possible costs of the repair, where c represents the cost of the repair in dollars. Which of the following are possible costs to repair Nina’s bike? Choose all that apply. A. $50.00

Proficient

Highly Proficient

Represent constraints by using equations or inequalities in one variable. Zara needs to create a portfolio with at least 15 drawings and no more than 25 drawings. She already has 6 drawings. Let n represent the number of drawings Zara still needs to create for the portfolio. Write an inequality that represents the number of drawings Zara still needs to create.

AD Indicators

B. $75.80 C. $89.75 D. $99.30 E. $125.60

Teacher Edition: Math 1, Module 1, Topic A

Topic A The Structure of Expressions In topic A, students solve a real-world problem by using their chosen solution path, describe patterns by using words or algebraic expressions, and generate different expressions to represent a situation. The activities allow students to recognize that they can represent a mathematical situation in several ways. Students share their reasoning for choosing a solution path with the class, justify their answers, and analyze different representations of the same situation. Students begin by recognizing how an everyday situation, such as deciding which checkout lane to choose when in the grocery store, incorporates statistical, geometric, and algebraic thinking. Students work in groups to choose a checkout lane in a grocery store that results in the fastest checkout time by analyzing diagrams, scatter plots, and data about the number of customers, the total number of items sold, and the total amount of time taken in a checkout lane. The context is designed to provide several viable solution paths that result in different checkout lane selections. Each student group presents and justifies their choice and reasoning to the class. Students analyze a pattern in successive geometric figures and use repeated reasoning to write an expression to determine the number of line segments in any figure in the pattern. They are encouraged to write their expression based on how they draw the pattern. This lesson allows students to generate different yet equivalent expressions that represent the same pattern. The variation leads to the question, How do we know these expressions represent the same value?

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Math 1 ▸ M1 ▸ TA

Students use their prior knowledge of properties of arithmetic to rewrite expressions and verify whether two expressions are equivalent. They use a flowchart or two-column table to show their thinking.

–8(–5b + 7) + 5b distributive property 40b – 56 + 5b commutative property of addition 40b + 5b – 56 addition of like terms

Expression

Property or Operation Used

−8(−5b + 7) + 5b 40b − 56 + 5b

Distributive property

40b + 5b − 56

Commutative property of addition

45b − 56

Addition of like terms

45b – 56 Students interpret the coefficients, terms, and factors of expressions that model a given situation. They justify how equivalent expressions can represent the same situation. They conclude the topic by writing a situation that can be modeled by different expressions. In module 1 topic B, students apply their knowledge of equivalent expressions to find and write the solution sets of one-variable equations and inequalities.

Progression of Lessons Lesson 1

A Powerful Trio

Lesson 2

Looking for Patterns

Lesson 3

The Commutative, Associative, and Distributive Properties

Lesson 4

Interpreting Linear Expressions

Teacher Edition: Math 1, Module 1, Topic A, Lesson 1 LESSON 1

A Powerful Trio Solve problems with a combination of tools used for statistics, geometry, and algebra.

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Name

Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Date

EXIT TICKET

Reflect on the lesson.

Lesson at a Glance This lesson begins with a video about the context of waiting in line at a grocery store. Students must choose which lane results in the fastest checkout time. Working in groups, they decide which lane has the fastest time by using given data and a combination of tools used for statistics, geometry, and algebra. After groups share their responses and compare solution paths, students watch the conclusion of the video and recognize ways to apply statistical, geometric, and algebraic thinking to solve problems. In an optional lesson activity, students consider another choice for a checkout line and use mathematical reasoning to convince their classmates which choice is better.

Key Question • How can we apply statistical, geometric, and algebraic thinking to solve problems?

Achievement Descriptors Math1.Mod1.AD1 Use units to understand and guide the solution

path of multi-step problems. (N.Q.A.1) Math1.Mod1.AD2 Define appropriate quantities for the purpose of descriptive modeling. (N.Q.A.2)

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Agenda

Materials

Fluency

Teacher

Launch 10 min

• Computer or device*

Learn 25 min • Waiting Is the Hardest Part

• Projection device* • Teach Book*

• Changing Lanes (Optional)

Students

Land 10 min

• Computers or devices (1 per student pair)* • Learn Book* • Paper or notebook* • Pencil* • Scientific calculator

Lesson Preparation • Gather a variety of materials, including (but not limited to) colored pencils and rulers, for students to use as they solve the problems. *These materials are only listed in lesson 1. Ready these materials for every lesson in this module.

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Fluency Determine Rates Students determine rates from different representations to prepare for solving a problem by using a combination of tools.

Fluency activities are short sets of sequenced practice problems that students work on in the first 3–5 minutes of class. Administer a fluency activity as a bell ringer or adapt the activity as a teacher-led Whiteboard Exchange or choral response. Directions for these routines can be found in the Resources.

Directions: Determine the rate 1.

Number of Minutes

54 9

Number of Miles

Teacher Note

minutes per mile

Number of Miles

4 3

_ 1 9

mile per minute

1 0

Number of Minutes

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

The runner jogs 5 miles in 45 minutes.

Let y represent the number of miles. Let x represent the number of minutes.

y = _1 x

minutes per mile

mile per minute

Launch

Students watch a video and identify what might affect how fast checkout lines move at a grocery store. Play part 1 of the Superheroes video. Then use the following prompt to begin a discussion. Describe what happened in the video. Someone is bored while waiting in line at the grocery store. They wonder how long it is going to take to check out their items. Invite students to describe their experiences when waiting in a line at a location such as a concession stand, an amusement park, or a movie theater.

Teacher Note The dialogue shown provides suggested questions and sample responses. To maximize student participation, facilitate discussion by using tools and strategies that encourage student-to-student discourse. For example, make flexible use of the Talking Tool, turn and talk, think–pair–share, and the Always Sometimes Never routine.

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Then display the image of 4 checkout lanes at a grocery store.

Lane 1

Lane 2

Lane 3

Lane 4 Imagine you are at a grocery store that has 4 checkout lanes. If you choose a lane randomly, then what is the probability of choosing the lane with the fastest checkout time? Explain. The probability is 25%. There are 4 lanes, and there is an equal probability of choosing each lane. Only 1 of the 4 lanes has the fastest time. What information can help you choose the lane with the fastest checkout time? Knowing which lane has the least number of people waiting in line Knowing the number of items that are in each shopping cart

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Display the image of shopping carts in each lane.

Lane 1

Lane 2

Lane 3

Lane 4 Ask students which lane they think has the fastest checkout time. Take a class poll, noting the number of students who choose each lane. Invite volunteers to explain their reasoning. Besides the number of shopping carts, what else could affect the speed of a checkout line at a grocery store? The number of items in each shopping cart; how fast the cashier scans items; how talkative the cashier and customers are; whether there is a price check; whether customers have coupons; whether customers have fruits and vegetables that must be weighed; whether customers or cashiers bag items; whether customers pay with cash, check, card, or mobile phone.

UDL: Engagement Time spent waiting in line at a grocery store provides a relatable context for students to apply statistical, geometric, and algebraic thinking. Consider asking students whether they have seen posted wait times when standing in a line or an estimated wait time when using the customer service chat feature on a website. Invite students to share their ideas about how the wait times might have been calculated.

Math 1 ▸ M1 ▸ TA ▸ Lesson 1

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If students do not offer a variety of ideas about what could affect the speed of a line, consider providing some. Then display the image that shows lane 4 is an express lane.

Lane 1

Lane 2

Lane 3

Lane 4 Express Lane 4 is an express lane. What does that typically mean? An express lane is for customers who have a limited number of items, such as 10 items or fewer. Does knowing lane 4 is an express lane change your answer about which lane you think has the fastest checkout time? Why? Yes. The shopping carts in lane 4 probably have fewer items than the shopping carts in the other lanes. No. Lane 4 has more shopping carts than the other lanes.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Display the image that shows what portion of each shopping cart is full.

Lane 1

Lane 2

Lane 3

Lane 4 Express Does knowing how full the shopping carts are change your answer about which lane you think has the fastest checkout time? Why? Yes. The shopping carts in lane 4 appear to have the fewest items. No. Lane 4 still has the most shopping carts. Invite students to turn and talk about the following question. What data could help you determine which lane is the fastest? Call the class back together. Today, we will use a combination of tools from statistics, geometry, and algebra to solve problems about waiting in line at the grocery store.

Teacher Note Queuing theory is the study of how people move through lines, or queues. This branch of mathematics dates to Danish mathematician A. K. Erlang, who worked for the Copenhagen Telephone Exchange in the early 20th century. Erlang analyzed operations and developed mathematical models to predict the number of switchboard operators needed to reduce the time people spent waiting in a telephone queue. His work led others to study and apply queuing theory to improve customer experiences in many different areas, such as the retail, airline, telecommunications, health care, and amusement park industries.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Learn

Waiting Is the Hardest Part Students use data and tools from statistics, geometry, and algebra to solve a problem. Direct students to problem 1. Have them read the problem and analyze the scatter plot and data. Then discuss the meaning of the following terms: • Items scanned: the number of items passed over a laser that reads barcodes and enters prices into the register • Items keyed: the number of items with prices or barcodes that must be typed into the register Have students turn and talk about whether they think it takes more time to scan or key an item and which types of items might need to be keyed rather than scanned. Bring the class back together to discuss the meaning of the following terms: • Register time: the time it takes cashiers to scan and key items • Exchange time: the time cashiers spend with customers separate from the register time, such as the time it takes for cashiers and customers to exchange greetings or for customers to pay and for cashiers to return receipts Invite students to share additional examples of ways the exchange time could be affected, such as by price checks, coupons, and conversations between cashiers and customers. Divide students into groups of 3 or 4 and continue to display the image that shows what portion of each shopping cart is full. Allow groups to work for about 15 minutes to solve problem 1 by using a strategy of their choice. Consider displaying a timer to help groups keep track of time. Have a variety of tools and materials available for use, including (but not limited to) calculators, colored pencils, and rulers, as students decide on a strategy.

Language Support Consider using strategic, flexible grouping throughout the module. • Pair students who have different levels of mathematical proficiency. • Pair students who have different levels of English language proficiency. • Join pairs to form small groups of four. As applicable, complement any of these groupings by pairing students who speak the same native language.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Expect groups to use a variety of strategies to decide how many items are in each shopping cart and to determine which lane is fastest. Circulate as students work, intervening only if a group needs support starting the problem. Refocus groups without leading them to a specific solution path. To advance student thinking, consider asking one or more of the following questions: • What are the important quantities? How are these quantities related? • How did you decide the number of items in each shopping cart? • What does each point on the scatter plot represent? • Are there quantities given in the checkout summary that you can use to find the average register time or average exchange time? If so, what are they? • What assumptions do you need to make to continue solving the problem? • Does the number of items in a customer’s shopping cart affect the register time or the exchange time? Why? Problem 1 can be solved in a variety of ways, which may lead to different responses. Two possible solution paths, each following a different strategy, are shown. Note that the scatter plot and checkout summary are based on two different data sets.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

1. There are 4 checkout lanes at a grocery store that have a different number of shopping carts in line. Use the data to determine which lane has the fastest checkout time. Show or explain your reasoning.

Checkout Times

Checkout Summary

250 225

Number of Customers

159

Items Scanned

2619

Items Keyed

477

Total Items Sold

3096

Total Time (seconds)

200 175 150 125 100 75 50 25 0

162.4 min

Exchange Time

98.4 min

Total Time

260.8 min

Language Support While students work in their groups to determine which lane is fastest, encourage them to use the Ask for Reasoning section of the Talking Tool with their group.

Differentiation: Challenge Encourage students to solve the problem a second time by using a different strategy than they used the first time. For example, if they used the scatter plot the first time, have them solve the problem again by using the checkout summary. If they determine on the second time that a different lane has the fastest checkout time, ask them to consider why they got a different answer when they used a different strategy.

Number of Items

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Sample:

Teacher Note

9 items

Lane 1

10 items

Lane 2

42 items

Lane 3

15 items

17 items

Lane 4

8 items

5 items

10 items

This sample student response shows how a student could use the scatter plot and fit a line to the data to solve the problem.

6 items

8 items

Checkout Times

y 250 225 Total Time (seconds)

200 175 150 125 100 75 50 25 0

Number of Items

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Lane 2

Lane 1 Number of Items

Predicted Time (seconds)

Number of Items

Predicted Time (seconds)

213

The total checkout time for lane 2 is about 213 seconds.

70 + 65 + 70 = 205 The total checkout time for lane 1 is about 205 seconds. Lane 3

Lane 4

Number of Items

Predicted Time (seconds)

Number of Items

Predicted Time (seconds)

100

92 + 100 = 192 The total checkout time for lane 3 is about 192 seconds.

60 + 45 + 50 + 60 = 215 The total checkout time for lane 4 is about 215 seconds. Lane 3 has the fastest checkout time. 26

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Sample:

14 items

Lane 1

12 items

Lane 2

45 items

Lane 3

20 items

18 items

Lane 4

5 items

2 items

Teacher Note

13 items

This sample student response shows how a student could use the checkout summary to solve the problem. Note that a student may use similar reasoning without writing an equation.

6 items

7 items

By using the data from the checkout summary, we can find the register time per item and exchange time per customer. Register time per item:

162.4 ÷ 3096 ≈ 0.0525

The rate is about 0.0525 minutes per item. Exchange time per customer:

98.4 ÷ 159 ≈ 0.6189

The rate is about 0.6189 minutes per customer.

Let n represent the number of items in a customer’s shopping cart, and let T represent the time in minutes it takes one customer to check out.

T = 0.0525n + 0.6189

Promoting the Standards for Mathematical Practice When students choose how to use the information provided in the scatter plot and checkout summary to determine the fastest checkout time, they are modeling with mathematics (MP4). Ask the following questions to promote MP4: • What key ideas in the grocery checkout scenario do you need to make sure to include in your model? • What equation can you write to represent the total amount of time it takes for a customer to check out, depending on the number of items in their cart?

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Lane 2

Lane 1 Number of Items, n

Time, T (minutes)

Number of Items, n

Time, T (minutes)

0.0525(12) + 0.6189 = 1.2489

0.0525(45) + 0.6189 = 2.9814

0.0525(14) + 0.6189 = 1.3539

0.0525(13) + 0.6189 = 1.3014

The total checkout time for lane 2 is about 3.0 minutes.

1.2489 + 1.3539 + 1.3014 = 3.9042 The total checkout time for lane 1 is about 3.9 minutes. Lane 3

Lane 4

Number of Items, n

Time, T (minutes)

Number of Items, n

Time, T (minutes)

0.0525(20) + 0.6189 = 1.6689

0.0525(5) + 0.6189 = 0.8814

0.0525(18) + 0.6189 = 1.5639

0.0525(2) + 0.6189 = 0.7239

0.0525(6) + 0.6189 = 0.9339

0.0525(7) + 0.6189 = 0.9864

1.6689 + 1.5639 = 3.2328 The total checkout time for lane 3 is about 3.2 minutes.

0.8814 + 0.7239 + 0.9339 + 0.9864 = 3.5256 The total checkout time for lane 4 is about 3.5 minutes. Lane 2 has the fastest checkout time. 28

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

When time is up, invite groups to share their responses and strategies for solving the problem. Encourage students to explain their thinking rather than just retelling their steps. Point out when groups apply the following: • Statistical thinking, such as representing the data in the scatter plot with a line or using the average register time and the average exchange time to approximate the total time for each customer • Geometric thinking, such as using spatial reasoning to decide the number of items in each shopping cart or drawing a line through the data and reading the coordinates of points on that line • Algebraic thinking, such as writing an equation that relates a customer’s checkout time to the number of items in the customer’s shopping cart As groups share, encourage students to record strategies that differ from their own. If time permits, consider asking one or more of the following questions and then directing students to discuss in groups before they share responses with the class: • What assumptions did your group make? How did that affect your results? • How did you decide the number of items in each shopping cart? Is it realistic to assume that all the shopping carts have the same number of items? • Is your group’s strategy the same as this group’s strategy, or is it different? • What are some questions you still have about this group’s strategy? After groups share, invite students to tell which strategy they like the most and why. Consider asking them how their responses for which lane has the fastest checkout time compare to their earlier predictions. Then continue the discussion by asking the following questions. What assumption did we make about the rate at which each cashier checks out customers? We assumed all cashiers check out customers at the same rate. Do all cashiers check out customers at the same rate? Why? No. Some cashiers might not scan as fast as others or may spend more time talking to customers. We assumed that all cashiers check out customers at the same rate because the data provided did not distinguish between cashiers. © Great Minds PBC

Teacher Note Mathematical modeling often requires making assumptions to model a problem and arrive at a solution. In problem 1, we assume all cashiers check out customers at the same rate. In contrast, in the practice set, students consider a scenario where cashiers check out customers at different rates. Acknowledging assumptions, how realistic they might be, and their effect on the model’s result is an important part of modeling. Once a model yields a solution, that solution must be interpreted in the context of the problem. If the answer is not reasonable, one must consider how assumptions can be changed.

Math 1 ▸ M1 ▸ TA ▸ Lesson 1

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Have students think–pair–share about the following question. In the video we watched earlier, a customer wondered how long it was going to take to check out at the grocery store. What does the customer need to know to figure that out? They need to know the number of shopping carts in line. They need to know the number of items in the shopping carts. They need to know how long it takes to scan each item. Play part 2 of the Superheroes video. Describe what happened in the video. The statistics, geometry, and algebra superheroes worked together to determine how long it would take to check out at the grocery store. What did each superhero do to help solve the problem? Statistics collected data and organized the data into a table. Geometry created a scatter plot and drew a line through the data. Algebra wrote an equation for the line and used the equation to answer the question. Have students turn and talk in their groups about how they applied statistical, geometric, and algebraic thinking in their strategies to solve problem 1.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Changing Lanes (Optional) Students make assumptions about a situation, answer questions, and defend their reasoning. Display the images of two scenarios: 4 lanes with 4 checkout lines and 4 self-checkout lanes with 1 line.

Lane 1

Self-Checkout Self-Checkout Self-Checkout Self-Checkout Lane 2 Lane 3 Lane 4 Lane 1

Lane 2

Lane 3

Lane 4 Express

Point to the picture that shows 4 self-checkout lanes with 1 line. Consider reminding students about the difference between lanes with cashiers and self-checkout lanes. Ensure that students understand that there is one self-checkout line where customers wait to access one of the several lanes, or registers. Once a customer gets to a register, they are no longer waiting in line. Facilitate a brief discussion about the goals of customers, such as checking out quickly and not being skipped in line, and the goals of the grocery store, such as keeping customers happy while maximizing profit and floor space.

Math 1 ▸ M1 ▸ TA ▸ Lesson 1

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Direct students to read problem 2. Then have students continue working in groups to complete the problem. Without providing any additional data, let students make assumptions about the two options. Circulate as students work, and ask the following questions as necessary to advance their thinking: • Do you think the average customer can scan items faster or slower than a cashier? Why? • Do you think the exchange time would increase, decrease, or stay the same if only 1 line was used for all 4 self-checkout lanes? Why? • How might having only 1 line for all 4 self-checkout lanes affect the register time per item and exchange time per customer? • How might the scatter plot in problem 1 look different if customers, rather than cashiers, scan their own items? Problem 2 can be answered in a variety of ways. Two sample responses are shown. 2. A grocery store has the following options: A. 4 lines for 4 checkout lanes that each have a cashier B. 1 line for all 4 self-checkout lanes Consider the options from the customer’s perspective and from the grocery store’s perspective. Which option is better? Why? Sample: I think option A is better. Most customers probably scan items slower than cashiers do, so option B might increase the time someone waits. If there is only one self-checkout line, a customer might have more people ahead of them than they would if there were multiple shorter lines. If the grocery store chooses option B, they might have fewer customers because customers don’t want to wait in a long line. If the grocery store has fewer customers, they might make less money. Sample: I think option B is better. With one self-checkout line, customers don’t have to choose which line they think is the fastest. They get in one line and check out in the order that they get in line. Also, with one self-checkout line, the grocery store might save some money because they don’t have to pay as many cashiers to work the registers. 32

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

When most groups are finished, invite them to share which option they think is better and why. Encourage friendly debate between groups as they use mathematical reasoning to defend their choice. As groups share, point out when they use statistical, geometric, and algebraic thinking in their reasoning.

Land Debrief 5 min Objective: Solve problems with a combination of tools used for statistics, geometry, and algebra. Facilitate a class discussion by asking the following questions. Encourage students to restate or build upon one another’s responses. Where did we apply statistical thinking in the problems we solved today? We represented the data on the scatter plot with a line. We used the average register time to approximate the register time for each customer. Where did we apply geometric thinking? We used spatial reasoning to decide the number of items in each shopping cart. We drew a line through the data in the scatter plot and found coordinates of the points on that line. We used the line we drew through the data to estimate the number of seconds it takes to check out depending on the number of items in each shopping cart.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Where did we apply algebraic thinking? We wrote an equation that relates the time it takes a customer to check out to the number of items in the customer’s shopping cart. We found an equation of the line we fit to the data and used it to determine the time it takes a customer to check out.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Students can solve the problems by using a variety of strategies.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

PRACTICE Name

Date

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

The number of shopping carts in each lane and number of items in each shopping cart are shown. a. Which lane has the faster checkout time? How do you know?

1. There are 2 checkout lanes at a grocery store that have different checkout data. The scatter plot and checkout summary for each lane are based on two different data sets. Lane 1 Checkout Times

110

100

Total Time (seconds)

110 Total Time (seconds)

120

80 70 60 50 40 30 20

Lane 2

The rate is about 0.0476 minutes per item.

Exchange time per customer:

The rate is about 0.6395 minutes per customer.

The rate is about 0.7049 minutes per customer.

Let n represent the number of items in a customer’s cart, and let T represent the time in minutes it takes one customer to check out.

Let c represent the number of items in a customer’s cart, and let M represent the time in minutes it takes one customer to check out.

24.3 ÷ 38 ≈ 0.6395

Lane 2 Checkout Summary

Number of Customers

Items Scanned

655

Items Scanned

628

Items Keyed

Total Items Sold

733

Total Items Sold

691

42.5 min

32.9 min

Exchange Time

24.3 min

Exchange Time

28.9 min

Total Time

66.8 min

Total Time

61.8 min

The rate is about 0.0580 minutes per item. x

Number of Items

Number of Customers

40 30

Number of Items

Lane 1 Checkout Summary

42.5 ÷ 733 ≈ 0.0580

Lane 2

Lane 1

10 x

Sample:

10 0

Lane 2 Checkout Times

120

Lane 1

T = 0.0580n + 0.6395 Substituting 1 for n:

32.9 ÷ 691 ≈ 0.0476 28.9 ÷ 41 ≈ 0.7049

M = 0.0476c + 0.7049 Substituting 1 for c:

T = 0.0580(1) + 0.6395 = 0.6975

M = 0.0476(1) + 0.7049 = 0.7525

The total checkout time for lane 1 is about 0.7 minutes.

The total checkout time for lane 2 is about 0.8 minutes.

Lane 1 has the faster checkout time.

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

b. Which lane has the faster checkout time? How do you know? Lane 1

Lane 2

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

d. Which lane has the faster checkout time? How do you know?

Lane 1

Lane 2

Sample:

Sample: Lane 1

Lane 2

Lane 1

T = 0.0580(8) + 0.6395 = 1.1035

M = 0.0476(8) + 0.7049 = 1.0857

The total checkout time for lane 1 is greater than 1.1 minutes.

The total checkout time for lane 2 is less than 1.1 minutes.

T = 0.0580(1) + 0.6395 = 0.6975

Lane 2 has the faster checkout time.

M = 0.0476(8) + 0.7049 = 1.0857

3(0.6975) = 2.0925

2(1.0857) = 2.1714

The total checkout time for lane 1 is about 2.1 minutes.

The total checkout time for lane 2 is about 2.2 minutes.

Lane 1 has the faster checkout time. c. Which lane has the faster checkout time? How do you know? Lane 1

Lane 2

2. Is it possible to apply a combination of statistical, geometric, and algebraic thinking to solve problem 1? If so, which combination did you use, and how?

Sample: Yes. I applied a combination of statistical thinking and algebraic thinking. I applied statistical thinking when I used the data to determine rates. I applied algebraic thinking when I used the rates to write equations.

3. Give two examples of where you apply statistics in your everyday life.

Sample:

Sample: using my batting average to predict how many hits I will get at my next baseball tournament; using the average summer temperature in my city to decide whether to go camping

Lane 2

Lane 1

T = 0.0580(8) + 0.6395 = 1.1035

M = 0.0476(16) + 0.7049 = 1.4665

2(1.1035) = 2.207

The total checkout time for lane 2 is about 1.5 minutes.

The total checkout time for lane 1 is about 2.2 minutes.

4. Give two examples of where you apply geometry in your everyday life. Sample: finding the area of my room; identifying shapes in architecture 5. Give two examples of where you apply algebra in your everyday life.

Lane 2 has the faster checkout time.

Sample: figuring out what score I need to make on my next test to improve my grade; figuring out whether it is faster to get to my friend’s house by bike or by bus

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

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Math 1 ▸ M1 ▸ TA ▸ Lesson 1

Remember For problems 6 and 7, evaluate. 6. −12.14 − 2.32

7. 9.9 + 11.25 + (−4)

−14.46

17.15

8. Which equations are true? Choose all that apply. A. 72 + 45 = 9(8 + 5) B. 80 + 60 = 8(10 + 60) C. 24 + 6 = 6(4 + 1) D. 39 + 27 = 3(19 + 9) E. 48 + 28 = 4(12 + 7)

9. Hana made a mistake solving w − 16 = −25.

w − 16 = −25 w − 16 + 16 = −25 − 16 w = −41 The solution is −41. a. Identify Hana’s mistake and explain what she should have done to solve the problem correctly. Hana added 16 to the left side of the equation, and she subtracted 16 from the right side of the equation. She should have added 16 to both sides of the equation.

b. Solve for w.

−9

P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic A, Lesson 2 LESSON 2

Looking for Patterns Extend patterns and write expressions to describe them.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Name

Date

EXIT TICKET

Examine the pattern of pentagons. The side length of each pentagon is 1 unit.

Figure 1

Figure 2

Figure 3

a. Describe the total number of 1-unit line segments for figure n.

Lesson at a Glance In this digital lesson, students write verbal and algebraic descriptions of visual patterns. Students begin by examining a hexagon pattern and describing how it changes with each step. They create representations and write expressions to describe hexagon patterns and compare their representations. Then students develop the formula for calculating the sum of the measures of the interior angles of a convex polygon by using patterns in polygon decompositions. This lesson introduces the term generalize. Use the digital platform to prepare for and facilitate this lesson. Students will also interact with lesson content and activities via the digital platform.

The total number of 1-unit line segments for any figure is 1 more than 4 times the figure number.

Key Questions • How can we write an expression to represent a situation?

b. Write an expression for your description in part (a).

4n + 1

• How do we know whether two different expressions are equivalent?

Achievement Descriptor Math1.Mod1.AD12 Write equivalent expressions for contexts that can be modeled by arithmetic sequences. (F.BF.A.1.a)

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Describing Patterns

• Computers or devices (1 per student pair)

• Sums of Interior Angle Measures

Lesson Preparation

Land 10 min

• None

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Fluency Find the Number of Segments and the Perimeter Students find the numbers of segments and the perimeters of figures to prepare for extending patterns and writing expressions to describe the patterns. Directions: Complete the table. The side length of each square is 1 unit. Figure

Number of 1-Unit Line Segments

Perimeter (units)

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Launch

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Students examine a visual pattern and describe how it changes. Students begin the lesson by noticing and wondering about a hexagon pattern. They continue the pattern and count the number of line segments required to create each figure. Students examine their counting strategies and look for patterns in the numbers of line segments. Describe how you counted the line segments so that someone else could count them in the same way you did. I counted the line segments across the top and bottom of each figure first. Then I counted the vertical line segments in the middle. I counted the line segments by moving in a clockwise direction around each hexagon, being careful not to count any line segments more than once.

Figure 1

Figure 2

Figure 3

Figure 4

Figure 5

Figure 6

What pattern do you notice for the number of line segments it takes to draw each figure? Each figure has 5 more segments than the one before it. The number of line segments always ends in 1 or 6.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Learn Describing Patterns

Students create visual or verbal representations and write expressions to describe a pattern. Students use their observations from Launch to describe the number of line segments in any figure. Then they compare their descriptions and look for patterns. How can you tell whether two different descriptions represent this hexagon pattern? Two descriptions represent the same pattern when the same number of new line segments is added each time, even if we draw or count the line segments in a different way. Students examine animations of distinct drawing and counting strategies for the same hexagon pattern. For the first animation, students determine whether a given verbal description of the pattern matches a given algebraic expression of the same pattern. For the other animations, Figure 1 students enter algebraic expressions to describe the patterns. Having identified different expressions for Figure 2 the same pattern, students conjecture about the expressions. They discuss Figure 3 and decide whether different expressions can describe the same pattern. Expression 1: 6 + 5(n − 1)

Figure 4

Promoting the Standards for Mathematical Practice Students look for and express regularity in repeated reasoning (MP8) when they examine the total number of line segments required to draw each figure before writing expressions for any figure. Ask the following questions to promote MP8: • What patterns did you notice when you counted line segments in each figure? How can that help you determine the number of line segments in any figure? • What new shape is added each time the figure number is increased? What repeated actions are you taking when you create new figures?

2(4) + 2(4) + 5

Expression 2: 1 + 5n Expression 3: 2n + 2n + (n + 1)

Figure 5

2(5) + 2(5) + 3

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Do you think different expressions can describe the same pattern? Explain. Yes. Different expressions can describe the same pattern if each expression evaluates to the same number for all values of the variable. Students then examine the algebraic expressions to determine whether they evaluate to the same number when a given figure number is substituted for n. To extend the examination, students choose several more figure numbers to substitute into each expression and then discuss the results. If two expressions evaluate to the same number for every value of the variable, then those two expressions are equivalent. Based on the table, do you think these three expressions are equivalent? Explain. Yes. For all the figure numbers we tested, all three expressions evaluated to the same number.

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

UDL: Engagement Digital activities align to the UDL principle of Engagement by including the following elements: • Opportunities to collaborate with peers. Students examine each other’s expressions to compare and refine them. • Immediate formative feedback. Students see their expressions evaluated. • Options that promote flexibility and choice. Students choose a personally suitable approach to examining patterns.

Does the information in the table prove that the expressions are equivalent? Explain. No. We would have to check every possible number to prove this. Students transition to thinking about the perimeter of each figure instead of the number of individual line segments needed to create it. After entering expressions to describe the perimeter of figure n, students again reason about the equivalence of the expressions. How can you determine the perimeter of figure n? Explain. I can use the expression 6 + 4(n − 1) since the perimeter grows by 4 units as the figure number increases by 1. I can take the expression that represents the total number of line segments, 5n + 1, and subtract n − 1. Do you think these expressions are equivalent? Explain. Yes. I think evaluating the expressions for any figure number would result in the same perimeter.

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

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Sums of Interior Angle Measures Students develop the formula for calculating the sum of the measure of the interior angles of a convex polygon by using patterns. Students examine a pattern of the sum of the measures of the interior angles of polygons. They begin by recalling that the sum of the measures of the interior angles of a triangle is 180°. Students examine other polygons and acknowledge that they may not know the sums of the measures of the interior angles for polygons with more than 3 sides. You might have known that the sum of the measures of the interior angles of any triangle is 180°. Let’s decompose the other polygons to determine the sums of the measures of their interior angles as well. After considering different approaches to decomposing polygons, students decide on one approach and use it to determine the sum of the measures of the interior angles of each given polygon. Students then write expressions for the sum of the measures of the interior angles of a polygon with n sides. What is your expression? Explain your thinking. My expression is 180(n − 2). I drew diagonals from one vertex of the polygon to each of the other vertices. The number of triangles inside the polygon is 2 less than the number of sides of the polygon, which is (n − 2). So the sum of the interior angle measures of a polygon with n sides can be represented by 180(n − 2). My expression is 180n − 360. I let each side of the polygon be one side of a triangle, and the center of the polygon was the third vertex of each of my triangles. I drew n triangles inside my polygon, but the third vertex of each triangle formed an angle at the center of the polygon. The sum of those angles is 360°.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Students then consider whether the expressions are equivalent and use their expressions to determine the sum of the measures of the interior angles of a given octagon. Do you think these expressions are equivalent? Why? Yes. Both expressions for each of the polygons we tested evaluate to the same number when we substituted the number of the polygon’s sides for n. What is the sum of the measures of the interior angles of the given octagon? How do you know? The sum of the measures of the interior angles of an octagon is 1080° because 180(8 − 2) = 180(6) = 1080. The sum of the measures of the interior angles of an octagon is 1080° because 180(8) − 360 = 1080. Draw on the polygons to decompose them if it helps you explain your thinking.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Land Debrief 5 min Objective: Extend patterns and write expressions to describe them. Display the three expressions from the hexagon patterns and facilitate a class discussion by using the following prompts. Encourage students to restate or build upon one another’s responses. These expressions all produced the same number of line segments in figure 10, but the expressions look different. Choose one of these expressions and explain how it represents the hexagon pattern. I chose the expression 6 + 5(n − 1). The pattern started with one hexagon, which is represented by the 6 in the expression. For each hexagon added to the figure, 5 line segments were added. The number of groups of 5 line segments added is equal to 1 less than the figure number, which is represented by 5(n − 1). I chose the expression 1 + 5n. The pattern started with 1 line segment, which is represented by the 1 in the expression. For each hexagon added to the figure, 5 line segments were added, which is represented by 5n. I chose the expression 2n + 2n + (n + 1). For each hexagon in the figure, there are 2 line segments across the top and two line segments across the bottom, which is represented by 2n + 2n. There is 1 more vertical line segment than the number of hexagons, which is represented by n + 1. How do we know if two expressions are equivalent? Two expressions are equivalent if they evaluate to the same number for every possible value of the variable.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

b. Complete the table. The side length of each pentagon is 1 unit. Only the line segments on the outside of the figure count toward the perimeter.

Looking for Patterns In this lesson, we •

examined and extended visual patterns.

•

wrote expressions to describe patterns.

Example 1. Examine the pattern of pentagons. Figure 1

Figure Number

Number of 1-Unit Line Segments

Perimeter (units)

1 + 4n

3n + 2

Figure 2

Figure 3

The number of line segments in each figure is 1 more than

4 times the figure number.

The perimeter of each figure is

2 more units than 3 times the figure number.

Figure 4

Figure 5

a. How many line segments are required to draw figure 6?

25 line segments are required to draw figure 6.

Each figure has 4 more line segments than the previous figure. Figure 5 has 21 line segments.

RECAP

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

PRACTICE Name

Date

d. Complete the table. The side length of each square is 1 unit.

1. Examine the pattern of squares. Figure 1

Figure 2

Figure 3

Figure 4

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Figure Number

Number of 1-Unit Line Segments

Perimeter (units)

6n − 2

Figure 5

a. Draw figure 4 and figure 5. (Note: The student response is shown in the original diagram.)

2. Examine the pattern of triangles.

b. Write an expression that can be used to find the number of squares in figure n.

Figure 1

2n − 1 c. Construct figure 10 and verify that the expression you wrote in part (b) gives the correct number of squares in the figure when n = 10.

Figure 2

Figure 3

Figure 4

Figure 5

Figure 6

a. Draw figure 5 and figure 6. (Note: The student response is shown in the original diagram.) b. Complete the table. The side length of each triangle is 1 unit. Figure Number

Number of 1-Unit Line Segments

Perimeter (units)

2n + 1

n+2

2n − 1 = 2(10) − 1 = 19 There are 19 squares in figure 10.

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

d. Complete the table. The side length of each polygon is 1 unit.

c. Verify that the expression you wrote for the perimeter of figure n gives the correct perimeter of figure 10. Sample:

Figure Number

Number of Line Segments

Perimeter (units)

2n + 1

n+2

n + 2 = (10) + 2 = 12 Figure 10 has a perimeter of 12 units. 3. Examine the pattern. Figure 1

Figure 2

Figure 3

Figure 4

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Figure 5

4. Examine the pattern of bees.

a. Draw figure 4 and figure 5.

Figure 1

Figure 2

Figure 3

b. Write an expression that can be used to find the number of triangles inside figure n.

n c. Construct figure 10 and verify that the expression you wrote in part (b) is valid when n = 10. Figure 10

a. How many bees will be in figure 4? There will be 17 bees in figure 4. The expression I wrote is n, so the number of triangles inside a figure is the same as the figure number. There are 10 triangles inside figure 10. © Great Minds PBC

P R ACT I C E

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

b. Describe how to find the number of mice in figure n.

b. Describe how to find the number of bees in figure n.

Sample: There are 4 mice in figure 1, and 3 mice are added to each new figure.

Sample: In each figure, there are 4 groups of bees plus 1 bee in the middle. The number of bees in each of the 4 groups is the figure number. Figure 1

Figure 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

Figure 3

Figure 1

Figure 2

Figure 3

I can represent the number of mice in figure 1 as 1 + 3, the number of mice in figure 2 as 1 + 3 + 3 or 1 + 3(2), and the number of mice in figure 3 as 1 + 3 + 3 + 3 or 1 + 3(3). So I can represent the number of mice in any figure as 3n + 1, where n is the figure number.

There are 4 groups of the figure number, plus 1. I can represent this as 4n + 1, where n is the figure number.

Remember For problems 6 and 7, evaluate. 6. 16.5(4.4)

5. Examine the pattern of mice.

72.6

Figure 1

Figure 2

7. 45 ÷ (−2.5)

−18

8. Evaluate 4x + 6y − 16 for x = 2 and y = 6.

Figure 3

a. How many mice will be in figure 4? There will be 13 mice in figure 4.

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 2

9. Find the solution to a = 8 by using the given method. g 7

a. Solve by using tape diagrams. ga7

8 g ga7

ga7

56 The solution is 56.

b. Solve algebraically.

ag = 8 7

ag a 7 = 8 a 7 7

g = 56

The solution is 56.

P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic A, Lesson 3 LESSON 3

The Commutative, Associative, and Distributive Properties Rewrite algebraic expressions in equivalent forms. Show the equivalency of two algebraic expressions by using properties and operations.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Name

Date

EXIT TICKET

Show that 6 + 5w − 3(w + 5) and 2w − 9 are equivalent expressions by writing the property or operation used in each step. Expression

Property or Operation Used

6 + 5w − 3(w + 5) 6 + 5w − 3w − 15

Distributive property

6 − 15 + 5w − 3w

Commutative property of addition

−9 + 2w

Addition of like terms

2w − 9

Commutative property of addition

Lesson at a Glance Students activate prior knowledge by matching simple equivalent expressions. They name the properties of arithmetic in a graphic organizer and use the properties to formalize their understanding of equivalent expressions. Students use two-column tables and flowcharts to formally show that two given expressions are equivalent. This lesson introduces a new formalization of the term algebraic expression.

Key Questions • What do the commutative, associative, and distributive properties allow us to do with expressions? • What strategy can we use to show that expressions are equivalent? • What strategy can we use to show that two expressions are not equivalent?

Achievement Descriptor Math1.Mod1.AD8 Explain each step in demonstrating that algebraic

expressions are equivalent. (A.REI.A.1)

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Defining Equivalent Expressions

• None

• Demonstrating Equivalence

Lesson Preparation

Land 10 min

• None

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Fluency Write Expressions Students write equivalent expressions to prepare for demonstrating the equivalency of two algebraic expressions by using properties. Directions: Write three different expressions that have the same value as the given expression. Use the numbers 1, 2, 3, 4; the variable x (as needed); and the operations +, −, ⋅, and ÷. For example, write 8 as 2 ⋅ 2 + 1 + 3. 1.

2x + 3

Sample:

1+2+3

2+4

4+3–1

Sample:

3⋅3+1

1+2+3+4

Sample:

3⋅x

(1 + 2)x

Sample:

2⋅x+3

3⋅4−2

2⋅3⋅2⋅x ________

3+2⋅x

(1 + 1)x + 2 + 1

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Launch

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Students use their prior knowledge to identify equivalent expressions. Direct students to problem 1. Have students work independently or with a partner to match each expression in table 1 to an expression in table 2. 1. Match each expression in table 1 to an expression in table 2. Table 1

Table 2

9+p+2

(9 + p) + 6

9(p + 2)

9p + 18

9⋅2⋅p

9 ⋅ (p ⋅ 6)

(9 ⋅ p) ⋅ 6

9(p + 6) 2⋅9⋅p

9 + (p + 6) 9p + 54

p+9+2

Once students have finished, ask a few students to share their matches and explain their reasoning. Examples: I matched 9 ⋅ 2 ⋅ p and 2 ⋅ 9 ⋅ p because they can both be written as 18p.

I matched 9(p + 2) and 9p + 18 because I can use the distributive property to rewrite 9(p + 2) as 9p + 18. Today, we will use properties to rewrite one expression to match another expression.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Learn Defining Equivalent Expressions Students define equivalent expressions based on the properties of arithmetic. Direct students to the Properties of Arithmetic tree map. Display the tree map and define each property of arithmetic. Then have students identify an example of each property in problem 1. For example, in the Distributive box, write the equation a(b + c) = ab + ac to complete the sentence. Then have students identify a pair of expressions from the matching activity that exemplifies the distributive property, such as 9(p + 2) = 9p + 18, and write them in the Example box.

UDL: Representation This lesson introduces the idea that the properties of arithmetic can be described by using variables. The tree map graphic organizer helps students build connections between this new, abstract idea of the properties of arithmetic and what they already know.

Distributive

9p + 54 = 9(p + 6)

9(p + 2) = 9p + 18

Example

a(b + c) = ab + ac.

If a, b, and c are real numbers, then

(9 · p) · 6 = 9 · (p · 6)

9 + (p + 6) = (9 + p) + 6

a · (b · c) = (a · b) · c.

Example

a + (b + c) = (a + b) + c.

Example

Multiplication

If a, b, and c are real numbers, then

Addition

If a, b, and c are real numbers, then

Associative

Properties of Arithmetic

Addition

Example 9·2·p=2·9·p

9+p+2=p+9+2

a · b = b · a.

Multiplication

If a and b are real numbers, then

Example

a + b = b + a.

If a and b are real numbers, then

Commutative

Complete the statements in the tree map, and provide examples of each property.

EUREKA MATH2 Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Differentiation: Support

If students need support making sense of the verbal descriptions of the properties, consider using visual models to aid their understanding. For example, the following image could represent 3 + p or p + 3.

3 p

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Direct students’ attention to problem 2, and introduce the definition for algebraic expression. Have students complete problem 2, and encourage them to refer to their work in problem 1 and the Properties of Arithmetic tree map. An algebraic expression is a number, a variable, or the result of placing previously generated algebraic expressions into the blanks of one of the four operators, ( ) + ( ), ( ) − ( ), ( ) ⋅ ( ), ( ) ÷ ( ), or into the base blank of an exponentiation with an exponent that is a rational number, ( )( ). 2. Write some examples of algebraic expressions. Sample:

6, 10, z, _1  w, 9 + p + 2, 9p + 54 2

Language Support Support understanding the formal definition of algebraic expression by using a simple example to demonstrate the description.

9, p, and 54 are all algebraic expressions because they are numbers or variables. We can create the algebraic expression 9p + 54 by multiplying the algebraic expressions 9 and p and adding the algebraic expression 54.

Have several students share their examples of algebraic expressions. Then debrief by asking the following questions. Based on the definition, is _1 w an algebraic expression? Why? 2

Yes. It is an algebraic expression because it is the product of a number and a variable. Based on the definition, is _1 wy an algebraic expression? Why? 2

Yes. It is the product of the algebraic expression _1 w and a variable, and the product 2

of an algebraic expression and a variable is an algebraic expression. Based on the definition, is _1 w + y an algebraic expression? Why? 2

Yes. It is the sum of the algebraic expression _1 w and a variable, and the sum of an 2 algebraic expression and a variable is an algebraic expression. Based on the definition, is 9p3 − 1 an algebraic expression? Why?

Yes. It is an algebraic expression because 9p3 is the product of 9, p, p, and p, all of which are algebraic expressions, and 9p3 − 1 is the difference of an algebraic expression and a number.

Direct students’ attention to problems 3 and 4, and review the definition for equivalent expressions, which was introduced in grade 6. Have students complete problems 3 and 4 and encourage them to refer to their work in the matching activity and the Properties of Arithmetic tree map. Ask the following questions to lead a class discussion about how 58

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

to establish whether two expressions are equivalent. Use student responses to problems 3 and 4 as examples during the discussion. How could we show that two algebraic expressions are not equivalent? We could find a value for the variable such that the expressions do not evaluate to the same number. Would this method also be a good way to show that two algebraic expressions are equivalent? Why? No. It would not be possible in most cases to use this method to check every possible value of the variable. How could we show that two algebraic expressions are equivalent? We could use the properties of arithmetic and the properties of exponents to rewrite one expression as the other. Consider using student responses to write a sentence explaining the equivalence of the algebraic expressions. For example, 2x is equivalent to x ⋅ 2 because of the commutative property of multiplication. It can be challenging to use operations and the properties of arithmetic and exponents to show that a given expression cannot be rewritten as another. Instead, a value can be found to substitute for the variable that results in a different number for each expression. This indicates that the expressions cannot be equivalent. Two expressions are equivalent expressions if both expressions evaluate to the same number for every possible value of the variables. If we can rewrite one expression as the other by using the commutative, associative, and distributive properties and the properties of exponents, then the two expressions are equivalent. 3. Write two equivalent expressions. Explain why the expressions are equivalent. Sample:

9(p + 2) and 9p + 18 are equivalent expressions because I can use the distributive property to rewrite 9(p + 2) as 9p + 18.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

4. Write a nonexample of two equivalent expressions. Explain why the expressions are not equivalent.

9 + p + 2 and 9 ⋅ (p ⋅ 6) are not equivalent expressions because they do not evaluate to the same number for every possible value of p. For example, for p = 1, the first expression evaluates to 12 and the second expression evaluates to 54. Sample:

Demonstrating Equivalence Students represent the equivalence of two algebraic expressions by using a flowchart or a two-column table. Work as a class to show the steps used to rewrite the expression −8(−5b + 7) + 5b as 45b − 56.

−8(−5b + 7) + 5b (1)

40b − 56 + 5b

(2)

40b + 5b − 56

(3)

(40b + 5b) − 56

(4)

(40 + 5)b − 56

(5)

45b − 56

Lead the class in identifying the property or operation used to rewrite the expression in each step. What property is used to rewrite −8(−5b + 7) + 5b as 40b − 56 + 5b? The distributive property

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Most students will skip steps (3) and (4). If they do, leave space to add these steps later and have the following discussion. When we rewrite 40b + 5b − 56 as 45b − 56, we usually call this combining like terms or addition of like terms. But this is actually an application of two properties. Display steps (3) and (4). What property is used to rewrite 40b + 5b − 56 as (40b + 5b) − 56? The associative property of addition What property is used to rewrite (40b + 5b) − 56 as (40 + 5)b − 56? The distributive property What we call addition of like terms or combining like terms is actually an application of the associative property of addition and the distributive property. We can include the steps for the associative and distributive properties. We can also combine these two steps into one step called addition of like terms. Model how to organize the information into a flowchart, and direct students to follow along in problem 5. Ask the following questions to discuss the important features of the flowchart. What information do you write on the lines next to the arrows? I write the property or operation used to rewrite the expression in each step. The arrows in our flowchart are double-ended. Why do you think this is important? The arrows show that the expressions are equivalent to one another. For example, 40b + 5b − 56 is equivalent to 45b − 56, and 45b − 56 is equivalent to 40b + 5b − 56. Model how to organize the information in a two-column table, and direct students to follow along in problem 5. Discuss the important features of the table by asking the following questions. In a two-column table, we write the property or operation used to rewrite each expression. Why is the Property or Operation Used column blank for the first expression? This is the original expression. I have not performed any operations or used any properties yet.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Where do we write the property or operation used to rewrite the original expression as the expression in step (1)?

Differentiation: Support

We write it in the space next to the expression in step (1). 5. Show that −8(−5b + 7) + 5b is equivalent to 45b − 56 by using a flowchart and a two-column table.

–8(–5b + 7) + 5b distributive property 40b – 56 + 5b commutative property of addition

Students may have difficulty justifying their steps when labeling the properties and operations in problem 9 and others like it. They may forget that they already know how to rewrite expressions from their work in prior grades. If students need support creating their own flowcharts or two-column tables, encourage them to rewrite the expression first and then label the properties and operations. By taking this approach, the problems will resemble the work done in problems 5–9.

40b + 5b – 56 addition of like terms

Promoting the Standards for Mathematical Practice

45b – 56

Expression

Property or Operation Used

−8(−5b + 7) + 5b

Ask the following questions to promote MP7:

40b − 56 + 5b

Distributive property

40b + 5b − 56

Commutative property of addition

45b − 56

When students use the properties of arithmetic to rewrite algebraic expressions, they are looking for and making use of structure (MP7).

Addition of like terms

• How can you use what you know about rewriting numerical expressions to help you rewrite expressions with variables? • How can you use what these expressions have in common to help you identify the property or operation used?

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Give students time to work independently or with a partner on problems 6−9. Circulate as students work, and confirm answers. 6. Show that 3(y + 2) + 9x is equivalent to 3(3x + y) + 6 by completing the flowchart. Write the property or operation used in each step on the line next to the double-ended arrow.

3(y + 2) + 9x distributive property

3y + 6 + 9x commutative property of addition 9x + 3y + 6 associative property of addition

(9x + 3y) + 6

distributive property 3(3x + y) + 6

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

7. Show that y(4 + y) − 12 − y2 is equivalent to 4y − 12 by completing the two-column table. Write the property or operation used in each step.

Expression

Property or Operation Used

y(4 + y) − 12 − y2

y ⋅ 4 + y ⋅ y − 12 − y2 4y + y ⋅ y − 12 − y2

Commutative property of multiplication

4y + y2 − 12 − y2

Property of exponents (ba ⋅ bc = ba + c )

4y − 12 + y2 − y2

Commutative property of addition

4y − 12

Distributive property

Addition of like terms

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

8. Show that 2(5a − 7 − 8 + 10a) is equivalent to 30(a − 1) by writing the property or operation used in each step on the line next to each double-ended arrow.

2(5a – 7 – 8 + 10a) commutative property of addition

distributive property

2(5a + 10a – 7 – 8)

10a – 14 – 16 + 20a commutative property of addition

addition of like terms 2(15a – 7 – 8)

10a + 20a – 14 – 16

addition

addition of like terms

2(15a – 15) distributive property

30a – 14 – 16 distributive property addition

2 · 15(a – 1)

30a – 30

multiplication

distributive property

30(a – 1)

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

9. Show that 2(3a) − a(b + 10) + 8(ab + 10) is equivalent to 7ab − 4a + 80. Organize your work in a two-column table or a flowchart. Expression

Property or Operation Used

2(3a) − a(b + 10) + 8(ab + 10)

(2 ⋅ 3)a − a(b + 10) + 8(ab + 10)

(2 ⋅ 3)a − a ⋅ b − a ⋅ 10 + 8 ⋅ ab + 8 ⋅ 10 6a − ab − a ⋅ 10 + 8ab + 80

Associative property of multiplication Distributive property Multiplication

6a − ab − 10a + 8ab + 80

Commutative property of multiplication

6a − 10a − ab + 8ab + 80

Commutative property of addition

−4a + 7ab + 80

Addition of like terms

7ab − 4a + 80

Commutative property of addition

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

2(3a) – a(b + 10) + 8(ab + 10) associative property of multiplication

(2 . 3)a – a(b + 10) + 8(ab + 10) distributive property

(2 . 3)a – a . b – a . 10 + 8 . ab + 8 . 10

UDL: Action & Expression To support the organization of reasoning in a flowchart, consider posting written directions for students to refer to as they work.

multiplication

6a – ab – a . 10 + 8ab + 80 commutative property of multiplication

6a – ab – 10a + 8ab + 80 commutative property of addition

6a – 10a – ab + 8ab + 80 addition of like terms

–4a + 7ab + 80

1. Write the first expression. 2. Apply one property of arithmetic or one operation, and write the new expression below the first expression. 3. Draw a double-ended arrow between the expressions to represent equivalence. 4. Write the property or operation used in the blank space next to the doubleended arrow. 5. Repeat steps 2 through 4 until you produce the second expression.

commutative property of addition

7ab – 4a + 80

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Land Debrief 5 min Objectives: Rewrite algebraic expressions in equivalent forms. Show the equivalency of two algebraic expressions by using properties and operations. Initiate a class discussion by using the prompts below. Invite students to restate their classmates’ responses. What do the commutative, associative, and distributive properties allow us to do with expressions? They allow us to create equivalent expressions or show that two expressions are equivalent. What strategy can we use to show that two expressions are equivalent? If we can apply some combination of the properties of arithmetic, arithmetic operations, and/or properties of rational exponents to rewrite one expression as the other, then the two expressions are equivalent. What strategy can we use to show that two expressions are not equivalent? If we can find a value for the variable where the two expressions evaluate to two different numbers, then the expressions are not equivalent.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

RECAP Name

Date

•

used the commutative, associative, and distributive properties to rewrite algebraic expressions.

•

applied properties and operations to show that two algebraic expressions are equivalent.

Examples 1. Show that 6 + 10x(2x − 3) − 20x 2 is equivalent to −30x + 6 by writing the property or operation used in each step next to each double-ended arrow.

The Commutative, Associative, and Distributive Properties In this lesson, we

6 + 10x (2x – 3) – 20x 2

Terminology

distributive property

An algebraic expression is a number, a variable, or the result of placing previously generated algebraic expressions into the blanks of one of the four operators, ( ) + ( ), ( ) − ( ),

(

)⋅(

), (

)÷(

), or into the base blank

6 + 20x 2 – 30x – 20x 2 The arrows are double-ended to show that the expressions

of an exponentiation with an exponent that is a rational number, ( )( ).

on each end are equivalent to each other.

The Properties of Arithmetic The Distributive Property

a(b + c) = ab + ac.

3(2 + 4) = 3 ⋅ 2 + 3 ⋅ 4

commutative property of addition

The Associative Property of Addition

The Associative Property of Multiplication

If a, b, and c are real numbers, then

(6 + t) + 5 = 6 + (t + 5)

–30x + 6

(a ⋅ b) ⋅ c = a ⋅ (b ⋅ c).

(3 ⋅ 2) ⋅ 4 = 3 ⋅ (2 ⋅ 4) (6 ⋅ t) ⋅ 5 = 6 ⋅ (t ⋅ 5)

The Commutative Property of Addition

The Commutative Property of Multiplication

If a and b are real numbers, then

a + b = b + a. 3+2=2+3 6+t=t+6 © Great Minds PBC

6 – 30x + 20x 2 – 20x 2

6 – 30x

6(t + 5) = 6t + 30

(a + b) + c = a + (b + c).

commutative property of addition

addition of like terms

If a, b, and c are real numbers, then

(3 + 2) + 4 = 3 + (2 + 4)

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

a ⋅ b = b ⋅ a. 3⋅2=2⋅3 6⋅t=t⋅6

RECAP

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

2. Show that 2(7a) − 3(4a − 1) is equivalent to 2a + 3 by writing the property or operation used in each step.

Expression

Leave this box blank because the expression in this row is the original expression.

Property or Operation Used

2(7a) − 3(4a − 1)

(2 ⋅ 7)a − 3(4a − 1)

(2 ⋅ 7)a − 3 ⋅ 4a − 3 ⋅ (−1) 14a − 12a + 3 2a + 3

Associative property of multiplication Distributive property Multiplication Addition of like terms

RECAP

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

3. Volume of the rectangular prism

For problems 1–4, write two equivalent expressions for each situation. 1. Total length of both segments

8 0

3 + 2; 2 + 3

(5 ⋅ 6) ⋅ 8; 8 ⋅ 5 ⋅ 6

4. Number of squares in the figure

2. Area of the rectangle

10 7

2 + 5; 5 + 2 For problems 5–10, identify the property that justifies why each pair of expressions is equivalent. 5. 3 + 5 + 2 and 3 + 2 + 5 Commutative property of addition

10 ⋅ 7; 7 ⋅ 10

7. 9(2 + 5) and 18 + 45 Distributive property 9. (y + 5)(3y) and (3y)(y + 5) Commutative property of multiplication

P R ACT I C E

6. 5 ⋅ (4 ⋅ 6) and (5 ⋅ 4) ⋅ 6

Associative property of multiplication

8. 2x + 3x2 − 5 and 3x2 + 2x − 5 Commutative property of addition 10. m2 + 2m − 6m − 12 and m2 + (2m − 6m) − 12 Associative property of addition

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

13. Show that −4a + (a + 2) ⋅ 3a − 3a2 is equivalent to 2a by writing the property or operation used in each step on the line next to each double-ended arrow.

11. Show that 2(3x + 2) − 4(3x + 2) is equivalent to −6x − 4 by writing the property or operation used in each step. Expression

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

n4a + (a + 2) n 3a n 3a2

Property or Operation Used

distributive property

2(3x + 2) − 4(3x + 2)

2 ⋅ 3x + 2 ⋅ 2 − 4 ⋅ 3x − 4 ⋅ 2

Distributive property

n4a + 3a2 + 6a n 3a2

6x + 4 − 12x − 8

Multiplication

6x − 12x + 4 − 8

Commutative property of addition

−6x − 4

commutative property of addition

Addition of like terms

n4a + 6a + 3a2 n 3a2 addition of like terms

12. Show that −8 + x(x + 3) + 4(y + 3) is equivalent to x2 + 3x + 4(y + 1) by writing the property or operation used in each step. Expression

Property or Operation Used

−8 + x(x + 3) + 4(y + 3)

−8 + x ⋅ x + 3x + 4y + 12 2

−8 + x + 3x + 4y + 12 x2 + 3x + 4y + 12 − 8 2

Distributive property

Property of exponents (ba ⋅ bc = ba+c) Commutative property of addition

x + 3x + 4y + 4

Addition of like terms

x2 + 3x + (4y + 4)

Associative property of addition

x2 + 3x + 4(y + 1)

Distributive property

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 3

14. Show that 2g + 8(g − 4) − 7g is equivalent to 3g − 32 by completing the flowchart. Write the property or operation used in each step on the line next to each double-ended arrow.

16. Show that 2(2 + n3) + n(5n2 + 2) is equivalent to 7n3 + 2n + 4. Write the property or operation used in each step.

2g + 8(g – 4) – 7g distributive property

Expression

2g – 7g + 8(g – 4) distributive property

commutative property of addition

4 + 2n3 + 5n3 + 2n

Distributive property

2n3 + 5n3 + 2n + 4

Commutative property of addition

7n + 2n + 4 2g + 8g – 7g – 32

Property or Operation Used

2(2 + n3) + n(5n2 + 2)

commutative property of addition

2g + 8g – 32 – 7g

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Math 1 ▸ M1 ▸ TA ▸ Lesson 3

Addition of like terms

2g – 7g + 8g – 32 Remember

addition of like terms

addition of like terms For problems 17 and 18, evaluate. 3 17. −2_ + _ 5 4

3g – 32

7 __ 20

15. Show that (5p + 3) − 4 − 9p is equivalent to −1 − 4p by using a flowchart or a two-column table. State the property or operation used in each step. Expression

Property or Operation Used

(3 − 4) + 5p − 9p −1 − 4p

1 _4 + − 2_ − __ 9 ( 3) 27 7 −__ 27

19. Evaluate.

_1 x + 1_ y + 9 for x = 8 and y = 27 2 3

(5p + 3) − 4 − 9p 5p + (3 − 4) − 9p

18.

Associative property of addition

For problems 20 and 21, solve the equation.

Commutative property of addition

20. −2(3x − 2) = 28

Addition of like terms

−4

P R ACT I C E

21. 5(2x + 4) − 12x = −6

Teacher Edition: Math 1, Module 1, Topic A, Lesson 4 LESSON 4

Interpreting Linear Expressions Interpret parts of linear expressions within a context.

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Name

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Date

EXIT TICKET

Computer World sells a certain laptop model for a markup price that is 15% more than its purchase price p. During a one-week sale, the markup price was discounted by $100. The expression 12(1.15p − 100) represents the total amount of money in dollars from the sale of this laptop model for that week. a. What does 1.15p represent in this situation?

1.15p represents the price that Computer World sells the laptop for with the 15% markup.

Lesson at a Glance In this lesson, students analyze which of two deals in a clothing store is better by using the Take a Stand routine to reflect on how to interpret an algebraic expression. Students work with partners to interpret algebraic expressions and to create situations that can be modeled with different expressions. As a class, students discuss equivalent expressions and what the coefficients and terms represent in relation to a context.

Key Questions • What can different parts of a linear expression tell us about a situation?

b. What does −100 represent in this situation?

• How can equivalent forms of an algebraic expression tell us different information?

−100 represents the $100 discount during the one-week sale.

Achievement Descriptor Math1.Mod1.AD3 Interpret parts of linear expressions, such as terms, factors, and coefficients in context. (A.SSE.A.1.a)

c. What does 12 represent in this situation?

12 represents the number of these laptops Computer World sold during the one-week sale.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Agenda

Materials

Fluency

Teacher

Launch 5 min

• Construction paper (3 sheets)

Learn 30 min • Interpreting Expressions

• Marker • Tape

• Coefficients and Terms

Students

• Creating a Situation for an Expression

• None

Land 10 min

Lesson Preparation • Prepare three signs with the following labels: $10 Off Before, $10 Off After, Doesn’t Matter. Post the signs around the classroom.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Fluency Write Equivalent Expressions Students write equivalent expressions to prepare for work adding and subtracting linear expressions. Directions: Add like terms to write an equivalent expression. 1.

2x + 5 + 4x − 3

6x + 2

6 − 8a + 8 + 4a

−4a + 14

Teacher Note Instead of this lesson’s Fluency, consider administering the Write Equivalent Expressions Sprint. Directions for administration can be found in the Fluency resource. EUREKA MATH2

Math 1 ▸ M1 ▸ Sprint ▸ Write Equivalent Expressions

Number Correct:

Rewrite each expression as an equivalent expression in standard form.

3. 4. 5. 6.

−4m + 10c + 5m − c 2(z − 12) + 6z 14 − 3(p + 4) −4(x − y) + y + 9x

m + 9c 8z − 24 −3p + 2 5x + 5y

1+1

23.

4x + 6x − 12x

1+1+1

24.

4x + 6x − 4x

(1 + 1) + 1

25.

(1 + 1) + (1 + 1)

26.

(4x + 3) + x

(1 + 1) + (1 + 1 + 1)

27.

(4x + 3) + 2x

x+x

28.

3x + (4x + 3)

x+x+x

29.

6x + (4x + 3)

(x + x) + x

30.

6x + (4 + 3x) 6x + (4 − 3x)

(x + x) + (x + x)

31.

10.

(x + x) + (x + x + x)

32.

11.

(x + x) + (x + x + x + x)

33.

2x + (4 − 3x)

12.

2x + x

34.

(3x + 9) + (3x + 9)

13.

3x + x

35.

(3x − 9) + (3x − 9)

14.

x + 4x

36.

15.

x + 7x

37.

(3x − 9) − (3x + 9)

16.

2x + 7x

38.

(11 − 5x) + (4x + 2)

17.

7x + 3x

39.

(5x + 11) + (2 − 4x)

4x + (4 − 3x)

(3x − 9) + (3x + 9)

10x − x

40.

(11 − 5x) − (2 − 4x)

19.

10x − 5x

41.

(2x + 3y) + (4x + y)

20.

10x − 10x

42.

(2x − 3y) + (4x − y)

21.

10x − 11x

43.

(2x − 3y) + (3y − 4x)

22.

10x − 12x

44.

(2x + 3y) − (2x − 3y)

18.

394

4x + 6x + 4

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Launch

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Students analyze a situation that can be interpreted in two different ways. Introduce the Take a Stand routine to the class. Draw students’ attention to the three signs posted in the classroom: $10 Off Before, $10 Off After, Doesn’t Matter. Direct students to complete problem 1. 1. A clothing store is having a sale where a 20% discount is applied at checkout. You also have a coupon for $10 off, which can also be applied at checkout. Is it a better deal to apply the $10 coupon before or after the 20% discount, or does it matter? Explain. Sample: Total Purchase Price: $120

$10 Off Before

$10 Off After

(120 − 10)0.8 = (110)0.8 = 88

120(0.8) − 10 = 96 − 10 = 86

Total Purchase Price: $20

$10 Off Before

$10 Off After

(20 − 10)0.8 = (10)0.8 = 8

20(0.8) − 10 = 16 − 10 = 6

UDL: Action & Expression To support planning and/or strategizing of work for problem 1, have students turn to a partner to restate the problem in their own words. Circulate as students work, and confirm that they understand what they need to do to answer the question. To provide further support, consider asking questions such as the following: • How do you know which deal is better? • What is the better deal if your total purchase price is $20? $100?

I think that the better deal is when the $10 coupon is applied after the 20% discount. I tested when my total purchase price was $120 and again when it was $20. In each case, I found that the better deal was when the $10 coupon was applied after the 20% discount. Invite students to stand near the sign that best describes their thinking. When all students are standing near a sign, allow 2 minutes for groups to discuss the reasons why they chose that sign. Then call on each group to share reasons for their selection. Invite students who change their minds during the discussion to join a different group.

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

EUREKA MATH2

Have students return to their seats. As a class, reflect on the strategies that students used to interpret each deal. How did your group decide which was the better deal? We assumed that our total purchase price was $120 and then calculated and compared the final cost if the coupon was applied before and if it was applied after the 20% discount. Then we repeated the process, assuming that the total purchase price was $20. We wrote expressions to represent each situation and compared the expressions by using the distributive property and the order of operations. Today, we will use algebraic expressions to model situations and interpret what expressions tell us about the situation.

Learn Interpreting Expressions Students interpret expressions that represent situations. Direct students to work with a partner on problem 2. 2. Let x represent the total purchase price in problem 1. a. What does the expression x − 0.2x − 10 represent in this situation? This expression represents that the 20% discount is applied first and then the $10 coupon is applied. b. What does the expression 0.8x − 10 represent in this situation? This expression represents that the 20% discount is applied first and then the $10 coupon is applied. c. What does the expression 0.8(x − 10) represent in this situation? This expression represents that the $10 coupon is applied first and then the 20% discount is applied. 78

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Confirm student responses when most of the pairs have completed the problem. Lead a discussion about their responses by using the following prompts. Generate an expression equivalent to x − 0.2x − 10 by combining like terms.

0.8x − 10 Generate an expression equivalent to 0.8(x − 10) by using the distributive property.

0.8x − 8 What does 0.8 represent in problems 2(b) and 2(c)?

0.8 represents 80% of the price that we pay after the 20% discount is applied.

Differentiation: Support If students need support understanding how 0.8 relates to the problem, consider asking questions such as the following: • What does it mean if the price of an item is 20% off? • If the price of an item is 20% off, then what percent of the original price do you pay?

Display the following equation.

0.8(x − 10) = 0.8x − 8 We know that these expressions are equivalent. However, these expressions tell two slightly different stories in relation to the context. Which of these expressions better tells the story of taking $10 off the total purchase price first and then applying the 20% discount? Why? The expression 0.8(x − 10) better tells the story because I can see that the $10 coupon is applied first and then the 20% discount is applied. Interpret 0.8x − 8 in this situation. The expression 0.8x − 8 represents a 20% discount being applied to the total purchase price and then $8 is subtracted. Recall that the expressions that represent the two deals are 0.8(x − 10) and 0.8x − 10. Which expression represents the better deal? Explain. The expression 0.8x − 10 represents a better deal. The expression 0.8(x − 10) is equivalent to 0.8x − 8. This means that the effect of taking $10 off first and then applying the 20% discount is the same as applying the 20% discount first and then taking $8 off the discounted value.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Does this information confirm your decision about the better deal, or does it change your mind? Why? This information confirms my decision because I thought that the better deal was for the 20% discount to be applied first. I see that applying the coupon first is equivalent to subtracting $8 from the cost, instead of subtracting $10. This information made me change my mind because I originally thought that the better deal was for the $10 coupon to be applied first. Now I see that applying the discount before applying the coupon means that I pay less money than if I apply the coupon first. Direct students to work independently on problem 3.

Language Support To support the context in problem 3, consider showing pictures or a video of people bowling in a bowling alley. Highlight the need for special shoes to protect the bowling lanes, prevent any foot injuries, and achieve smoother motion that results in having more control of the ball.

3. On Monday nights, a bowling alley charges $10 per person and $3.80 for each person’s shoe rental. On Wednesday nights, the same bowling alley charges $12 per person with no charge for shoe rental. a. Write an expression that represents the total cost for p people to bowl on a Monday night.

10p + 3.8p b. Write an expression that represents the total cost for p people to bowl on a Wednesday night.

12p c. Which expressions represent the total cost for p people to bowl if each person bowls on both a Monday night and a Wednesday night? Choose all that apply. A. 10p + 3.8p + 12p B. 10p + 3.8p C. 25.8p D. 3.8p + 22p E. 12p

Promoting the Standards for Mathematical Practice When students interpret expressions or parts of expressions in a given context and use those interpretations to determine the better deal, they are reasoning abstractly and quantitatively (MP2). Ask the following questions to promote MP2: • What does 22 in the expression in answer choice D tell us about the situation? • How does 25.8p represent this context?

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Ask a few students to share their responses to part (c) and to explain their reasoning. Then display all five expressions from part (c) and ask the following questions. Examine these expressions from problem 3(c). What does each expression tell us about the situation? The expressions 10p + 3.8p + 12p, 25.8p, and 3.8p + 22p tell us the total cost for p people to bowl on both a Monday night and a Wednesday night. The expression 10p + 3.8p tells us the total cost for p people to bowl on just a Monday night. The expression 12p tells us the total cost for p people to bowl on just a Wednesday night. How did you find all the expressions that represent the total cost for p people to bowl if each person bowls on both a Monday night and a Wednesday night? I added my answer from problem 3(a) to my answer from problem 3(b) and then combined like terms. Each time I applied different properties to the expression, I checked to see whether it was one of the answer choices.

Coefficients and Terms Students interpret coefficients and terms of a linear expression that represents a situation. Continue having students work with their partner to complete problems 4 and 5. While students are working on problem 4, they might question if Zara is continuing to receive messages. Acknowledge the students’ observation but encourage them to complete problem 4. Revisit their question after problem 5 is completed. 4. Zara works as a customer service representative for an online clothing store. At the beginning of the day, Zara begins responding to messages in the company’s email inbox. The expression 105 − 15n models the number of messages remaining in the inbox n hours after Zara began responding to the messages. a. What does 105 represent in this situation? The company email inbox had 105 messages at the beginning of the day. b. What does −15 represent in this situation? The number of messages in the inbox decreases by 15 messages each hour that Zara responds to them. © Great Minds PBC

Differentiation: Challenge For an additional challenge, have students answer the following question: Mason begins responding to messages in the company’s email inbox 2 hours after Zara began responding. The expression 105 + 5n − 20n − 9(n − 2) models the number of messages remaining n hours after Zara began responding to them. Interpret −9 and n − 2 in the expression.

−9 tells us that the number of messages remaining decreases by 9 messages each hour that Mason responds to them.

n − 2 tells us that Mason began responding to messages 2 hours after Zara began responding to them.

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

EUREKA MATH2

5. The expression 105 + 5n − 20n also models the number of messages remaining in the inbox n hours after Zara began responding to them. a. Is 105 + 5n − 20n equivalent to the expression given in problem 4? Explain. Yes. If we combine like terms in 105 + 5n − 20n, the result is 105 − 15n. b. Based on the expression 105 + 5n − 20n, what might be happening in this situation? The company’s inbox could receive 5 more messages each hour, and Zara responds to 20 messages each hour. Ask a few students to share their answer to problem 4(b) and to explain their reasoning. What does n represent in this situation? It represents the number of hours that Zara responds to messages in the inbox. By using this information, what does the coefficient of −15 in −15n tell us about this situation? It tells us that there are 15 fewer messages in the inbox every hour Zara responds to them. So we can say that −15 represents that the number of messages in the inbox decreases by 15 messages per hour that Zara responds to them. Does this information change your answer for problem 4(b)? Why? This does not change my answer because I originally thought that the messages in the inbox were decreasing by 15 messages per hour. This does change my answer because I originally interpreted the expression in relation to the numbers of messages Zara responded to or received each hour. Direct students to work on problem 6 with a partner. Circulate as students work, and confirm their answers.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

6. Nina and Fin are raising money for charity by asking for cash donations and selling T-shirts. The expression 250 + 8n models the total amount of money Nina raised, where n represents the number of T-shirts Nina sold. The expression 175 + 8f represents the total amount of money Fin raised, where f represents the number of T-shirts Fin sold. a. Who raised more money in cash donations? Explain. Nina raised more money in cash donations with $250, compared to Fin who raised $175 in cash donations. b. What is the price that Nina and Fin sold each T-shirt for? How do you know? Nina and Fin sold each T-shirt for $8. The coefficient 8 represents the money raised in dollars for each T-shirt sold. c. What does the expression (250 + 8n) + (175 + 8f ) represent? The expression represents the total amount of money both Nina and Fin raised for charity. d. Show that 8(n + f ) + 425 is equivalent to the expression given in part (c).

(250 + 8n) + (175 + 8f ) = 250 + 8n + 175 + 8f = 8n + 8f + 250 + 175 = 8(n + f ) + 425 e. What does the expression (n + f ) represent in part (d)? The expression (n + f ) represents the total number of T-shirts Nina and Fin sold. f. What does 425 represent in part (d)?

425 represents the total amount in cash donations Nina and Fin collected.

Creating a Situation for an Expression Students create a situation that can be modeled by a given expression. Present problem 7 to the class. Have students work with a partner, and use the Co-construction routine to have pairs create a context that could apply to problem 7.

Language Support While working in pairs during the Co-construction routine, consider directing students to use the Share Your Thinking section of the Talking Tool.

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

EUREKA MATH2

7. Write a situation that can be modeled by the expression 50x + 100. Samples: Ana opens a savings account with $100 and then deposits $50 into the account each month for x months. Mason reads 100 pages of a novel at school and then reads 50 pages each night for x nights. Once pairs finish, have them compare the situations they construct with those of other pairs. Then invite a few pairs to share their situations with the class.

Land Debrief 5 min Objective: Interpret parts of linear expressions within a context. Facilitate a class discussion by asking the following questions. Encourage students to restate or build upon one another’s responses. What can different parts of a linear expression tell us about a situation? Use an example to explain. Different parts of a linear expression can tell us what is constant and what is changing in a situation and at what rate it is changing. For example, in the situation modeled by the expression 250 + 8n, we see that Nina raised $250 in cash donations and also that she sold each T-shirt for $8.

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

How can equivalent forms of an algebraic expression reveal different information about a situation? Use an example to explain. One algebraic expression can sometimes represent a situation more accurately than its equivalent expression. For example, the expression 105 + 5n − 20n tells us the rate that new messages were coming in and the rate that Zara was responding to the messages. But the expression 105 − 15n only told us the rate that the messages in the inbox were decreasing.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Recap

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

2. Write a situation that can be modeled by the expression 1.25p + 7.50. Sample: Huan paid $7.50 for a notebook and bought p pencils that cost $1.25 each.

Interpreting Linear Expressions In this lesson, we •

described what the coefficients and terms of a linear expression represent in a given situation.

•

interpreted linear expressions that represent different situations.

•

created a situation that could be modeled by a given expression.

When deciding on a context that can be modeled by a specific expression, identify the part of the context that is constant and the part that changes.

Examples 1. At a local pizza restaurant, the price of one slice of pizza is $2.25 more than the price of one soda, which is x dollars. Bahar and Angel get dinner at the pizza restaurant. The expression 2x + 3(x + 2.25) models the total price of their dinner. a. What does the expression 2x represent in this situation? The expression 2x represents the price of 2 sodas. b. What does the expression x + 2.25 represent in this situation? The expression x + 2.25 represents the price of 1 slice of pizza. c. What does the expression 3(x + 2.25) represent in this situation? The expression 3(x + 2.25) represents the price of 3 slices of pizza. d. Write an expression that represents the total price of their dinner if they instead purchased 1 soda and 4 slices of pizza.

x + 4(x + 2.25)

To write a new expression, consider how the coefficients and terms in the original expression change based on the context.

e. Write an equivalent expression to your answer for part (d). Sample: 5x + 9

RECAP

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

PRACTICE Name

Date

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

EUREKA MATH2

3. On Fridays, a water park charges $25 per person and $8 to rent an inner tube. On Saturdays, the same water park charges $26 per person, and there is no charge to rent an inner tube. a. Write an expression that represents the total cost for n people to go to the water park on Friday if each person rents an inner tube.

1. Mason played 4 games at the arcade on his own, and then he played g more games after meeting up with his friends. Mason paid $0.75 to play each game.

25n + 8n

a. What does the expression 4 + g represent in this situation? The expression 4 + g represents the total number of games Mason played at the arcade.

b. Write an expression that represents the total cost for n people to go to the water park on Saturday.

26n

b. What does the expression 0.75(4 + g) represent in this situation? The expression 0.75(4 + g) represents the total cost of playing 4 + g games at the arcade.

c. Suppose n people go to the water park on both Friday and Saturday. Write an expression that represents the total cost if each person rents an inner tube both days.

26n + 25n + 8n

c. What does the expression 3 + 0.75g represent in this situation? The expression 3 + 0.75g represents the total amount of money Mason spent to play games at the arcade.

d. Write an equivalent expression to your answer for part (c).

59n 2. Li Na watched 3 videos on her phone before school and then watched v more videos after school. Each video was 2 minutes long. The expressions 6 + 2v and 2(3 + v) both model this situation. a. What does each part of the expression 6 + 2v represent in this situation? The 6 represents that Li Na watched 6 minutes of videos before school, and the 2v represents that she watched 2v more minutes of videos after school.

4. A bookstore sells each book at a 40% markup. The expression 0.4b + b models the selling price of each book, where b represents the manufacturer’s price of a book in dollars. a. What does 0.4 represent in this situation?

0.4 represents the 40% markup of the manufacturer’s price of each book.

b. What does each part of the expression 2(3 + v) represent in this situation?

b. The expression 1.4b also models the selling price of each book. Is this expression equivalent to the expression in problem 4? Explain.

The factor 3 + v represents that Li Na watched 3 + v videos, and the factor 2 represents that each video was 2 minutes long.

Yes, 1.4b is equivalent to 0.4b + b. If we rewrite 0.4b + b by combining like terms, we get 1.4b.

P R ACT I C E

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

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Math 1 ▸ M1 ▸ TA ▸ Lesson 4

6. Write a situation that can be modeled by the expression 3n + 25.

5. An indoor gym charges rock climbers the following fees: •

An hourly harness rental fee

•

An hourly shoe rental fee, which is twice as much as the harness rental fee

•

A monthly fee for using the facilities

EUREKA MATH2

Math 1 ▸ M1 ▸ TA ▸ Lesson 4

Sample: Riku’s family of n people paid $25 to visit a trampoline park and $3 for each person’s pair of trampoline socks.

Angel spends the same number of hours h rock climbing each time he goes to this gym. The following expressions each model the total amount of money Angel spent on rock climbing during the same month at this gym:

5(h(1 + 0.50)) + 15

7. Write a situation that can be modeled by the expression 6a + 4b.

5(h + 0.50h) + 15

Sample:

5(1.50h) + 15

The drama club sells a total of a adult tickets for $6 each and b child tickets for $4 each.

a. What does 5 represent in this situation?

5 represents the number of times Angel rock climbs at this gym in the same month. b. What does 1 represent in this situation?

Remember

1 represents the cost of the shoe rental per hour.

For problems 8 and 9, evaluate. 8. − 2_ n 5_ 3 6

c. What does 0.50 represent in this situation?

0.50 represents the cost of the harness rental per hour.

− 5_

7 ÷ − 2_ 9. − __ 10 ( 9)

63 __ 20

d. What does 1.50 represent in this situation?

1.50 represents the total cost of the shoe rental and harness rental per hour. e. What does 15 represent in this situation?

10. Evaluate 0.2x + 11 − 0.5y when x = 16 and y = 6.

15 represents the constant monthly fee for using the facilities. f.

11.2

What does the expression 1.50h represent in this situation? The expression 1.50h represents the total cost of the shoe rental and harness rental each time Angel goes rock climbing at this gym.

11. Solve for x in the equation −4(x − 6) = −2(3x + 2) − 8.

−18

g. What does the expression h + 0.50h + 15 represent in this situation? The expression h + 0.50h + 15 represents the total cost Angel pays to go rock climbing at this gym one time in a month.

P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic B

Topic B Solving Equations and Inequalities in One Variable In grades 6, 7, and 8, students solve one-variable equations and inequalities of increasing complexity. Topic B builds on that foundation by formalizing the properties of equality and inequality. When solving equations, students justify each step of the solution path. Their work with equations is further extended to include equations with coefficients represented by letters and the concept of rearranging a formula to highlight a particular quantity. Students solve multi-step linear inequalities, write the solution sets by using set-builder notation, and graph the solution sets on the number line. Students are introduced to solving problems in real-world and mathematical contexts by creating equations and inequalities in one variable, which helps them understand how equations and inequalities can be used to represent constraints of a given situation. Students begin the topic by exploring a problem about printing presses that can be solved in many ways. Students are invited to solve the problem by using any method they believe is best. The class generalizes the different methods that can be used by developing an algebraic equation. Students learn that an algebraic approach is sometimes the most efficient way to solve a problem.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB

Number of Printing Presses

Number of Novels

Number of Cookbooks

10(1) + 5(1), or 15

5(2) + 1(2) + 1(2), or 14

12(1) + 6(1), or 18

6(2) + 1(2) + 1(2), or 16

8(1) + 4(1), or 12

4(2) + 1(2) + 1(2), or 12

x + _x

1 2

1 x (2) + 2 + 2 2

Because the print shop must print the same number of copies of each book, the expression representing the total number of novels printed must equal the expression representing the total number of cookbooks printed. Solution set is defined in terms of both equations and inequalities. Students are introduced to set-builder notation which allows them to represent solution sets with formality and precision. Students use their understanding of properties of arithmetic and equality to justify each step used in solving a linear equation and to evaluate different solution paths that yield the same result.

3 + _ = __ − 2 g 7

3 + _ = __ − 2

g 14

g 7

42 + 2g = g − 28

multiplication property of equality

42 + g = −28

addition property of equality

g = −70

addition property of equality

g 14

_ __

g g = −5 7 14

g = −5 14

g = −70

addition property of equality addition property of equality multiplication property of equality 91

EUREKA MATH2

Math 1 ▸ M1 ▸ TB

Students explore actions that may or may not preserve a solution set. This lesson is designed to deepen students’ understanding of properties of equality, in particular the multiplication property of equality, and to explore the effect of squaring both sides of an equation. These explorations set the stage for equations seen in later courses that have extraneous solutions. Students return to concepts found in the opening lesson, exploring problems that can be modeled and solved by using a one-variable equation. The lesson includes an engaging digital activity where students use equations to find missing numbers in a puzzle that would be cumbersome to solve by using non-algebraic methods. Students use their knowledge of solving linear equations with known coefficients by using the properties of equality in more abstract contexts, as they rearrange formulas and solve linear equations with coefficients represented by letters. Topic B concludes with students solving linear inequalities, a process introduced in grade 7. Students solve linear inequalities, express the solution sets by using set-builder notation, and graph the solution sets on the number line. While finite solution sets can often be written concisely by using set notation (e.g., {2}), set-builder notation represents the solution sets of linear inequalities with precision (e.g., {x | x > 2}). The lesson closes by summarizing and comparing properties of equality and properties of inequality.

50 – 2n

50 – n

50 – n + 1

Understanding how to find and represent solution sets of linear equations and inequalities in one variable prepares students for topic C, which focuses on finding, writing, and graphing solution sets to compound statements involving equations or inequalities.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB

Progression of Lessons Lesson 5

Printing Presses

Lesson 6

Solution Sets of Equations and Inequalities in One Variable

Lesson 7

Solving Linear Equations in One Variable

Lesson 8

Some Potential Dangers When Solving Equations

Lesson 9

Writing and Solving Equations in One Variable

Lesson 10 Rearranging Formulas Lesson 11 Solving Linear Inequalities in One Variable

Teacher Edition: Math 1, Module 1, Topic B, Lesson 5 LESSON 5

Printing Presses Investigate a problem that can be solved by reasoning quantitatively or algebraically.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Name

Date

EXIT TICKET

Evan is 7 years younger than his sister, Danna. The sum of their ages is 37. Using any method, find their ages. Let x represent Danna’s age in years. Then x − 7 represents Evan’s age in years. So the expression x + (x − 7) represents the sum of their ages.

x + (x − 7) = 37 2x − 7 = 37 2x = 44 x = 22

Substituting 22 for x:

Lesson at a Glance In this lesson, students work in groups to find an entry point to solve a problem about printing presses. Because this problem can be solved by using various methods, it provides an opportunity for both collaboration and student choice in a solution path. This lesson introduces the Five Framing Questions routine as students analyze a variety of strategies, building connections between quantitative reasoning and the process of writing and solving an equation in one variable. Before beginning the problem, students analyze a picture of and read some information about the Gutenberg printing press to connect art and history with the problem they are solving in the lesson.

Key Questions

x − 7 = 22 − 7 = 15

• What are some strategies for solving a mathematical problem? How are these strategies related?

Danna is 22 years old, and Evan is 15 years old.

• When is writing an equation helpful in solving a problem?

Achievement Descriptors Math1.Mod1.AD4 Create equations and inequalities in one variable

and use them to solve problems. (A.CED.A.1) Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3)

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Printing Presses

• None

• Sharing Work Samples

Lesson Preparation

Land 10 min

• None

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Fluency Write Algebraic Expressions Students write algebraic expressions to prepare for writing equations in one variable. Directions: Write an algebraic expression to represent the verbal description.

A number x increased by 5

x+5

Two less than a number n

n−2

Ten times a number y

10y

Four less than 3 times a number c

3c − 4

Six times the sum of a number p and 7

6(p + 7)

One less than two times the quantity 5 less than a number z

2(z − 5) − 1

EUREKA MATH2

Launch

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Students recognize the historical significance of the printing press by discussing a photograph and related written information. Display the photograph of the Gutenberg Press taken in the Gutenberg Museum in Mainz, Germany.

Allow 2 minutes for students to study the photograph and to read the related information.

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

EUREKA MATH2

Read the following excerpt. At the start of the 15th century, Johannes Gutenberg developed the printing press. Although Chinese monks had developed a printing technique around 600 CE, the technology was not used widely until the Gutenberg printing press was introduced. It could print much faster than other presses, which meant that publishers could produce more books and pamphlets for a lower cost. Thanks to Gutenberg’s printing press, more people could now afford books. New ideas in science and mathematics spread quickly. Invite students to notice and wonder. Select a few volunteers to share what they notice and wonder, and facilitate a discussion of the historical significance of the Gutenberg printing press. If students do not wonder about how the use of the Gutenberg printing press impacted the number of books that could be printed in a specific amount of time, guide students to think about this. I wonder how the use of the Gutenberg printing press impacted the number of books that could be printed each year. I also wonder how much faster modern printing presses print compared to the Gutenberg printing press. Conclude the discussion by transitioning to the Printing Presses problem. Today, we will solve a problem related to the use of modern printing presses.

EUREKA MATH2

Learn Printing Presses Students select a strategy to solve a problem in context. Direct students’ attention to the Printing Presses problem. Consider using the Three Reads routine to provide students with a structure for deconstructing this problem. Complete a first read of the problem. In your groups, answer the question, What is the problem about? Complete a second read of the problem. In your groups, answer the question, What is the important information? Complete a third read of the problem. In your groups, answer the question, What is the question?

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

UDL: Action & Expression Consider supporting students in organizing information. To accompany the Three Reads routine, have students organize their page into four sections: three small sections and one large section. The labels of the smaller sections should correspond to the questions from the Three Reads strategy. The larger section is reserved for students to solve the problem. What is the problem about?

What is the important information?

What is the question?

Workspace:

A print shop needed to print the same number of copies of a novel as copies of a cookbook. • The novel has twice as many pages as the cookbook. • On day 1, all of the printing presses made copies of the novel. • On day 2, the printing presses were split into two groups of equal size. ▸ The first group continued to print copies of the novel. The last novel was finished at the end of the day. ▸ The second group printed copies of the cookbook. There were still cookbooks left to print at the end of the day. • On day 3, one printing press continued to print copies of the cookbook. • On day 4, one printing press continued to print copies of the cookbook and finished at the end of the day. • All of the printing presses printed pages at the same constant rate. How many printing presses are at the print shop? There are 8 printing presses at the print shop. (Note: Sample student solution paths are shown later in the lesson.) © Great Minds PBC

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Have students solve the problem in their groups by using a solution path of their choice. Circulate as students work, and intervene only if a group needs support starting the problem. Refocus groups without leading them to a specific solution path. Expect students to employ strategies such as creating tape diagrams, guess and check, or writing equations. As students reason abstractly and quantitatively, they may make assumptions about a specific number of pages in one of the books or a constant rate at which the presses print. Allow this to happen naturally. This is part of the guess, check, generalize, and verify technique that allows students to build toward the creation of a one-variable equation. If students need support making sense of the problem, consider asking one of the following questions and modeling a think-aloud.

Promoting the Standards for Mathematical Practice

The number of pages in each cookbook, the number of pages in each novel, the rate at which each printing press prints, and the total number of cookbooks and novels that must be printed.

Students demonstrate perseverance and sense-making (MP1) when they make sense of the printing presses problem, look for entry points to its solution, and evaluate their own progress, changing course as necessary.

What assumptions can we make?

Ask the following questions to promote MP1:

Because the printing presses all print at the same constant rate, we can simplify the calculations by assuming each printing press completes exactly one copy of the novel per day or exactly two copies of the cookbook per day.

• What information or facts do you not have that you need to find the number of printing presses at the print shop?

What information remains constant?

How are the numbers of novels printed and cookbooks printed related? There are the same number of novels printed as cookbooks printed. What other quantities are relevant in the problem? The number of printing presses and the number of days they each printed are also relevant quantities. If groups are attempting to use guess and check, recommend organizing their work to see the flow of their calculations. This will help them generalize their calculations when writing an equation later in the lesson. Consider posing one the following questions: • What number should you start with? What does this number represent? • How can you verify that you have guessed the correct number of printing presses?

100

• How can you use the fact that the printing presses all print at the same constant rate to simplify the problem? • Does your final answer make sense? Why?

Differentiation: Support If groups need additional prompting, suggest that they assume a specific quantity for the rate at which a press prints. They can generalize after making reasonable calculations.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Sharing Work Samples Students make connections between various representations and solution paths by sharing work samples. Ask volunteers to share their group’s strategy for solving this problem. Encourage students to take notes on the various strategies presented. As groups present, display a list of strategies used, and indicate when a strategy is repeated. Record your notes and comments about work samples from other students. Consider having groups that used models such as tape diagrams and numerical methods share first, followed by groups that wrote and solved an equation in one variable. Discuss other solution strategies, including those that led to incomplete or incorrect solutions. Encourage students to explain their thinking rather than retelling the steps. Call attention to groups that defined and used variables, solved the problem by using different representations and made connections between those representations, or showed detailed and accurate work. As groups share their solution paths, use the Five Framing Questions routine to invite students to analyze work samples. Ask a question, and then direct students to discuss in groups before they share responses with the class.

Notice and Wonder What do you notice about this work? From your observations, what do you wonder?

Organize What steps did this group take? How do you know? Use any of the following questions to advance the discussion.

Language Support If students require a high level of support, direct them to use the Ask for Reasoning section of the Talking Tool to support their communication with their groups.

What could this group do next? How is your group’s method similar to this group’s method? How is it different? What connections do you see between this solution path and others that were shared? © Great Minds PBC

101

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Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Which strategy do you like the most? Why? What are some questions you still have about this process? This problem can be solved in a variety of ways. Two possible solution paths are shown. This solution path uses a tape diagram.

Teacher Note

1 unit (Number of printing presses in the shop) Day 1

Novels

Day 2

Cookbooks 1 unit 2

In Eureka Math2, students begin using tape diagrams in A Story of Units to model word problems involving all four operations with concrete numbers. In A Story of Ratios, students discover that tape diagrams are useful tools for solving equations and making sense of application problems.

1 unit 2

Day 3 ? unit Day 4 ? unit All of the presses printed novels on day 1. We labeled the tape that represents the presses printing novels on day 1 as 1 unit. This unit represents the number of printing presses in the shop. On day 2, _1 of the presses printed novels and _1 of the presses 2

printed cookbooks, so we labeled each section of the tape as _1 unit. These two sections 2

represent all of the printing presses. On day 3 and day 4, only 1 press is printing to finish the cookbooks. Because we don’t know what fraction of the presses this 1 press represents, we don’t know how long the tapes should be. So we just drew shorter tapes. 102

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

The tapes representing the number of presses printing cookbooks on day 3 and day 4 must be the same size.

1_1 units in the tapes for days 1 and 2 represent the number of presses that printed all 2

the pages for the novel. We don’t know the number of presses that printed all the pages for the cookbooks. We

know the cookbooks have _1 as many pages as the novels. Because the presses print both 2

books at the same constant rate, it would take _1 of the printing presses used for novels 2

to print the cookbooks. So we need to multiply _1 by the 1_1 units representing the presses that printed the novels.

3 1 3 ⋅ units = units 2 2 4

_ _

So _3 units represent the number of presses that printed all the cookbook pages. 4

We want to determine the fraction of the presses represented by the 1 press used on days 3 and 4. To find that fraction, we can subtract the number of units representing the presses used on day 2 from _3 and then divide the result by 2. On day 2, _1 unit 2 4 represents the number of printing presses that printed cookbooks.

_3 − _1 = _1 4

Together, the number of presses used on days 3 and 4 represent _1 unit. 4

1 ÷2=1⋅1=1 4 4 2 8

_ _ _

The tapes for day 3 and day 4 should each be labeled as _1 unit. We know that _1 unit represents 1 printing press. 8 8 _ Therefore, units represent 8 printing presses.

Because 1 unit represents the total number of printing presses in the print shop, there are 8 printing presses in the print shop.

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

This solution path follows a guess-and-check approach.

On day 2, _1 of the presses print cookbooks, and _1 of the presses print novels. We need 2

to pick an even number for the total number of printing presses to avoid fractions when divided by 2. We picked 10 because it is an even number and calculations would be manageable. Because the presses all print at the same rate, we assumed each press can print 1 novel per day or 2 cookbooks per day. Number of Printing Presses

Number of Novels

Number of Cookbooks

10(1) + 5(1), or 15

5(2) + 1(2) + 1(2), or 14

Numbers of each book are not equal.

12(1) + 6(1), or 18

6(2) + 1(2) + 1(2), or 16

Numbers of each book are not equal.

8(1) + 4(1), or 12

4(2) + 1(2) + 1(2), or 12

Numbers of each book are equal.

When we tried 10 printing presses, we calculated that 15 novels were printed, but only 14 cookbooks were printed. So we tried 12 printing presses. When we tried 12 printing presses, we calculated that 18 novels and 16 cookbooks were printed. This did not produce the same number of copies of each book either. The number of copies printed were actually not as close as our last guess. Next, we tried fewer printing presses. We tried 8. When we tried 8 printing presses, we calculated 12 novels and 12 cookbooks. The same number of copies of each book were printed. There must be 8 printing presses. Display sample student work that resembles the guess-and-check approach, or use the sample work provided. Work with the class to generalize by writing an equation. Using the following questions, advance the discussion to focus on writing an equation, and encourage student thinking that makes connections to the other methods. 104

Teacher Note If student groups used an algebraic approach to solve the problem, display this work along with work showing an arithmetic strategy. Point out parallels between the work samples while guiding the class in representing the situation with an equation.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

The table shows a solution path that uses a guess-and-check approach. Number of Printing Presses

Number of Novels

Number of Cookbooks

10(1) + 5(1), or 15

5(2) + 1(2) + 1(2), or 14

12(1) + 6(1), or 18

6(2) + 1(2) + 1(2), or 16

8(1) + 4(1), or 12

4(2) + 1(2) + 1(2), or 12

When using guess and check, how do we know that a value we picked did not work? The number of copies of novels printed was not the same as the number of copies of cookbooks printed. Why was the next number chosen [smaller/larger], and not [larger/smaller]? The next number chosen was smaller because, when we used a larger number of printing presses, the difference between the number of novels printed and the number of cookbooks printed was larger than the original guess. We want the difference between the two numbers of copies of each book to be smaller, not larger.

Reveal Let’s focus on writing an equation. Where can we generalize our findings in this work? We can look at the work for the guesses to see which parts stayed the same. Which parts stayed the same? When calculating the number of novels printed, the 1’s stayed the same because we assumed that each press printed 1 novel per day. When calculating the number of cookbooks printed, the 2’s stayed the same. If a press can print 1 novel, it can print 2 cookbooks in the same amount of time because the cookbook has half the number of pages as the novel. The 1 press that was used on day 3 and day 4 also stayed the same.

105

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Can we replace specific values from the table with more general variables? What variables could we define? Yes. We could define p as the total number of printing presses in the shop. If we let p represent the total number of printing presses in the shop, how could we represent the total number of novels printed? Explain. Assuming that each printing press prints 1 novel or 2 cookbooks each day, we could represent the total number of novels printed as p + _1 p. In each row, the number of novels 2 printed on day 1 is equal to the number of printing presses, and the number of novels printed on day 2 is equal to half the number of printing presses. How could we represent the total number of cookbooks printed? Explain. We could represent the total number of cookbooks printed as p + 4. In each row, the number of cookbooks printed on day 2 is equal to the number of printing presses, the number of cookbooks printed on day 3 is 2, and the number printed on day 4 is 2. Have students think–pair–share about the following prompt. Write and solve an equation to represent this situation. Be ready to explain your reasoning. Because the print shop must print the same number of copies of each book, the expression representing the total number of novels printed must equal the expression representing the total number of cookbooks printed.

p + _1 p = p + 4 2 1 p=4 2

p=8 Solving the equation confirmed the answer we found through guess and check, but we assumed that the printing rates were 1 novel per day or 2 cookbooks per day. If we had chosen different values for the rates, would our result change?

106

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Allow students time to make a conjecture in their groups about whether choosing different rates would change the answer. Then have them test the conjecture by selecting different rates, creating a new equation, and solving the equation. Consider assigning each group a different rate to test. For example, What is the solution if a press can print 2 novels per day or 4 cookbooks per day?

p(2) + _1 p(2) = _1 p(4) + 4 + 4 2

2p + p = 2p + 8 p=8 Distill What effect does changing the rates have on this work? Why? Changing the rates has no effect. All the rates we chose resulted in the solution 8.

Differentiation: Challenge If students are ready for an additional challenge, have them generalize further by using a rate of y novels per day to solve the problem.

1 py = py + 2y + 2y py + __ 2

1 p = y(p + 4) y p + __

(

2 )

Because y is a factor in both the left and right sides of the equation, divide both sides by y.

1p = p + 4 p + __ 2

__3 p = p + 4 2 __1 p = 4 2

p=8

Further challenge students to consider why it is allowable to consider dividing both sides by y. Highlight that this can be done only because, in this context, y represents the rate of novels per day that a press can print, and therefore y cannot equal 0.

107

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Land Debrief 5 min Objective: Investigate a problem that can be solved by reasoning quantitatively or algebraically. Facilitate a brief class discussion by asking the following questions. What are some strategies for solving a mathematical problem? We can use an arithmetic approach by either guessing and testing values for an unknown quantity or by working backward to find the unknown value. We can draw a picture like a tape diagram to represent what is known and what is unknown. We can use an algebraic approach by writing and solving an equation. How are these strategies related? The algebraic approach generalizes the arithmetic approach because a variable is assigned to represent the unknown quantity.

Know When is writing an equation helpful in solving a problem? Using equations to solve problems may be more efficient than guess and check or tape diagrams if the solutions are not numbers that are easy to guess or figure out, if we need to make repeated calculations, or if the relationships between quantities are complicated.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

108

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

RECAP Name

Date

Printing Presses

Solve Algebraically Let x represent the number of student tickets sold. Then 2x + 5 represents the number of adult tickets sold. The sum of the number of student tickets sold and the number of adult tickets sold is the total number of tickets sold.

In this lesson, we •

solved a problem by reasoning quantitatively, by using a variety of strategies such as guess and check or creating tape diagrams.

•

solved a problem algebraically by creating an equation in one variable.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

x + (2x + 5) = 98 3x + 5 = 98 3x = 93

Example

x = 31

The number of adult tickets sold at a basketball game is 5 more than 2 times the number of student tickets sold. A total of 98 tickets are sold. How many adult tickets are sold?

There are 31 student tickets sold.

Solve with a Tape Diagram

Substitute 31 for x in 2x + 5 to find the number of adult tickets sold.

This section is 1 unit and represents the number of student tickets sold.

2(31) + 5 = 62 + 5 = 67

98 is the total Number of Student Tickets Sold

number of tickets sold.

Number of Adult Tickets Sold

These sections are 2 units and represent 2 times the number

This section represents

There are 67 adult tickets sold.

5 tickets.

of student tickets sold.

98 − 5 = 93 3 units = 93 1 unit = 31 There are 31 student tickets sold.

98 − 31 = 67 There are 67 adult tickets sold. © Great Minds PBC

RECAP

109

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Math 1 ▸ M1 ▸ TB ▸ Lesson 5

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

PRACTICE Name

Date

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

EUREKA MATH2

3. A printing company will print equal numbers of textbooks and novels to complete an order. a. A textbook has 3 times as many pages as a novel. Write expressions to represent the number of pages in each type of book.

1. The larger of two numbers is 16 more than 3 times the smaller number. The sum of the two numbers is 62.

Let n represent the number of pages in the novel. Then the number of pages in the textbook can be represented by 3n.

Nina used a tape diagram to find the two numbers. Find the error in Nina’s work and correct it. Then find the numbers. Smaller Number

Larger Number

16 62

b. Write an expression for the total number of pages if the order is for 2560 copies of each type of book.

The error is that Nina marked the larger number as 62 in her diagram instead of marking the sum as 62.

An expression that represents the total number of pages is 2560(4n).

The two numbers are 50 _ and 11 _ . 1 2

1 2

2. Mason read _ of his book on Monday. On Tuesday, he read half of the remaining pages. 1 3

On Wednesday, he read 54 pages to finish the book. What is the total number of pages in Mason’s book? Use a tape diagram to solve.

Total number of pages

c. The printing company needs to print a total of 1,925,120 pages to complete the order. Use your expression from part (b) to write and solve an equation to determine the number of pages in each type of book.

Number of Number of pages read pages read on Monday on Tuesday

There are 188 pages in a novel and 564 pages in a textbook.

3(54) = 162 The total number of pages in Mason’s book is 162.

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P R ACT I C E

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 5

For problems 4–7, use any method to solve.

Remember

4. Levi is 4 years older than twice his younger brother’s age. The sum of their ages is 25. How old is each boy?

For problems 8 and 9, solve the equation. 2 8. _ + x = 4 5

Levi is 18 years old, and his brother is 7 years old.

18 5

5. Three times the difference of a number and 5 is equal to the sum of the number and 12. What is the number?

_x = − 2.1 3

−6.3

10. Show that 7 + 2m + 6m(m + 3) and 6m2 + 20m + 7 are equivalent expressions. Write the property or operation used in each step.

The number is 13 _ . 1 2

7 + 2m + 6m(m + 3) = 7 + 2m + 6m2 + 18m

6. Danna, Evan, and Mason combine their money to buy their friend a present. Evan spends $5.50 more than Danna. Danna spends $3.00 less than Mason. They spend a total of $25. How much money does each person spend?

Distributive property

= 7 + 2m + 18m + 6m2

Commutative property of addition

= 7 + 20m + 6m2

Addition of like terms

= 6m2 + 20m + 7

Commutative property of addition

Mason spends $8.50. Danna spends $5.50. Evan spends $11.00.

For problems 11 and 12, solve the inequality. 11. x − 5 > −13

x > −8

7. Ji-won has $68, and Emma has $18. How much money will Ji-won have to give Emma so that Emma will have 4 times as much money as Ji-won?

12. 2 ≤ x + 7

−5 ≤ x

Ji-won will have to give Emma $50.80.

P R ACT I C E

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Teacher Edition: Math 1, Module 1, Topic B, Lesson 6 LESSON 6

Solution Sets of Equations and Inequalities in One Variable Find values to assign to the variables in equations or inequalities that make the statements true. Describe a solution set in set notation, in words, and on a graph.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Name

EXIT TICKET

Date

Solve each equation or inequality. Write the solution set, and graph the solution set on a number line. 1. 8 + 2p = 2(p + 4)

8 + 2p = 2(p + 4) 8 + 2p = 2p + 8 8 + 2p = 8 + 2p The solution set is ℝ.

p –10

–8

–6

–4

–2

In this lesson, students formalize their understanding of solution sets of equations and inequalities. Students test possible solutions by substituting values and determining the truth-value of resulting number sentences. This lesson emphasizes the meaning of solution sets rather than the process of solving equations or inequalities, which is covered in subsequent lessons. Students complete a graphic organizer to record various representations of solution sets by using set notation, words, and graphs. This lesson introduces the terms elements, empty set, and set-builder notation and the symbol for the set of real numbers, ℝ.

Key Questions • What are characteristics of real numbers that are in the solution set of an equation or inequality?

2. 4 + x ≤ 7

• What is the difference between using set notation for finite solution sets and using set notation for infinite solution sets?

4+x≤7

x≤3

The solution set is {x | x ≤ 3}.

Achievement Descriptors Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3)

x –10

Lesson at a Glance

–8

–6

–4

–2

Math1.Mod1.AD11 Solve linear inequalities in one variable. (A.REI.B.3)

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Agenda

Materials

Fluency

Teacher

Launch 5 min

• Chart paper

Learn 30 min

• Marker

• Make It True

Students

• Never True and Always True

• None

• Inequality

Lesson Preparation

Land 10 min

• Use the chart paper to create an anchor chart with the same table as the Solution Set graphic organizer.

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Fluency Substitute Values to Determine Truth-Values Students substitute a value for a variable and determine whether the resulting number sentence is true or false to prepare for describing solution sets of equations and inequalities. Directions: Determine whether the equation or inequality is true or false for the given value of x. 1.

2x + 7 = −1 when x = −4

x > −2 when x = −5

5x − 5 = 5 − 5x when x = 0

3≥x+5

when x = −2 5.

x2 = 9 when x = −3

2(2x − 6) = x − 3 when x = 3

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True

False

True

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Launch

Students apply the concept of truth-values to number sentences. Direct students’ attention to problems 1–10. Give students a minute to work independently to answer as many problems as they can. Encourage students to use mental math. For problems 1–10, determine whether the sentence is true or false. 1. I am in math class.

True

2. The president of France is also the president of Mexico.

False

3. 4 + 1 = 3 + 2 True 4. 3 + 2 = 4 − 9 False 5. 3 + 2 > 4 − 9 True

4 1 5 6. _ + _ = _ 3 3 3 1 1 1 7. _ + _ = _ 2 3 5

True False

8. (7 + 9)2 = 162

True

9. 32 + 42 = 72

False

10. 32 × 42 = 122

True

Facilitate a brief discussion with students about why each number sentence is true or false, assigning each statement a truth-value by asking the following question. Can a number sentence be both true and false? Why? No. A number sentence is a statement of equality or inequality between two numerical expressions. That statement is true or false, but it cannot be both. Today, we will explore solution sets of equations and inequalities and how the solution sets are represented.

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Learn Make It True Students interpret solution sets and represent them by using set notation and graphs on a number line. Direct students to problem 4 and ask the following question. Why is the number sentence 3 + 2 = 4 − 9 false? The expression on the left side of the equal sign is equivalent to 5. The expression on the right side of the equal sign is equivalent to −5. Those numbers are not equal, so the number sentence is false. Display problem 11 and ask the following question.

Language Support This lesson uses terminology covered in previous grades. Consider reviewing each of these terms with students. Ask students about the similarities between two of the terms, such as number sentence and equation, and the differences between two other terms, such as expression and equation.

3(2) + 4 ≤ 10

Consider the equation 3 + x = 4 − 9. How is this equation similar to the number sentence in problem 4? How is it different?

Number sentence: 3(2) + 4 = 10 or

The equation and the number sentence have the same expression on the right side. The expressions on the left sides are both the sum of 3 and another value. The value in the number sentence is 2, and the value in the equation is x.

Expression: 3(2) + 4 or 3n + 4

Direct students to problem 11. Have them record their calculations when substituting 0 or 1 for x in the middle column of the table. Have students record whether the resulting number sentence is true or false in the last column. When they finish, ask the following questions.

Equation: 3(2) + 4 = 10 or 3n + 4 = 10

Inequality: 3(2) + 4 ≤ 10 or 3n + 4 ≤ 10

Is the equation true when x = 0? Why? No. The expression on the left side of the equal sign is equivalent to 3 when x = 0. The expression on the right side is equivalent to −5. The expressions do not evaluate to the same number.

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Is the equation true when x = 1? Why? No. The expression on the left side of the equal sign is equivalent to 4 when x = 1. The expression on the right side is still equivalent to −5. The expressions do not evaluate to the same number. Have students think–pair–share about the following question. Is there a value of x that makes this equation true? How do you know? Yes. When x = −8, the equation is true because the expression on each side of the equation is equal to −5. Have students record −8 as a value of x in the third row of the table and complete the rest of the row if they have not done so already. Because −8 makes the equation true, we can say that −8 is a solution to the equation. The collection of all the solutions to an equation or inequality is called the solution set. Before we can say that −8 is the only solution, let’s try substituting a few more values for x. Choose a value and write it in the last row of the table. Determine whether your chosen value is a solution to the equation. Encourage students to test negative numbers, fractions, or irrational numbers in the equation.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

For problems 11–15, complete the table by using the provided equation or inequality to determine whether the number sentence is true or false for each value of x. 11. 3 + x = 4 − 9 Value of x

Number Sentence

3+0=4−9

3 = −5 3+1=4−9

4 = −5 3 + (−8) = 4 − 9

−8

−5 = −5

1 Sample: 3

3 + 1_ = 4 − 9 3

3 _1 = −5

True or False False

False

True

False

When students finish, have them reveal their answer for the tested value. Then ask the following question. A number that makes the equation a true statement is a solution to the equation. If the equation is false, what does that tell us about the value of x ? A value of x that makes an equation false is not a solution to the equation. Pair students and assign each pair one problem from problems 12–15. Have pairs complete the table for the given values of the variable for their assigned problem. Students will create the last row of the tables in problems 13 and 14 later in the lesson.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

12. a2 = 25 Value of a

−5

Number Sentence

(−5)2 = 25 25 = 25 02 = 25 0 = 25 52 = 25 25 = 25 82 = 25 64 = 25

True or False True

False

True

False

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

13. 3g = 2(g + 1) + g Value of g

Number Sentence

True or False

3(−5) = 2(−5 + 1) + (−5) −15 = 2(−4) − 5

−5

False

−15 = −8 − 5 −15 = −13 3(0) = 2(0 + 1) + 0

0 = 2(1) + 0

False

0=2 3(5) = 2(5 + 1) + 5 15 = 2(6) + 5

False

15 = 12 + 5 15 = 17 3(8) = 2(8 + 1) + 8 24 = 2(9) + 8

False

24 = 18 + 8 24 = 26 3(1_) = 2(1_ + 1) + 1_ 3

1 Sample: 3

1 = 2(4_) + 1_ 3

8 1 = _ + 1_ 3

False

1=3

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

14. 2(x − 6) = −12 + 2x Value of x

Number Sentence

True or False

2(−5 − 6) = −12 + 2(−5) −5

2(−11) = −12 − 10

True

−22 = −22

2(0 − 6) = −12 + 2(0) 0

2(−6) = −12 + 0

True

−12 = −12

2(5 − 6) = −12 + 2(5) 5

2(−1) = −12 + 10

True

−2 = −2

2(8 − 6) = −12 + 2(8) 8

2(2) = −12 + 16

True

4=4

1 Sample: _ 2

2(1_ − 6) = −12 + 2(1_) 2

__ = −12 + 1 2(− 11 2)

True

−11 = −11

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

15. x + 1 ≥ 6

Value of x

Number Sentence

True or False

−5

(−5) + 1 ≥ 6

0+1≥6

False

5+1≥6

True

8+1≥6

True

−4 ≥ 6

1≥6 6≥6 9≥6

False

When students finish, choose a pair for each problem and invite them to display their table. Ask students to record True or False in the last column of the tables for the problems they were not assigned. Lead a class discussion by asking the following questions. What do you notice about the number of true statements in each table? What do you wonder? I notice that all the number sentences in problem 13 are false, and all the number sentences in problem 14 are true. I notice that problems 12 and 15 have two true number sentences. I wonder why there are some problems with more than one true number sentence and others with no true number sentences. Direct students back to problem 11.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Before we resolve any of our wonderings, let us look back at the first equation, 3 + x = 4 − 9. We determined −8 is a solution. Do you think we can assign any other values to x to make the equation true? Why? No. I couldn’t find any other values of x that make this equation true. Is it possible to check whether every real number is a solution to an equation by substituting in values for the variable? Why? No. There are infinitely many real numbers and it would take an infinite amount of time. To solve an equation means to find its solution set. Have students solve the equation 3 + x = 4 − 9 with their partner if they have not done so already. They may include their work below the table in problem 11.

Teacher Note

3+x=4−9 3 + x = −5 x = −8 The solution is −8. Once students finish, ask them to share how they solved the equation. Then use the following prompts to introduce set notation. So our solution set is the set of −8. A set is essentially a collection of things. In this case, the things are numbers that are solutions to the equation. Display the set {−8}. This notation is called set notation. The curly braces indicate that we are denoting a set. Each item in the set is called an element. In this example, the only element in our set is −8. No other elements belong in this particular set because no other numbers are solutions to 3 + x = 4 − 9. How might we graph the solution set of this one-variable equation? We can draw a number line and plot a point at −8.

Sets are denoted in various ways in mathematical literature. Choosing a notation is simply a matter of convention. The conventions used in Eureka Math2 will be specifically defined within a lesson. Sample notations follow. • Empty set Eureka Math2

{}

Other notations

• Set of all real numbers Eureka Math2 Other notations

ℝ

{x | x ∈ ℝ}

While all of these are acceptable, Eureka Math2 elects to use the conventions outlined within the body of the lesson.

When graphing the solution set of a one-variable equation, draw a number line and label the number line with the variable, in this case x, on the right. For every solution in the solution set, plot a point on the number line that corresponds to that solution.

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Direct students to record the solution set {−8} under their work in problem 11. Have students annotate that the number listed inside the curly braces is called an element. Then have them include the graph of the solution set.

element

{−8} x –10

–8

–6

–4

–2

Do you think all solution sets will contain only one element? Explain. No. Problems 12, 14, and 15 have more than one value that makes each of their statements true. Direct students to problem 12. Which given values are solutions to a2 = 25? How do you know?

−5 and 5 are solutions to the equation because both values make the equation true. Do you think we can assign any other values to a to make the equation true? Why? No. −5 and 5 are the only two values that have a square of 25. With your partner, write and graph the solution set for a2 = 25. Have students include their solution set and graph below the table in problem 12.

{−5, 5}

a –10

–8

–6

–4

–2

Invite students to share how they represented their solution sets and graphs. Then reveal and confirm answers.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Although it is common to write the elements of a solution set in increasing order when using set notation, {5, −5} represents the same solution set. Direct students to the Solution Set graphic organizer. Model how to complete the organizer by adding only the information about finite sets shown in the green color to the anchor chart. The information shown in the other colors will be added throughout the lesson. Consider having students put a sticky note on this page in their books so they can easily access the graphic organizer during the rest of the lesson.

Solution Set Graphic Organizer

Type of Solution Set

Sample Equation or Inequality

Solution Set Set Notation

In Words

a2 = 25

{−5, 5}

The set of −5 and 5

3g = 2(g + 1) + g

{}

Empty set

2(x − 6) = −12 + 2x

ℝ

The set of all real numbers

x+1≥6

{x | x ≥ 5}

The values of x such that x is greater than or equal to 5

Finite sets

Infinite sets

Graphically a –5

5 g

–5

5 x

–5

5 x

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Never True and Always True Students use set notation to represent the solution sets of equations with no solution or infinitely many solutions and graph the solution sets on a number line. Direct students to problem 13. For problem 13, how many of the values in the table make the equation true? Are there any values that make this equation true? None of the values in the table make the equation true. There might be values that make the equation true that aren’t listed in our tables. With your partner, choose another value for x. Complete the blank row of the table to determine whether your chosen value of x is a solution to the equation. Encourage students to test negative numbers, fractions, or irrational numbers in the equation. When students are finished, invite a few pairs to each share their value of x and whether that value is a solution to the equation. So far we have not found a value of x that makes the equation true. How many solutions do you think the equation has? I think the equation has no solution. Solve the equation 3g = 2(g + 1) + g with your partner. Then write and graph the solution set. Students may include their work below the table in problem 13. Expect them to realize there are no elements to list inside the set’s curly braces and to wonder whether there is a special symbol for denoting this case.

3g = 2(g + 1) + g 3g = 2g + 2 + g 3g = 3g + 2 0=2 The equation has no solution.

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{}

g –10

–8

–6

–4

–2

Invite students to share their answers. Accept all reasonable answers and encourage students to explain their reasoning for their solution sets and graphs. Ask the following question to guide students to interpret the solution set. What does this result tell us about the solution set? Explain. It tells us that there are no elements in the solution set of this equation. When we arrive at a false statement when solving an equation, we know that there is no value that will make the equation true. When a set has no elements, we call it the empty set. Consider having students turn and talk about why it’s called the empty set. Then display the symbol { } and the graph used to represent an empty set. Have students revise their answers for problem 13 as needed. Direct students to the Solution Set graphic organizer. Model how to complete the organizer by adding only the information about the empty set shown in the blue color on the anchor chart. Then say the following statement before asking students to add the term finite set to their graphic organizer. If you can list all the solutions or if there is no solution to an equation, then the solution set is a finite set, meaning there is a limited number of elements in the set. A finite set is represented by curly braces, which enclose the list of elements. Direct students to problem 14. For problem 14, how many of the values in the table make the equation true? Are there any values that make this equation false? All the values in the table make the equation true. Maybe the values that make the equation false aren’t listed in our tables. With your partner, choose another value for x. Complete the blank row of the table to determine whether your chosen value of x is a solution to the equation. © Great Minds PBC

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Encourage students to test negative numbers, fractions, or irrational numbers in the equation. When students finish, invite a few pairs to each share their value of x and whether that value is a solution to the equation. So far, all the values we have tried have made the equation true. How many solutions do you think the equation has? I think the equation has infinitely many solutions. With your partner, solve the equation 2(x − 6) = −12 + 2x. Then write and graph the solution set. Students may include their work below the table in problem 14. Expect them to realize that the set has infinitely many elements to list inside the curly braces and to wonder whether there is a special symbol for denoting this case.

To help students develop understanding of the concept of set notation, consider comparing a set to a familiar object by using a suitcase analogy. Explain that a set is a collection of numbers called elements. A suitcase is a case used to carry the personal objects of a person traveling. A set is like a suitcase. Each object in the suitcase is called an element. The empty set is like an empty suitcase.

The symbol ℝ represents the set of all real numbers, so all the numbers students currently know are in that suitcase. The incorrect notation {ℝ} would indicate a suitcase containing another suitcase that contains all the real numbers.

2(x − 6) = −12 + 2x 2x − 12 = −12 + 2x 2x = 2x The equation has infinitely many solutions.

UDL: Representation

ℝ

x –10

–8

–6

–4

–2

Invite students to share their answers. Accept all reasonable answers and encourage students to explain their reasoning for their solution sets and graphs. Guide students to interpret the solution set by using the following prompts. The solution set of the equation 2(x − 6) = −12 + 2x is the set of all real numbers. We cannot possibly list all the solutions inside the curly braces. We denote the set of all real numbers in set notation as a symbol.

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EUREKA MATH2

Display the symbol for the real numbers, ℝ. Then draw the symbol by hand: This symbol represents the set of all real numbers. We do not write {ℝ} because that would mean “the set of the set of all real numbers,” which is not the same as ℝ. Can we represent infinitely many solutions on a number line? Explain your reasoning. No. We cannot plot infinitely many points on a number line. Yes. We can use a line to represent infinitely many solutions. When graphing the set of all real numbers, we indicate that all points are included in the solution set of the equation by drawing a solid line. We also shade the arrows to indicate that the solution set extends infinitely to the right and to the left. Display the graph of the set of all real numbers. Have students revise their answers for problem 14 as needed.

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Promoting the Standards for Mathematical Practice Students attend to precision (MP6) when they understand and use symbols to communicate solution sets. Ask the following questions to promote MP6: • How can we express the empty set by using the new symbols we have learned? • Where is it easy to make mistakes when finding the solution set of an equation or inequality?

Direct students to the graphic organizer. Model how to complete the organizer by adding only the information about the set of real numbers shown in the purple color to the anchor chart.

Inequality Students use set notation to represent the solution to an inequality and graph the solution set on a number line. Direct students to problem 15. How many of the values in the table make the inequality true? Two of the values in the table made the inequality true. How do we know whether these values are the only solutions to the inequality? We can solve the inequality to see if there are other solutions. When we solve an inequality, we find the solution set, and this set contains all the values that are solutions to the inequality. With your partner, solve the inequality x + 1 ≥ 6. Then write and graph the solution set.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Students may include their work below the table in problem 15. Expect them to write x ≥ 5 as the solution set.

x+1≥6

The solution set of the inequality is x ≥ 5.

x≥5

{x | x ≥ 5}

–10

–8

–6

–4

–2

x 2

Invite students to share their answers. Encourage students to explain their reasoning for their solution sets and graphs. Introduce set-builder notation by asking the following questions. What values of x make this inequality a true statement? All numbers greater than or equal to 5 make this inequality a true statement. Does this inequality have a finite number of solutions or infinitely many solutions? Why? This inequality has infinitely many solutions because there are infinitely many values that are greater than or equal to 5. An infinite set is not a finite set, so we cannot list all the elements. Because we cannot list every element, we cannot represent an infinite set using the same notation we use to represent a finite set. Since we cannot list the elements, we describe them.

Differentiation: Support If students do not recall that many inequalities have infinitely many solutions, ask them to generate a list of values that make the inequality true by expanding the table from problem 15 to include more values. Record and display this expanding list of values. Elicit a few rational values that lie between a pair of consecutive integers to help students conceptualize the infinitely many solutions to the inequality.

Write the solution set for the class by using set notation while narrating with the appropriate words. Annotate what each part of the notation means while explaining the following information.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

x x >5

x is greater than the set of all x such that or equal to 5 When writing solution sets for inequalities, we traditionally use set-builder notation. We can build the set together by describing what it is. We are saying that the solution set is the set of all values of x such that x ≥ 5.

We denote a set by using the curly braces. We identify our variable, in this case x, inside the opening curly brace. Then we draw a vertical bar. In this notation, this vertical bar is typically read as so that or such that. Now we describe our constraints by using symbols, in this case x ≥ 5. We read this as the set of all values of x such that x ≥ 5.

Teacher Note If students ask why they cannot simply write the solution set as {x ≥ 5}, remind them that the elements of a solution set are solutions. As written, this set has a single element, the inequality x ≥ 5. The inequality x ≥ 5 is not a solution. It represents the constraints on the solution set, hence the need to write x | in the solution set. Writing {x | x ≥ 5} provides a method for building the solution set of all values of x that satisfy the constraint that x is greater than or equal to 5. As they explore new concepts, students need to identify the elements of the set before defining the constraints on the set.

Earlier, we graphed an infinite set of solutions, more specifically the set of all real numbers. How is this solution set different from the set of all real numbers? There are infinitely many solutions, but not all real numbers are solutions to the inequality. What is the least possible value of x that makes this inequality true? The least possible value of x that makes this inequality true is 5. How can we identify the least possible value in the solution set on the graph? The least possible value is an endpoint on the graph of the solution set.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

How do we mark the endpoint for an inequality with the greater than or equal to symbol when graphing on a number line?

Teacher Note

We mark the endpoint with a closed circle. Are the numbers to the right or to the left of the endpoint in the solution set of the inequality? How do you know? The numbers to the right of 5 are in the solution set because the numbers greater than or equal to 5 make the inequality true.

Students use the term endpoint in grades 6 and 7 to describe the circle used when graphing inequalities.

Have students record the set-builder notation for the inequality in problem 15 in their books. Then direct students to the graphic organizer. Model how to complete the organizer by adding only the information about infinite sets shown in the red color to the anchor chart. Have students complete problem 16 with their partner. 16. Solve the inequality. Write the solution set by using set-builder notation. Then graph the solution set on the number line.

b − 6 < −3 b − 6 < −3 b<3

Solution set: {b | b < 3}

–10

–8

–6

b –4

–2

Discuss the answers as a class. Ensure that students write the solution set by using set-builder notation and that they label the graph properly. Debrief problems 15 and 16 by asking the following questions. What is similar about the graphs of the solution sets in problems 15 and 16? What is different? The graphs are both rays. The endpoint in problem 15 is a closed circle, and the endpoint in problem 16 is an open circle.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Why do we use an open circle for the endpoint on the graph in problem 16? We use an open circle because the endpoint’s value is not a solution to the inequality. Inequalities such as b − 6 < −3 in problem 16 are often called strict inequalities because they do not allow the two expressions to be equal to each other. Inequalities such as x + 1 ≥ 6 in problem 15 are often called not strict because the two expressions can be equal to each other. The solution set of a strict inequality does not include the value of the endpoint, but the solution set of an inequality that is not strict does include the value of the endpoint. How can you determine by looking at an inequality if it is strict or not strict? A strict inequality uses the less than symbol or the greater than symbol. An inequality that is not strict uses the greater than or equal to symbol or the less than or equal to symbol.

Land Debrief 5 min Objectives: Find values to assign to the variables in equations or inequalities that make the statements true. Describe a solution set in set notation, in words, and on a graph. Initiate a class discussion by asking the following questions. What are the characteristics of real numbers that are in the solution set of an equation or inequality? The real numbers that are in the solution set of an equation or inequality are values that make the equation or inequality a true statement.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

How do we find the solution set of a linear equation or inequality? We can solve the linear equation or inequality by isolating the variable. Have students answer each of the following questions with a thumbs-up or thumbs-down. Invite a few students to share their reasoning before asking the next question. Do equations always have a solution? No. When we arrive at a false number sentence after solving an equation, we know that there is no value that will make the equation true. Do equations always have solution sets? Yes. Some equations don’t have elements in their solution set. These equations don’t have solutions, but they still have a solution set. The solution set is the empty set. Do equations always have finite solution sets? No. Some equations have infinitely many solutions, which indicates a solution set with infinitely many elements. How is using set notation for finite solution sets different from using set notation for infinite solution sets? We use set notation to list each element in a finite solution set. We use set-builder notation for infinite solution sets except the set of real numbers because we cannot list all the individual elements. We use a shorthand symbol, ℝ, for the set of all real numbers.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

RECAP Name

Date

•

determined whether equations and inequalities are true for different values of the variable.

•

represented solution sets in set notation, in words, and on graphs.

For problems 3 and 4, solve the equation. Write the solution set, and then graph the solution set on a number line. 3. 6x − 12 = 6(x − 12)

Solution Sets of Equations and Inequalities in One Variable In this lesson, we

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

6x − 12 = 6(x − 12) 6x − 12 = 6x − 72

Terminology

The solution set of this problem is the empty set. This notation means “a set with no elements.”

−12 = −72 {}

An element of a set is an item in the set. An empty set is a set that has no elements.

Set-builder notation is a representation of a set that builds the set by describing what is in it. The set is denoted by curly braces containing a variable on the left followed by a vertical bar and constraints listed on the right of the bar.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

A graph of the empty set does not show any points or shading.

Examples 1. The graph represents the solution set of an equation or inequality. Write the solution set.

a –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

{a | a ≤ 0}

4. 6x − 12 = 6(x − 2)

6x − 12 = 6(x − 2)

9 10

The solution set of this problem contains all values of x. This notation

6x − 12 = 6x − 12

The solution set of this problem is infinite, so we use set-builder notation. This solution set means “the set of all values of a such that a ≤ 0.”

means “the set of all real numbers.”

ℝ

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

2. Solve the equation 9 + 3x = 6x. Write the solution.

9 10

9 + 3x = 6x 9 = 3x

A graph of the set of all real numbers shows shading on the entire number line, including both arrows.

3=x {3}

The solution set of this problem is finite, so we use set notation. This solution set means “the set consisting of the number 3.”

RECAP

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Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

For problems 5–8, each graph represents the solution set of an equation or inequality. Write the solution set.

For problems 1–4, each set is the solution set of an equation or inequality. Graph the solution set.

–10

1. {8}

–8

–6

–4

–2

–8

–6

–4

–2

–8

–6

–4

–2

{n | n ≥ −5}

–6

–4

–2

{−3}

–10

–8

–6

–4

–2

10 p

–10 {−6, 6}

3 2. {− _} 4

–2

–1

–10 {} 3. ℝ

–10

–8

–6

–4

–2

–3

–2

–1

–10

1 4. {x x < _} 2

–5

136

–8

9. What is the solution set of an equation? Explain in your own words.

x –4

The solution set of an equation is the set of all values of the variable that make the equation true.

P R ACT I C E

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

For problems 10–13, write a sentence that interprets the solution set of the equation or inequality. 10. 19 + x = 10; {−9}

For problems 20–23, solve the equation or inequality. Write the solution set, and then graph the solution set on the number line.

11. c 2 = 16; {−4, 4}

20. x 2 = 36

The equation c2 = 16 is only true when c = −4 or c = 4.

The equation 19 + x = 10 is only true when x = −9.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

{−6, 6} x –10

12. x − 3 = x − 5; { }

13. −6b + 4 < 10; {b | b > −1}

There is no solution to x − 3 = x − 5.

–8

–6

–4

–2

–8

–6

–4

–2

–4

–2

–4

–2

21. x − 1 ≤ 3

Only values of b greater than −1 make the inequality −6b + 4 < 10 true.

{x | x ≤ 4} –10

For problems 14–19, find the solutions to the equation. Write the solution set. 14. x − 10.5 = 12

15. 4x = 6 + 2x

{22.5}

{3} 22. 5(x − 2) = 5x − 2

{}

16.

_ __ 3 = x 4 16

{12}

17.

_ ____ 3 x+1 = 4 16

–10

–8

–6

{11}

23. 5(x − 2) = 5x − 10

ℝ 18. x 2 = −9

{}

19. 3x + 5 = 6x + 5

{0}

–10

P R ACT I C E

–8

P R ACT I C E

–6

137

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

24. The equation A = l ⋅ w gives the area of a rectangle in square units with length l units and width w units.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 6

Remember For problems 28 and 29, solve the equation.

a. Find A when l = 10 and w = 15.5.

28.

A = 155

2 = −4a 3 9 −3 2

3 1 29. a + (− _) = − _ 4 8 1_ − 8

b. Find l when A = 26 and w = 2.

l = 13

30. A theater charges $8 for an adult ticket and $6 for a senior citizen ticket. The theater sells a total of 150 tickets. Write an expression that represents the total amount of money the theater makes when n adult tickets are sold.

c. Find w when A = 9 and l = _ .

w = 18

1 2

8n + 6(150 − n)

25. Create an equation in one variable with one solution. Write the solution set. Sample: 2a + 4 = a − 8, {−12} For problems 31 and 32, solve the inequality. 31. x − 5 > −13

26. Revise one expression from the equation you created in problem 25 so the solution set of the new equation is all real numbers.

x > −8

Sample: 2a + 4 = 2(a + 2)

32. 2 ≤ 7 + x

−5 ≤ x

27. Revise one expression from the equation you created in problem 25 so that the new equation has no solution. Sample: 2a + 4 = 2a − 8

138

P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic B, Lesson 7 LESSON 7

Solving Linear Equations in One Variable Justify each step in solving a linear equation.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Name

Date

EXIT TICKET

Solve the equation 2x − 3(x − 2) = 4x + 34. Choose the appropriate reason from the answer choices to justify each step, and write the solution set by using set notation. Answer Choices Commutative property of addition

Commutative property of multiplication

Associative property of addition

Associative property of multiplication

Distributive property

Addition of like terms

Addition property of equality

Multiplication property of equality

2x − 3(x − 2) = 4x + 34 2x − 3x + 6 = 4x + 34 −x + 6 = 4x + 34 −5x + 6 = 34 −5x = 28

__ x = − 28 5

Distributive property Addition of like terms Addition property of equality

Lesson at a Glance In this lesson, students explore whether rewriting equations by using properties of arithmetic and equality results in equations that have the same solution sets as the original equations. They conclude that rewriting equations by adding the same expression to both sides of an equation preserves their solution sets, as long as the expressions can be evaluated for every number. Students also conclude that rewriting equations by multiplying both sides of the equation by the same number preserves their solution sets, as long as that number is not 0. Students apply this understanding to solve equations and justify solution paths by using properties and operations. They build flexible thinking by critiquing the reasoning of their classmates as they compare different solution paths with corresponding justifications.

Addition property of equality Multiplication property of equality

Key Questions

The solution set is {− __}. 28 5

• What does it mean to rewrite an equation in a way that preserves its solution set? • What properties can we use to rewrite an equation in a way that preserves its solution set?

Achievement Descriptors Math1.Mod1.AD9 Explain steps required to solve linear equations

in one variable and justify solution paths. (A.REI.A.1) Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3) © Great Minds PBC

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Properties of Arithmetic

• None

• Properties of Equality

Lesson Preparation

• Justifying Steps

• None

Land 10 min

141

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Fluency Solve One- and Two-Step Equations Students solve one- and two-step equations to prepare for solving multi-step equations and justifying each step in the solution path. Directions: Solve each equation for x. 1.

x + 12 = 2

−10

Teacher Note Instead of this lesson’s Fluency, consider administering the Solve One- and Two-Step Equations Sprint. Directions for administration can be found in the Fluency resource. EUREKA MATH2

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Equations

Number Correct:

Solve each equation.

−9 = 3 + x

− 4 x = −6 5 27 = 4x + 11 14 − x = 20

−12

__ 15 2

4 −6

x+2=6

23.

x+4=6

24.

2x + x = 18

x+6=6

25.

2x + x + 3 = 18 2x + x + 6 = 18

x − 3 = 12

26.

x − 5 = 12

27.

3x + 6 = 18

x − 7 = 12

28.

18 = 3(x + 2)

8x = 16

29.

4x = 16

30.

12 = −x − x − x

2x = 16

31.

−x − 3 − x − x = 12

10.

142

32.

−3x − 3 = 12

33.

12 = −3(x + 1)

_1 x = 4

12.

8 + x = −4

13.

−8 + x = −4

35.

14.

x − 8 = −4

36.

−2 = 1_ x − 1

15.

x − (−8) = 4

37.

3 x+6=9 2

16.

−5x = 10

38.

17.

10x = −5

39.

18.

2x = −6

40.

19.

−6x = 2

41.

20. 21.

382

2(x − 6) = −8

_1 x = 4

12 = −3x

11.

22.

2x = 18

34.

−1_ (x − 4) = 2 4

−1_ x + 1 = 2 4

_ _

3 (x + 4) = 9 2

9 = _ (x + 8) 3 4

−9 = _ (x + 8) 3 4

−_ (x + 8) = −9 3 4

−2_ x + 1 = 1

1 x = −6 2

42.

43.

−2_ (x − 1) = 1

44.

−1_ x = −6 −1_ x = 2

1 = −2_ x + 2_ 3

EUREKA MATH2

Launch

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Students use prior knowledge to solve linear equations and to explain their solution paths. Have students complete problems 1 and 2 individually. Then have them compare their solution sets and solution paths with those of a partner. For problems 1 and 2, identify the solution set of the equation. Check your solution and describe your solution path. 1. 5(v − 4) = 100

5(v − 4) = 100 v − 4 = 20 v = 24 Solution set:

{24}

Check:

Because 5(24 − 4) = 100 is a true number sentence, I know that 24 is a solution to 5(v − 4) = 100.

Solution path:

First, I divided both sides of the equation by 5. Then I added 4 to both sides of the equation.

2. 3z + 1.6 = 2z

3z + 1.6 = 2z 3z − 3z + 1.6 = 2z − 3z −1(1.6) = −1(−z) −1.6 = z Solution set:

{−1.6}

Check:

Because 3(−1.6) + 1.6 = 2(−1.6) is a true number sentence, I know that −1.6 is a solution to 3z + 1.6 = 2z.

Solution path:

First, I subtracted 3z from both sides of the equation. Then I multiplied both sides of the equation by −1.

143

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

EUREKA MATH2

Once students have completed the problems, reveal the solution set of each equation and have several students share their solution paths. Encourage students with different solution paths for the same problem to share. Also, invite students who used a similar solution path but described their path differently to share. For example, some students may refer to specific if–then moves, while other students may mention properties of arithmetic and equality. Debrief student thinking by asking the following questions. Some of you applied the same steps in your solution paths but described these steps differently. For example, some of you said that you subtracted 3z from both sides of the equation as your first step in solving the equation in problem 2. What property states that the left and right sides are still equal when we subtract 3z from both sides of the equation? The addition property of equality We needed to use the addition property of equality. Did we also need to use the multiplication property of equality? If so, where? Yes. We used the multiplication property of equality when we divided both sides of the equation in problem 1 by 5 and when we multiplied both sides of the equation by −1 in problem 2. Some of you used different solution paths to find the solution set of an equation. How can you know that your solution path produces the correct solution set of an equation? I can substitute the solution into the original equation to make sure that it creates a true number sentence. When a step in a solution path results in an equation that has the same solution set as the original equation, we say that the solution set is preserved. In previous courses, you used properties of arithmetic and properties of equality to solve equations. Today, we will explore these properties in more detail and use them to justify the steps we take when solving an equation.

144

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Learn Properties of Arithmetic Students confirm that using the properties of arithmetic to rewrite expressions preserves the solution set of an equation. Direct students’ attention to problem 3. What do you notice about the two equations? The right side of each equation is 100. The factors on the left sides of the equations are written in a different order. The expressions on the left sides of the equations are equivalent. Can we be sure that the expressions on the left sides of the equations are equivalent? If so, how? Yes. We can be sure that the expressions are equivalent because we can use the commutative property of multiplication to write 5(v − 4) as (v − 4)5. Because 5(v − 4) is equivalent to (v − 4)5, what can we say about the solution sets of the two equations? Explain. We can say that the solution sets of the two equations must be the same because 5(v − 4) and (v − 4)5 are equivalent expressions, and they are both set equal to 100. Have students work in pairs to complete problems 3–5.

145

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

EUREKA MATH2

For problems 3–5, without solving, write the property that justifies why the two equations must have the same solution set. 3. 5(v − 4) = 100 and (v − 4)5 = 100 Commutative property of multiplication 4. 5(v − 4) = 100 and 5v − 20 = 100 Distributive property

5. 2[5(v − 4)] = 2(100) and (2 ⋅ 5)(v − 4) = 2(100) Associative property of multiplication

Debrief student thinking by asking the following questions. We used the commutative, associative, and distributive properties in lesson 3 to demonstrate that two expressions are equivalent. Why can we use these properties to justify relationships between two equations? Since equivalent expressions evaluate to the same number for every value of the variable, replacing an expression in an equation with an equivalent expression will produce an equation that has the same solution set as the original equation. We can say that using these properties of arithmetic to rewrite an expression on one side of an equation preserves the solution set of the original equation. Besides the properties of arithmetic, what other properties were used in solving the equations in problems 1 and 2? The addition property of equality and the multiplication property of equality Next, we’ll explore whether applying the addition and multiplication properties of equality to rewrite an equation preserves the solution set of the original equation.

146

EUREKA MATH2

Properties of Equality Students identify whether rewriting an equation by using the properties of equality preserves the solution set of the equation. Display the addition property of equality and the multiplication property of equality: Addition property of equality: For expressions a, b, and c, if a = b, then a + c = b + c. Multiplication property of equality: For expressions a, b, and c, if a = b, then ac = bc. Facilitate a brief discussion with the following prompts.

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Teacher Note In grade 8, the properties of equality were defined for numbers a, b, and c. This lesson expands the properties to include expressions that are defined for all possible values of the variable. If an expression that is not defined for one or more values of the variable is added to both sides of an equation, the new equation

You were introduced to these properties in grade 8. For both properties, what do we assume to be true?

may not have the same solution set as the

We assume that a = b.

1 1 the resulting equation is x + 1 + __ = 1 + __ , x x

1 is added to original equation. For example, if __ x

both sides of an equation such as x + 1 = 1,

Describe these properties in your own words.

which does not have the same solution set as

If you have an equation and you add the same expression to both sides of the equation, the two sides of the new equation are still equal.

x + 1 = 1.

If you have an equation and you multiply both sides of the equation by the same expression, the two sides of the new equation are still equal. Let’s explore whether adding or multiplying by the same expression on both sides of an equation preserves its solution set. Have students continue working in pairs to complete problem 6.

In lesson 8, students explore whether multiplying both sides of an equation by a variable expression preserves the solution set. In later courses, students will refine their understanding to recognize that although the addition property of equality and the multiplication property of equality apply when adding or multiplying by any expression, adding or multiplying by an expression might change the solution set of the original equation.

147

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Addition property of equality: For expressions a, b, and c, if a = b, then a + c = b + c. Multiplication property of equality: For expressions a, b, and c, if a = b, then ac = bc. 6. Complete the table by performing the indicated operation on both sides of the given equation to write a new equation. Then write the solution set of the new equation. Original Equation

Solution Set of Original Equation

Operation

New Equation

Solution Set of New Equation

5+x=3

{−2}

Add −5

x = −2

{−2}

3x = 2x + 6

{6}

Subtract 2x

x=6

{6}

8=x

{8}

Add 0

8=x

{8}

5 x = 20 2

{8}

2 Multiply by _ 5

x=8

{8}

5=x

{5}

Multiply by 0

0=0

ℝ

Debrief student responses and facilitate a discussion by asking the following questions. What do you notice about the solution sets of the equations you wrote in problem 6? What do you wonder? I notice that all the equations have the same solution set as the solution sets of the original equations except for the last one when I multiplied both sides of the equation by 0. I notice that all the equations where I added the same expression to both sides of the equation have the same solution set. I wonder why the solution set was not preserved in the last row but it was in the other rows.

148

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

The addition property of equality states that if you add the same expression to both sides of an equation, the left and right sides of the new equation are still equal. It does not state that the new equation has the same solution set as the original equation. Do you think that adding the same number to both sides of an equation preserves its solution set? Explain. Yes. I have solved many equations by adding the same number to both sides of the equation, and the solutions to the new equations are always the same as the solutions to the original equations. Adding the same number to both sides of an equation preserves its solution set. Do you think adding the same expression to both sides of an equation preserves its solution set? Explain. Yes. An expression represents a number, so adding the same expression to both sides of an equation is just like adding the same number to both sides of an equation. Adding the same expression to both sides of an equation preserves its solution set, as long as the expression is defined for all possible values of the variable. The multiplication property of equality states that if you multiply both sides of an equation by the same expression, the left and right sides of the new equation are still equal. It does not state that the new equation has the same solution set as the original equation. Do you think multiplying both sides of an equation by the same expression preserves the solution set? Explain. Yes. When we multiplied both sides of the original equation in problem 6 by _, the 2 5

solution set of the new equation is the same as the solution set of the original equation. Yes. I have multiplied both sides of equations by the same number many times before, and the solutions to the new equations are the same as the solutions to the original equations. I am not sure. When we multiplied both sides of the equation in problem 6 by 0, the new equation has a different solution set from that of the original equation. Multiplying both sides of an equation by a number that is not 0 preserves its solution set.

149

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Direct students to the last row of the table in problem 6. What is the resulting equation in problem 6 when you multiply both sides of the original equation by 0? The resulting equation is 0 = 0. What is the solution set of the equation 0 = 0? Explain. The solution set is all real numbers. No matter the value of x, the equation 0 = 0 is always true. The multiplication property of equality states that when we multiply both sides of an equation by 0, the resulting equation is also true. However, multiplying both sides of an equation by 0 almost never preserves the solution set of an equation. Have students think–pair–share about what properties or steps they can apply to rewrite an equation so that it has the same solution set as the original equation. As students share, emphasize the following points:

Differentiation: Challenge Invite students to find an equation that has the same solution set when both sides are multiplied by 0.

• The properties of arithmetic can be used to rewrite equations so that the solution set of the original equation is preserved. • Adding the same expression that is defined for all possible values of the variable to both sides of an equation preserves the solution set of the original equation. • Multiplying both sides of an equation by the same number that is not 0 preserves the solution set of the original equation. Ask students to complete problem 7 with their partner.

150

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

7. Tiah says 11 − 9q = −9 has the same solution set as 1 − 9q = −19 and shows the following work:

11 − 9q = −9 11 − 9q − 10 = −9 − 10 11 − 10 − 9q = −9 − 10 1 − 9q = −19 Do you agree with Tiah? Explain. Yes. Tiah subtracts 10 from both sides of 11 − 9q = −9 to produce 1 − 9q = −19. I know that subtracting 10 from both sides of an equation preserves the solution set of the original equation. Debrief problem 7 by having students share their reasoning. Then ask the following questions. What property did Tiah use when she subtracted 10 from both sides of the equation? Addition property of equality Did subtracting 10 from both sides of the equation help Tiah solve the equation? Why? No, because that step did not help her to isolate q. When we solve equations, it is important to consider both whether the steps we take preserve the solution set of the original equation and whether these steps move us closer to finding the solution set.

Justifying Steps Students solve equations and justify solution sets by using properties of arithmetic and equality. Display problem 8 and solve as a class by asking students to suggest a solution path. Display each step and record the property or operation justification next to it. Probe student thinking by eliciting the justification and asking for alternate steps students might have taken to get to the same solution.

151

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

For problems 8–10, find the solution set of the equation. Choose the appropriate reason from the answer choices to justify each step. Write the solution set by using set notation. Answer Choices Commutative property of addition

Commutative property of multiplication

Associative property of addition

Associative property of multiplication

Distributive property

Addition of like terms

Addition property of equality

Multiplication property of equality

g g 8. 3 + _ = __ − 2 7 14

3 + _ = __ − 2 g 7

g 14

UDL: Engagement Building a classroom culture that supports a growth mindset includes sharing efficient work samples and work samples that may include unnecessary steps or repeated steps. Emphasize that learning is most effective if students evaluate their work, identify their errors, and adjust their strategies accordingly to ensure future success.

UDL: Action and Expression

_g = __g − 5

Addition property of equality

__g = −5 14

Addition property of equality

g = −70

Multiplication property of equality

Solution set: {−70}

Consider scaffolding practice. While students complete this problem, display a worked example. The example should match the written instructions by including the solution steps, the justification for each step, and the solution set written in set notation.

Ask students to complete problems 9 and 10. Monitor students as they work, and look for a variety of solution paths to display during the debrief. 9. 3x + 7 = 3 + 8x − 16

3x + 7 = 3 + 8x − 16 3x + 7 = 8x + 3 − 16 3x + 7 = 8x − 13 7 = 5x − 13 20 = 5x 4=x Solution set: 152

{4}

Commutative property of addition Addition of like terms Addition property of equality Addition property of equality Multiplication property of equality

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

10. 6(2 + x) + 4x − 19 = −18x − 3

6(2 + x) + 4x − 19 = −18x − 3 12 + 6x + 4x − 19 = −18x − 3 6x + 4x + 12 − 19 = −18x − 3 10x − 7 = −18x − 3 28x − 7 = −3 28x = 4

Solution set:

1 {7 }

x = _1 7

Distributive property Commutative property of addition Addition of like terms Addition property of equality Addition property of equality Multiplication property of equality

Promoting the Standards for Mathematical Practice Students construct viable arguments and critique the reasoning of others (MP3) when they justify each step in solving an equation by using the associative, commutative, and distributive properties and the properties of equality and then peer-review their classmates’ solution paths and corresponding justifications. Ask the following questions to promote MP3:

Display the solution sets of both problems. Then have students exchange their work with a partner and conduct a peer review. Students should briefly discuss their solution path with their partner. Close the activity by displaying selected student work samples and by addressing any outstanding questions or disputes. Debrief by eliciting the following points:

• Why does your solution path work? Convince your partner.

• There are infinitely many ways to create a new equation that has the same solution set as the original equation.

• Is your partner’s solution path valid? How do you know?

• What questions can you ask your partner to make sure you understand their justifications?

• When solving an equation, there are many solution paths that will lead to the correct solution as long as the properties of arithmetic and equality are applied correctly.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Land Debrief 5 min Objective: Justify each step in solving a linear equation. Facilitate a brief discussion by asking the following questions. What does it mean to rewrite an equation in a way that preserves its solution set? Rewriting an equation in a way that preserves its solution set means that we create a new equation that has the same solution set as the original equation. What properties can we use to rewrite an equation in a way that preserves its solution set? We can use the associative property of addition, the associative property of multiplication, the commutative property of addition, the commutative property of multiplication, the distributive property, the addition property of equality with expressions that are defined for all possible values of the variable, and the multiplication property of equality with a nonzero number.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

154

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

I can use the commutative property of addition to rewrite 16m + 48 as 48 + 16m. This preserves the solution set.

−6(m − 5) = 48 + 16m

Solving Linear Equations in One Variable

The original equation is

−3(m − 5) = 8(m + 3).

In this lesson, we •

solved linear equations.

•

verified that applying the properties of arithmetic preserves the solution set of an equation.

•

observed that adding the same expression to both sides of an equation or multiplying each side of an equation by the same nonzero number preserves its solution set.

•

justified each step in solving an equation by applying properties and operations.

The equations −3(m − 5) = 8(m + 3) and −6(m − 5) = 48 + 16m have the same solution set because each step I used preserved the original equation’s solution set.

For problems 2–4, find the solution set. Write the property or operation used in each step. 2. 9x − 5 = 10x − 5

Examples

9x − 5 = 10x − 5 −5 = x − 5

1. Explain why the equations −3(m − 5) = 8(m + 3) and −6(m − 5) = 48 + 16m have the same solution set.

0=x

I can multiply both sides of the first equation by 2. This preserves the solution set because I multiplied by a nonzero number.

Addition property of equality Addition property of equality

The solution set is {0}.

2(−3(m − 5)) = 2(8(m + 3))

The solution set {0} means that the equation is true only when x equals 0. The solution set {0}

I can use the associative property of multiplication to regroup the factors. This preserves the solution set.

(2(−3))(m − 5) = (2 ⋅ 8)(m + 3)

is not the same as { }.

−6(m − 5) = 16(m + 3)

3. 11x + 7(2 − x) − 8 = 4x + 6 The associative property of multiplication states that you can regroup the factors to multiply 2 and −3 first.

11x + 7(2 − x) − 8 = 4x + 6 11x + 14 − 7x − 8 = 4x + 6 11x − 7x + 14 − 8 = 4x + 6 4x + 6 = 4x + 6

I can use the distributive property to rewrite 16(m + 3) as 16m + 48. This preserves the solution set.

Distributive property Commutative property of addition Addition of like terms

The solution set is ℝ.

−6(m − 5) = 16m + 48

The resulting expressions are identical. Therefore, the expressions evaluate to the same number for all values of x.

105

106

RECAP

155

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

− 1) x − 2 4(5x ____ = _______ 3

x − 2 4(5x−1) ____ = ______ 3

24 4(5x−1)

x−2 = 24(______) 24(____ 3 ) 24

8(x − 2) = 4(5x −1) 8x − 16 = 20x − 4 −16 = 12x − 4 −12 = 12x −1 = x

Multiplication property of equality Multiplication Distributive property Addition property of equality Addition property of equality Multiplication property of equality

The solution set is {−1}.

156

RECAP

107

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

PRACTICE Name

Date

6. 3w + (8 + 6w) + 2w = 180 and 11w = 172 Because of the associative and commutative properties of addition, I can regroup and change the order of the terms in the left-side expression, resulting in the equation 3w + 6w + 2w + 8 = 180. Using these properties to rewrite the equation preserves its solution set. By adding like terms, the resulting equation is 11w + 8 = 180. I can then use the addition property of equality. I can add −8 to both sides of the equation to obtain 11w = 172. This means that the two equations have the same solution set.

For problems 1–4, without solving, write all properties that justify why the two equations must have the same solution set. 2. 6(5x) = 90 and (6 ⋅ 5)x = 90

1. x − 4 = 10 and −4 + x = 10 Commutative property of addition

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

Associative property of multiplication 7. For each step, write the property or operation that was used.

9x − 3 + 2x + 4 = 8(x − 3) + 4 9x − 3 + 2x + 4 = 8x − 24 + 4

4. 4(x ⋅ 5) = 10 and (4 ⋅ 5)x = 10

3. 2x + 10 = 40 and 2(x + 5) = 40 Distributive property

Distributive property

9x + 2x − 3 + 4 = 8x − 24 + 4

Commutative property of multiplication and Associative property of multiplication

Commutative property of addition

11x + 1 = 8x − 20

Addition of like terms

3x + 1 = −20

Addition property of equality

3x = −21

Addition property of equality

x = −7

Multiplication property of equality

For problems 8–13, find the solution set of the equation.

For problems 5 and 6, without solving, explain why the two equations have the same solution set. Include references to properties in your answer.

8. 4x − 9 = 5

5. 3(a + 5) = −4a + 8 and 9(a + 5) = −12(a − 2)

_7 {2}

Multiplying both sides of the first equation by 3 preserves the solution set.

x x 9. 3 + _ = __ + 2 6 18

{−9}

3(3(a + 5)) = 3(−4a + 8) Because of the associative property of multiplication, rewriting 3(3(a + 5)) as 9(a + 5) preserves the solution set. Because of the distributive property, rewriting 3(−4a + 8) as −12a + 24 preserves the solution set.

9(a + 5) = −12a + 24 I can apply the distributive property again to rewrite −12a + 24 as −12(a − 2), resulting in 9(a + 5) = −12(a − 2). This means that the two equations have the same solution set.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

10. 3m − 5 = 2(m + 6) − 17 + m

ℝ

3 11. 5(2x + 1) − 2x = 10x − 2 (x − _) 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

For problems 14 and 15, solve the equation. Write the property or operation used in each step. Write the solution set by using set notation.

{}

14. 5x + 7 = 4x + 7

5x + 7 = 4x + 7 x+7=7 x=0

Addition property of equality Addition property of equality

{0}

12.

2(3x + 8) x + 6 _______ ____ = 5

{4}

13.

__ ____ 2q 7 − q = 4 6

{3}

15. 3(m − 2) = 5m + 4(2m + 1)

3(m − 2) = 5m + 4(2m + 1) 3m − 6 = 5m + 8m + 4 3m − 6 = 13m + 4 −6 = 10m + 4 −10 = 10m −1 = m

Distributive property Addition of like terms Addition property of equality Addition property of equality Multiplication property of equality

{−1}

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

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Math 1 ▸ M1 ▸ TB ▸ Lesson 7

16. Nina and Angel both solved an equation. They compared their solution paths. The beginning steps of each path are shown. Nina’s Work

1 (x + 3) = − 45 x + 53 + x 5

1 x + 35 = − 45 x + 35 + x 5

Remember For problems 17 and 18, solve the equation.

Angel’s Work

17. 1.6 − m = 14.2

1 (x + 3) = − 45 x + 35 + x 5

_ _

1 x + 35 = 15 x + 35 5

ℝ

18. −5m = −22.5

−12.6

5(1_5 (x + 3)) = 5(− 4_5 x + 3_5 + x)

4.5

1(x + 3) = 5 ⋅ (− 4_5 x) + 5 ⋅ 3_5 + 5 ⋅ x

a. Complete Nina’s work and Angel’s work. Write each solution set by using set notation. Nina’s work:

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 7

19. Mr. Wu’s age is 7 years more than twice Riku’s age. The sum of their ages is 58. How old is Riku? Riku is 17 years old.

Angel’s work:

x + 3 = −4x + 3 + 5x x+3=x+3 ℝ

For problems 20 and 21, solve the inequality. 20. −3x ≤ 42

x ≥ −14

b. Explain why Nina and Angel will both arrive at the same solution set. Use properties or operations to justify your response.

21.

_x > −5 2

x > −10

Nina used the distributive property to rewrite the equation. Angel used the multiplication property of equality to multiply both sides of the equation by 5. Then Angel used the distributive property to rewrite the expression on the right and the associative property to rewrite the expression on the left. Angel also added like terms. Using any of these properties or operations to rewrite the original equation creates an equation with the same solution set as the original equation. So both students should arrive at the same solution set.

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Teacher Edition: Math 1, Module 1, Topic B, Lesson 8 LESSON 8

Some Potential Dangers When Solving Equations Explore steps in solving an equation that are not guaranteed to preserve the solution set.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Name

Date

EXIT TICKET

Consider the equation x + 3 = 10, which has a solution set of {7}. a. Add x to both sides of x + 3 = 10, and verify that the solution set of the new equation is {7}.

x + 3 + x = 10 + x 2x + 3 = 10 + x 2x + 3 − x = 10 + x − x x + 3 = 10 x=7 The solution set of the new equation is {7}.

b. Multiply both sides of x + 3 = 10 by x and verify that 7 is a solution to the new equation.

Lesson at a Glance In this lesson, students expand their knowledge of ways to solve an equation by exploring steps beyond those they have used previously. Students independently explore the effects of squaring both sides of an equation and multiplying both sides of an equation by a variable expression. Exploring these steps not only solidifies students’ understanding of the properties applied in the previous lessons of this topic but also gives students the freedom to experiment. Students conclude the lesson by solving a problem written by Muhammed ibn Mūsā al-Khwārizmī, who is known as the “father of algebra.” Students solve the problem by writing and solving a simple rational equation.

x(x + 3) = x(10)

Key Questions

Substitute 7 for x. Left side: 7(7 + 3) = 7(10) = 70

• What are some potential dangers when solving an equation by taking steps such as multiplying both sides of the equation by a variable expression?

Right side: 7(10) = 70 Because 7(7 + 3) = 7(10) is a true number sentence, I know that 7 is a solution to x(x + 3) = x(10).

• While it is a good habit to always check solutions, when must we check solutions to an equation?

c. Verify that 0 is also a solution to the equation from part (b).

x(x + 3) = x(10) Substitute 0 for x.

Achievement Descriptors

Left side: 0(0 + 3) = 0(3) = 0 Right side: 0(10) = 0

Math1.Mod1.AD9 Explain steps required to solve linear equations

Because 0(0 + 3) = 0(10) is a true number sentence, I know that 0 is a solution to x(x + 3) = 10.

in one variable and justify solution paths. (A.REI.A.1) Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3)

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Agenda

Materials

Fluency

Teacher

Launch 10 min

• None

Learn 25 min

Students

• Squaring Both Sides

• Personal whiteboard

• Multiplication Property of Equality

• Dry-erase marker

• Algebra Begins

• Personal whiteboard eraser

Land 10 min

Lesson Preparation • Review the Math Past resource.

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Fluency Solve Equations by Inspection Students solve equations by using mental math to prepare for exploring steps in solving an equation that are not guaranteed to preserve the solution set. Directions: Solve each equation by using mental math. Write the solution set by using set notation. 1.

x + 5 = 12

{7}

2x + x + 5 = 2x + 12

{7}

x 2 = 25

{−5, 5}

x 2 + 2 = 27

{−5, 5}

5x = 10

{2}

5x − 2 8 = 3 3

_____ _

{2}

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Teacher Note If time permits, ask students to identify all properties that explain why the two equations in each pair (problems 1 and 2, problems 3 and 4, and problems 5 and 6) have the same solution set.

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Launch

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Students summarize what steps are guaranteed to preserve a solution set and explore steps that do not preserve a solution set. Ensure students have their personal whiteboards, dry-erase markers, and personal whiteboard erasers. Initiate a class discussion by asking the following questions. Have students use their whiteboards to show work and make notes about their observations. What are the properties we used when solving equations in previous lessons? Commutative, associative, and distributive properties; addition property of equality; and multiplication property of equality by a number that is not 0 What does it mean if we say that using a property to rewrite an equation preserves its solution set? If we use a property to rewrite an equation, the new equation has the same solution set as the original equation.

Display the equation __ = _ . Direct students to find the solution set. x 12

1 4

What is the solution set of the equation, and how did you determine it? The solution set is {3}. I determined the solution set by multiplying both sides of the equation by 12. The properties of equality are sometimes summarized by stating that whatever is done to one side of the equation has to be done to the other. Applying this idea, Nina decides to erase both denominators. Does that step preserve the solution set of the original equation? Explain. No. Erasing both denominators gives her the equation x = 1, which has the solution set {1}. However, this is not the solution set of the original equation. When we say we can do the same thing to both sides of an equation, we are referring to addition of any expression that is defined for all possible values of the variables or multiplication by a nonzero number. We have verified that these steps preserve the solution set, making them valid mathematical steps to solve an equation. Nina’s action wasn’t a valid mathematical step to solve an equation. © Great Minds PBC

Promoting the Standards for Mathematical Practice When students test conjectures, examine counterexamples, and justify their conclusions, they are constructing viable arguments (MP3). Ask the following questions to promote MP3: • Is it true that using the properties of equality to rewrite an equation will always preserve its solution set? How do you know? • Can you find a situation where using the multiplication property of equality to rewrite an equation does not preserve its solution set?

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Display the equation 3x + 4 = 4x − 5. Direct students to find the solution set. Then confirm that the solution set is {9}. Huan decides to add x + 2 to both sides of the equation. Does this step preserve the solution set of the original equation? Explain. Yes. He added the same expression to both sides of the equation, so the new equation will have the same solution set as the original equation. Confirm by solving the problem as a class or by having students work individually. Display the equation x 2 − 4 = 5. Direct students to find the solution set. Then confirm that the solution set is {−3, 3}. Lyla decides to ignore the exponent on the left side of the equation, resulting in the equation x − 4 = 5. Does this step preserve the solution set of the original equation? Explain. No. The new equation has the solution set {9}. Have students summarize their key takeaways from this sequence of problems. Lead them to the idea that although they can try various steps to solve an equation, not all steps are guaranteed to result in an equation with the same solution set as the original equation. Today, we will explore steps for solving equations that may not preserve solution sets but may still help us find the solution set of our original equation.

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Learn

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Squaring Both Sides Students explore how squaring both sides of an equation may not preserve the solution set of the original equation. Allow students to work on problems 1–3 either independently or with a partner. 1. Consider the equations x + 1 = 4 and (x + 1)2 = 16. a. What is the solution set of x + 1 = 4? The solution set is {3}. b. Verify that both 3 and −5 are solutions to (x + 1)2 = 16. Substitute 3 for x.

(3 + 1)2 = 42 = 16 Because (3 + 1)2 = 16, we know that 3 is a solution to the equation. Substitute −5 for x.

(−5 + 1)2 = (−4)2 = 16 Because (−5 + 1)2 = 16, we know that −5 is a solution to the equation. 2. Consider the equations x − 2 = 0 and (x − 2)2 = 0. a. What is the solution set of x − 2 = 0? The solution set is {2}. b. Verify that 2 is a solution to (x − 2)2 = 0. Substitute 2 for x. Because (2 − 2)2 = 0, we know that 2 is a solution to the equation. c. Are there any other solutions to (x − 2)2 = 0? How do you know? No. For the equation to be true, x − 2 must be equal to 0, which is only true when x = 2. © Great Minds PBC

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3. Based on your results from problems 1 and 2, do you think squaring both sides of an equation preserves the solution set of the original equation? Explain. Sometimes. For problem 2, squaring both sides of the equation preserves the solution set of the original equation, but for problem 1, the solution set is not preserved. Once students are finished, review the answers to problems 1 and 2, and have students share their responses to problem 3. Display the first equation from Launch:

__ _

x =1 12 4

If we square both sides of the equation, do you think the new equation will have the same solution set as the original equation? Explain. No. I think 3 will still be a solution, but there will probably be a second solution to the new equation. Have students work on their whiteboards to square both sides of the equation and find the solution set of the new equation.

__x 2 1_ 2 (12) = (4) x2 1 ___ = __ 144

x2 = 9 x=3

x = −3

The solution set is {−3, 3}. Once students finish, confirm solutions by using the following prompt. The new equation’s solution set consists of the original solution, 3, and an additional solution, −3. Squaring both sides of the equation did not preserve the solution set of the original equation.

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Teacher Note The work done in this lesson will help students understand why extraneous solutions sometimes arise when solving radical and rational equations in Mathematics III.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Multiplication Property of Equality Students explore how multiplying both sides of an equation by an expression may not preserve the solution set of the original equation. Allow students time to work on problems 4 and 5 either independently or with their partner. 4. Consider the equation x − 3 = 5. a. Multiply both sides of the equation by a nonzero constant, and verify that the solution set of the new equation is the same as the original equation. Sample:

7(x − 3) = 7(5) 7x − 21 = 35 7x = 56 x=8 The solution set of both the original equation and the new equation is {8}. b. Based on your results, does multiplying both sides of an equation by a nonzero constant preserve the solution set of the original equation? Explain. Yes. Multiplying both sides of an equation by a nonzero constant produces an equation that has the same solution set as the original equation. 5. Consider again the equation x − 3 = 5. a. Multiply both sides of the equation by x, and verify that 8 is a solution to the new equation.

x(x − 3) = 5x Substitute 8 for x. Left side: 8(8 − 3) = 8(5) = 40 Right side: 5(8) = 40 Because 8(8 − 3) = 5(8), we know that 8 is a solution to x (x − 3) = 5x.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

b. Verify that 0 is also a solution to the equation from part (a). Substitute 0 for x. Left side: 0(0 − 3) = 0 Right side: 5(0) = 0 Because 0(0 − 3) = 5(0), we know that 0 is a solution to x (x − 3) = 5x. c. Based on your results from problems 4 and 5, does multiplying both sides of an equation by an algebraic expression always preserve the solution set of the original equation? Explain. No. Multiplying both sides of an equation by an expression could produce an equation with a solution set that is different from that of the original equation. The solution set of the new equation contains the solution to the original equation, but it also might contain more than one solution.

Teacher Note Based on the results of problem 4, students will likely say that multiplying both sides of an equation by an expression will always result in a new equation that has an additional solution. If time permits, show an example where this is not the case. For example, if we multiply both sides of the equation x − 2 = 0 by (x − 2), we have (x − 2)2 = 0, and we saw in problem 2 that the solution set of this new equation and the original equation is {2}.

Debrief as a class by having students share their responses to problems 4(b) and 5(c). Then facilitate a discussion about the multiplication property of equality. The multiplication property of equality does not make statements about the solution sets of equations. We know that multiplying both sides of an equation by a nonzero constant preserves the solution set of the equation. But multiplying both sides of an equation by an algebraic expression may or may not preserve the solution set because the expression could be equal to zero. Display the second equation from Launch.

3x + 4 = 4x − 5 We saw earlier that adding x + 2 to both sides of the equation preserved the solution set of the original equation. What do you think multiplying both sides of the equation by x + 2 will do? I think that multiplying both sides of the equation by x + 2 will create a new equation that has a different solution set than the original equation.

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Display (x + 2)(3x + 4) = (x + 2)(4x − 5). Then have students think–pair–share about the following questions. Is 9 a solution to this new equation? Explain. Yes. When I substitute 9 into the expressions on each side of the equal sign, I get the same value of 341. Is −2 a solution to the new equation? How can you tell? Yes. When x is −2, the expression x + 2 is 0, so when I substitute −2 for x, both sides of the equation equal 0. Did multiplying both sides of the equation by x + 2 preserve the solution set? How do you know? No. The solution set of the original equation is {9}. When we multiply both sides of the equation by x + 2, the solution set of the resulting equation also includes −2. Direct students to problems 6 and 7 and ask the following question. How might you solve problem 6? Since both denominators are the same, I can set the numerators equal to each other and solve. I can multiply both sides of the equation by x − 2. Give students time to complete problems 6 and 7 either independently or with their partner. For problems 6 and 7, solve for x and then check the solution.

Differentiation: Support

5 x−4 6. ____ = ____ x−2 x−2

x−4=5 x=9 The solution is 9.

Because ____ = ____ is a true number sentence, I know that 9 is a solution to ____ = ____ . 9−4 9−2

5 9−2

x−4 x−2

5 x−2

If students need support solving the rational equations in problems 6 and 7, remind them of work they did with equivalent fractions in previous grades. Example: x __ If __ = , then what value must x have for those 7

5 7

fractions to be equivalent? What about if x − 4 __ ____ = 5? x

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

x+3 5 7. ____ = ____ x−2 x−2

x+3=5 x=2

2+3 5 When x = 2, ____ = _ , which is undefined. The equation has no solution. 2−2 0

Debrief the problems by asking the following questions. How did you solve the equations in problems 6 and 7? I ignored the denominators because they were the same. Then I set the numerators equal to each other and solved the resulting equations. I multiplied both sides of the equations by (x − 2), and then I solved the resulting equations. What happened when you checked the solutions to the equations in problems 6 and 7? For problem 6, the solution was verified when I checked it. For problem 7, I ended up with 0 in the denominator when I checked the solution. How could your solution path for problem 7 result in an equation with a different solution set than the solution set of the original equation? To solve the equation in problem 7, I multiplied both sides by (x − 2). Multiplying both sides of an equation by an algebraic expression does not always preserve the solution set of the original equation. Although multiplying both sides of an equation by a variable expression does not always preserve the solution set of the original equation, it can be a helpful strategy for solving some equations. When using this strategy, it is important to check all solutions to make sure they are also solutions to the original equation.

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Algebra Begins Students represent a historical algebra problem with an equation and solve it. Display problem 8 and provide time for students to read it. This problem appeared about 1200 years ago in a book that represents the beginning of algebra. The author of that book was Persian scholar Muhammad ibn Mūsā al-Khwārizmī (780–850 CE). Al-Khwārizmī lived in the city of Baghdad as a member of the House of Wisdom, a center of learning. His book was published in the year 820. The title of his book means “Book of Restoration and Balancing.” The word al-jabr in the title gives us the modern word algebra. Historians call al-Khwārizmī “the father of algebra” because he was the first mathematician to establish systematic procedures for solving problems that we represent by equations.

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Math Past Consider extending the discussion about the history of algebra by referring to the Math Past resource. The resource explains that al-Khwārizmī invented procedures to simplify any linear equation to the form bx = c, procedures that are still used today to solve equations.

Let’s see whether we can use algebra to solve the problem that al-Khwārizmī wrote. The first part of the problem says, “You have separated ten into two parts.” How can we state this in a different way? The sum of two numbers is 10. What else do we know about the numbers in this problem? When you divide one number by the other, the result is four. The quotient of the numbers is four. How can we represent the numbers in this problem by using algebraic expressions? We can let x represent one number and 10 − x represent the other number. What is an equation that represents the problem? 10 − x _____ x =4 Have the students write their equation for problem 8(a). Then allow a few minutes for students to solve the equation they wrote with a partner. If the partners wrote different equations, encourage them to solve their own equation and compare their answers.

UDL: Representation Consider highlighting different portions of the problem and writing the corresponding algebraic expressions or equations next to them.

x, 10 − x You have separated ten into two parts , and you have divided one by the other ; the obtained quotient is four .

10 − x _____ x

10 − x _____ =4 x

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8. The following problem was written by Muhammad ibn Mūsā al-Khwārizmī, known as the father of algebra: You have separated ten into two parts, and you have divided one by the other; the obtained quotient is four. Find the two parts. a. Write an equation that represents the problem. Let x represent one number. Then 10 − x represents the other number. 10 − x Sample: _____ x =4

b. Solve the equation to answer the problem.

_____

10 − x x =4

Teacher Note Students may instead write the equation

x _____ = 4 and solve by multiplying both sides 10 − x

of the equation by 10 − x. Some students may represent the situation with a system of two equations in two variables. If they do, encourage them to use substitution to rewrite the system as an equation in one variable.

x(____ x ) = x(4) 10 − x

10 − x = 4x 10 = 5x When x = 2, _____ = 4. The solution is 2.

2=x

10 − 2 2

10 − 2 = 8 The two parts are 2 and 8.

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When students have completed the problem, invite a few to share their solution paths. Ask students the following questions as they share to support them in explaining their reasoning. What steps did you take to solve the equation? I multiplied both sides of the equation by x, resulting in the equation 10 − x = 4x. Then I solved the equation by using properties of equality to find the first part. Finally, I subtracted the first part from 10 to determine the second part.

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Teacher Note The student responses provided to the questions in the debrief are based on the solution path shown for problem 8.

Did multiplying both sides of the equation by a variable expression guarantee that the solution set of the original equation would be preserved? Why? No. Multiplying both sides of the equation by a variable expression did not guarantee that the original equation’s solution set would be preserved because the expression could have had a value of 0. How did you make sure the solution you found was the solution to the original equation? I substituted the value I got for x back into the original equation I wrote.

Land Debrief 5 min Objective: Explore steps in solving an equation that are not guaranteed to preserve the solution set. Ask students to summarize their key takeaways from the lesson. During the discussion, pose the following questions. What are some potential dangers when solving an equation by taking steps such as multiplying both sides of the equation by a variable expression? We might get an equation with a solution set that is different from that of the original equation.

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While it is a good habit to always check solutions, when must we check solutions to an equation? We must check solutions if we do any step other than applying a property of arithmetic, the addition property of equality for expressions that are defined for all possible values of the variable, or the multiplication property of equality with a nonzero constant.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

Recap

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

c. Verify that multiplying both sides of 4x − 3 = x + 9 by x + 5 results in a new equation with a different solution set that includes −5.

(4x − 3)(x + 5) = (x + 9)(x + 5)

Some Potential Dangers When Solving Equations

Substitute −5 for x to verify that −5 is also a solution to the new equation. Left side: (4(−5) − 3)(−5 + 5) = −23(0) = 0

In this lesson, we •

discovered that not all steps taken when solving an equation preserve the solution set of the original equation.

•

explored steps that are not guaranteed to preserve the solution set, such as squaring both sides of an equation and multiplying both sides of an equation by a variable expression.

Right side: (−5 + 9)(−5 + 5) = 4(0) = 0

The solution set of the new equation is {−5, 4}. So multiplying both sides of this equation by an algebraic expression resulted in a new equation with a different solution set.

2. Consider the equation 3x = x − 4. a. Find the solution set of 3x = x − 4.

Examples

3x = x − 4 2x = −4

1. Consider the equation 4x − 3 = x + 9.

x = −2

a. Find the solution set of 4x − 3 = x + 9.

The solution set is {−2}.

4x − 3 = x + 9 3x − 3 = 9 3x = 12

b. Square both sides of 3x = x − 4. Verify that −2 is a solution to the resulting equation.

x=4

(3x)2 = (x − 4)2

The solution set is {4}.

Substitute −2 for x to verify that −2 is a solution. Left side: (3(−2))2 = (−6)2 = 36 Right side: (−2 − 4)2 = (−6)2 = 36

b. Verify that adding x + 5 to both sides of 4x − 3 = x + 9 preserves the solution set.

c. Verify that 1 is also a solution to the resulting equation.

4x − 3 + (x + 5) = x + 9 + (x + 5)

(3x)2 = (x − 4)2

5x + 2 = 2x + 14

Substitute 1 for x to verify that 1 is also a solution.

3x + 2 = 14 3x = 12 x=4 The solution set is {4}.

Left side: (3(1))2 = (3)2 = 9

Adding an expression to both sides of an equation preserves the solution set.

Right side: (1 − 4)2 = (−3)2 = 9

The solution set of the resulting equation is {−2, 1}. So squaring both sides of this equation resulted in a new equation with a different solution set.

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

c. Verify that multiplying both sides of x + 4 = 3x + 2 by x − 4 creates a new equation that has a different solution set that includes 4.

(x − 4)(x + 4) = (x − 4)(3x + 2)

1. Consider the equation a = 3. Create a new equation by performing the step described. Then write the solution set of the new equation.

Substitute 4 for x. Left side: (4 − 4)(4 + 4) = 0

a. Square both sides of a = 3.

Right side: (4 − 4)(3(4) + 2) = 0

a2 = 9, {−3, 3}

Because (4 − 4)(4 + 4) = (4 − 4)(3(4) + 2), we know that 4 is a solution to (x − 4)(x + 4) = (x − 4)(3x + 2).

3. Consider the equation x + 2 = 2x.

b. Cube both sides of a = 3.

a. Find the solution set of x + 2 = 2x.

a = 27, {3}

{2}

b. Square both sides of x + 2 = 2x, and verify that your solution satisfies this new equation.

2. Consider the equation x + 4 = 3x + 2.

(x + 2)2 = (2x)2

a. Find the solution set of x + 4 = 3x + 2.

Substitute 2 for x.

{1}

Left side: (2 + 2)2 = 16

Right side: (2 · 2)2 = 16

Because (2 + 2)2 = (2 · 2)2, we know that 2 is a solution to (x + 2)2 = (2x)2. b. Verify that adding x − 4 to both sides of x + 4 = 3x + 2 creates a new equation that has the same solution set.

c. Verify that − _ is also a solution to the new equation. 2 3

2 Substitute − _ for x. 3 2 2 4 2 16 Left side: − _ + 2 = _ = __

(x − 4) + x + 4 = (x − 4) + 3x + 2 2x = 4x − 2

( 3

2 = 2x

2 3

176

4 2 3

16 9

Because (− _ + 2) = (− _ · 2) , we know that − _ is a solution to (x + 2)2 = (2x)2.

The solution set is {1}.

(3 )

)

Right side: (2(− _)) = (− _) = __

1=x

2 3

123

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Math 1 ▸ M1 ▸ TB ▸ Lesson 8

For problems 4–7, solve the equation for x.

Remember

x 1 4. _ = __ 4 1 3

For problems 8 and 9, solve the equation.

5 8. − 3_ p = __ 12 4 5_ − 9

x + 3 ____ ____ = 5 x+1

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 8

9. p − 4.7 = −12.2

−7.5

10. Solve the equation. Write the solution set by using set notation, and graph the solution set on the number line.

x+1

2x + 5 = 5 + 2x ℝ x 6.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

x + 3 _____ ____ = 5 x−2

9 10

x−2

No solution

For problems 11 and 12, solve the inequality. 11. 5 − n > 35

n < −30

x−5 7. ____ = 1

3 12. − _ n ≤ −9 8

n ≥ 24

x−5

All real numbers except 5

P R ACT I C E

125

126

P R ACT I C E

177

Teacher Edition: Math 1, Module 1, Topic B, Lesson 9 LESSON 9

Writing and Solving Equations in One Variable

Create equations in one variable and use them to solve problems. EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

Name

Date

EXIT TICKET

Nina sells tickets to the school play. Adult tickets are $8 each, and student tickets are $5 each. She sells 15 fewer student tickets than adult tickets. If the total amount of Nina’s ticket sales are $315, how many student tickets does she sell? Let x represent the number of adult tickets Nina sells. Then x − 15 represents the number of student tickets Nina sells.

8 x + 5( x − 15 ) = 315 8 x + 5 x − 75 = 315 13 x − 75 = 315 13 x = 390 x = 30

Lesson at a Glance In this digital lesson, students visualize connections between a given context and the equation they write to represent it. Empowered by this experience, students transition to writing and solving equations for other real-world and mathematical contexts. Through discussion, students connect the process of creating and solving equations from real-world and mathematical problems. Use the digital platform to prepare for and facilitate the digital portion of this lesson. Students will also interact with lesson content and activities via the digital platform.

Substituting 30 for x, the number of student tickets Nina sells is 15 because 30 − 15 = 15. Nina sells 15 student tickets.

Key Questions • When we write an equation to represent a situation, why is it important to define the variable as specifically as possible? • When we write an equation in one variable to represent a situation, what information is important to consider?

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Achievement Descriptors Math1.Mod1.AD4 Create equations and inequalities in one variable and use them to solve problems. (A.CED.A.1) Math1.Mod1.AD6 Interpret solutions to equations and inequalities in one variable as viable or nonviable options in a modeling context. (A.CED.A.3) Math1.Mod1.AD9 Explain steps required to solve linear equations in one variable and

justify solution paths. (A.REI.A.1) Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3)

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Polyomino Squares

• Computers or devices (1 per student pair)

• Creating and Solving Equations

Lesson Preparation

Land 10 min

• None

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Fluency Solve Multi-Step Equations Students solve multi-step equations to prepare for using equations in one variable to solve problems. Directions: Solve each equation for m. 1.

2(3m + 4) − 3m = 35

0 = 7 − 1_ ( 8 − 5 m )

−4

10 − 6m = 2m + 58

−6

−2(4m − 1) = 9(2m − 2)

10 13

Launch

Students identify patterns in a given context to determine an unknown number.

Students look for the difference between the numbers inside two vertically adjacent squares, or a domino, of a 3 × 3 grid. Students continue completing the same exercise for dominoes inside a grid of increasing size. Students observe patterns between grid 180

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size and the difference of the numbers inside a domino. These patterns help students determine the difference between numbers inside any two vertically adjacent squares on a 10 × 10 grid. Students discuss their strategies for answering a final challenge involving a 7 × 7 grid. What pattern do you see between the grid size and the difference of the numbers inside a domino on the grid? The difference of the numbers inside a domino on a 3 × 3 grid is 3. The difference of the numbers inside a domino on a 4 × 4 grid is 4 and so on. The difference between the two numbers inside a domino corresponds to the dimension of the square grid.

4 8 12

5 10

13 14 15 16 20 21 22 23 24 25

20 40 50 60 70 80 90

91 92 93 94 95 96 97 98 99 100

Learn Polyomino Squares

Students write algebraic equations to generalize relationships. Students continue their exploration of polyominoes. Students are introduced to a new shape, the triomino, on the 7 × 7 grid from Launch. They encounter a variety of problems where they must find one or more missing numbers in a triomino. Strategies using algebraic expressions or equations to solve for the unknown values are highlighted. Students transition to problems where they must write an equation to determine grid dimensions when given information about specific squares of the polyomino.

Promoting the Standards for Mathematical Practice When students use polyomino squares to write algebraic equations and generalize relationships, they are making sense of problems and persevering in solving them (MP1). Ask the following questions to promote MP1:

50 ̶ 2n

50 ̶ n

50 ̶ n + 1

• What information can you use to write an equation for the sum of the numbers inside the triomino? • Does your value of n make sense? Why?

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Does it make sense to use algebraic equations to solve polyomino problems instead of using guess and check? Why? Yes. It makes sense because using equations can be faster than using guess and check. Algebraic equations help us describe general relationships that exist. We can use equations to efficiently determine unknowns in a variety of situations.

Creating and Solving Equations Students create and solve equations from real-world and mathematical problems. Consider allowing students to work in groups. Read problem 1 to the class. This problem has two unknowns: the number of dimes and the number of quarters. Share your thinking aloud with the class as you model how to generate an equation by letting q represent the number of quarters. Invite students to write an expression in terms of q for the number of dimes. Organize the information to scaffold writing an equation by documenting what each expression represents. Allow students to write an equation that represents the situation before offering support. Verify that students have written an accurate equation. Then pose the following question.

UDL: Engagement Digital activities align to the UDL principle of Engagement by including the following elements: • Immediate formative feedback. The activity reveals select information once a student submits a response. • Options that promote flexibility and choice. Students choose which polyomino they use to write and solve an equation.

How could we define a different variable? We could define a variable to represent the number of dimes. Have students write an equation to represent the situation where d represents the number of dimes. Display the different equations side by side. Approach 1:

Approach 2:

Let q represent the number of quarters.

Let d represent the number of dimes.

Then q − 5 represents the number of dimes.

Then d + 5 represents the number of quarters.

Equation: 0.25q + 0.10(q − 5) = 5.45

Equation: 0.1d + 0.25(d + 5) = 5.45

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Direct students to write both equations. Have half of the class solve the equation for approach 1 and the other half of the class solve the equation for approach 2. Ensure that students use their solutions to find the number of quarters and the number of dimes, regardless of how they define the variables. Solve each problem by writing and solving an equation. 1. Angel has $5.45 in quarters and dimes. He has 5 fewer dimes than quarters. How many of each coin does Angel have? Approach 1:

Approach 2:

Let q represent the number of quarters.

Let d represent the number of dimes.

The expression q − 5 represents the number of dimes.

The expression d + 5 represents the number of quarters.

Then 0.25q represents the amount of money in dollars from the quarters.

Then 0.10d represents the amount of money in dollars from the dimes.

Then 0.10(q − 5) represents the amount of money in dollars from the dimes.

Then 0.25(d + 5) represents the amount of money in dollars from the quarters.

0.25 q + 0.10( q − 5 ) = 5.45

0.10 d + 0.25( d + 5 ) = 5.45

0.25 q + 0.10 q − 0.50 = 5.45

0.10 d + 0.25 d + 1.25 = 5.45

0.35 q − 0.50 = 5.45

0.35 d + 1.25 = 5.45

0.35 q = 5.95

0.35 d = 4.20

q = 17

d = 12

Substituting 17 for q, the number of dimes is 12 because 17 − 5 = 12.

Substituting 12 for d, the number of quarters is 17 because 12 + 5 = 17.

Angel has 17 quarters and 12 dimes.

Angel has 12 dimes and 17 quarters.

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Debrief student work by using the following prompt. The solution in approach 1 is different from the solution in approach 2. How are these approaches different? In approach 1, solving for q tells us the number of quarters. Then we can substitute the number of quarters to find the number of dimes. In approach 2, solving for d tells us the number of dimes. Then we can substitute the number of dimes to find the number of quarters. The only difference between the two approaches is the order in which we find the number of dimes and the number of quarters. Direct students to complete problems 2 and 3 in groups. Circulate as students work, and offer support as appropriate. Ensure that students attend to the following details:

Differentiation: Challenge

• Define variables. • Document each expression created as steps toward creating the equation. • Substitute the solution to the equation into each expression to answer any multipart questions. If students need support writing the related expressions for either problem, encourage them to use a tape diagram or to guess, check, generalize, and verify to make sense of the problem. Students can use the tape diagram or the repeated reasoning used in guess and check to generalize by creating an equation. Students can use colors to highlight the portion of each equation that changes with each new guess. The following is an example of the guess, check, and generalize approach from problem 2.

In problem 2, consider asking students who are ready for a challenge to define the variable t to represent Tiah’s age 12 years ago or to represent Tiah’s age 15 years from now. Ask students to create an equation by using this new relationship. Students create expressions depending on how the variable is defined. Follow up by asking students why the equations they create have different solution sets even though the equations all model Tiah’s age.

Let 10 be Tiah’s current age. Tiah’s age in 15 years:

Twice Tiah’s age 12 years ago:

10 + 15 = 25

2(10 − 12) = − 4

An age of − 4 years doesn’t make sense, so Tiah must be older than 10. Let 20 be Tiah’s current age. Tiah’s age in 15 years:

Twice Tiah’s age 12 years ago:

20 + 15 = 35

2(20 − 12) = 16

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The numbers 35 and 16 are pretty far apart, so 20 is not a good guess. However, these numbers are closer together than the numbers I got when I guessed Tiah was 10, so my next guess should be greater than 20. Let 40 be Tiah’s current age. Tiah’s age in 15 years:

Twice Tiah’s age 12 years ago:

40 + 15 = 55

2(40 − 12) = 56

The numbers 55 and 56 are very close together, so 40 is not correct but it is a close guess. Generalize by writing an equation: Let t represent Tiah’s current age in years.

t + 15 = 2(t − 12) Circulate as students work, and look for solution paths such as those where one group may have defined a variable differently from the other groups. Invite these students to share their solution path with the class. 2. Fifteen years from now, Tiah’s age will be twice her age 12 years ago. What is Tiah’s current age? Let t represent Tiah’s current age in years. Then t + 15 represents Tiah’s age 15 years from now, and t − 12 represents Tiah’s age 12 years ago.

t + 15 = 2( t − 12 ) t + 15 = 2 t − 24 15 = t − 24 39 = t Tiah’s current age is 39 years.

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3. Bahar and Huan filled bags of popcorn to sell at a baseball game. • Huan filled 25% more bags of popcorn than Bahar. • After the game, 15% of the total number of their bags was not sold. • The bags sold for $0.75 each. • Bahar and Huan made a total of $114.75 selling bags of popcorn. How many bags of popcorn did each person fill? Let b represent the number of bags of popcorn Bahar filled. Then b + 0.25b, or 1.25b, represents the number of bags of popcorn Huan filled. Together they filled b + 1.25b bags of popcorn. If 15% of the total number of bags was not sold, Bahar and Huan sold 85% of the total number of bags. Then 0.85(b + 1.25b) represents the total number of bags sold. Therefore, 0.75(0.85(b + 1.25b)) represents the total amount of money in dollars they made by selling bags of popcorn.

0.75( 0.85( b + 1.25b )) = 114.75 0.75( 0.85( 2.25b )) = 114.75 1.434375 b = 114.75 b = 80 Substituting 80 for b, the number of bags of popcorn Huan filled is 100 because 1.25(80) = 100. Bahar filled 80 bags of popcorn, and Huan filled 100 bags of popcorn.

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Land Debrief 5 min Objective: Create equations in one variable and use them to solve problems. Facilitate a discussion by using the following prompts. When we write an equation to represent a situation, why is it important to define the variable as specifically as possible? For example, in problem 2, why shouldn’t we define t to represent Tiah’s age? Problem 2 mentioned three different ages for Tiah: her current age, her age 12 years ago, and her age 15 years from now. If we define the variable to represent her age, it is unclear what our result means. If we define t to represent Tiah’s current age, we can solve the equation and answer the question without further calculations. Why does it make sense that the solution set of the equation in problem 2 has only one element? Tiah cannot be more than one age at any given time, whether that is now, 12 years ago, or 15 years from now. When can we write a one-variable equation to solve a problem in a real-world situation? We can write a one-variable equation to solve a problem in a real-world situation when we have enough information to create expressions for any related quantities in terms of the defined variable. When writing an equation in one variable to represent a situation, what information is important to consider? Explain. It is important to consider how to define the variable. The structure of the equation will depend on how the variable is defined.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem. © Great Minds PBC

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

2. The sum of Angel’s age now and his sister’s age now is 25. Three years ago, Angel’s age was 1 less than 3 times his sister’s age at that time. How old are Angel and his sister now? Let a represent Angel’s age now. Then 25 − a represents his sister’s age now.

Writing and Solving Equations in One Variable

Also, a − 3 represents Angel’s age 3 years ago, and 22 − a represents his sister’s age 3 years ago.

In this lesson, we •

wrote equations in one variable.

•

used equations in one variable to solve problems. Angel’s age

Examples

3 years ago

1. A bowling alley offers two options for the cost of bowling: •

$5.75 per game plus a fee of $3.50 for each player’s shoe rental or

•

$32.25 per hour with no fee for shoe rental.

a − 3 = 3(22 − a) − 1 a − 3 = 66 − 3 a − 1 a − 3 = 65 − 3 a 4 a − 3 = 65 4 a = 68 a = 17

Three years ago, Angel’s sister was

(25 − a) − 3, or 22 − a.

One less than 3 times his sister’s age 3 years ago

Angel is 17 years old now. Substitute 17 for a in 25 − a to find his sister’s age now.

How many games must one player bowl in one hour for the costs of the two options to be the same?

25 − (17) = 8

Let g represent the number of games.

Angel’s sister is 8 years old now.

5.75 g + 3.50 = 32.25 5.75 g = 28.75 g=5 A player must bowl 5 games in one hour for the costs of the two options to be the same.

188

131

132

RECAP

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

PRACTICE Name

Date

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

d. Use your expression from part (b) to find n, the number covered by the center of the pentomino when the sum of the five covered numbers is 118. Write an equation that verifies your placement.

If 5n = 118, then n = 23 _3 . The center of the pentomino cannot be placed on a mixed number, 5 so it is not possible for the sum of the five numbers to be 118.

1. The pentomino can be placed on the grid to cover a set of numbers.

EUREKA MATH2

2. Consider two integers. The first integer is 3 more than twice the second integer. Adding 21 to 5 times the second integer will give us the first integer. Find the two integers. The first integer is −9 and the second integer is −6.

a. Where should the center of the pentomino be placed so that the sum of the five covered numbers is 55? Write a number sentence that verifies your placement of the pentomino. The center of the pentomino should be placed at 11 for the sum of the five numbers to be 55.

11 + 17 + 5 + 12 + 10 = 55 b. Write an expression to represent the sum of any five numbers covered by the pentomino on the grid. Let n represent the number covered by the center of the pentomino.

3. When the measures of the interior angles of a pentagon are listed in order from least to greatest, each successive angle measure increases by 20°. Write the measures of the angles.

(n + 6) + (n − 6) + n + (n − 1) + (n + 1)

68°, 88°, 108°, 128°, and 148° c. Use your expression from part (b) to find n, the number covered by the center of the pentomino when the sum of the five covered numbers is 75. Write an equation that verifies your placement. The center of the pentomino should be placed over the number 15 for the sum of the five numbers to be 75.

21 + 9 + 15 + 14 + 16 = 75

133

134

P R ACT I C E

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EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

4. Ana sells T-shirts and sweatshirts for the soccer team. The price of one sweatshirt is $10 more than the price of one T-shirt. She sells 50 T-shirts and 40 sweatshirts for a total of $1480. What is the price of one T-shirt? What is the price of one sweatshirt? The price of one T-shirt is $12, and the price of one sweatshirt is $22.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 9

Remember For problems 7 and 8, solve the equation. 7. −6r + 8 = −28

8. 32 = 4 − 9r + 6r

__ − 28 3

5. In a triangle, the measure of the first angle is half the measure of the second angle, and the measure of the third angle is 6° more than the measure of the first angle. Find the measures of the angles of the triangle. The measures of the angles of the triangle are 43.5°, 87°, and 49.5°.

9. Solve the inequality. Write the solution set by using set notation, and graph the solution set on the number line.

{x | x S 7} –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 6. A pet rescue center is creating a monthly food budget. The center currently has two-thirds as many dogs as cats. Each animal is fed 2 cans of food a day. Dog food costs $0.68 per can, while cat food costs $0.65 per can. The rescue center has budgeted a total of $2581.80 for food for this 30-day month. How many dogs does the rescue center have?

x−6S1

x 0

9 10

10. Which equation correctly models the statement that −30 is 30 units from 0 on a number line? A. (−30) = 30

The rescue center has 26 dogs.

B. −30 = 30

C. |30| = −30

D. |−30| = 30

190

P R ACT I C E

135

136

P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic B, Lesson 10 LESSON 10

Rearranging Formulas Rearrange formulas to highlight a quantity of interest.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 10

Name

Date

EXIT TICKET

energy is K = _1 mv 2, where v ≠ 0. Rearrange the formula to isolate m.

The kinetic energy K of an object depends on its mass m and its velocity v. The formula for kinetic 2

K = 1_ mv 2 2

2 K = mv 2 2K ___ =m v2

Lesson at a Glance Students work in groups to rearrange formulas by applying properties of equality to solve for a specific unknown quantity. Through discussion, they connect the process of rearranging a formula or equation to solving one-variable equations. Students share and compare different representations to verify their equivalence. This lesson promotes flexible thinking by encouraging students to choose their own solution paths.

Key Questions • How is rearranging a formula or equation to highlight a variable of interest similar to solving an equation in one variable? • When can it be useful to rearrange a formula or equation?

Achievement Descriptor Math1.Mod1.AD7 Rearrange formulas to highlight a quantity

of interest. (A.CED.A.4)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Width of Any Rectangle

• None

• Use a Formula to Make a Formula

Lesson Preparation

• Numbers Versus Letters

• None

Land 10 min

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Fluency Solve Proportions Students solve proportions to prepare for rearranging formulas. Directions: Solve each equation.

c 3 = 4 1

_ _

_h = 7_

3 6 y = 22

_ __

3 w = 5 8

3 8 = 5 k

x =3 4

194

_ __ _ _

__ 24 5

__ 40 3

EUREKA MATH2

Launch

Math 1 ▸ M1 ▸ TB ▸ Lesson 10

Students substitute known quantities into a formula and solve for the unknown quantity. Direct students to problem 1. Allow them to solve the problem with a partner by using any strategy they choose. Monitor how students solve this problem. Find two student work samples to display: one where the student first distributed the 2 and one where the student 1 first multiplied both sides of the equation by _ . 2

1. A rectangular portion of a park will be fenced off to create a playground. A construction company donated a total of 117.5 feet of fencing. Based on the space provided to create the playground, the length of the playground must be 35.5 feet. Use the formula for the perimeter of a rectangle to determine what the width of the playground must be to use all the fencing.

P = 2(l + w) 117.5 = 2(35.5 + w) 117.5 = 71 + 2w 46.5 = 2w 23.25 = w The width of the playground must be 23.25 feet. In previous lessons, we used properties and operations to rewrite equations in one variable. Today, we will rewrite equations with more than one unknown value.

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Learn Width of Any Rectangle Students use properties to rearrange a formula and solve for a specific unknown quantity. Introduce problem 2 with the following prompt. In problem 1, there were restrictions on the total amount of fencing provided and on the length of the playground. Let’s assume now that there are no restrictions on the amount of fencing or on the dimensions of the rectangular playground. How could we represent the width of the playground by using the formula for the perimeter of a rectangle?

UDL: Action & Expression As students work, provide worked examples of analogous problems. For example, display a solution path for an analogous one-variable equation and include justifications for each step.

We could rearrange the formula to isolate w. Have students complete problem 2 with their partner. Circulate as students work and give the following guidance as needed: • Recall that the variables are placeholders for unknown quantities. • There will be no numeric calculations. • When solving for w, the goal is to isolate that variable. Monitor how students approach this problem. 2. Rearrange the perimeter formula P = 2(l + w) to isolate w.

P = 2(l + w) P = 2l + 2w P − 2l = 2w P − 2l _____ =w 2

Select two student work samples to debrief different solution paths: one where students 1 first distribute the 2 and one where students first multiply by _ . Display the most used 2 solution path first. After displaying both solution paths, ask the following question. 196

UDL: Engagement Consider providing mastery-oriented feedback by focusing attention on students’ effort, the application of learned strategies, and the value in understanding errors. For example, if students incorrectly apply the distributive property, first acknowledge their decision to apply the distributive property to eliminate the grouping symbols. Then direct them to verify that they distributed to every term within the grouping symbols. When students discover their mistake, discuss the advantages of finding errors and using this knowledge to ensure success with similar problems in the future.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 10

Both of these formulas can be used to find the width of a rectangle. Recall that a formula is an equation often used to determine something. In this lesson, we rearrange formulas to isolate a variable so we can see how it relies on the other variables. How can we determine whether these two formulas represent the same relationship?

Promoting the Standards for Mathematical Practice

If we can rewrite the expression for the width in one formula to look like the other expression for width by using the commutative, associative, and distributive properties, the two formulas represent the same relationship.

When students apply what they know about solving one-variable equations to rearrange formulas by using mathematical rules such as the arithmetic properties, they are making use of structure (MP7).

Below the equation that you created, rearrange the formula to resemble one of the two work samples just shared.

Ask the following questions to promote MP7:

P − 2l _____ =w

• How can the properties of arithmetic and properties of equality help you rewrite formulas?

1 (P − 2l) = w 2 1 P−l=w 2 P −l=w 2

• How is figuring out the next step in isolating a variable in a formula similar to figuring out the next step in solving a one-variable equation?

Emphasize that these formulas might look different, but they both find the width of a rectangle in terms of perimeter and length. Address any outstanding questions. Revisit the problem from Launch with the known perimeter of 117.5 feet and the length of P − 2l 35.5 feet. Ask half of the class to substitute these known values into the formula _____ =w 2

and the other half of the class to substitute these values into the formula _ − l = w. Have P 2

them record their work below their answer to problem 1. Compare student responses to the calculations from Launch. Consider displaying each student work sample from Launch next to its analogous approach.

Use a Formula to Make a Formula Students rearrange well-known formulas for a specified quantity.

Teacher Note Watch for students who rewrite

P − 2l _____ = w as 2

P − l = w because they incorrectly divided the numerator and denominator by 2. Consider having students compare the calculations from Launch with known values for P and l to help them conceptualize that they divided the difference P − 2l by 2, not just the expression 2l.

Transition students to problems 3–5. Allow students to work in groups to complete the problems. Focus their attention on using the properties of equality to rearrange the formulas. © Great Minds PBC

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3. The formula for the area of a rectangle is A = lw, where l represents the length and w represents the width. a. Rearrange the formula to isolate l.

A = lw

__A = l

b. Rearrange the formula to isolate w.

A = lw

A =w l

4. To find the volume of a cylinder with radius r and height h, we use the formula V = πr 2h. Rearrange the formula to isolate h.

V = 𝜋r 2h

V ___ =h 𝜋r 2

5. Angel decided to memorize three formulas that relate the volume V, area of the base B, and the height h of a right rectangular prism. V B = __ h

V h = __ B

V = Bh

Does he need to memorize all three formulas, or can he memorize just one? Explain your reasoning. Angel does not need to memorize all three formulas. He can memorize just one of them and rearrange the formula to isolate any variable that he needs. Review the answers with students. Then debrief by asking the following question. Why must the variables in these formulas have positive values? The variables must have positive values because they represent measures of figures, which cannot be negative or 0.

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Facilitate a brief discussion about why it might be helpful to rearrange a formula by asking the following question. Why might you want to rearrange a formula like the one for volume in problem 5? I might want to rearrange a formula if I need to perform repeated calculations. For example, if I need to compare several different volumes, I might want to rearrange the formula to isolate V. If I need to compare several different heights, I might want to rearrange the formula to isolate h. If I need to compare several different base areas, I might want to rearrange the formula to isolate B. Allow groups to complete problems 6–8. Circulate as students work, and look for interesting solution paths that showcase the variety of equivalent representations. Once groups finish, debrief student responses by displaying student work samples. If student work lacks variety, consider introducing your own equivalent representation. Invite students to compare work samples and build connections between solution paths. 6. The formula for the equation of a line in standard form is Ax + By = C, where A, B, and C are constants. Rearrange the formula to isolate y. Assume B ≠ 0.

Ax + By = C

By = −Ax + C

__ x + __ y = −A B

C B

7. The formula for the surface area of a rectangular prism with a square base is SA = 2w2 + 4hw, where h represents the height of the prism and w represents the side length of the square base, with w ≠ 0. Rearrange the formula to isolate h.

SA = 2w2 + 4hw

SA − 2w2 = 4hw SA − 2w2 _______ =h 4w

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8. If F represents temperature in degrees Fahrenheit and C represents the same temperature in degrees Celsius, then the formula relating F and C is 5F − 9C = 160. a. Rearrange the formula so it is more convenient for the following situations. Situation 1: A traveler to the United States from a country that measures temperature in degrees Celsius

5F − 9C = 160 −9C = 160 − 5F −9C = −5(F − 32)

_ _ (−9) (−9C) = (−9)(−5)(F − 32) 1

C = _ (F − 32) 5 9

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Situation 2: A traveler from the United States to a country that measures temperature in degrees Celsius

5F − 9C = 160 5F = 9C + 160 F = _ C + 32 9 5

b. Why would the two travelers prefer different formulas? The travelers would prefer different formulas because they are performing opposite conversions. The traveler to the United States wants to convert Fahrenheit into the more familiar Celsius; the traveler from the United States wants to do the reverse.

Numbers Versus Letters Students compare solving equations containing more than one variable with related one-variable equations. Present problem 9 to the class. Have students solve the one-variable equation and try to use their work to solve the equation with more than one variable. When the struggle is no longer productive, guide students through the problem. While working through the problem, highlight the following: • The distributive property is used to isolate x. • Precise language is used to describe steps. For example, justify the step from

d−b x (a + c) = d − b to x = ____ a + c by saying that students will multiply the expressions on both

sides of the equation by the reciprocal of the sum of a and c.

• Division by 0 or having a denominator equal to 0 is avoided. Emphasize the similarities of solving each type of equation, and inform students that variables are simply placeholders for unknown numbers.

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Ask students to complete problem 10. For each pair of equations in problems 9 and 10, first solve the equation that contains one variable. Then solve the equation that contains more than one variable. Equation Containing One Variable 9. Solve 3x + 4 = 6 − 5x for x.

Related Equation Containing More Than One Variable Solve ax + b = d − cx for x.

3x + 4 = 6 − 5x 3x + 5x + 4 = 6

ax + b = d − cx ax + cx + b = d

3x + 5x = 6 − 4

ax + cx = d − b

x (3 + 5) = 6 − 4

x (a + c) = d − b

____

6−4 x= 3+5 x=2 8 x=1 4

_ _

10. Solve __ + _ = 3 for x. 2x 5

x 7

__ _

x = ____ a+c

where a + c ≠ 0

d−b

Solve __ + __ = f for x. ax b

cx d

__ __

2x x + =3 5 7 2x x (5 ⋅ 7)( + ) = (5 ⋅ 7)(3) 5 7

ax cx + =f b d ax cx bd( + ) = bd( f ) b d

14x + 5x = 105

adx + bcx = bdf

x (14 + 5) = 105

x (ad + bc) = bdf

__ _

x = _____ 105 14 + 5

x = ___ 105 19

__ __

x = ______

where ad + bc ≠ 0

bdf ad + bc

Debrief problem 10 by inviting students to display their work and share their thinking. 202

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Land Debrief 5 min Objective: Rearrange formulas to highlight a quantity of interest. Facilitate a discussion by asking the following questions. Why can we use the properties of arithmetic and properties of equality to highlight a variable of interest? We can use the properties to solve equations to highlight a variable of interest because variables are just placeholders for unknown numbers. How is rearranging a formula or equation to highlight a variable of interest similar to solving an equation in one variable? The properties and reasoning used to solve equations apply regardless of the number of variables that appear in the equation or formula. When can it be useful to rearrange a formula or equation? Rearranging formulas or equations to isolate or solve for a specific variable can be useful when solving applied problems or when performing the same calculations repeatedly with different values of the variables.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

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RECAP Name

Date

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Math 1 ▸ M1 ▸ TB ▸ Lesson 10

b. Using the formula SA = 2lw + 2(l + w)h, rearrange the variables to isolate h. Recall that SA = 108, l = 6, and w = 4.

SA = 2lw + 2( l + w ) h

Rearranging Formulas

SA − 2lw = 2( l + w ) h

The rearranged formula is equivalent to the given formula.

SA − 2lw _______ =h

In this lesson, we •

compared solving equations with more than one variable to solving equations with one variable.

•

rearranged formulas to solve for a specific quantity.

2(l + w)

108 − 2(6) (4) __________ =h

Generally assume that shapes like rectangular prisms have positive side lengths. So the possibility of dividing by 0 is not a concern in the

Examples

2( 6 + 4 )

108 − 48 _______ =h 20

rearranged formula.

1. One formula for the surface area of a rectangular prism with length l, width w, and height h is SA = 2lw + 2(l + w)h. Suppose the surface area is measured in square inches and the length and width are each measured in inches.

60 __ =h 20

3=h The height of the rectangular prism is 3 inches.

a. Find the height of a rectangular prism when SA = 108, l = 6, and w = 4.

choose.

SA = 2lw + 2(l + w)h 108 = 2 ( 6 ) ( 4 ) + 2 ( 6 + 4 ) h 108 = 48 + 20 h 60 = 20 h

We get the same answer regardless of which arrangement of the formula we

2. Solve each equation for n. Substitute the known values into the formula. Then solve the equation.

Equation Containing One Variable n−6 ____ =4 10

3=h

n−6 ____ =4 10

The height of the rectangular prism is 3 inches.

n − 6 = 40

Related Equation Containing More Than One Variable n−a ____ = c where b ≠ 0 b

Apply the same properties of equality to solve for n.

n = 46

n−a ____ =c b

n − a = bc n = bc + a

3. Rewrite the equation in slope-intercept form, y = mx + b.

_2 x − 3y = 24 3

2_ x − 3 y = 24 3

− 3 y = − 2_ x + 24 3

y = 2_ x − 8 9

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RECAP

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TB ▸ Lesson 10

PRACTICE Name

Date

For problems 3 and 4, solve the equation for the given variable. Equation Containing One Variable

1. The formula for the perimeter of a rectangle with length l and width w is P = 2(l + w). a. Find the width of a rectangle when P = 92 and l = 13. The measurements are in centimeters. The width of the rectangle is 33 cm.

3. Solve 3 = _____ for m.

b. Using the formula P = 2(l + w), isolate w first. Then find the width of the rectangle when P = 92 and l = 13. P __ − l = w 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 10

m+2 5

4. Solve 13 − 5x = 3x − 3 for x.

The width of the rectangle is 33 cm.

Related Equation Containing More Than One Variable

Solve t = _____ s for m where s ≠ 0. m+n

m = ts − n

Solve b − ax = cx − d for x. b+d x = ____ where c + a ≠ 0 c+a

2. The formula for the volume of a cylinder with radius r and height h is V = πr 2h. Suppose the volume is measured in cubic inches and the radius and height are each measured in inches. a. Find the height of a cylinder when r = 4 and V = 112π.

5. Use problems 3 and 4 to compare each equation containing one variable to its related equation containing more than one variable. Explain how they are similar and how they are different.

The height of the cylinder is 7 inches.

The equations are similar because I use the same properties and operations to solve them. The equations are different because in the equation with one variable I ended up with a solution, but in the related equation I don’t know the value of any variable. Another difference is that with one-variable equations, I don’t need to make sure I don’t divide by 0. When solving equations with more than one variable, I need to know what values are excluded to make sure I don’t divide by 0.

b. Using the formula V = πr 2h, isolate h first. Then find the height of the cylinder when r = 4 and V = 112π. V ___ =h 𝜋r 2 The height of the cylinder is 7 inches.

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11. The formula for the area of a trapezoid is A = (____)h. 2 Rearrange the formula to isolate b. a+b

For problems 6–11, rearrange the formula by isolating the given variable. 6. The formula for the perimeter of a square is P = 4s, where s is the side length. Rearrange the formula to isolate s.

2A ___ − a = b h

P __ = s 4

a For problems 12–15, rewrite the equation in slope-intercept form, y = mx + b.

7. The formula for the circumference of a circle with radius r is C = 2πr. Rearrange the formula to isolate r. 2𝜋

C __ =r

8. The formula for the force on an object with mass m and acceleration a is F = ma. Rearrange the formula to isolate m.

12. 2x + 4y = 8 y = − _1 x + 2 2

13. −3x + 7y = 14 y = 3_ x + 2 7

14. 6 x + 1_ y = 2 3

3 15. _ x − 5 y = 20 4 3 x−4 y = __ 20

y = −18x + 6

__F = m a

16. A snow cone consists of flavored ice that fills a right circular paper cone and sits on top of the cone in the shape of a hemisphere. The volume of a snow cone is estimated by the formula

V = 1_ 𝜋r 2h + 2_𝜋r 3.

9. The formula for the perimeter of a triangle with side lengths a, b, and c is P = a + b + c. Rearrange the formula to isolate b.

a. Rearrange the formula to isolate h.

P−a−c=b

3V − 2𝜋r 3 ________ =h 𝜋r 2

10. The formula for the surface area of a prism is SA = 2B + ph, where B represents the area of the base, p represents the perimeter of the base, and h represents the height of the prism. Rearrange the formula to isolate h. SA − 2B _______ =h p

b. Use the formula you wrote in part (a) to estimate the height of the paper cone of a snow cone that has an estimated volume of 17.08 cubic inches and a diameter of 3 inches. The height of the paper cone of the snow cone is approximately 4.25 inches.

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17. When is it useful to rearrange a formula?

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Math 1 ▸ M1 ▸ TB ▸ Lesson 10

21. Use the number line to answer each question.

It might be useful to rearrange a formula if you need to make repeated calculations for the same variable or if the calculations are difficult to make when solving an equation.

–11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

a. Which points, if any, correspond to a number with an absolute value of 3? Explain. Points C and E each correspond to a number with an absolute value of 3 because they are each 3 units from 0 on the number line.

Remember For problems 18 and 19, solve the equation. 19. 4_ t − 2_ t = − 2 3 9

18. 9 − 1.6t = −11

− 9_

12.5

b. Which points, if any, correspond to a number with an absolute value that is greater than 3? Explain.

Points A, B, and F each correspond to a number with an absolute value that is greater than 3 because they are each more than 3 units from 0 on the number line.

20. Write the property or operation used for each step in solving the equation 5(x + 4) + 6x = 11 − 3x.

5(x + 4) + 6x = 11 − 3x 5x + 20 + 6x = 11 − 3x

Distributive property

5x + 6x + 20 = 11 − 3x

Commutative property of addition

11x + 20 = 11 − 3x

Addition of like terms

14x + 20 = 11

Addition property of equality

14x = − 9

Addition property of equality

9 x = − __ 14

c. Which points, if any, correspond to a number with an absolute value of −7? Explain. No points correspond to a number with an absolute value of −7. The absolute value of a number is the distance of that number from 0 on the number line, and distance cannot be negative.

Multiplication property of equality

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Teacher Edition: Math 1, Module 1, Topic B, Lesson 11 LESSON 11

Solving Linear Inequalities in One Variable Solve inequalities and graph the solution sets on the number line.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

Name

EXIT TICKET

Date

Find the solution set of each inequality. Write the solution set by using set-builder notation, and then graph the solution set on the number line. 1. 6x − 5 < 7x + 4

6x − 5 < 7x + 4 −5 < x + 4 −9 < x {x | x > −9} x –20 –18 –16 –14 –12 –10

–8

–6

–4

–2

Key Questions • How can we determine what conditions make an inequality true?

10 − 3 x ≤ − 6( x − 2 ) 10 − 3 x ≤ − 6 x + 12 10 + 3 x ≤ 12 3x ≤ 2

• How are the properties of equality and the properties of inequality similar? How are they different?

x ≤ 2_

Achievement Descriptors

2_ {x x ≤ 3 }

Through a cycle of independent think time, collaboration, and discussion, students classify a set of inequalities as always, sometimes, or never true. Students determine the conditions for which each inequality is true, which provides them the opportunity to formalize the properties of inequality. Students organize the properties of inequality in a tree map and then use the properties to solve inequalities and graph the solution sets on the number line. By the end of the lesson, students work in groups to write and solve inequalities to represent a real-world context.

• What does it mean if an inequality is sometimes true?

2. 10 − 3x ≤ −6(x − 2)

–2

Lesson at a Glance

–1

Math1.Mod1.AD4 Create equations and inequalities in one variable and use them to solve problems. (A.CED.A.1)

x 1

Math1.Mod1.AD11 Solve linear inequalities in one variable. (A.REI.B.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Naming the Properties of Inequality

• None

• Solving Inequalities

Lesson Preparation

• Making the Grade

• None

Land 10 min

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Fluency

Teacher Note

Solve One- and Two-Step Inequalities Students solve one- and two-step inequalities to prepare for solving multi-step inequalities and graphing the solution sets on a number line.

Instead of this lesson’s Fluency, consider administering the Solve One- and Two-Step Inequalities Sprint. Directions for administration can be found in the Fluency resource. EUREKA MATH2

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Inequalities

Directions: Solve each inequality for n. 1.

n+8>6

n > −2

4 ≤ −3 + n

7≤n

5n > 25

n>5

n ≤9 2

24 ≥ 6n − 12

n ≤ 18

4(n + 1) < −48

n < −13

210

Number Correct:

Solve each inequality for x. 1.

x + 2 > 10

17.

5x < −20

x + 4 > 10

18.

−4x < −20

x + 6 > 10

19.

2x < −20

x−3<9

20.

−2x < −20

x−5<9

21.

−x < −20

x−7<9

22.

1 x < −2 2

2x < 18

23.

−1_ x < 3

3x > 18

24.

−1_ x > −2

10 + x > 15

25.

3(x + 3) > 18

10.

10 + x > −15

26.

−4(x + 3) > 24

11.

−10 + x > −15

27.

−4x − 12 > 24

12.

−10 + x > 15

28.

2 x − 6 < 12 3

13.

4x > −4

29.

−6 − 2_ x < 12

14.

4x > 8

30.

2 (x + 9) < −12 3

15.

4x > −16

31.

−12 − _ x < −6

16.

4x > 32

32.

3 (x + 20) < 6 5

386

3 5

6≥n

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

Students classify inequalities as always, sometimes, or never true. Display problem 1. Use the Always, Sometimes, Never routine to engage students in constructing meaning and discussing their ideas. Give students 1 minute of silent think time to evaluate whether the statement is always, sometimes, or never true. Have students discuss their thinking with a partner. Circulate as students talk, and identify a few students to share their thinking. Repeat the routine for problems 2–4. For problems 1–4, let x be a real number. Determine whether the statement is always, sometimes, or never true. 1. 3x + 1 > 3x + 5 This statement is never true. 2. 3x + 8 > 3x + 5 This statement is always true.

Teacher Note In grade 7, students learn that solving an inequality reveals the constraints on the values that make the inequality a true statement.

3. 4x − 12 > 3x + 5 This statement is sometimes true.

1 4. _ x + 3 > x + 8 2

This statement is sometimes true. Invite identified students to share their answers with the class. Encourage them to provide examples and nonexamples to support their claims. Elicit the following details through the class discussion: • Problem 2 is always true because any value of x makes the statement true.

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• Problem 3 is sometimes true. For example, the statement is true when x is 20 but false when x is 0. • Problem 4 is sometimes true. For example, the statement is true when x is −20 but false when x is 0. Today, we will solve a variety of inequalities and graph their solution sets on number lines.

Learn Naming the Properties of Inequality Students formalize the properties of inequality and use them to solve inequalities. Have students solve the inequalities for problems 1–3 if they did not do so during Launch. Circulate as students work, and provide support for problem 3 as needed by using the following prompts: • The statement 4x − 12 > 3x + 5 is sometimes true. If 4x − 12 is sometimes greater than 3x + 5, then what happens the other times? • For what values of x do you think 4x − 12 might be greater than 3x + 5? Once students finish, confirm answers and debrief each problem by asking the following questions: • What steps did you take to solve the inequality? • How would we write the solution set by using set-builder notation? Have students turn and talk about whether they are convinced that adding the same values to both sides of the inequality will preserve the solution set of the original inequality. Encourage them to justify their reasoning.

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There is a property of equality that tells us that adding the same value to both sides of an equation ensures its solution set is preserved. However, because this is an inequality, we cannot apply a property of equality. We now need to define the addition property of inequality. Have students record the description and an example of the addition property of inequality in the Naming the Properties of Inequality tree map. The multiplication property of inequality will be addressed later in this lesson. Complete the statements in the tree map, and provide examples of each property.

For expressions

a and b and real number c: If a > b and c > 0, then ac > bc.

Example

If –5 > 12 x, then

2(–5) > 2( 12 x).

Language Support

Multiplication

a + c > b + c.

4x – 12 – 3x > 3x + 5 – 3x.

Ask the following questions to promote MP7:

• How can what you know about the properties of equality help you with the properties of inequality?

Addition For expressions a, b, and c: If a > b, then

Example

Students look for and make use of structure (MP7) when they use the properties of equality to formalize and use the properties of inequality.

• How are equations and inequalities related? How can that relationship help you find the solution set of an inequality?

Properties of Inequality

If 4x – 12 > 3x + 5, then

Promoting the Standards for Mathematical Practice

For expressions

a and b and real number c: If a > b and c < 0, then ac < bc.

Example If – 12 x > 5, then

–2(– 12 x) < –2(5).

The consequence of the multiplication property of inequality when multiplying by negative values can be difficult to communicate. When displaying solution steps, consider annotating solution steps to emphasize reversing the inequality symbol by drawing an arrow. Use of the annotation may be easier for students to understand than a verbal explanation.

1x+3>x+8 2 – 12 x + 3 > 8 – 12 x > 5

(–2) · (– 12 x) < (–2) · 5 x < –10

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To support students with remembering when to reverse the inequality symbol, display each of the following statements and have students give a thumbs-up or thumbs-down to indicate whether the statement is true or false. • −3 = 5 • −3(1) = 5(1) • −3(−1) = 5(−1) • −3 < 5 • −3(1) < 5(1)

UDL: Representation The tree map graphic organizer highlights the relationships associated with the addition property of inequality and the multiplication property of inequality. Throughout the lesson, assist students in completing each section of the graphic organizer by adding examples.

• −3(−1) < 5(−1) Display −3(−1) ______ 5(−1). Have students use their hands to make a greater than or less than symbol to correctly complete the statement. When multiplying both sides of an inequality by a negative number, reverse the inequality symbol to maintain a true inequality statement. Direct students to problem 4. Ask half of the class to solve by first adding −_  1 x to both sides 2 of the inequality, and ask the other half of the class to solve by first adding −x to both sides. Compare the different solution paths. Formalize by writing the solution set of problem 4 by using set-builder notation.

What is the solution set of the inequality _1 x + 3 > x + 8? Write the solution set 2 by using set-builder notation.

{x | x < −10}

Everyone added some value in the first step and likely in the first few steps. Besides addition, what other steps did you and your classmates take to solve the inequality? Some of us multiplied both sides by 2 and some of us multiplied both sides by −2 in our final step in solving the inequality. Those two different steps may not lead to the solution the same way that they would when solving an equation. Why? Multiplying both sides of an equation by any value besides 0 preserves the solution set. Inequalities do not always behave in the same way. I can multiply both sides of an

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inequality by a positive value and create a new inequality with the same solution set. I cannot multiply by a negative value to produce a new inequality with the same solution set as the original inequality without reversing the direction of the inequality symbol. Have students revisit the Naming the Properties of Inequality tree map. Name the multiplication property of inequality. Specify that there are conditions to consider before summarizing the property: one for multiplying by a positive value and another for multiplying by a negative value. Encourage students to consider a third condition where the value is neither positive nor negative with the following question. Consider the inequality −3 < 5 again. If we multiply both sides of the inequality by 0, is the resulting inequality true? No. The result is 0 < 0, which is not true. In general, multiplying both sides of an inequality by 0 may not preserve the solution set. Direct the students to the box labeled Multiplication in the tree map, and have students summarize the properties of inequality and record the descriptions and examples. Once students finish recording their descriptions and examples, invite them to summarize the properties of inequality in their own words. Ask them to cite evidence from the tree map as they summarize. Elicit the following details: • The addition property of inequality states that adding an expression to both sides of an inequality creates a new inequality. If we can evaluate the expression for every value of the variable, then the solution set of the new inequality is the same as the original inequality. • The multiplication property of inequality states that multiplying both sides of an inequality by the same number creates a new inequality. If the number is positive, then the new inequality has the same solution set as the original inequality. If the number is negative, then reversing the inequality symbol of the new inequality results in an inequality with the same solution set as the original inequality. • We can always rearrange the inequality by using the addition property of inequality so that the variable term is positive.

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Solving Inequalities Students practice solving inequalities and representing their solution sets. Transition students to problems 5–10. Allow students to work in pairs to complete the problems. Circulate as students work, paying careful attention to how students graph the solution sets and how they apply the properties of inequality. If students need support to graph the correct interval, inform them that they can find the endpoint of the graph of the solution set, test a value on an interval by using the original inequality, and shade the interval if the inequality is true. For problems 5–9, find the solution set of the inequality. Write the solution set by using set-builder notation, and then graph the solution set on the number line.

Teacher Note If students have difficulty applying the multiplication property of inequality for negative values, recommend that they use the addition property of inequality instead.

4 − 3 x > 10 4 > 10 + 3 x −6 > 3x −2 > x This eliminates the need to reverse the inequality symbol.

5. 3(x − 4) > 18

3 ( x − 4 ) > 18 3 x − 12 > 18 3 x > 30 Solution set:

{x | x > 10}

x > 10

x –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

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6. 6 − 2x ≥ −4

Solution set:

6 − 2x ≥ −4

− 2 x ≥ − 10 x≤5

{x | x ≤ 5}

x –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 5 2 7. _ g − 3 ≥  1_ g +  _ 3 3 3

2_ g − 3 ≥ 1_ g + _ 3

5 3

g ≥ 14

__ 2_ g ≥ 1_ g + 14

__ 1_ g ≥ 14 Solution set:

{g | g ≥ 14}

g –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

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Math 1 ▸ M1 ▸ TB ▸ Lesson 11

8. −6(x − 1) < 6 − 6x

Differentiation: Support

−6(x − 1) < 6 − 6x −6x + 6 < 6 − 6x 6<6 Solution set:

{}

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 9. 2(x − 3) + x < 3(x + 4)

2 ( x − 3 ) + x < 3( x + 4 )

If students need support identifying the solution sets of the inequalities in problem 8 or 9, ask them to determine whether the inequality in the final step in each problem is true. Summarize how to determine if the solution set of an inequality is the empty set or the set of all real numbers. Emphasize that students interpret the final step of solving an inequality the same way that they interpreted the final step of solving an equation.

2 x − 6 + x < 3 x + 12 3 x − 6 < 3 x + 12 − 6 < 12 Solution set:

ℝ

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

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Math 1 ▸ M1 ▸ TB ▸ Lesson 11

10. Fin and Bahar each solved the inequality 8 − 5x > −10. They found different solution sets. Fin’s Work

Bahar’s Work

8 − 5 x > − 10

8 − 5 x + 5 x > − 10 + 5 x

8 − 8 − 5 x > − 10 − 8

8 + 10 > − 10 + 10 + 5 x 18 > 5 x

_ _ ( )

( )

__  > x 18 5

{x __   > x} 5 18

− 5 x > − 18

(− _1)(−5x ) > (− _1)( − 18 ) 5

x > __ 

18 5

{x x > __ } 18 5

a. Whose solution set is correct? Fin’s solution set is correct. b. Explain the mistake. Bahar multiplied each side of the inequality by a negative number but did not reverse the inequality symbol. When most students have completed the problems, reveal the answers. During the debrief, ask students to justify their steps. Display −6(x − 1) ≤ 6 − 6x. Assess student understanding by asking the following question. Would your answer change if the inequality symbol in problem 8 was replaced with the less than or equal to symbol? Why?

Yes. If the inequality symbol was replaced with the less than or equal to symbol, my final step would result in 6 ≤ 6. This statement is always true because this new inequality symbol allows the expression on the left to be equal to the expression on the right. The solution set of this new inequality would be all real numbers.

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Making the Grade Students create and solve an inequality in context. Present problem 11, and read the prompt aloud. Allow students to work in groups to complete the problem. Circulate as students work, monitoring their ability to decontextualize to write and solve an inequality and then recontextualize to interpret the solution set. Debrief student responses. Showcase any variety present in student work. 11. Every student in Zara’s science class will have 5 test scores by the end of the grading period. Zara sets a goal to have at least a 90 average for the grading period. Her first four test scores are 97, 85, 96, and 89. Write and solve an inequality to find the lowest score Zara must earn on the fifth test to meet her goal.

Differentiation: Support If students need support with writing an inequality from the context, ask them to first write an expression for Zara’s average score on the 5 science tests. Then, ask them to use this expression to write an equation to represent an average score of exactly 90. Finally, they can transition to writing an inequality.

Let x represent Zara’s fifth test score.

________________   ≥ 90 97 + 85 + 96 + 89 + x 5

97 + 85 + 96 + 89 + x ≥ 450 367 + x ≥ 450 x ≥ 83

Zara must earn a score of at least 83 on the fifth test to meet her goal.

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UDL: Representation

Debrief 5 min Objective: Solve inequalities and graph the solution sets on the number line. Use the following prompts to guide a discussion about solving inequalities. What does it mean if an inequality is sometimes true? How can we determine what conditions make the inequality true? If an inequality is sometimes true, some values will make the inequality a true statement when they are substituted for the variable. The conditions that make the inequality true can be determined by solving the inequality. Problems 1 through 4 required us to identify whether each inequality is always true, sometimes true, or never true when values are substituted for the variable. How do we represent inequalities that are always, sometimes, or never true on number lines? If the entire number line is shaded, then the inequality is always true and the solution set is all real numbers.

As the class summarizes solving inequalities, consider using a graphic organizer to compare the graphic representations of inequalities. Provide a three-column chart with the labels Always True, Sometimes True, and Never True. Students can record a description and include a sketch for each type of graph. Inequalities

Always True

Sometimes True

Never True

The solution set is the set of all real numbers because every value makes it true.

Only shade the portion of the number line where the values make the inequality true.

The solution set is the empty set because the graph of the empty set has no shading.

{x 1 x ≥ 5}

{}

If the graph has no shading, then the inequality is never true and the solution set is the empty set. For inequalities that are sometimes true, only the portion of the number line that includes the values that make the inequality true is shaded. The marking used to indicate the endpoint of the graph of the solution set tells us whether the value of the endpoint of the graph is included in the solution set.

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How are the properties of equality and the properties of inequality similar? How are they different? The addition property of equality and the addition property of inequality are the same. We can add the same expression to both sides of an equation or inequality and preserve its solution set. The multiplication property of equality and the multiplication property of inequality are the same only if we multiply both sides of the equation or inequality by a positive number. They are different, because multiplying both sides of an inequality by a negative number requires reversing the inequality symbol to create a new inequality that has the same solution set as the original inequality. Neither multiplication property is guaranteed to preserve the solution set when multiplying both sides by 0.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

222

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

2. One number is 5 less than half another number. The numbers have a sum of at most −68. What are the largest numbers that satisfy these conditions? Let x represent the first number. Then _1 x − 5 represents 2 the second number.

Solving Linear Inequalities in One Variable

x + 1_ x − 5 ≤ − 68

In this lesson, we •

formalized the addition and multiplication properties of inequality.

•

solved inequalities and graphed the solution sets on the number line.

_3 x − 5 ≤ − 68 2

_3 x ≤ − 63 2

Examples

1. Find the solution set of the inequality 10 − 9w ≥ 28. Write the solution set by using set-builder notation, and then graph the solution set on the number line.

10 − 9 w ≥ 28 − 9 w ≥ 18

w ≤ −2

10 ≥ 9 w + 28

10 − 9 w ≥ 28

Reverse the inequality symbol

− 18 ≥ 9 w

when applying the multiplication property of inequality with a negative value.

−2 ≥ w

{w | w ≤ −2} –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

Apply the addition property of inequality so that the coefficient of the variable term is positive.

Method 2:

Method 1:

Substitute −42 for x in _1 x − 5.

x ≤ − 42

The phrase at most −68 indicates that the sum of the numbers must be less than or equal to −68.

The largest number less than or equal to −42 is −42.

_1 ( − 42 ) − 5 = − 21 − 5 = − 26 2 The largest numbers that satisfy the given conditions are −42 and −26.

w 4

9 10

161

162

RECAP

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Math 1 ▸ M1 ▸ TB ▸ Lesson 11

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

PRACTICE Name

Date

6. Why is graphing solution sets of one-variable inequalities on number lines more helpful than graphing solution sets of one-variable equations on number lines? We graph solution sets of inequalities on number lines because they’re usually more difficult to visualize than solution sets of equations. A graph is a convenient way to represent the entire solution set of an inequality. Many equations generally have a limited number of solutions, and listing them is an efficient way of representing the solution set of an equation.

For problems 1 and 2, graph the solution set on the number line. 1. {g | g > 4}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 2. {m | 3 ≤ m}

–10

–8

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

For problems 7–10, place the correct inequality symbol in the box. Write the property used in each case. 7. If y + 3 > 9, then y + 3 − 3

m –6

–4

–2

m , then 3 ( 17 ) 9. If 17 > __

{p | p < 0}

–4

–3

–2

–1

n –6

–5

{n | n > −5}

–4

–3

–2

–1

12. h − 7 < 3 Solution set:

h –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

224

≤

3(_1 k). 3

Multiplication property of inequality

–4

c –3

–2

–1

{h | h < 10} h

{}

−2

1 {c c > − 4 2 }

–5

10. If − 9 ≤ _1 k, then 3(−9)

m 3(__ ).

11. 2c > −9 Solution set:

8 ___ .

For problems 11–14, find the solution set of the inequality. Write the solution set by using set-builder notation, and then graph the solution set on the number line.

p –5

≥

Multiplication property of inequality

For problems 3–5, write the solution set that represents the graph shown on the number line.

–6

−2

Addition property of inequality

− 2x 8. If −2x ≤ 8, then ___

9 − 3.

–5 –4 –3 –2 –1

163

164

P R ACT I C E

9 10 11 12 13 14 15

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Math 1 ▸ M1 ▸ TB ▸ Lesson 11

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

13. 54 ≥ −6v

For problems 16–18, find the solution set of the inequality. Write the solution set by using set-builder notation, and then graph the solution set on the number line.

{v | v ≥ −9}

Solution set:

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

16. −6b + 4 < 10

v –15 –14 –13 –12–11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

{b | b > −1}

Solution set:

b –10

14. −9 ≤ 3_ f 4

2 17. _ u − 1_ > 2 5 4

{ f | f ≥ −12}

Solution set:

–8

Solution set:

f –20

–18 –16

–14

–12 –10

–8

–6

–4

–2

–6

–4

–2

| u > 5 5_8}

u 6

18. 2(g − 8) > 2g − 16 15. Write and solve an inequality that represents the statement that a number increased by 4 is greater than 16.

{}

Solution set:

Let x represent a number.

g x + 4 > 16 x > 12

–10

–8

–6

–4

–2

Any number greater than 12 is a solution to this statement. 19. Evan buys 3 yards of fabric that is the same price per yard. He needs to have at least $5 left after making his purchase. He has a $50 bill. What’s the greatest price per yard Evan can pay for the fabric? The greatest price per yard Evan can pay for the fabric is $15.

P R ACT I C E

165

166

P R ACT I C E

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Math 1 ▸ M1 ▸ TB ▸ Lesson 11

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

For problems 20–22, find the solution set of the inequality. Write the solution set by using set-builder notation, and then graph the solution set on the number line.

EUREKA MATH2

Math 1 ▸ M1 ▸ TB ▸ Lesson 11

23. One number is 8 more than twice the other number. If the numbers have a sum of at least −22, what are the smallest numbers that satisfy these conditions?

20. 2g + (g + 5) ≥ 6g − 10

The smallest numbers are −10 and −12.

{g | g ≤ 5}

Solution set:

g –10

3 21. _ ( 10 b − 6 ) ≥ 16 b 5

Solution set:

–8

–6

–4

–2

2__ q 2q + 4 < __ − 9 5 3

Solution set:

9 | b ≤ −__ 25 }

For problems 25 and 26, solve the equation.

25. − 3x + 21 − 8x = − 34

24. Twelve more than four times a number is at most 42 more than twice the number. Find the solution set.

Remember

– 9

{x | x ≤ 15}

–1

22.

26. 10x − 3 − 4x = − 1 _1 3

27. In △ ABC, the measure of ∠B is twice the measure of ∠A, and the measure of ∠C is 40° less than the measure of ∠A. What are the measures of the three angles? The measure of ∠A is 55°, the measure of ∠B is 110°, and the measure of ∠C is 15°.

q > 48 _3} 4 3

48 4 47

q 49

51 28. Order the following values from least to greatest.

| |

− 7, − 1_ , 0, |0.6|, − 3_ , |− 4|, 5, |6| 4

226

P R ACT I C E

167

168

P R ACT I C E

| |

|0.6|, − 1_4 , 5, 0, |6|, |− 4|, − 7, − 3_2

Teacher Edition: Math 1, Module 1, Topic C

Topic C Compound Statements Involving Equations and Inequalities in One Variable In topic C, students build upon the work done with equations and inequalities in topic B by examining and creating compound statements involving two linear equations or inequalities and by interpreting their solution sets. Students solve a variety of compound statements and graph the solution sets on a number line. Students then apply that concept to solving absolute value equations and inequalities. To begin topic C, students examine real-world situations with multiple constraints, requiring students to represent the situations with compound inequalities. The goal is for students to develop an understanding of mathematical compound statements, emphasizing the difference between statements connected by and and those separated by or. Students then participate in a digital lesson in which they first solve compound inequalities through a trial-and-error process. Students independently find solutions that are then compiled as a class to produce the solution set on a number line. They then find the solution sets algebraically. Students work through a set of compound inequalities, searching for the integer that satisfies all of the restrictions of the inequalities.

Clue B When I subtract 2 from my integer and then triple that diﬀerence, the result is strictly between –21 and 12.

–21 < 3(x – 2) < 12 –16 –14 –12 –10 –8 –6 –4 –2

228

10 12 14 16

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Math 1 ▸ M1 ▸ TC

Students use their knowledge of solving compound statements to solve absolute value equations and inequalities. A large number line placed on the classroom floor is used to connect students’ previous understanding of the definition of absolute value, from grade 6, to finding the solution sets of absolute value equations or inequalities. Students solve absolute value equations and determine when an absolute value equation has no solution, one solution, or two solutions. Students also solve absolute value inequalities by rewriting them as compound inequalities connected by and or or, with an emphasis on understanding when each word is used to connect the inequalities. In an optional lesson, students write and solve absolute value equations and inequalities that represent real-world situations.

−2|6x − 8| = −40 |6x − 8| = 20

6x − 8 = 20 or −(6x − 8) = 20 −(6x − 8) = 20 6 x = 28 14 x = __ 3

6x − 8 = −20 6x = −12 x = −2

14 {−2, __ } 3

The work in topic C prepares students for module 2, as they transition from working with systems of linear equations and inequalities in one variable to those in two variables.

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Progression of Lessons Lesson 12 Solution Sets of Compound Statements Lesson 13 Solving and Graphing Compound Inequalities Lesson 14 Solving Absolute Value Equations Lesson 15 Solving Absolute Value Inequalities Lesson 16 Applying Absolute Value (Optional)

230

Teacher Edition: Math 1, Module 1, Topic C, Lesson 12 LESSON 12

Solution Sets of Compound Statements Describe the solution set of two equations or inequalities joined by and or or and graph the solution set on a number line. Write a compound statement to describe a situation.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Name

EXIT TICKET

Date

1. Match each compound statement to the graph of its solution set on the number line. A. x = 5 or x = −10 B. x < 5 and x > −10 C. x > 5 or x < −10 D. x ≥ 5 or x ≤ −10

E. x ≤ 5 and x ≥ −10

x –15

–10

–5

–15

–10

–5

x –15

–10

–5

–15

–10

–5

–15

–10

–5

Students extend their study of inequalities and equations by exploring those that can be written as compound mathematical statements. Students build a need for mathematical statements connected by and or or by using a familiar context: height requirements for amusement park rides. Then they use the truth value of compound English sentences to identify the truth values for compound mathematical statements. Partners work to solve and graph compound mathematical statements to summarize how the words and or or affect the solution set. Students end the lesson by representing real-world constraints with compound statements. This lesson formalizes the terms statement and compound statement.

Key Questions

• What is required for a compound statement connected by the word and to be true?

• What is required for a compound statement connected by the word or to be true?

Achievement Descriptors

2. The acceptable range of chlorine levels in swimming pools is at least 1 part per million and no more than 3 parts per million. Write a compound inequality for the acceptable range of chlorine levels in a swimming pool.

Math1.Mod1.AD5 Represent constraints by using equations and inequalities. (A.CED.A.3)

Let x represent the chlorine level in a swimming pool in parts per million. The acceptable range of chlorine levels is 1 ≤ x ≤ 3.

Lesson at a Glance

Math1.Mod1.AD6 Interpret solutions to equations and inequalities

in one variable as viable or nonviable options in a modeling context. 179

(A.CED.A.3)

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Determining Truth

• Colored pencil set, 3 different colors

• Graphing Solution Sets

Lesson Preparation

• Let’s Apply It

• None

Land 10 min

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Fluency Truth Values Students determine whether equations and inequalities are true or false to prepare for describing the solution sets of compound statements. Directions: Decide whether each statement is true or false. 1.

4 · 23 = 32

True

6 < −3(−2)

False

8 + x = 12 when x = −4

False

9 ≥ a − 6 when a = 15

True

14 = 2(5 − m) when m = 12

False

3y + 1 < −2 when y = −2

True

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Launch

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Students write inequality statements for a real-world situation. Use the following prompt to introduce the context in this activity. Epic Park is a theme park that is home to the Caterpillar Coaster. It is a roller coaster made for small children. Then have students turn and talk about the following question. What is a reasonable prediction for the height requirement in inches to ride the Caterpillar Coaster? Display the Epic Park interactive. Then display the description of the Caterpillar Coaster. As a class, discuss an inequality that would represent the height of a rider. What does Height Requirement: 36 inches actually mean? It means that a rider has to be at least 36 inches tall. How could we symbolically write that the height of a rider must be at least 36 inches? Let’s let h represent the height of a rider in inches.

h ≥ 36

Direct students to problem 1 and have them record the answer. Display the descriptions for the rides listed in problems 2–4 one at a time, ensuring that the height requirements are visible. Have students write inequalities to represent the height requirements for the rides.

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EUREKA MATH2

For problems 1–4, use one or more inequalities to represent the height requirements of a rider for the rides at Epic Park. Let h represent the height of a rider in inches.

h ≥ 36

1. Caterpillar Coaster

h ≥ 54

2. Mobius Loop

h ≥ 36 and h ≤ 54

3. Roaring River 4. Lil’ Dipper

h < 48 and h > 0 Debrief as a class by asking students to share the inequalities they wrote for each ride, and discuss and revise them as a class. Consider asking the following questions to guide the discussion. There are two height requirements, or conditions, for Roaring River. How did you approach writing an inequality for this ride? I wrote two inequalities because the rider must be at least 36 inches tall but can’t be taller than 54 inches. Would h < 48 be a precise description for the height requirement for the Lil’ Dipper ride? Explain. Heights cannot be negative, so we know h > 0. There is no minimum height listed, but it isn’t realistic that a very small child would be able to ride. Even a very small child would not have a height close to 0 inches. What would be an appropriate minimum height requirement for the Lil’ Dipper ride? Allow students to answer intuitively with a height that seems appropriate. Today, we will analyze statements with more than one condition, such as the height requirements for the Roaring River and Lil’ Dipper rides. 236

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Learn Determining Truth Students determine when mathematical statements connected by the words and or or are true by comparing them to English sentences with the same connecting words. Have students complete problems 5–8 and then read the definitions for statement and compound statement. For problems 5–13, determine whether each statement is true or false. 5. Right now, I am in math class, and I am in English class. False

Language Support Consider creating a tree map graphic organizer for compound statements to support comprehension of truth values. Include the rules for truth values and the examples of compound statements. The definition for statement and compound statement could be included at the top of the map.

6. Right now, I am in math class, or I am in English class.

Compound Statements

True (assuming students answer this question in math class or English class) 7. Ice is cold, and fire is hot. True 8. Ice is cold, or fire is hot. True A statement is a sentence that is either true or false, but not both. A compound statement consists of two or more statements connected by logical modifiers, like and or or.

and

True if both statements are true.

True if at least one statement is true.

Examples Ice is cold and fire is hot. TRUE

Examples Ice is cold or fire is hot. TRUE

3 + 5 = 8 and 5 < 7 – 1

TRUE

3 < 5 + 4 or 6 + 4 = 9

TRUE

10 + 2

FALSE

16 – 20 > 1 or 5.5 + 4.5 = 11

FALSE

12 and 8 – 3 > 0

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Once students finish, lead a class discussion by asking the following questions. Problems 5 and 7 both use the connecting word and. How do we determine whether the statement is true or false for these problems? For problem 5, I am only in math class. I can’t be in two different classes at the same time, so the statement in problem 5 is false. For problem 7, I know that both are true, so the statement is true. These are examples of compound statements. When two statements in a sentence are connected by the word and, how many statements have to be true to make the compound statement true? Both statements have to be true for the compound statement to be true. Problems 6 and 8 both use the connecting word or. How do we determine whether the compound statement is true or false for problem 6? For problem 6, I am in math class. Because I am in one of the classes listed, the statement is true. For problem 8, how many of the statements are true? Both statements are true. Do you think this makes the entire statement true or false? Turn and talk with a partner. When two true statements are separated by or, then the entire statement is true. When two statements in the sentence are connected by the word or, how many statements have to be true to make the compound statement true? At least one of the statements has to be true for the compound statement to be true. Encourage students to use their understanding of truth value in English sentences to complete problems 9–13.

238

Differentiation: Support If students are confused about the truth value for problem 8, consider using the following discussion for clarification: • Suppose we went to an ice cream store, and I said, “I can’t decide if I want chocolate or vanilla. Surprise me!” Getting one scoop of vanilla, one scoop of chocolate, or one scoop of each flavor are all actions that would fulfill my request.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Consider pairing students to discuss their reasoning. 9. 3 + 5 = 8 and 5 < 7 − 1 True

10. 10 + 2 ≠ 12 and 8 − 3 > 0 False

11. 3 < 5 + 4 or 6 + 3 = 9 True 12. 16 − 20 > 1 or 5.5 + 4.5 = 11 False 13. 16 + 20 > 1 or 5.5 + 4.5 = 11 True Confirm answers and initiate a class discussion to summarize the following truth value requirements for compound statements connected by and or or: • The word and has the same meaning in a compound mathematical statement as it does in an English sentence. • For a compound statement connected with and to be true, both statements in the sentence have to be true. • The word or has a similar meaning in a compound mathematical statement as it does in an English sentence, with one important difference. In an English sentence, the word or commonly means that if one, not both, of the statements is true, the compound statement is true. In mathematics, the word or means the compound statement is true if either statement is true or if both statements are true. • For a compound statement connected by or to be true, at least one statement in the sentence needs to be true.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Graphing Solution Sets Students graph the solution sets of one-variable compound equations and inequalities on the number line. Guide students to problem 14. Have students discuss with a partner their approach for solving the problem. Inform students that they do not actually solve the problem yet. Invite one or two pairs to share their approaches for solving problem 14. The compound statement has or, so at least one part of the statement needs to be true. The solution set consists of all values of y, where y + 8 = 3 is true or y − 6 = 2 is true, so we need to find the solution to both equations to find the possible values of y. Direct students to work with their partners to complete problem 14. After students have finished, call the class back together to discuss the solution set and graph. Direct students to continue their work through problem 16. Then display the correct answers, and summarize the process of writing the solution sets by using set notation and graphing the solution sets. For problems 14–16, solve each compound statement. Write the solution set by using set notation. Then graph the solution set on the given number line. 14. y + 8 = 3 or y − 6 = 2

y = −5 or y = 8; {−5, 8}

y –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

240

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

15. d − 6 = 1 and d + 2 = 9

d = 7 and d = 7; {7}

d –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 16. 2w − 8 = 10 and w > 9

w = 9 and w > 9; { }

w –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 Give each student three different colored pencils. Ask students to complete problems 17(a) through 17(d). 17. x < 3 and x > −1

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 a. Using a colored pencil, graph the inequality x < 3. b. Using a different colored pencil, graph the inequality x > −1. c. Using a third colored pencil, darken the section of the number line where x < 3 and x > −1.

{x | x < 3 and x > −1}

d. Write the solution set by using set-builder notation.

241

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Once students have finished, discuss the results. For the compound statement x < 3 and x > −1 to be true, what has to be true about x? The value of x has to be both greater than −1 and less than 3. For which part of this problem did you color the solution set of the compound statement? I colored the solution set of the compound statement in part (c). Draw a new number line below the one you completed that displays only the solution set. List some of the solutions to the compound statement. 3 Sample: 0, 0.25, 2, −_ 7 If students do not offer any examples of non-integer solutions, prompt them to do so. Then direct them to answer part (e). e. How many solutions to this compound inequality are there? Explain. There are infinitely many solutions to this compound inequality because there are infinitely many real numbers between −1 and 3. For example, 0.1, 0.01, 0.001, 0.0001, … f. Write the given compound inequality in a more efficient way.

−1 < x < 3

UDL: Representation Consider annotating the number line to draw students’ attention to the section of the line that represents the solution. For example, display problem 17 as the results are discussed. Annotate the number line by darkening the section where x < 3 and x > −1, and label that section as the solution set. Encourage students to make the same notation on their number lines. Then draw a new number line that displays only the solution set.

Discuss parts (e) and (f) as a class. Notice in part (e) that the statement x < 3 and x > −1 is referred to as a compound inequality. A compound inequality is a type of compound statement that involves two inequalities. How many solutions to this compound inequality are there? Explain your answer. There are infinitely many solutions to this compound inequality because there are infinitely many real numbers between −1 and 3. For example, 0.1, 0.01, 0.001, 0.0001, …

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Display the following:

x < 3 and x > −1

−1 < x < 3

These two statements both represent the same compound inequality. Reading the second statement, we say that x must be greater than − 1 and x must be less than 3. As you read the statement, point to each part of the compound inequality. Point out that when the compound inequality is written as − 1 < x < 3, the x is between the − 1 and the 3, just as students saw on the graph of the solution set of x < 3 and x > − 1. Note that − 1 < x < 3 is a strict inequality because it is a representation that x is strictly between − 1 and 3 but cannot be equal to − 1 or 3. Display the following:

3 > x > −1 Have students read the statement with a partner. This statement also represents the same compound inequality. Reading the statement, we say that x must be less than 3 and x must be greater than − 1. Writing a compound inequality like this is valid. However, it is convention to write the numbers from least to greatest by using less than symbols. Have students write the answer for part (f), and then direct them to work on problem 18. 18. x < − 1 or x > 3

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 a. Using a colored pencil, graph the inequality x < − 1. b. Using a different colored pencil, graph the inequality x > 3.

{x | x < − 1 or x > 3}

c. Write the solution set by using set-builder notation.

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When students have finished, debrief with the following discussion. What values of x make the compound inequality x < − 1 or x > 3 true? The compound inequality is true when the value of x is less than − 1 or greater than 3. Remind students that for a compound statement separated by or to be true, at least one of the statements must be true. So if both statements are true, then the compound statement is true. Is it possible for a value of x to make both inequalities of the compound inequality x < − 1 and x > 3 true? Explain. No. There is no value of x that is both less than − 1 and greater than 3. Display the following:

x < − 1 or x > 3

3 < x < −1

Have students think–pair–share about the following question. Do you think the second statement makes sense? Why? No. It seems to be stating that x is greater than 3 and less than − 1, which is impossible. Cross out the second statement. When we write a compound inequality such as − 1 < x < 3 there is an implied and. Compound inequalities separated by or cannot be written this way. Give students time to work on problems 19–22, and then allow students to compare answers with a partner. Encourage students to continue using colored pencils if needed. Circulate as students work, and listen for opportunities to guide student thinking by asking questions such as the following: • Is there a more efficient way to write your solution set? • How can you use your graph to verify your solution set?

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For problems 19–22, write the solution set of the compound statement by using set-builder notation. Then graph the solution set on the number line. 19. m > 2 or m > 6

{m | m > 2}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 20. m > 2 and m > 6

{m | m > 6}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10

21. x ≥ − 5 or x ≤ 2

ℝ

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 22. x > − 5 and x < 2

{x | − 5 < x < 2}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 Confirm answers by comparing the differences in the solution sets when the inequalities are separated by or or and.

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Let’s Apply It Students represent situations with compound statements. Have students think independently about the situations given in problems 23–25. Give students time to discuss the problems with their partner. Encourage them to define variables and then write compound statements that represent each situation. Call the class back together. Then display the answers and discuss the problems. 23. Each student has to present a speech in English class. The guidelines state that the speech must be at least 7 minutes, but it must not exceed 12 minutes. Write a compound inequality for the possible lengths of the speech. Let s represent the length of a speech in minutes.

7 ≤ s ≤ 12

24. The element mercury has a freezing point of −37.9°F and a boiling point of 673.9°F. Mercury is liquid between these temperatures. Write a compound inequality for the temperatures at which mercury is a liquid. Let m represent temperature in degrees Fahrenheit.

−37.9 < m < 673.9 25. The internal temperature of a cooked steak must be at least 145°F when warm or below 40°F when refrigerated. At any other temperature, bacteria can grow and make the steak unsafe to eat. Write a compound inequality for the internal temperature of a cooked steak that is safe to eat. Let t represent temperature in degrees Fahrenheit.

t ≥ 145 or t < 40

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Promoting the Standards for Mathematical Practice When students decontextualize problems by using compound inequalities, they are reasoning abstractly (MP2). Ask the following questions to promote MP2: • What does this context tell you about whether to use strict or not strict inequalities to write your compound statements? • What does this context tell you about whether to connect your inequalities with and or or ? • Does your answer make sense in the context of this problem?

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Land Debrief 5 min Objectives: Describe the solution set of two equations or inequalities joined by and or or and graph the solution set on a number line. Write a compound statement to describe a situation. Use the following prompts to guide a discussion about compound statements. What is required for a compound statement connected by the word and to be true? Both statements must be true for the compound statement to be true. What is required for a compound statement connected by the word or to be true? At least one statement must be true for the compound statement to be true. Some inequalities are actually compound statements without including the words and or or. What are some examples of these inequalities, and how can we rewrite them by using and or or ?

−3 < x ≤ 5 means x > −3 and x ≤ 5. 6 ≥ k means k < 6 or k = 6.

7 < m < 15 means m > 7 and m < 15.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

RECAP Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

3. c ≥ −7 and c ≤ 2

Solution Sets of Compound Statements

The connecting word is and. Solutions must make both parts of the compound statement true.

In this lesson, we

Terminology

•

described solution sets of two equations or two inequalities joined by and or or.

•

graphed solution sets of two equations or two inequalities joined by and or or on a number line.

•

wrote compound statements to describe contexts.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

A statement is a sentence that is either true or false, but not both.

For problems 1–3, graph the solution set of the compound statement.

9 10

The graph of the solution set is the region where the graphs of c ≥ −7 and c ≤ 2 overlap.

A compound statement consists of two or more statements connected by logical modifiers, like and or or.

Examples

4. Write a compound inequality for the graph.

m –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

−4 ≤ m ≤ 6

1. x + 1 = 3 or x − 2 = 5

9 10

Another way to write this compound inequality is m ≥ −4 and m ≤ 6.

5. Consider the compound inequality −5 < x ≤ 5.

The connecting word is or. Solutions must make at least one part of the compound statement true.

a. Rewrite as a compound statement separated by and or or.

x > −5 and x ≤ 5

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

b. Graph the compound inequality.

2. w > 5 or w < −4

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 w

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

There are infinitely many solutions to the compound inequality. 6. Write a compound inequality to represent the situation. The car repair is expected to cost at least $100 but no more than $150. Let r represent the cost for the car repair in dollars.

100 ≤ r ≤ 150

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c. How many solutions are there to the compound inequality?

9 10

The graph of the solution set consists of both regions.

181

182

RECAP

The phrases at least and no more than indicate that the inequality should include the values 100 and 150. © Great Minds PBC

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

For problems 7–16, graph the solution set of the statement on the given number line. 7. x = −2 or x = 8

For problems 1–6, write whether the compound statement is true or false.

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

1. 6 + 1 = 7 and 8 − 6 = 2

9 10

True 8. w > 5

2. 5 > 3 or 7 < 1

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

True

9. g ≤ 2

1 3. 3 · 4 = 12 and _ · 7 = 4 2 False

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 21 4. 9 · 1 < 22 or __ > 6 3 True

10. k > 5 or k < 2

k 5.

_1 · 9 ≠ 3 and 0.2 · 10 > 5 3

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 11. m ≤ −8 or m ≥ −1

False

1 6. 4 − 10 > 6 or _ · 16 ≤ 3 4 False

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

12. d < 9 and d > 7

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

For problems 17 and 18, write a compound inequality for each graph.

d –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

17.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

b < 1 or b > 3 13. v > −3 and v < 5

v –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

–4

p < 2 or p > 2

c –3

–2

–1

–4

19. Consider the compound inequality 0 < x < 3.

a. Rewrite the compound inequality as a compound statement connected by and or or.

x > 0 and x < 3

1 1 15. q ≤ −2 _ or q ≥ 2 _ 4 4

–5

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

3 1 14. c ≤ _ or c > _ 2 4

–5

18.

b. Graph the compound inequality on a number line.

q –3

–2

–1

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

c. How many solutions to the compound inequality are there? There are infinitely many solutions to the inequality.

16. x = −2 and x = 8

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

20. Consider the compound inequality −1.75 ≤ u < 4.

a. Rewrite the compound inequality as a compound statement connected by and or or.

u ≥ −1.75 and u < 4

b. Graph the compound inequality on a number line.

u –5

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P R ACT I C E

185

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P R ACT I C E

–4

–3

–2

–1

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

21. Consider the inequality n ≤ 6.

26. Children and senior citizens receive a discount on tickets at the movie theater. The discount applies to children who are 2 to 12 years of age as well as adults who are 60 years of age or older.

a. Rewrite the inequality as a compound statement connected by and or or.

Let a represent the age of a person in years.

2 ≤ a ≤ 12 or a ≥ 60

n < 6 or n = 6

Note: If students define age continuously, they may also write the inequality 2 ≤ a < 13 or a ≥ 60.

b. Graph the inequality on a number line.

n –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

9 10

For problems 22–26, write a single or compound inequality for the situation. 22. The scores on the last test ranged from 65% to 100%.

27. Consider the following compound statements.

Let s represent a score on the last test, as a percent.

65 ≤ s ≤ 100

x < 1 and x > −1

x < 1 or x > −1

Does changing the word and to or change the solution set? Explain and graph the solution sets of both statements.

23. To ride the roller coaster, a person must be at least 48 inches tall. Let h represent the height of a person in inches.

h ≥ 48

For the first statement, both inequalities must be true, so x can only equal values that are both greater than −1 and less than 1. For the second statement, only one inequality must be true, so x must be greater than −1 or less than 1. This means x can equal any number on the number line.

24. Unsafe body temperatures are those lower than 96°F or above 104°F. Let t represent body temperature in degrees Fahrenheit.

t < 96 or t > 104

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

25. For a shark to survive in an aquarium, the water in its tank must be at least 18°C and no more than 22°C. Let x represent water temperature in degrees Celsius.

18 ≤ x ≤ 22

P R ACT I C E

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

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Math 1 ▸ M1 ▸ TC ▸ Lesson 12

31. A soccer team’s coach made a dot plot showing the number of goals scored in each game this season.

Remember For problems 28 and 29, solve the equation. 28. −9n + 5n − 10 + 14n = 30

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 12

3 1 29. 12 − __ n − _5 n = 26 10

Goals Scored in Games This Season

−20

30. Without solving, explain why the equations 5x + 6 = 3 − 4 x and 18 + 15x = 9 − 12x have the same solution set. Include properties in your answer.

Number of Goals

I can apply the multiplication property of equality to the first equation. Multiplying both sides of the first equation by 3 creates an equation that has the same solution set as the original equation.

a. How many games did the soccer team play this season? The soccer team played 15 games this season.

5x + 6 = 3 − 4 x 3(5x + 6) = 3(3 − 4 x) 15x + 18 = 9 − 12 x

b. In how many games this season did the soccer team score at least 2 goals? The team scored at least 2 goals in 10 games this season.

I can apply the commutative property of addition to rewrite the equation 15x + 18 = 9 − 12x as 18 + 15x = 9 − 12x.

c. What is the total number of goals the team scored this season? The team scored a total of 29 goals this season.

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Teacher Edition: Math 1, Module 1, Topic C, Lesson 13 LESSON 13

Solving and Graphing Compound Inequalities Find the solution sets of compound inequalities in one variable and graph the solution sets on the number line.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Name

EXIT TICKET

Date

Solve each compound inequality. Write the solution set by using set-builder notation. Then graph the solution set on the number line. 1. 9 + 2x > 15 or 7 + 4x < −9

{x | x > 3 or x < −4}

9 + 2 x > 15 2x > 6 x>3

7 + 4 x < −9 4 x < −16 x < −4

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

Lesson at a Glance In this digital lesson, students solve compound inequalities and graph their solution sets on the number line as they progress through a series of clues to solve a riddle. They use digital tools to visualize key characteristics of compound inequalities and make connections to the graphs of their solution sets. Students have opportunities to interact with technology independently and as a class. Use the digital platform to prepare for and facilitate this lesson. Students will also interact with lesson content and activities via the digital platform.

Key Questions x 2. 6 ≤ _ ≤ 11 2

6 ≤ _2 x

{x | 12 ≤ x ≤ 22}

_x ≤ 11 x ≤ 22

and

12 ≤ x

• How do the inequality symbols of a compound inequality affect the graph of the solution set?

• How do the words and and or affect the solution set of a compound inequality? x

–2

Achievement Descriptors Math1.Mod1.AD5 Represent constraints by using equations and

inequalities. (A.CED.A.3) Math1.Mod1.AD11 Solve linear inequalities in one variable. (A.REI.B.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Solving Compound Inequalities

• Computers or devices (1 per student pair)

• What’s the Mystery Integer?

Lesson Preparation

• They Look the Same

• None

Land 10 min

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Fluency Solve Inequalities Students solve inequalities and graph the solution sets to prepare for solving compound inequalities and graphing the solution sets.

Teacher Note Students may use the Number Lines removable.

Directions: Solve each inequality. Write the solution set by using set-builder notation, and then graph the solution set on a number line.

−3 + x ≤ −5

− 2_ x > 2 7

18 ≥ −2(2 x − 1)

{x | x ≤ −2}

–10 –8 –6 –4 –2

8 10

{x | x < −7}

–10 –8 –6 –4 –2

x 4

8 10

{x | x ≥ −4}

–10 –8 –6 –4 –2

x 4

8 10

{x | x > 8}

8x + 27 < 12 x − 5

–10 –8 –6 –4 –2

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x 4

8 10

EUREKA MATH2

Launch

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Students interpret and graph solution sets of simple compound inequalities on the number line.

Students explore two compound inequalities in the digital platform.

−5 < x < 4 x < − 5 or x > 4 Students identify values of x that make the inequality true, plot the endpoints, and graph the solution sets on the number line. Students recall the differences between and and or when working with inequalities. What differences do you notice about the graphs of the solution sets of the compound inequality −5 < x < 4 and the compound inequality x < −5 or x > 4?

Promoting the Standards for Mathematical Practice

For −5 < x < 4, the solutions must satisfy both −5 < x and x < 4. The graph of the solution set is a line segment between, but not including, the endpoints. For x < −5 or x > 4, the solutions satisfy either inequality. The graph of the solution set consists of two non-overlapping rays that face opposite directions. Each ray starts at, but does not include, the different endpoints.

When students compare their predictions with the graphed class data and use these data to determine the solution set of a compound inequality and the endpoints of its graph, they are making sense of a problem and persevering in solving it (MP1).

Learn Solving Compound Inequalities Students solve compound inequalities and graph their solution sets on the number line.

Ask the following questions to promote MP1:

• What steps can you take to determine whether the endpoints you chose are correct? • Does your solution set make sense? Why?

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Each student identifies four solutions to −5 < 2x + 1 < 4. As students enter unique solutions, the graph of the solution set builds. Students study the graph and discuss its endpoints. What do you notice about the endpoints of the graph of the solution set? I notice that the endpoints are not in the compound inequality. I notice that we can solve two inequalities to find the endpoints of the graph of the solution set.

What’s the Mystery Integer? Students find solution sets of a series of compound inequalities. This digital activity presents students with a riddle. Characteristics of an unknown integer are presented in a series of clues written in sentence form and as a compound inequality. Students use these clues to practice solving and graphing the solution sets of compound inequalities of varying complexity. As students graph the solution sets, they gather information to solve the riddle.

2x + 1 ≤ −3 or 2x + 1 ≥ 9

Clue B When I subtract two from the number and then triple that difference, the result is strictly between –21 and 12. –21 < 3(x – 2) < 12

x ≤ −2 or x ≥ 4

Digital activities align to the UDL principle of Action & Expression by including the following elements:

−21 < 3(x − 2) < 12 −5 < x and x < 6

_____

• Embedded prompts that promote strategic planning. Students receive immediate feedback and hints.

_____

−x + 5 −x + 5 < 3 or >4 2 2

x > −1 or x < −3

− 4 ≤ x − _1 x ≤ 2

−6 ≤ x and x ≤ 3 3

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UDL: Action & Expression

–16 –14 –12 –10 –8 –6 –4 –2

10 12 14 16

• Multiple exemplars of work that support students in monitoring their own progress. The activities model the correct use of strategy within the animations.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

What’s the only integer found in all the solution sets? How do you know? The only integer found in all the solution sets is −4 because that is the only integer found in every graph. Have students think–pair–share about the following questions to discuss how to solve compound inequalities. When solving compound inequalities, what effect does the word and or the word or have on the solution set? When the word and is in the compound inequality, the solution set is the set of all values of the variable that make both statements true. When the word or is in the compound inequality, the solution set is the set of all values that make at least one of the statements true. What effect does the word and or or have on the graph of the solution set? When the word and is in a compound inequality, the graph of the solution set consists of all points that lie in the region where the graphs of both inequalities intersect. When the word or is in a compound inequality, the graph of the solution set consists of all points that lie on the graph of either inequality or on the graphs of both inequalities. For each clue, complete the following steps to determine the mystery integer. • Find the solution set of the compound inequality. • Write the solution set in set-builder notation. • Graph the solution set by using the tools on the corresponding slide. • Guess what the mystery integer is. Write your guess in the table and enter it in the box on the corresponding slide.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Clue A. When I double the mystery integer and then add 1, the result is less than or equal to −3 or greater than or equal to 9.

2x + 1 ≤ − 3 or 2x + 1 ≥ 9

2 x + 1 ≤ −3 2 x ≤ −4

x ≤ −2

2x + 1 ≥ 9 2x ≥ 8 x≥4

B. When I subtract 2 from the mystery integer and then triple that difference, the result is strictly between −21 and 12.

− 21 < 3(x − 2) < 12 − 21 < 3(x − 2)

and

3(x − 2) < 12

−7 < x − 2

Set-Builder Notation

Guess

{x | x ≤ − 2 or x ≥ 4}

Sample: − 2

{x | − 5 < x < 6}

Sample: − 2

{x | x < − 3 or x > − 1}

Sample: 4

x−2<4

−5 < x

x<6

C. Half the sum of the opposite of the mystery integer and 5 is less than 3 or greater than 4. −x + 5 −x + 5 ______ < 3 or ______ >4 2 2 −x + 5 − x+5 _____ _____ <3 or >4 2

−x + 5 < 6

−x + 5 > 8

−x < 1

−x > 3

x > −1

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x < −3

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Clue

Set-Builder Notation

Guess

D. The difference between the mystery integer and _1 of the integer is at 3 least − 4 and at most 2. − 4 ≤ x − _1 x ≤ 2 3 and x − _1 x ≤ 2 x − _1 x ≥ − 4 3 3 _2 x ≥ − 4 _2 x ≤ 2 3 3

{x | − 6 ≤ x ≤ 3}

−4

x ≥ −6

x≤3

What is the mystery integer? The mystery integer is − 4.

They Look the Same Students identify characteristics that make the graphs of the solution sets of similar compound inequalities different. In this digital activity, students match the following compound inequalities to their graphs.

1 + x ≥ −4 or 3x − 6 > −12 1 + x ≥ −4 or 3x − 6 < −12

1 + x ≥ −4 and 3x − 6 > −12

1 + x ≥ −4 and 3x − 6 < −12

Which characteristics of compound inequalities are important to pay attention to when graphing their solution sets? It is important to pay attention to whether the compound inequality includes and or or, as well as whether it includes inequalities that are strict or not strict.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Land Debrief 5 min

Objective: Find the solution sets of compound inequalities in one variable and graph the solution sets on the number line. Facilitate a class discussion by inviting students to compare the key characteristics of each inequality in the They Look the Same section. Consider using the following questions to assess students’ understanding of how to solve and graph the solution sets of compound inequalities. How do the inequality symbols in a compound inequality affect the graph of the solution set? The inequality symbols determine which part of the graph is shaded. If x < a, then the graph of the solution set shows that the values to the left of the endpoint a are shaded. If x > a, then the graph of the solution set shows that the values to the right of the endpoint a are shaded. How do the words and and or affect the solution set of a compound inequality? When the word and is in a compound inequality, the solution set is the set of all values of the variable that make both inequalities true. When the word or is in the compound inequality, the solution set is the set of all values of the variable that make one or both inequalities true.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

262

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

RECAP Name

Date

1 2. 1 ≤ _ w + 5 ≤ 8 2

−4 ≤ 1_ w

•

graphed solution sets of compound inequalities.

{w | −8 ≤ w ≤ 6}

Examples

Rewrite as a compound statement by using and.

1 w≤3 2

The notation −8 ≤ w ≤ 6 means w ≥ −8 and w ≤ 6.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

1. 5x − 2 > 8 + 3x or 5 − 2x > 3

5x − 2 > 8 + 3x 5x > 10 + 3x 2x > 10 x>5

w≤6

For problems 1 and 2, find the solution set of the compound inequality. Write the solution set in set-builder notation. Then graph the solution set on the number line.

Solve each part of the compound inequality individually.

−8 ≤ w

1 w+5≤8 2

and

In this lesson, we found solution sets of compound inequalities.

1 ≤ 1_ w + 5 2

Solving and Graphing Compound Inequalities •

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

9 10

5 − 2x > 3 −2x > −2 x<1

{x | x < 1 or x > 5} x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

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RECAP

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

PRACTICE Name

Date

and

2. x > 4

3. x + 1 > 3

_1 x > 0 3

1 2

x < 12 For problems 8–10, find the solution set of each compound inequality. Write the solution set by using set-builder notation. Then graph the solution set on the number line.

x < −1

x 8. x − 2 < 6 or _ > 4 3

{x | x < 8 or x > 12}

x + 1 < −3

and

1 7. x ≥ _ and x < 1 2 1_ ≤ x<1 2

The value of x can be any number greater than or equal to _ and less than 1.

For problems 1–4, write and or or to complete the compound inequality so that the solution set is neither the empty set nor the set of all real numbers. 1. x > −2

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

_1 x < 4 3

x 5

–2

–1

9. 5y + 2 ≥ 27 and 3y − 1 < 29

{y | 5 ≤ y < 10}

For problems 5–7, write each compound inequality as one statement without the word and. Then write a sentence describing all possible values of x that satisfy the compound inequality. 5. x > −6 and x < −1

−6 < x < −1

y 2

The value of x can be any number greater than −6 and less than −1. 6. x ≤ 5 and x > 0

10. 2w > 8 or −2w < 4

{w | w > −2}

0<x≤5

The value of x can be any number greater than 0 and less than or equal to 5.

–5

264

199

200

–4

P R ACT I C E

w –3

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

11. Consider the compound inequality 4p + 8 > 2p − 10 or _ p − 3 < 2. 1 3

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

13. Consider the compound inequality −1 < g − 6 < 1. a. Rewrite the compound inequality as a compound statement with the word and or or.

a. Find the solution set. Then graph the solution set on the number line.

g − 6 > −1 and g − 6 < 1

ℝ p 0

13 b. Solve each inequality from part (a) for g.

b. If the inequalities are joined by and instead of or, what is the solution set? Graph the solution set on the number line.

g > 5 and g < 7

1 4p + 8 > 2p − 10 and _3 p − 3 < 2

{p | p > −9 and p < 15} –10

–8

–6

c. Write the solution set by using set-builder notation.

{g | 5 < g < 7}

–4

–2

12. Consider the compound inequality 7 − 3x < 16 and x + 12 < −8.

d. Graph the solution set on the number line.

a. Find the solution set. Then graph the solution set on the number line.

{} –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

x 2

9 10

b. If the inequalities are joined by or instead of and, what is the solution set? Graph the solution set on the number line.

7 − 3x < 16 or x + 12 < −8 {x | x > −3 or x < −20}

–22 –21 –20 –19 –18 –17 –16 –15 –14 –13 –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

P R ACT I C E

201

202

P R ACT I C E

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

14. Consider the compound inequality −1 < __ ≤ 3. h 2

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

For problems 15–17, solve the compound inequality. Then graph the solution set on the number line. 15. 0 ≤ 4x − 3 ≤ 11

a. Rewrite the compound inequality as a compound statement with the word and or or.

{x 4 ≤ x ≤ 2 }

_h > −1 and h_ ≤ 3 2 2

|_ 3

b. Solve each inequality from part (a).

h > −2 and h ≤ 6

16. −8 ≤ −2(x − 9) ≤ 8

c. Write the solution set by using set-builder notation.

{h | −2 < h ≤ 6}

{x | 5 ≤ x ≤ 13} 1

d. Graph the solution set on the number line.

x 3

h –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10 x+1 17. −6 < ____ < 3 4

{x | −25 < x < 11} –27 –25 –23 –21 –19 –17 –15 –13 –11 –9 –7 –5 –3 –1

266

P R ACT I C E

203

204

P R ACT I C E

x 1

9 11 13 15

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Math 1 ▸ M1 ▸ TC ▸ Lesson 13

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 13

Remember For problems 18 and 19, solve the equation. 18. 3(x + 5) = −18

19. 7 = − 4(2 x − 1) 3 − _8

−11

20. The formula for the volume V of a right cone is V = _ πr 2h, where r represents the radius of the 3 base of the cone and h represents the height of the cone. Rearrange the formula to isolate h. 1

3V ___ =h πr 2

21. These data represent the heights of the vertical jumps in inches of 18 players on a youth basketball team.

Make a dot plot of the data. Vertical Jump Heights

Vertical Jump (inches)

P R ACT I C E

205

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Teacher Edition: Math 1, Module 1, Topic C, Lesson 14 LESSON 14

Solving Absolute Value Equations Write absolute value equations in one variable as compound statements and solve.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 14

Name

EXIT TICKET

Date

Consider the equation |2x + 5| = 3.

a. Rewrite the equation as a compound statement.

2x + 5 = 3 or −(2x + 5) = 3

b. Use the compound statement you wrote in part (a) to find the solution set of the original equation.

2x + 5 = 3

− (2 x + 5) = 3

2x = −2

2x + 5 = −3

x = −1

2x = −8

Lesson at a Glance Through discussion, partner work, and physical movement on a large number line in the classroom, students recognize absolute value equations as compound statements connected with the word or. Students discuss solutions to absolute value equations to refine their thinking about the solution set. They apply the properties of equality to isolate the absolute value expressions. Then they write compound statements to solve the absolute value equations. This lesson reminds students of the definition of absolute value and introduces the term absolute value equation.

Key Questions

x = −4

• How can we predict the number of solutions to an absolute value equation before we solve it?

{−4, −1} c. Use the original equation to check the solutions you found in part (b).

Substitute −4 for x. Because |2(−4) + 5| = |−3| = 3, we know that −4 is a solution to the equation.

• Why do we rewrite absolute value equations as compound statements?

Substitute −1 for x. Because |2(−1) + 5| = |3| = 3, we know that −1 is a solution to the equation.

Achievement Descriptor Math1.Mod1.AD10 Solve linear equations in one variable. (A.REI.B.3)

d. How many solutions does the equation |2x + 5| = −3 have? Explain.

The equation has no solutions because the absolute value of a number cannot be negative. It cannot be negative because absolute value represents the distance from 0 on the number line. Distance is never negative. Therefore, there is no value of x that produces a value of −3 on the left side of the equation, so it is impossible to create a true number sentence.

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• Painter’s tape

Learn 30 min

• Index cards (4)

• Absolute Value Equations

Students

• More Than Just Absolute Value

• Personal whiteboard

Land 10 min

• Dry-erase marker • Personal whiteboard eraser

Lesson Preparation • Prepare a large number line on the classroom floor by using painter’s tape. (See figure 14.1.) Use floor tiles or a meter stick to create 21 equidistant tick marks. This number line will be used again in lesson 15. • Write a large 0 on an index card, and place it below the middle tick mark on the floor number line. • Write x, k, and y on the three other index cards (one letter per card).

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Fluency Evaluate Absolute Values Students evaluate numerical expressions containing absolute value symbols to prepare for solving absolute value equations. Directions: Evaluate each expression. 1. 2. 3. 4. 5. 6.

|−3|

|9 − 14|

|2(−3) + (−3)(−4)| 6 − |−10|

|5(−5)| − 8

−|−3 ⋅ 6| + |2 − 20|

3 5

Teacher Note Instead of this lesson’s Fluency, consider administering the Evaluate Expressions with Absolute Values Sprint. Directions for administration can be found in the Fluency resource.

Number Correct:

Evaluate the expression.

1. 2. 3. 4. 5.

−4 17

6. 7.

23.

|−4|

25.

|4|

|−5| |5| |6|

|−6|

24.

26. 27.

31.

−|−20|

32.

11.

−|20|

33. 34. 35.

14.

−|−4(−5)|

36.

15.

−|−4(5)|

37.

16. 17. 18. 19.

|−4(−5)| |−4(5)|

−|2(4)(5)| |2(4)(−5)|

38. 39.

5 + |−6| 5 − |6|

|6| + 5

|−6| − 5

−5 + |2 + 3| −5 − |3 − 2|

|−3 + 2| + 5 |−3 − 2| − 5

|−3( − 2)| − 5 |3(−2)| + 5

40.

−|−3(−2)| + 5

41.

−|−3(2)| − 5

−|−2(4)(−5)|

42.

21.

2|−4(−5)|

43.

22.

−2|−4(−5)|

44.

370

−|5 + 1|

−|−5 − 1|

10.

|4(−5)|

|−5 − 1| −|−5 + 1|

30.

|4(5)|

|5 − 1|

|−5 + 1|

29.

−|6|

12.

|5 + 1|

28.

−|−6|

13.

|3|

20.

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Math 1 ▸ M1 ▸ Sprint ▸ Evaluate Expressions with Absolute Values

|−3(2) + 5|

|−3(−2) − 5|

−|3(2) + 5| + 5

EUREKA MATH2

Launch

Math 1 ▸ M1 ▸ TC ▸ Lesson 14

Students recall the definition of absolute value and apply the definition in an activity. Direct students to the number line on the classroom floor. (See figure 14.1.) Identify 0 on the number line. Figure 14.1

0 Ask a pair of students to go to the number line. Have one student stand at 4 and the other stand at −4. Place the index card labeled x near the right arrow of the number line.

UDL: Representation Students engage in the kinesthetic activity of using a life-size number line, which is another format for presenting the concept of absolute value as the distance from zero on the number line.

0 Have students think–pair–share about the following question. Expect some students to say 4 or − 4 rather than providing a compound statement. Encourage these students to revoice their response as a compound statement. The students standing at the number line represent the graph of what compound statement? They represent the graph of the compound statement x = 4 or x = − 4. These students represent the graph of the compound statement x = 4 or x = −4. Give students 1 minute of think time to examine the points on the graph where their peers are standing. Facilitate student conversation by asking the following questions. What do you notice about the placement of your peers in relation to 0? One is to the left of 0. The other is to the right of 0. They are both 4 units away from 0.

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Help students notice that the points on the graph represent points that are each a distance of 4 units from 0 on the number line. What do we call a number’s distance from 0? We call it the absolute value. Even if students do not use precise language, encourage them to discuss their ideas. Have the standing students return to their seats.

Discuss that the equation |x| = 4 means that the value of x is a distance of 4 units from 0. This means that either x is 4 or the opposite of x is 4. Write the equation as a compound statement.

x = 4 or −x = 4

Explain that the equation |x| = 4 can be rewritten as a compound statement because there are two values of x that are a distance of 4 units from 0. Therefore, there are two solutions to the equation, −4 and 4. Today, we will use our knowledge of absolute value and of writing and solving compound statements to write and solve absolute value equations.

Learn Absolute Value Equations Students solve absolute value equations by solving an equivalent compound statement. Direct students to problems 1–5. Give students 1 minute of silent time to think about how to solve the equations. Then have students discuss their thinking with a partner. For problems 1–5, write the equation as a compound statement, if applicable. Use the compound statement to find the solution set of the equation. Check your solutions in the original equation. Then graph the solution set on the given number line. 272

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

1. |x| = 5

x = 5 or −x = 5

{−5, 5}

Substitute 5 for x. Because |5| = 5, we know that 5 is a solution to the equation.

Substitute −5 for x. Because |−5| = 5, we know that −5 is a solution to the equation.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

x 8

9 10

2. |x| = −5

The equation has no solution. The expression |x| represents a distance, and distance is never negative.

{}

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

3. |k − 2| = 7

k − 2 = 7 or −(k − 2) = 7

{−5, 9}

Substitute −5 for k. Because |−5 − 2| = |−7| = 7, we know that −5 is a solution to the equation.

Substitute 9 for k. Because |9 − 2| = |7| = 7, we know that 9 is a solution to the equation.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

k 9 10

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

4. |4y + 2| = 3

4y + 2 = 3 or −(4y + 2) = 3

_5 1_

{− 4 , 4 }

Substitute −_5 for y. Because 4(−_5) + 2 = |− 3| = 3, we know that −_5 is a solution to 4

the equation.

Substitute _1 for y. Because 4(_1) + 2 = |3| = 3, we know that _1 is a solution to 4

the equation.

y –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

5. |2x − 3| = 0

9 10

2x − 3 = 0 or −(2x − 3) = 0

{2 } 3

Substitute _3 for x. Because 2(_3) − 3 = |3 − 3| = 0, we know that _3 is a solution to 2 2 2 the equation.

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

Direct students to problem 1, and facilitate a discussion by asking the following questions. We can interpret |x| = 5 to mean that the value of x is a distance of 5 units from 0 on the number line.

How can we interpret the equation |x| = 5?

Ask a pair of students to go to the number line on the classroom floor and stand on the marks that represent the values of x that are 5 units from 0 on the number line. They represent the values of x that make the equation |x| = 5 true. What do these students represent?

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Look at where the students are standing. How can we describe the compound statement that is equivalent to | x| = 5? Either x is 5 or the opposite of x is 5. Ask students to write this compound statement symbolically and then find the solution set. Have the standing students return to their seats, and ask their classmates to share what they have written for problem 1. Ask students to check their solutions in the original equation. Review the solution set and its graph with the class. Guide students to problem 2. How can we interpret the solutions to the equation |x| = −5? We can interpret the solutions as the values of x, that are each a distance of −5 units from 0 on the number line. Ask a pair of students to go to the number line on the classroom floor and stand on the marks that represent the values of x that are −5 units from 0. Students should realize that, because distance is never negative, it is impossible for them to stand on the marks on the number line.

Teacher Note In preparation for the next lesson, encourage students to write the compound statement as x = 5 or −x = 5. Once students have mastered this conceptual foundation, writing x = 5 or x = −5 without first writing the compound statement x = 5 or −x = 5 is acceptable.

What does this distance mean for the solution set of this equation? There is no solution, so the solution set is the empty set. Ensure the class records their answer and a brief explanation for problem 2. Guide students to problem 3. How is the equation for problem 3 different from the equations for problems 1 and 2? We are given the absolute value of (k − 2), not just k. What does the equation |k − 2| = 7 mean? The equation means that k − 2 represents a distance of 7 units from 0 on the number line.

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Have two students bring their whiteboards and markers and stand on the number line at the marks that each represent a distance of 7 units from 0 on the number line. Ask the students on the number line to write k − 2 on their whiteboards. Are the numbers the students are standing on solutions to the equation |k − 2| = 7? Explain.

No. The solutions to the equation are the numbers that can each replace the variable and make the equation a true statement. Look at where the students are standing. How can we explain the compound statement that is equivalent to |k − 2| = 7?

Either k − 2 is 7, or the opposite of k − 2 is 7.

Have students write this compound statement symbolically and then find the solution set. Direct the students standing on the number line to revise their whiteboards so they have only k written on them. Where should your classmates stand now that they are representing the solutions to the equation? They should stand at 9 and −5. Ask the students at the number line to move to these new locations. Direct students to check their solutions in the original equation. Review the solution set by using set notation and its graph with the class.

Differentiation: Support If students write k − 2 = −7 without conceptual understanding of −(k − 2) = 7, consider concentrating on the idea that the definition of absolute value says the expression and its opposite are the same distance from 0 on the number line. • k − 2 equals 7. • The opposite of k − 2 equals 7. In addition, thinking about absolute value as a distance will help students avoid procedural misconceptions such as removing all of the negative signs.

Language Support To support the term absolute value equation, consider completing a Frayer model graphic organizer to add to the definition of absolute value from grade 6.

Continue this discussion pattern for problems 4 and 5. Have students read the description of an absolute value equation and discuss as a class, addressing any questions. Students should note that problems 1–5 are all examples of absolute value equations that were already written in the form |bx − c| = d.

An equation that can be written in the form |bx − c| = d, where b, c, and d are real numbers, is an example of an absolute value equation. Solutions to the equation |bx − c| = d are values of x that make the value of bx − c a distance of d units from 0 on the number line. The equation has the same solution set as the compound statement bx − c = d or −(bx − c) = d.

276

Teacher Note In Mathematics I, absolute value equations are limited to those that have linear expressions within the absolute value. Therefore, absolute value equations are described but not formally defined in this manner. In subsequent courses, students may encounter other types of absolute value equations.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

Facilitate a class discussion to summarize this section. Consider using the following statements to address some common student misconceptions: • It is possible for the solution to an absolute value equation to be a negative value. • It is not possible for an absolute value expression to be equal to a negative value. • An absolute value equation may have zero, one, or two solutions. Language Support

More Than Just Absolute Value Students isolate the absolute value expression to solve more complex absolute value equations. Guide students to problem 6. Have them turn and talk about the following question. Based on the description of an absolute value equation, is this an absolute value equation? Circulate as students discuss, and listen for those who mention the part of the description that states can be written in the form. Then ask the following question.

To support the process of isolating an absolute value expression, consider emphasizing that solving for the value of a variable in an algebraic equation is equivalent to isolating that variable. Therefore, isolating the absolute value expression means using operations and properties so that the absolute value expression is alone on one side of the equation.

The equation is not written in the form |bx − c| = d as it is in the previous problems. But can it be written in that form? If so, how? Yes, it can, by subtracting 2 from both sides. Why do you think it is useful to isolate the absolute value expression first? By isolating the absolute value expression first, I can rewrite the equation as a compound statement to find the solutions. Direct students to write the compound statement for problem 6 and to find the solution set. Then have students complete problem 7 independently. Have students turn and talk to share their solution paths for problems 6 and 7. Facilitate student conversation by asking the following questions. How are the solution paths for problems 6 and 7 similar to the solution paths for problems 1 through 5? How are the solution paths different? The solution paths are similar after we isolate the absolute value expression. The expression within the absolute value bars and its opposite expression are equal to the same number.

Differentiation: Support If students isolate the absolute value expression to get |bx − c| = d and they automatically conclude the solution set of the equation is {−d, d}, guide them to see the absolute value expression as a single object that is composed of different parts. First, physically cover the expression within the absolute value bars and ask students to analyze what the absolute value means. Then uncover the expression and ask students to write the compound statement with the expression inside the absolute value bars.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

The solution paths are different because we had to use the properties of equality to isolate the absolute value expressions in problems 6 and 7. The absolute value expressions were already isolated in problems 1 through 5. Have students continue working with their partners to complete problems 8–10. Display the answers to problems 8–10. Then discuss solution pathways. Students may want to use the distributive property as they would with equations that have parentheses. Emphasize to students that, just like with absolute value problems involving addition and subtraction, it is best to isolate the absolute value expression first and then rewrite the equation as a compound statement. For problems 6–10, solve the equation. Write the solution set by using set notation. 6. |b − 5| + 2 = 10

b−5=8

When students apply their previous understanding of solution sets and absolute values to solving absolute value equations, they are making use of structure (MP7). Ask the following questions to promote MP7: • How can what you know about absolute value expressions help you find the solution set of this absolute value equation? • What is another way you could write this equation that would help you find the solution set?

|b − 5| + 2 = 10 |b − 5| = 8

Promoting the Standards for Mathematical Practice

−(b − 5) = 8 b − 5 = −8

b = 13

b = −3 {−3, 13}

7. 14 = 3 + |m + 6|

11 = m + 6 5=m

14 = 3 + |m + 6| 11 = |m + 6|

11 = −(m + 6) −11 = m + 6 −17 = m

{−17, 5}

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

8. 36 = 4|2 − h|

36 = 4|2 − h| 9 = |2 − h|

9=2−h

9 = − ( 2 − h) −9 = 2 − h

7 = −h

− 11 = − h

−7 = h

11 = h {−7, 11} 9. −2|6x − 8| = −40

− 2|6x − 8| = −40 6x − 8 = 20 6x = 28 14 x = __ 3

|6x − 8| = 20 or

−(6x − 8) = 20 6x − 8 = −20 6x = −12 x = −2

{− 2, 3 } 14

279

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

10. 8 + 2|3w + 8| = 4

8 + 2|3w + 8| = 4 2|3w + 8| = −4 |3w + 8| = −2

The equation has no solution. The expression |3w + 8| represents a distance, and distance is never negative.

{}

Land Debrief 5 min Objective: Write absolute value equations in one variable as compound statements and solve. Facilitate a brief discussion by asking the following prompts. Compare and contrast solving linear equations and solving absolute value equations. We can use the properties of equality for solving both linear equations and absolute value equations. We isolate the variable to solve a linear equation. We isolate the absolute value expression and then write and solve a compound statement to solve an absolute value equation. Can we predict the number of solutions an absolute value equation has before solving it? If so, how? Yes. An absolute value equation has two distinct solutions if the absolute value expression is equal to a positive number. It has one solution if the absolute value expression is equal to 0. It has no solutions if the absolute value expression is equal to a negative number.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

Why do we rewrite absolute value equations as compound statements for nonnegative values of d?

When d is not negative, we can write absolute value equations as compound statements because |bx − c| = d means that bx − c is distance d from 0 on the number line. So (bx − c) = d or − (bx − c) = d.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 14

RECAP Name

Date

Solving Absolute Value Equations In this lesson, we •

wrote absolute value equations as compound statements.

•

solved absolute value equations.

•

graphed solution sets of absolute value equations on the number line.

b. Solve each linear equation.

z−6=2

z=4

An equation that can be written in the form |bx − c| = d, where b, c, and d are real

c. Write the solution set in set notation.

numbers, is an example of an absolute value equation. Solutions to the equation

{4, 8}

|bx − c| = d are values of x that make the on the number line. The equation has the same solution set as the compound statement bx − c = d or −(bx − c) = d.

2. |m| = −9

{}

{−7, 7} The compound statement means that

x is 7 or the opposite of x is 7. If the opposite of x is 7, then x is −7.

d. Graph the solution set on the number line.

z –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

4|6 − 4x|+ 9 = 81 |6 − 4x| = 18

4|6 − 4x| = 72

distance is never negative. This equation is not true for any value of m.

6 − 4x = 18

9 10

Apply properties of equality to first isolate the absolute value expression.

− (6 − 4x) = 18

− 4x = 12

6 − 4x = − 18

x = −3

− 4x = − 24 x=6

a. Rewrite the absolute value equation as a compound statement.

{−3, 6}

z − 6 = 2 or −(z − 6) = 2

282

4. Solve the absolute value equation 4|6 − 4x| + 9 = 81. Write the solution set in set notation. Absolute value represents distance from 0 on the number line, and

3. Consider the absolute value equation |z − 6| = 2.

z − 6 = −2

Terminology

For problems 1 and 2, write the equation as a compound statement, if applicable. Then write the solution set in set notation.

x = 7 or −x = 7

− (z − 6) = 2

z=8

value of bx − c a distance of d units from 0

Examples

1. |x| = 7

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

215

216

RECAP

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 14

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TC ▸ Lesson 14

5. Consider the absolute value equation |x + 7| = 3.

a. Rewrite the absolute value equation as a compound statement.

x + 7 = 3 or −(x + 7) = 3

For problems 1–4, write the equation as a compound statement, if applicable. Then graph the solution set of the equation on the number line. 1. |p| = 10

b. Solve the compound statement. Write the solution set by using set notation.

p = 10 or −p = 10

{−10, −4}

p –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

2. |x| = 0

9 10 c. Graph the solution set on the given number line.

x = 0 or −x = 0 x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

3. |t| = 1.5

9 10

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

|2 |

a. Rewrite the absolute value equation as a compound statement. d _ − 4 = 1 or − d_ − 4 = 1 (2 ) 2

t 4. |y| = −3

d 6. Consider the absolute value equation __ − 4 = 1.

t = 1.5 or −t = 1.5 –5

–4

–3

–2

–1

b. Solve the compound statement. Write the solution set by using set notation.

{6, 10}

Distance is never negative, so this equation has no solution.

c. Graph the solution set on the given number line.

y –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

217

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For problems 7–11, solve the absolute value equation. Write the solution set by using set notation.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 14

12. Why do we rewrite an absolute value equation as a compound statement?

7. |x + 2| − 4 = 6

We rewrite an absolute value equation as a compound statement because absolute value represents the distance from 0 on the number line. The two statements in the compound statement represent the two values that are the same distance from 0 on the number line.

{−12, 8}

13. Why is there no solution to an absolute value equation when the absolute value expression is equal to a negative number? Absolute value represents a number’s distance from 0 on the number line. Because distance is always nonnegative, an absolute value cannot equal a negative number. So there is no solution.

8. 2|2p − 11| = 66

{−11, 22} Remember For problems 14 and 15, solve the equation.

9. 8 + |5x − 2| − 2 = 6 2_ {5 }

14. 2_ (− 6r + 9) = − 12 3

r 15. − 5(__ + 2) = 15 10

_9 2

− 50

10. −2|2x + 25| = 14

{}

16. Find the solution set of the inequality. Write the solution set by using set-builder notation, and graph the solution set on the number line.

5x − 3 > 6x + 2 x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

11. 8|3 − 3x| − 5 = 91

{x | x < − 5}

{−3, 5}

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17. Emma surveyed her friends and recorded the number of states each friend has visited. The numbers of states each friend visited are 3, 8, 7, 6, 5, 4, 4, 5, 6, 4, and 5. a. Find the mean number of states visited by Emma’s friends. Round to the nearest tenth. The mean number of states visited by Emma’s friends is about 5.2.

b. Find the median number of states visited by Emma’s friends. The median number of states visited by Emma’s friends is 5.

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Teacher Edition: Math 1, Module 1, Topic C, Lesson 15 LESSON 15

Solving Absolute Value Inequalities Write absolute value inequalities in one variable as compound statements joined by and or or. Solve absolute value inequalities and graph the solution sets on the number line.

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

Name

EXIT TICKET

Date

Consider the inequality |2x + 5| ≤ 3.

a. Rewrite the inequality as a compound statement.

2x + 5 ≤ 3 and −(2x + 5) ≤ 3

b. Find the solution set.

2x + 5 ≤ 3

2x ≤ − 2 x ≤ −1

and

Students work with partners to match equations and inequalities to solution sets. Through continued use of the classroom floor number line, this lesson guides students to summarize when an absolute value inequality corresponds to a compound statement with the word and versus the word or. Partners solve absolute value inequalities by writing and solving compound statements. Students synthesize their learning through the Always, Sometimes, Never routine, working with partners to include examples or counterexamples to support their claims. This lesson introduces the term absolute value inequality.

Key Question

− (2x + 5) ≤ 3

2x + 5 ≥ − 3

• Do the solution sets of absolute value equations differ from the solution sets of absolute value inequalities? If so, how?

2x ≥ − 8 x ≥ −4

{x | −4 ≤ x ≤ −1}

Lesson at a Glance

Achievement Descriptor Math1.Mod1.AD11 Solve linear inequalities in one variable. (A.REI.B.3)

c. Graph the solution set on a number line.

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• Painter’s tape

Learn 25 min

• Index cards (4)

• Between or Not Between

Students

• More Than Just Absolute Value

• None

Land 15 min

Lesson Preparation • Prepare a large number line, used in lesson 14, on the classroom floor by using painter’s tape (see figure 14.1). Use floor tiles or a meter stick to create 21 equidistant tick marks. • Write a large 0 on an index card, and place it below the middle tick mark of the floor number line. • Write x, k, and a on the three other index cards (one letter per card).

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Fluency Graph Compound Inequalities Students graph compound inequalities to prepare for solving absolute value inequalities and graphing the solution sets on the number line.

Teacher Note Students may use the Number Lines removable.

Directions: Graph the solution set of each compound inequality.

x < −5 or x > 6

x ≥ −1 and x < 4

x > 8 or x ≤ −3

−9 < x < −2

x ≥ 0 or x < 3

288

x –10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

Teacher Note

Students use the structure of equations and inequalities to match the equations and inequalities to the graphs of their solution sets. Direct students to the matching problem set. Have students work with a partner on the matching problem set. Circulate as students work, and listen for different strategies.

The matching problem set includes problems with absolute value inequalities. Students may have difficulty with those problems, but they should be able to use the process of elimination to assist them with correctly matching those statements to the graphs of their solution sets.

Write the letter of the equation or inequality next to the graph of its solution set. Statement A. |x| = 7 B. x > 7 C. |x| < 7

Language Support

Graph

x –10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10 x

–10 –8 –6 –4 –2

8 10 x

D. x = 7

–10 –8 –6 –4 –2 E. |x| > 7

8 10 x

–10 –8 –6 –4 –2

8 10

The definition and language support ideas for the term absolute value can be found in lesson 14.

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Explain how you matched each equation or inequality to the graph of its solution set.

Sample: I knew that the graph of x = 7 would have a single point at 7 and that the graph of |x| = 7 would have two points at −7 and 7. I knew that the graph of x > 7 would have an open circle at 7 and then the values to the right would be shaded. For |x| < 7 and |x| > 7, I tried a test value of 0. Because |0| < 7 is true, I knew that the graph of |x| < 7 had to contain 0.

Invite pairs of students to share the different strategies they used. Then use the following statement to transition to the next activity. Today, we will combine our knowledge of absolute value with our knowledge of solving inequalities to solve absolute value inequalities.

Learn Between or Not Between Students write absolute value inequalities as compound statements by using and or or. Ask the following questions to introduce problem 1.

How can we interpret the solutions to the equation |x| = 4?

The solutions can be interpreted as the values of x that are each a distance of 4 units from 0 on the number line. How do you think we can interpret the solutions to the inequality |x| < 4?

I think we can interpret the solutions as the values of x that are each a distance of less than 4 units from 0 on the number line. Direct students’ attention to the number line on the classroom floor. Identify 0 on the number line. Ask a pair of students to go to the number line and stand on the marks that represent the values of x that are each exactly 4 units from 0. Ask students the following questions to discuss the placement of the students on the number line in relation to the inequality. 290

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What do these students represent in relation to our inequality?

Language Support

They represent the endpoints for the graph of the solution set. Are 4 or −4 solutions to the inequality? How do you know?

Asking students whether the inequality they are describing is a strict inequality will help students begin to use the terminology in their own discussions when describing solution sets. It is not required that students master the use of this terminology.

No. This is a strict inequality, so neither 4 nor −4 is part of the solution set. Ask another student to stand at a point on the number line that is less than 4 units from 0. Ask five more students to do the same. Then ask the following question. What can you conclude about all the points in the solution set of this inequality? All the points in the solution set of this inequality are between 4 and − 4. Ask the standing students to be seated, and have all students complete problem 1 with a partner. Circulate as students work, and ask the following questions: • Should the compound statement have and or or between the two inequalities? Explain. • Did the representation of the absolute value inequality shown by using the number line on the classroom floor and your classmates help you find and graph the solution set? If so, how? • Can you write your compound statement as one inequality? Explain how you know. Review with students how to write the solution set by using set-builder notation. For problems 1–4, write the inequality as a compound statement. Find the solution set and write it by using set-builder notation. Then graph the solution set on the given number line. 1. |x| < 4

x < 4 and −x < 4

{x | −4 < x < 4}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

x 0

9 10

Direct students to problem 2. Have students discuss with a partner and predict what the solution set and graph will look like for this inequality. Ensure that students conclude that the solutions to the inequality are the values of x that are each a distance of more than 4 units from 0 on the number line. © Great Minds PBC

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Invite a few students to represent the absolute value inequality by using the number line on the classroom floor. Once students accurately represent the inequality, have them return to their seats. Then have all students complete problem 2 with a partner. Circulate as students work, and support their understanding by asking the same questions provided for problem 1. When students finish, have them compare problems 1 and 2. 2. |x| > 4

x > 4 or −x > 4

{x | x < −4 or x > 4}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

Have students solve problems 3 and 4 with a partner. Circulate as students work, and listen for students experiencing challenges, such as inaccurately identifying endpoints and having difficulty writing a compound statement. Consider asking any of the following questions to advance students’ thinking:

Differentiation: Support

• How do you think we can interpret the solutions to the inequality |k + 2| ≤ 7?

If necessary, use the number line on the floor to solidify understanding. For example, for problem 3, invite two students to stand at − 9 and 5, and ask what these values represent, eliciting the answer that they are the least and greatest possible values of k. Then ask students to stand at other values that are in the solution set.

• How can we rewrite this inequality as a compound statement? • How is this problem like problem 1? Problem 2? 3. |k + 2| ≤ 7

k+2≤7 k≤5

|k + 2| ≤ 7

− (k + 2) ≤ 7

k + 2 ≥ −7

and

k ≥ −9

{k | −9 ≤ k ≤ 5}

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

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k 0

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4. |a − 5| > 3

|a − 5| > 3

a−5>3

− (a − 5) > 3

a − 5 < −3

a>8

a<2

{a | a < 2 or a > 8} 0

a 2

When students finish, invite a few pairs to display their answers and share their reasoning. Then lead a class discussion by asking the following questions. When rewriting an absolute value inequality as a compound statement, when do we use the word and ? Explain. We use the word and when an absolute value expression is less than or less than or equal to a positive number. When an inequality states that an absolute value expression is less than a positive number, solutions on the number line lie between the endpoints. For example, all solutions to |x| < 4 are less than 4 and greater than − 4. When an inequality states that an absolute value expression is less than or equal to a positive number, solutions on the number line lie between the endpoints and include the endpoints. When rewriting an absolute value inequality as a compound statement, when do we use the word or ? Explain. We use the word or when an absolute value expression is greater than or greater than or equal to a positive number. When an inequality states that the absolute value expression is greater than a positive number, solutions on the number line are to the left of the lesser endpoint and to the right of the greater one. For example, all solutions to |x| > 4 are values less than − 4 or greater than 4. When an inequality states that the absolute value expression is greater than or equal to a positive number, solutions on the number line are the lesser endpoint and all values to the left of that endpoint and the greater endpoint and all values to the right of that endpoint.

Promoting the Standards for Mathematical Practice When students rewrite absolute value inequalities as compound statements and determine whether to connect individual inequalities with and or or, they are looking for regularity in repeated reasoning (MP8). Ask the following questions to promote MP8: • What patterns did you notice when you solved an absolute value inequality and joined the statements with and ? • What patterns did you notice when you solved an absolute value inequality and joined the statements with or ? • What patterns did you notice in the absolute value inequalities’ solution sets and their graphs?

When solving an absolute value inequality, whether the compound statement is joined by and or or depends on the direction of the inequality symbol. © Great Minds PBC

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Have students read the description of an absolute value inequality and discuss as a class, addressing any questions. In particular, note the portion of the description that says using one of the four inequality symbols to ensure that students understand that this means the inequality could be >, <, ≥, or ≤. Have students complete the statements after the absolute value inequality description and then verify the correct responses.

Teacher Note

Ask students to work on problems 5 and 6 with a partner.

A statement that compares the expressions |bx − c| and d, where b, c, and d are real numbers, using one of the four inequality symbols is an example of an absolute value inequality.

In Mathematics I, absolute value inequalities are limited to those with linear expressions within the absolute value. Thus, absolute value inequalities are described, but not formally defined, in this manner. In subsequent courses, students may encounter other types of absolute value inequalities.

Complete each statement with either more or less based on the absolute value inequality.

The inequality |bx − c| > d means that the value of the expression bx − c is more than d units from 0.

The inequality |bx − c| < d means that the value of the expression bx − c is than d units from 0.

less

For problems 5 and 6, write the inequality as a compound statement. Find the solution set and write it by using set-builder notation. Then graph the solution set on the given number line. 5. |6y − 2| ≥ 5

6y − 2 ≥ 5 6y ≥ 7

|6y − 2| ≥ 5

− (6y − 2) ≥ 5

6y − 2 ≤ − 5

6y ≤ − 3

y≥_ 7 6

y ≤ − 1_ 2

_ _ {y y ≤ − 2 or y ≥ 6 }

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

6. |10x + 1| < 11

|10x + 1| < 11

10x + 1 < 11

− (10x + 1) < 11

and

10x < 10

10x + 1 > − 11

x<1

10x > − 12 x > −_ 6 5

_ {x − 5 < x < 1} 6

x –2

–1

Once students finish, display correct answers and address any questions. Debrief by having students compare and contrast solving absolute value equations with solving absolute value inequalities. Address the following points if students do not raise them: • Both are solved by rewriting as a compound statement. • When solving an absolute value equation, the compound statement is always joined by or. When solving an absolute value inequality, whether the compound statement is joined by and or or depends on the direction of the inequality symbol. • An absolute value equation can have zero, one, or two solutions, while an absolute value inequality can have infinitely many solutions.

UDL: Representation Consider using a graphic organizer to compare solving absolute value equations with solving absolute value inequalities. For example, create a three-column table and solve |2x + 1| = 9, |2x + 1| ≤ 9, and |2x + 1| ≥ 9 in each column.

Have students turn and talk about the following question. Do you think an absolute value inequality can have no solution? Only one solution? A solution of all real numbers?

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Give students 1 minute to discuss with a partner, and then direct them to work on problems 7–10. Tell the class before they start that instead of solving algebraically, they should use what they know about absolute value expressions to find the solution sets. For problems 7–10, find the solution set. Write a brief justification for your answer. 7. |m| > −2

ℝ

Distance is never negative, so the value of |m| will always be greater than or equal to 0. Therefore, the value of |m| will always be greater than −2.

8. |m| < −2

{}

Distance is never negative, so there is no value of m that will make the value of |m| less than −2.

9. |g − 4| ≥ 0

ℝ

Distance is never negative, so the value of |g − 4| will always be greater than or equal to 0 for any value of g.

10. |g − 4| ≤ 0

{4}

Distance is never negative, so the value of |g − 4| can never be less than 0. The value of |g − 4| can equal 0 when g = 4, so that is the only solution to the inequality.

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After students have completed these problems, review the answers as a class. Debrief by asking the following questions and discussing any misconceptions or errors. How can we interpret the solutions to the inequality |m| > −2?

The solutions are the values of m that are each a distance of more than −2 units from 0 on the number line. What values of m will make the inequality |m| > −2 true? Why?

Any value of m will make the inequality true because the absolute value will always be greater than or equal to 0. How can we interpret the solutions to the inequality |m| < −2? The solutions are the values of m that are each a distance of less than −2 units from 0 on the number line. What values of m will make the inequality |m| < −2 true? Why? No value of m will make the inequality true because the absolute value will never be negative, so it can never be less than a negative number. If time permits, have students explore how the solution sets to problems 9 and 10 change if the inequality symbols are changed to make each problem a strict inequality.

Teacher Note Some students may need support with the idea that |m| > −2 has a solution set of all real numbers, when previously an absolute value equation with a negative number had no solution. Clarify for students that it does not matter that |m| will never take on a value between 0 and −2. Because |m| will always have a nonnegative value, it will always be greater than −2. This clarification could lead students to the generalization that, for a negative number b, |x| > b will have a solution set of all real numbers.

More Than Just Absolute Value Students solve absolute value inequalities by first isolating the absolute value expression. Direct students to problem 11 and ask the following questions. Based on the description of an absolute value inequality that we read earlier, is this an absolute value inequality? How do you know? Yes. It can be written in the form described. What should the first step be in solving the inequality? We should add 8 to both sides to isolate the absolute value expression.

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Have students work problems 11 and 12 with their partners. For problems 11 and 12, solve the inequality. Write the solution set by using set notation.

11. |c + 4| − 8 < 20

c + 4 < 28 c < 24

|c + 4| − 8 < 20 |c + 4| < 28 and

c + 4 > −28 c > −32

{c | −32 < c < 24}

12. −3 | w + 12| + 10 ≤ −5

−(c + 4) < 28

−3|w + 12| + 10 ≤ −5 − 3|w + 12| ≤ −15

w + 12 ≥ 5

w ≥ −7

{w | w ≤ −17 or w ≥ −7}

|w + 12| ≥ 5 or

−(w + 12) ≥ 5

w + 12 ≤ −5

w ≤ −17

When students finish, display the answers and discuss as a class. Debrief by asking the following questions. When rewriting an absolute value inequality as a compound statement, when should we decide whether to use the word and or or ? We should decide whether to use the word and or or after isolating the absolute value expression.

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For problem 11, do we use and or or when rewriting the inequality as a compound statement? Why? We use and. Once we isolate the absolute value expression, the inequality has a less than symbol. For problem 12, do we use and or or when rewriting the inequality as a compound statement? Why? We use or. When isolating the absolute value expression, we reverse the inequality symbol to be a greater than or equal to symbol when we multiply by − _1 . 3

Land Debrief 10 min Objectives: Write absolute value inequalities in one variable as compound statements joined by and or or. Solve absolute value inequalities and graph the solution sets on the number line. Use the statements provided to engage students in the Always, Sometimes, Never routine. For each statement, give students 30 seconds of thinking time followed by another 30 seconds to share their thinking with a partner. Ask pairs to share whether they think the statement is always, sometimes, or never true, encouraging examples or counterexamples to support their claim. Come to a consensus that the statement is [always/sometimes/never] true [because…]. Repeat the routine for each statement selected. • Absolute value inequalities have infinitely many solutions.

Sometimes, because there are times when there will be no solution or a finite number of solutions. For example, |x| > −5 has infinitely many solutions, but |x| < −5 has no solution.

Sometimes, because inequalities such as |x| ≤ 0 have one solution, while other inequalities such as |x| > 4 have more than one solution.

• Absolute value inequalities have one solution.

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• Absolute value expressions are less than a negative number. Never, because distance is never negative, so absolute value expressions will always be greater than a negative number. • Absolute value expressions are greater than a negative number. Always, because distance is always greater than a negative number. • Absolute value equations have two solutions.

Sometimes, because there are equations that can have one solution, no solution, or two solutions. For example, |x| = 2 has two solutions, while |x| = 0 has one solution and |x| = −7 has no solution.

Conclude the discussion by asking the following question.

Do the solution sets of absolute value equations differ from the solution sets of absolute value inequalities? If so, how? Yes. The solution sets of absolute value equations can have one solution, no solution, or two solutions. The solution sets of absolute value inequalities can have one solution, no solution, or infinitely many solutions.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

RECAP Name

Date

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

2. Consider the absolute value inequality 2|6a − 10| ≥ 4. a. Isolate the absolute value expression.

Solving Absolute Value Inequalities

2|6a − 10| ≥ 4 |6a − 10| ≥ 2

Terminology

In this lesson, we •

wrote absolute value inequalities as compound inequalities.

•

solved absolute value inequalities.

•

graphed solution sets of absolute value inequalities on the number line.

A statement that compares the expressions |bx − c| and d,

b. Rewrite the absolute value inequality as a compound statement.

where b, c, and d are real

If 6a − 10 and its opposite are at least 2 units from 0, then the value of 6a − 10 must be less than or equal to −2 or greater

6a − 10 ≥ 2 or −(6a − 10) ≥ 2

numbers, using one of the four inequality symbols is an example of an absolute value inequality.

than or equal to 2. Use or.

Examples

1. Consider the absolute value inequality |4r + 3| < 7.

c. Solve each inequality.

a. Rewrite the inequality as a compound statement.

6a − 10 ≥ 2

6a ≥ 12

If 4r + 3 and its opposite are each less than 7 units from 0

4r + 3 < 7 and −(4r + 3) < 7

a≥2

on the number line, then the value of 4r + 3 must be

− (6a − 10) ≥ 2

6a − 10 ≤ − 2

6a ≤ 8 a ≤ 4_ 3

between −7 and 7. Use and.

b. Solve each inequality.

4r + 3 < 7 4r < 4

d. Write the solution set by using set-builder notation. 4_ {a a ≤ 3 or a ≥ 2}

− (4r + 3) < 7

and

4r + 3 > − 7

r<1

4r > − 10 r > − 5_ 2

e. Graph the solution set on the number line.

c. Write the solution set by using set-builder notation.

5_ {r − 2 < r < 1}

–4

–3

–2

–1

d. Graph the solution set on the number line.

r –5

–4

–3

–2

–1

231

232

RECAP

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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PRACTICE Name

Date

1. Consider the inequality |x + 1| < 2.

For problems 3–9, find the solution set of the inequality. Write the solution set by using set-builder notation. Then graph the solution set on the given number line. 3. |w + 2| < 10

{w | −12 < w < 8}

a. Rewrite the inequality as a compound statement.

x + 1 < 2 and −(x + 1) < 2

{x | −3 < x < 1}

2. Consider the inequality |2b − 1| ≥ 6.

–5

9 10 11 12 13

–4

p –3

–2

–1

9 10

1 6. _ |4d − 2| < 1 2

c. Graph the solution set on the number line provided.

{d | 0 < d < 1}

302

13 __ {p − 3 ≤ p ≤ 1}

5_ 7_ {b b ≤ − 2 or b ≥ 2 }

–7 –6 –5 –4 –3 –2 –1

b. Solve the compound statement. Write the solution set by using set-builder notation.

–1

5. |3p + 5| + 20 ≤ 28

2b − 1 ≥ 6 or −(2b − 1) ≥ 6

–2

9 10

a. Rewrite the inequality as a compound statement.

–3

{m | m < −6 or m > 13}

x –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

4. |2m − 7| > 19

c. Graph the solution set on the number line provided.

–4

–12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

b. Solve the compound statement. Write the solution set by using set-builder notation.

–5

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–10 –9 –8 –7 –6 –5 –4 –3 –2 –1

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7. |2r − 3| ≥ 0

Math 1 ▸ M1 ▸ TC ▸ Lesson 15

10. Compare and contrast the solution set of |k − 2| = 3 with the solution set of |k − 2| ≥ 3.

ℝ

The solution set of the inequality has infinitely many solutions, while the solution set of the equation has only two solutions. Both solution sets contain −1 and 5. Those are the only solutions to the equation, while the solution set of the inequality also includes all values greater than 5 or less than −1.

r –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

9 10

8. −5|4c + 25| + 7 ≥ −8

For problems 11–14, create an equation or inequality for the situation.

{c − 7 ≤ c ≤ − 2 }

11 __

11. An absolute value equation that has two solutions Sample: |x − 3| = 5

c –10

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–9

–8

–7

–6

–5

–4

–3

–2

–1

9. |5 − x| − 3 > 7

12. An absolute value equation that has one solution Sample: |x − 3| = 0

{x | x < −5 or x > 15} –5 –4 –3 –2 –1

x 0

9 10 11 12 13 14 15 13. An absolute value inequality that has a solution of all real numbers Sample: |x − 3| > −2

14. An absolute value inequality that has no solution Sample: |x − 3| < −2

P R ACT I C E

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

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Math 1 ▸ M1 ▸ TC ▸ Lesson 15

18. A lifeguard counted and recorded the number of children using the splash pool at each half hour. The data are listed in ascending order.

Remember For problems 15 and 16, solve the equation. 15. 0 = − 4(−6t + 3) 1_ 2

16. 2.5(8 − 2t) = 2

3.6

n ≤ 4 and n ≥ −2

n = 4 or n = −2

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Splash Pool Usage

n –8

–6

–4

–2

n –8

–6

–4

–2

–8

–6

–4

–2

–8

–6

–4

–2

–8

–6

–4

–2

Number of Children

−2 < n < 4

n ≥ 4 or n ≤ −2

Make a box plot of the data set.

17. Match each compound statement to the graph of its solution set on the number line.

n > 4 or n < −2

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Teacher Edition: Math 1, Module 1, Topic C, Lesson 16 LESSON 16

Applying Absolute Value (Optional) Use absolute value equations and absolute value inequalities to solve real-world problems.

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Name

Date

EXIT TICKET

A team uses volleyballs that have certain weights for games. Each volleyball must not be more than 10 grams from the desired weight of 270 grams. Let w represent the weight of each volleyball in grams. Write and solve an absolute value inequality to find the range of acceptable weights of a volleyball.

w − 270 ≤ 10

w ≤ 280

|w − 270| ≤ 10 and

−(w − 270) ≤ 10

w − 270 ≥ −10 w ≥ 260

Lesson at a Glance In this optional lesson, students watch a video to make sense of a context about a car’s distance from a roadside attraction. They work in pairs to answer questions related to a similar situation. Through class discussion and teacher guidance, students write absolute value equations to model situations. Building from that experience and what they know about solution sets of absolute value inequalities, students continue working in pairs to write and solve absolute value inequalities that represent real-world situations.

The range of acceptable weights of a volleyball is 260 grams to 280 grams.

Key Question • What types of situations can be modeled with an absolute value equation or an absolute value inequality?

Achievement Descriptors Math1.Mod1.AD4 Create equations and inequalities in one variable and use them to solve problems. (A.CED.A.1) Math1.Mod1.AD5 Represent constraints by using equations and

inequalities. (A.CED.A.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• The Road Trip

• None

• Applying Absolute Value Inequalities

Land 10 min

Lesson Preparation • None

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Fluency Graphs of Solution Sets of Equations and Inequalities Students match a solution set graphed on a number line to its equation or inequality. Directions: Match each equation or inequality to its corresponding solution set graphed on the number line. A. D.

|x| = 4

B. |x| = −4 E. |x| ≥ 4

x=4

x –8 –6 –4 –2

C. |x| ≤ 4

F. −x = 4 2.

–8 –6 –4 –2

x –8 –6 –4 –2

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x –8 –6 –4 –2

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Launch

Math 1 ▸ M1 ▸ TC ▸ Lesson 16

Students collect information about a situation to prepare to model a similar situation with an absolute value equation. Play the Road Attractions video. Then play the video a second time and direct students to problem 1. Have them record important information about the situation shown in the video. Pause or replay the video as needed. 1. Record important information from the video. Then draw and label a sketch of the road.

UDL: Representation Presenting the roadside attraction situation in video format supports students in understanding the problem context by removing barriers associated with written and spoken language.

The trip odometer is set to 0 miles at the beginning of the trip. The driver saw the first sign for World Famous Tacos after driving 239 miles. The distance between the first sign and World Famous Tacos is 100 miles. After driving 301 miles, World Famous Tacos is 38 miles away.

While replaying the video, consider pausing it and pointing out the odometer. Highlight that an odometer is an instrument used to measure distance traveled.

After driving 332 miles, World Famous Tacos is 7 miles away. After driving 346 miles, the driver pulls off the road and stops.

332 mi Mile 0

Language Support

World Famous 7 mi Tacos

7 mi 346 mi

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Elicit the important information depicted in the video by asking the following questions. What did you notice? The driver drove a total of 346 miles. World Famous Tacos is 339 miles from where the trip began. The driver passed World Famous Tacos without stopping. The driver was daydreaming when they were within 7 miles of World Famous Tacos. What do you wonder? I wonder whether the person drove at a constant speed. I wonder how long it will take for the driver to turn around and arrive back at World Famous Tacos. I wonder how many miles past World Famous Tacos the driver is. I wonder how much time it would take for the person to drive to the end of the road. What questions might we want to answer? How far past World Famous Tacos is the driver now? What is the driver going to do since they passed World Famous Tacos? If the driver turns around, how far would they have to drive to get to World Famous Tacos? Was the person driving at a constant speed? If so, how fast was the car traveling? How long would it take for the person to drive to the end of the road? How far away is World Famous Tacos when the driver stops the car? Explain. World Famous Tacos is 7 miles away. Based on the signs and the trip odometer, it is located 339 miles from where the trip began, and 346 − 339 = 7. Earlier in the trip, there was a sign that said that World Famous Tacos was 7 miles away. How can the driver be 7 miles away at two different places on this road? The driver can be 7 miles before World Famous Tacos or 7 miles beyond it. We can use absolute value equations or inequalities to answer questions related to situations like this one. Today, we will write and solve absolute value equations and inequalities to solve real-world problems.

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Learn The Road Trip Students write and solve absolute value equations based on a real-world situation. Have students work in pairs to complete problem 2(a). Circulate as students work, and confirm answers. 2. Mrs. Allen takes a road trip. She sets her trip odometer to 0 as she starts her trip. • Mrs. Allen plans to stop at Mystery Dino Land, which is 84 miles from her starting location. • She then plans to continue driving in the same direction and stop for lunch at Crystal Trees Park, which is 142 miles from her starting location.

UDL: Representation Consider highlighting relationships between the location of the car and specific locations along the road mentioned in problem 2. Have students sketch the road and label, color-code, or highlight important details.

• After lunch, Mrs. Allen plans to continue driving in the same direction. a. How far has Mrs. Allen driven from her starting location when she is 20 miles from Mystery Dino Land?

20 miles before Mystery Dino Land: 84 − 20 = 64 20 miles beyond Mystery Dino Land: 84 + 20 = 104 Mrs. Allen has driven either 64 miles or 104 miles from her starting location when she is 20 miles from Mystery Dino Land. Call the class back together and ask the following question. We found that Mrs. Allen is 20 miles from Mystery Dino Land when she has driven 64 miles and again when she has driven 104 miles. Why are there two answers? There are two answers because Mrs. Allen can be located either 20 miles before Mystery Dino Land or 20 miles beyond Mystery Dino Land.

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Have students continue working in pairs to complete part (b). Circulate as students work, and ask them whether their expressions make sense for both answers in part (a). Identify one or two pairs to share their expressions during the debrief. b. Write two or more expressions that each represent the distance between Mrs. Allen on her drive and Mystery Dino Land. Let x represent the number of miles Mrs. Allen has driven from her starting location.

x − 84

84 − x

|x − 84|

|84 − x|

Debrief by having student pairs share their expressions for part (b). Facilitate a brief discussion by asking the following questions. If x represents the number of miles Mrs. Allen has driven, what does x − 84 represent? It represents the difference between the number of miles Mrs. Allen has driven and the number of miles to Mystery Dino Land. Does the expression x − 84 represent the distance between Mrs. Allen and Mystery Dino Land? Explain. No. That expression can give us negative numbers for some values of x, and distance is always positive. Display the expressions x − 84 and 84 − x. Are these expressions equivalent? How do you know?

If students need support identifying how x − 84 and 84 − x are related, consider having them rewrite the expression 84 − x showing the x-term first to compare the two expressions:

84 − x = 84 + (−x) = −x + 84 Then ask the following questions: • What do you notice about x and −x ?

No. If x = 64, for example, the two expressions evaluate to two different numbers.

• What do you notice about −84 and 84?

These expressions are not equivalent, but they are related. How are they related?

• How do the expressions x − 84 and −x + 84 compare?

The expressions are opposites of each other.

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Have students think–pair–share to answer the following question. How can we algebraically represent that 84 − x is the opposite of x − 84? We can write 84 − x as −(x − 84). If Mrs. Allen drives less than 84 miles, the expression x − 84 would evaluate to a negative number. If Mrs. Allen drives more than 84 miles, the expression x − 84 would evaluate to a positive number. What single expression would tell us the distance between Mrs. Allen’s location and Mystery Dino Land, regardless of the number of miles she has driven?

|x − 84|

Direct students to work with a partner to complete part (c). Circulate as students work, and ensure they write the correct absolute value equation before they solve the equation. c. Write and solve an absolute value equation that models the situation from part (a). Let x represent the number of miles Mrs. Allen has driven.

x − 84 = 20 x = 104

|x − 84| = 20 or

−(x − 84) = 20 x − 84 = −20

Teacher Note Even though students can answer problem 2(a) without writing an equation, this problem helps reinforce the concept of absolute value.

x = 64 When Mrs. Allen is 20 miles from Mystery Dino Land, she has driven either 64 miles or 104 miles. Once students finish, have them verify their answers from part (a). Invite students to share their work. Then facilitate the following discussion to help students interpret absolute value equations in the form |x − a| = b, where a is a real number and b is a nonnegative number. Why do you think that we are asked to write an absolute value equation in this situation?

I think that we are asked to write an absolute value equation because the absolute value expression is used to represent distance. I think that we are asked to write an absolute value equation because this situation is about a specific distance on either side of Mystery Dino Land.

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How can we interpret the expression |x − 84| in terms of the context?

We can interpret it as the distance between Mrs. Allen’s location and Mystery Dino Land. How can we interpret the equation |x − 84| = 20 in terms of the context?

The distance between Mrs. Allen’s location and Mystery Dino Land is 20 miles.

How do 64 and 104 relate to the equation |x − 84| = 20? They are the solutions to the equation.

Display the equation |x − 84| = 20 and a blank number line.

How do we graph the solution set of |x − 84| = 20 on the number line?

We plot points at 64 and 104.

Draw and label tick marks at 64 and 104. Plot points at 64 and 104. Then annotate the number line as shown and ask the following questions.

20 x

104

What do you notice? I notice 84 is in the middle of 64 and 104. I notice 84 is 20 units from both 64 and 104. I notice this number line reminds me of the situation from part (a). How does this number line model the situation from part (a)? The number line shows the two values of x that are each 20 units from 84.

The solutions to |x − 84| = 20 are 64 and 104 because they are each 20 units from 84.

In general, an absolute value equation in the form |x − a| = b, where a is a real number and b is a nonnegative number, can be interpreted as the values of x that are each b units from a on a number line.

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Have students work in pairs to complete problem 3. 3. Mrs. Allen is 40 miles from Crystal Trees Park. a. Write and solve an absolute value equation to find the number of miles Mrs. Allen has driven.

x − 142 = 40 x = 182

|x − 142| = 40 or

−(x − 142) = 40 x − 142 = −40 x = 102

When Mrs. Allen is 40 miles from Crystal Trees Park, she has driven either 102 miles or 182 miles. b. How far is Mrs. Allen from Mystery Dino Land?

182 − 84 = 98 102 − 84 = 18 When Mrs. Allen is 40 miles from Crystal Trees Park, she is either 18 miles or 98 miles from Mystery Dino Land. Confirm student answers. Have students think–pair–share about the following questions. How can we interpret the equation |x − 142| = 40 in terms of the context?

The equation models the numbers of miles Mrs. Allen drives x that are each 40 miles from Crystal Trees Park, which is 142 miles from her starting point. We can also think of the equation |x − 142| = 40 as representing the values of x that are each 40 units from 142 on the number line. How else have we interpreted this equation?

We have interpreted this equation as the values of x that make the expression x − 142 evaluate to a distance of 40 units from 0 on the number line. What is similar about these interpretations? What is different? Both interpretations have to do with a distance of 40 units on the number line. One measures distance from 0 and the other measures distance from 142.

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Have student pairs complete problem 4. Circulate as students work, and look for student work that is similar to Angel’s and Levi’s work shown in the discussion that follows. 4. Suppose Mrs. Allen drives at a constant rate of 50 mph. After how many hours is Mrs. Allen 80 miles from Crystal Trees Park? Let x represent the number of miles Mrs. Allen has driven.

x − 142 = 80

|x − 142| = 80

x = 222

−(x − 142) = 80 x − 142 = −80 x = 62

Let d represent the number of miles Mrs. Allen has driven, let r represent the rate of travel in miles per hour, and let t represent the number of hours Mrs. Allen has driven.

d = rt

Differentiation: Support If students need support to complete problem 4, consider asking the following questions to encourage them to write and solve an absolute value equation: • Is there more than one place on the road where Mrs. Allen is 80 miles from Crystal Trees Park? • Does relating the location of Crystal Trees Park on the road to 0 on the number line help us write an equation? • Can this situation be modeled with an absolute value equation? Why?

d = 50t After driving 222 miles:

222 = 50t

4.44 = t

After driving 62 miles:

62 = 50t

1.24 = t

Mrs. Allen is 80 miles from Crystal Trees Park after driving for 1.24 hours and again after driving for 4.44 hours.

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When student pairs finish, display either the selected student work samples or Angel’s and Levi’s work samples.

If students are ready for a challenge, consider asking the following question:

Angel’s Work

• How many miles has Mrs. Allen driven when she is twice as far from Mystery Dino Land as she is from Crystal Trees Park?

Let x represent the number of miles Mrs. Allen has driven.

x − 142 = 80

|x − 142| = 80

x = 222

Differentiation: Challenge

−(x − 142) = 80 x − 142 = −80

2 122 __ miles or 200 miles 3

x = 62 Let d represent the number of miles Mrs. Allen has driven, let r represent the rate of travel in miles per hour, and let t represent the number of hours Mrs. Allen has driven.

d = rt d = 50t After driving 222 miles:

After driving 62 miles:

222 = 50t

62 = 50t

4.44 = t

1.24 = t

Mrs. Allen is 80 miles from Crystal Trees Park after driving for 1.24 hours and again after driving for 4.44 hours.

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Levi’s Work Let t represent the number of hours Mrs. Allen has driven.

50t − 142 = 80 50t = 222

|50t − 142| = 80 or

−(50t − 142) = 80 50t − 142 = −80

t = 4.44

50t = 62 t = 1.24

Mrs. Allen is 80 miles from Crystal Trees Park after driving for 1.24 hours and again after driving for 4.44 hours.

Then ask the following questions to facilitate a discussion. What do you notice about Angel’s and Levi’s work samples? I notice that the solutions to Angel’s absolute value equation are the numbers of miles Mrs. Allen has driven and the solutions to Levi’s absolute value equation are the numbers of hours Mrs. Allen has driven. I notice that Levi’s sample has fewer steps. What do you wonder? I wonder if there are any other ways to solve this problem. Why did Angel and Levi find different solutions to their absolute value equations? The solutions to Angel’s absolute value equation correspond to the numbers of miles Mrs. Allen has driven, and the solutions to Levi’s absolute value equation correspond to the numbers of hours Mrs. Allen has driven. What does 50t represent in Levi’s equation? It represents the distance traveled after driving t hours at 50 mph.

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How does 50t in Levi’s equation relate to x in Angel’s equation? How do you know? The value of 50t in Levi’s equation is equal to the value of x in Angel’s equation because the distance Mrs. Allen drove after t hours of driving at 50 mph corresponds to the number of miles she drove. The questions presented in this lesson so far have two answers that make sense in context. Now, let’s consider questions where the answers include a range of values that make sense in context.

Applying Absolute Value Inequalities Students write and solve absolute value inequalities based on real-world situations involving restrictions or error in measurement. Consider asking students whether they are aware of the terms margin of error or range of acceptable measurement. If students are, allow a few to share what the terms mean with the class, and then read problem 5 aloud to the class. If they are not familiar with the terms, direct students to read problem 5, and then summarize the definitions. Guide students through writing an absolute value inequality for the given situation by using the following prompts. What is the target weight for a bag of flour? The target weight is 5 pounds. How much error is allowed? In other words, how much greater than or less than the target weight of 5 pounds can the weight of the bag of flour be? The weight of the bag of flour can be 0.04 pounds greater than or less than the target weight. If we let w represent the weight of a bag of flour in pounds, what does w − 5 represent? It represents the difference between the actual weight of the bag of flour and the target weight. What does |w − 5| represent?

It represents the absolute error that the weight of a bag of flour is from the target weight of 5 pounds. Turn to a partner and discuss the process for writing an inequality that represents this problem. © Great Minds PBC

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Direct students to complete problem 5. When they finish, invite a few students to share their inequalities and reasoning. Direct students to complete problem 6 with their partners. Review the correct answer, and let student understanding dictate the amount of discussion needed. 5. A machine weighs bags of flour. The weight of a filled bag of flour should be no more than 0.04 pounds from the target weight of 5 pounds. Write and solve an absolute value inequality to find the range of acceptable weights of a bag of flour. Let w represent the weight of a bag of flour in pounds.

w − 5 ≤ 0.04 w ≤ 5.04

|w − 5| ≤ 0.04 and

−(w − 5) ≤ 0.04 w ≥ 4.96

The range of acceptable weights of a bag of flour is 4.96 pounds to 5.04 pounds. 6. An engineer designs a sensor for an airplane. The sensor’s height can be no more than 0.001 inches from the desired height of 6 inches. Write and solve an absolute value inequality to determine the range of acceptable heights for a sensor. Let h represent the height of a sensor in inches.

h − 6 ≤ 0.001 h ≤ 6.001

|h − 6| ≤ 0.001 and

Promoting the Standards for Mathematical Practice Students reason quantitatively and abstractly (MP2) as they write and solve absolute value inequalities based on real-world situations. Ask the following questions to promote MP2:

−(h − 6) ≤ 0.001

h ≥ 5.999

• What real-world situations are modeled by absolute value expressions? • How does an absolute value inequality represent the range of acceptable values?

The range of acceptable heights for a sensor is 5.999 inches to 6.001 inches.

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Land Debrief 5 min Objective: Use absolute value equations and absolute value inequalities to solve real-world problems. Facilitate a class discussion by using the following questions. Encourage students to restate or build upon one another’s responses. In what ways can we interpret the equation |x − 5| = 2?

We can interpret the equation as x is a distance of 2 units from 5 or that x − 5 is a distance of 2 units from 0 on the number line. In the previous topic, we used linear equations and inequalities in one variable to solve real-world problems. Why is the situation in problem 2(a) modeled with an absolute value equation? The situation in problem 2(a) is modeled with an absolute value equation because the unknown values are equal distances from a specific location. Why is the situation in problem 5 modeled with an inequality instead of an equation? The situation in problem 5 is modeled with an inequality instead of an equation because there is a range of values that are acceptable solutions in this context. What types of situations can be modeled with an absolute value equation or an absolute value inequality? Absolute value equations are used to represent situations where two unknown values are the same distance from a given value on the number line. Absolute value inequalities can represent unknowns that cover a range of values either within or outside of a given distance from a fixed value on the number line.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem. © Great Minds PBC

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

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Math 1 ▸ M1 ▸ TC ▸ Lesson 16

RECAP Name

Date

2. Professional baseball players typically use a bat that weighs up to 2 ounces greater than or less than 34 ounces. a. Write an absolute value inequality that represents the typical range of weights of a bat that professional baseball players use.

Applying Absolute Value (Optional) In this lesson, we •

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Math 1 ▸ M1 ▸ TC ▸ Lesson 16

Let w represent the weight of a bat in ounces.

|w − 34| ≤ 2

wrote and solved absolute value equations and absolute value inequalities that represent real-world situations.

This situation is modeled with an inequality because there is a range of values that make sense in context.

Examples 1. The thickness of a steel pipe can be up to 0.01 inches greater than or less than the desired thickness of 3.5 inches when it is manufactured. Let x represent the thickness of a steel pipe in inches. Match each situation to the equation or inequality that represents it. Situation

b. Solve the absolute value inequality you wrote in part (a) to find the typical range of weights of a bat that professional baseball players use.

Equation or Inequality

All thicknesses that indicate the steel pipe is not manufactured within the acceptable limits All thicknesses that indicate the steel pipe is manufactured within the acceptable limits All thicknesses that indicate the steel pipe is manufactured at the acceptable limits

|x − 3.5| = 0.01

w − 34 ≤ 2

w ≤ 36

|x − 3.5| > 0.01

|w − 34| ≤ 2 and

−(w − 34) ≤ 2

w − 34 ≥ −2 w ≥ 32

The typical range of weights of a bat that professional baseball players use is 32 ounces to 36 ounces.

|x − 3.5| ≤ 0.01

The situation is modeled by an absolute value equation because the unknown values are the same distance from

3.5 on the number line.

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245

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RECAP

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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PRACTICE Name

Date

Math 1 ▸ M1 ▸ TC ▸ Lesson 16

3. A candy manufacturer typically fills each bag with 362 pieces of candy. The number of pieces of candy in a bag can be up to 12 pieces greater than or less than 362. a. Write an absolute value inequality that represents the acceptable range of the number of pieces of candy in a bag.

1. For each situation, the temperature of the water inside a water heater x in degrees Fahrenheit can be up to 5°F greater than or less than the 120°F setting. Match each situation to the equation or inequality that represents it.

|x − 120| ≤ 5

All temperatures of the water x that indicate that the water heater needs repairs

b. Solve the absolute value inequality you wrote in part (a) to find the acceptable range of the number of pieces of candy in a bag.

|x − 120| > 5

All temperatures of the water x that indicate that the water heater is functioning at its limits

The acceptable range of the number of pieces of candy in a bag is 350 pieces to 374 pieces.

2. A decorator wants to hang a shelf in the center of a wall. She wants to mark the wall for placement of the anchors that will support the weight of the shelf. The decorator has the following information: •

|c − 362| ≤ 12

|x − 120| = 5

All temperatures of the water x that indicate that the water heater is functioning properly

•

Let c represent the number of pieces of candy in a bag.

Equation or Inequality

Situation

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4. A typical body temperature of an adult is 98.6°F, but adult body temperatures can vary by up to 1°F and still be considered normal.

The center of the wall is 25 _ inches from the left edge.

a. Write an absolute value inequality that represents the range of normal body temperatures of an adult in degrees Fahrenheit.

5 8

The anchors should be placed exactly 3 inches from the center of the wall.

Let t represent the body temperature of an adult in degrees Fahrenheit.

|t − 98.6| ≤ 1

a. Write an absolute value equation to find where the anchors should be placed on the wall. Let m represent the distance from the left edge of the wall in inches where the anchors should be placed. m − 25 5_ = 3 8

b. Solve the absolute value inequality you wrote in part (a) to find the range of normal body temperatures of an adult in degrees Fahrenheit.

b. Solve the absolute value equation you wrote in part (a) to find where the anchors should be placed on the wall.

The range of normal body temperatures of an adult in degrees Fahrenheit is 97.6°F to 99.6°F.

The anchors should be placed 22 _ inches and 28 _ inches from the left edge of the wall. 5 8

5 8

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c. Write an absolute value inequality that represents the range of abnormal body temperatures of an adult in degrees Fahrenheit. Let t represent the body temperature of an adult in degrees Fahrenheit.

|t − 98.6| > 1

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Math 1 ▸ M1 ▸ TC ▸ Lesson 16

8. During one day, managers of two grocery stores tracked how long each of their customers spent shopping. One random sample was taken from all shoppers at store A, and another random sample was taken from all shoppers at store B. The box plots show the distributions of the lengths of time in minutes shoppers spent at each store. Which statements must be true based on the data distributions shown in the box plots? Choose all that apply. Length of Time Spent in Grocery Store

Store A

d. Solve the absolute value inequality you wrote in part (c) to find the abnormal body temperatures of an adult in degrees Fahrenheit.

Store B

The abnormal body temperatures of an adult in degrees Fahrenheit are below 97.6°F or above 99.6°F.

12 18 24 30 36 42 48 54 60 66 72 78 Time (minutes)

A. The ranges of the two data sets are the same. B. The median for store A is less than the median for store B.

Remember For problems 5 and 6, solve the equation. 5. −2(d − 6) + 5d = 12

C. The interquartile range for store B is greater than the interquartile range for store A.

D. The difference in the medians of the two data sets is _ of the interquartile range for store A. 1 2

_ 6. −42 = 8d − 9 + d 4

E. The difference in the medians of the two data sets is 2 times the interquartile range for store B.

−4

7. The acceptable temperature of a freshwater aquarium is at least 72°F and no more than 82°F. Write a compound inequality for the acceptable range of temperatures for a freshwater aquarium. Let x represent the temperature of a freshwater aquarium in degrees Fahrenheit. The acceptable range of temperatures is 72 ≤ x ≤ 82.

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Teacher Edition: Math 1, Module 1, Topic D

Topic D Univariate Data Beginning in grade 6, students identify statistical questions for which variation in the responses is anticipated. They collect data on a single variable and display it in dot plots, histograms, and box plots. Students identify the shapes of data distributions from these graphs. Additionally, students calculate statistical measures of center and spread to describe data distributions, and they analyze how the shapes of data distributions affect which measures of center and spread most accurately represent the data sets. In grade 7, students use measures of center and variability to make informal comparative inferences about univariate data distributions. In Mathematics I, students build on their middle school experiences to describe and compare univariate data distributions by using shape, center, and spread, noting the effect of outliers and interpreting comparisons in context. Topic D begins by reintroducing students to collecting and analyzing univariate data. In a digital lesson, students participate in a fun activity in which they make predictions, generate data, and analyze dot plots of pooled class data to informally compare the shapes of data distributions. This lesson provides an access point for all students to informally compare univariate data distributions. The activity also provides an opportunity to formatively assess what students recall from middle school about univariate data analysis. Following the digital lesson, students use measures of center to compare univariate data distributions. They compare the centers of data sets by using mean and median and by discussing the relationship between the shape of a data distribution and the mean and median of the data set. Students use measures of center to compare data distributions represented in dot plots, and they estimate means and medians of distributions displayed in histograms. Students then use the medians to compare distributions represented in box plots. Throughout these lessons, students are encouraged both to use precision when describing data distributions and to discuss comparisons in the context in which the data are collected. The goal is for students to strategically choose which measures of center are the most appropriate to compare data distributions based on the shapes of the distributions.

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Students are introduced to standard deviation to measure and compare the spreads of two univariate data distributions. The lessons emphasize understanding how the standard deviation measures a typical distance between a value in the distribution and the mean of the distribution, as well as how the standard deviation is affected by outliers. The usefulness of standard deviation as a measure of variation for normal distributions is addressed in Mathematics III. Students estimate measures of center and spread to compare univariate distributions displayed by using histograms. The goal is to enable students to approximate the center and typical variation in a distribution to facilitate comparison. Students also use the interquartile range to compare sets of data for distributions displayed in box plots as well as for skewed distributions represented in dot plots.

Maximum Speeds of Roller Coasters

Wooden Steel

100

110

120

130

Speed (miles per hour)

Topic D concludes with a culminating activity as students decide which measures of center and spread to use to compare two or more sets of univariate data. Students present their comparisons, justifying their choices of statistics for comparing the distributions and interpreting similarities and differences between the distributions in context. Throughout the topic, a principal goal is to build students’ capacity to explain what statistical measures indicate about a data set and how statistics can be used to answer questions. The skills students develop in module 1 will be applied in module 2 when they analyze bivariate data, in module 6 when they solve real-world problems by using statistical models, and in Mathematics III when they engage in formal comparative inference.

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Progression of Lessons Lesson 17 Distributions and Their Shapes Lesson 18 Describing the Center of a Distribution Lesson 19 Using Center to Compare Data Distributions Lesson 20 Describing Variability in a Univariate Distribution with Standard Deviation Lesson 21 Estimating Variability in Data Distributions Lesson 22 Comparing Distributions of Univariate Data

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Teacher Edition: Math 1, Module 1, Topic D, Lesson 17 LESSON 17

Distributions and Their Shapes Informally describe a data distribution displayed in a dot plot.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 17

Name

Date

EXIT TICKET

The dot plot shows the delay times rounded to the nearest five minutes for 60 Big Air flights in 2019. Delay Times of Big Air Flights from 2019

10 20 30 40 50 60 70 80 90 100 110 120 Delay Time (minutes)

In this digital lesson, students use digital tools to efficiently collect and analyze data via a memory game. The aggregate data provide students the unique opportunity to dynamically interact with class-generated data distributions as students approximate and compare a typical measure of two data sets. Students answer statistical questions related to data type and the shapes, centers, and spreads of data distributions on dot plots. Several terms used to describe the shapes of data distributions are reviewed in this lesson, including symmetric, right-skewed, and left-skewed. This lesson also formalizes the terms univariate quantitative data, mound-shaped distribution, and uniform distribution. Use the digital platform to prepare for and facilitate this lesson. Students will also interact with lesson content and activities via the digital platform.

a. What shape is the distribution? The distribution is right-skewed.

Key Question

b. Was 70 minutes a typical delay time for these Big Air flights? Justify your answer. No, a typical delay time was not 70 minutes. Only 12 of the 60 flights had delays of 70 minutes or more, and more than half the flights had delays of less than 40 minutes.

Lesson at a Glance

• What information can we use from a dot plot to help describe a set of univariate quantitative data?

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Achievement Descriptors Math1.Mod1.AD13 Represent data with dot plots, histograms, or box plots. (S.ID.A.1) Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers,

and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the context of the

data sets. (S.ID.A.3)

Agenda

Materials

Fluency

Teacher

Launch 10 min Learn 25 min

• None

Students

• Comparing Dot Plots

• Computers or devices (1 per student pair)

• Typical Values

Lesson Preparation

• Putting It All Together

• None

Land 10 min

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Fluency Read a Dot Plot Students read a dot plot to prepare for describing shapes of distributions represented in dot plots. Directions: The dot plot shows the number of siblings each student in one class has. Use the dot plot to answer each question. Students’ Siblings

Number of Siblings

How many students are in the class?

What is the greatest number of siblings any student in the class has?

What is the least number of siblings any student in the class has?

What number of siblings is the most common for students in the class?

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Launch

Teacher Note

Students play a game to collect data that will be analyzed informally.

Students play a memory game to collect data about the number of letters students can remember in a given time frame. The data collected are analyzed in subsequent activities in the lesson.

Learn

Comparing Dot Plots

Students compare data distributions and acquire formal language to describe their shapes. Students compare the distributions from two trials of a memory game. Students recognize that each of the two dot plots tells its own story due to the differences in the data distributions. Students answer a series of questions, through which they progress from noticing and wondering to formally describing distribution shape. Students summarize the information by using the Univariate Quantitative Data graphic organizer. They will continue to add information to this graphic organizer throughout the topic.

The purpose of the Test Your Memory game is to collect data for students to analyze later in the lesson. In case it is preferred that students do not participate in the game or analyze their own data, prepopulated data sets are provided. Use the provided data on the first slide of the lesson and consider having students move beyond the data collection slides. For classes with few students, consider selecting the Use Combined Data choice to combine students’ collected data with the provided data sets.

Your Class

Trial 1

Trial 2 0 5 10 15 20 25 30 35 40 45 50 55 60 65

Number of Letters Remembered

Teacher Note Consider having students mark the location of the Univariate Quantitative Data graphic organizer with a sticky note so they can easily locate it as they progress through the work in this topic.

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How would you describe the shape of the distribution from class A? The distribution is roughly symmetric. The distribution appears to have a peak in the middle. When the values in a data distribution look like they are close together in the center, like a mound of dirt, we refer to the distribution as mound shaped. A mound-shaped distribution is a symmetric distribution that has exactly one peak. A distribution is considered approximately uniform when most of the values have about the same frequency.

Promoting the Standards for Mathematical Practice When students collect and analyze data to estimate the typical length of a memorized sequence of letters in a trial and compare distributions across two trials, they are modeling with mathematics (MP4). Ask the following questions to promote MP4:

Which of the four classes, if any, have an approximately uniform distribution? How do you know?

• What statistical questions can you ask about these data sets?

Class B’s distribution is approximately uniform because all the data values have frequencies that are approximately the same value.

• What key ideas are represented in these dot plots? What key ideas come from comparing these dot plots?

Typical Values

Class A

Students informally explore the meaning of a typical value in a data distribution. Students estimate a typical value for the two data distributions resulting from the memory game. They recognize that it is easier to agree on a typical value for a distribution that is symmetric and mound-shaped than for one that is not. Students informally discuss how shape affects the variability in their estimates of a typical value.

UDL: Engagement Digital activities align to the UDL principle of Engagement by including the following elements:

Trial 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13

• Opportunities to collaborate with peers. Students collaborate with peers about data sets with different shapes and variabilities.

Trial 2 0 5 10 15 20 25 30 35 40 45 50 55 60 65

Number of Letters Remembered

334

• Immediate formative feedback. Students compare their answers with those of their classmates.

• Options that promote flexibility and choice. Students choose the sequence of letters that is easier for them to remember.

EUREKA MATH2

What do you notice about the estimates the class made for the typical score for trial 1? I notice that all of the estimates are clumped together.

Math 1 ▸ M1 ▸ TD ▸ Lesson 17 EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 17

LESSON Name

Date

Distributions and Their Shapes Comparing Dot Plots

What do you notice about the estimates the class made for the typical score for trial 2? I notice that the estimates are more spread out than the estimates for trial 1.

Putting It All Together

Typical Values

Putting It All Together

Students describe the shapes of the distributions shown in the dot plots and informally discuss typical values and spread. Students match dot plot distributions to descriptions. They analyze each dot plot and choose which of four descriptions best represents the dot plot. Students discuss what it means for data distributions to be symmetric, right-skewed, or left-skewed, as well as what it means for a value to be typical of a data distribution.

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Land Debrief 5 min Objective: Informally describe a data distribution displayed in a dot plot. Facilitate a brief class discussion by asking the following question. What information can we use from a dot plot to help us describe a set of univariate quantitative data? We can look at the shape of the distribution to determine whether it is approximately symmetric or skewed, for example. We can also look at the distribution in the dot plot to estimate a typical value and determine how spread out the data are in the distribution.

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Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

Recap

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

RECAP Name

Date

Distributions and Their Shapes •

described shapes of data distributions displayed in dot plots by using the terms symmetric, right-skewed, left-skewed, and mound-shaped.

•

compared data distributions displayed in dot plots.

•

estimated typical values for data distributions.

The distribution is right-skewed. c. Use the dot plot to estimate the typical number of grams of protein per serving of one of these cereals. The typical number of grams of protein per serving is about 3 grams.

The data value 3 grams falls in the middle of the data set, so it is a good estimate of a typical data value.

A mound-shaped distribution is a symmetric distribution that looks like a smooth, round hill. A mound-shaped distribution has exactly one peak.

The numbers of grams of protein per serving in 20 different varieties of breakfast cereals are listed.

A data distribution

1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8 a. Create a dot plot showing the distribution of the number of grams of protein per serving in the 20 breakfast cereals.

EUREKA MATH2

b. Describe the shape of the distribution.

Univariate quantitative data means there is one variable, and the data values are numerical.

Example

Terminology

In this lesson, we

Each dot represents one value in the data set.

Math 1 ▸ M1 ▸ TD ▸ Lesson 17

is approximately uniform when most of the values have about the same frequency.

Protein per Serving of Cereal

Protein (grams)

A skewed distribution has a tail on one side. This distribution has a tail on the right side.

259

260

RECAP

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

PRACTICE Name

Date

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

2. Ana recorded the daily high temperature in degrees Fahrenheit at her house for 10 days.

77, 85, 77, 79, 82, 84, 79, 78, 79, 80 1. Match each dot plot to the description of its shape.

a. Create a dot plot showing the distribution of high temperatures in degrees Fahrenheit for the 10-day period.

Number of Pets Owned by Students

Daily High Temperatures

Approximately symmetric

75 76 77 78 79 80 81 82 83 84 85 86 87 0

Temperature (degrees Fahrenheit)

Number of Pets

Measurements of Length of East Hallway

b. What is the shape of the distribution? The distribution is right-skewed.

Right-skewed

8.2

8.3

8.4

8.5

8.6

8.7

8.8

8.9

Length (meters)

Ages of Cars Driven by Teachers

Left-skewed

Age (years)

338

261

262

P R ACT I C E

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

4. A random sample of 80 viewers of a television show was selected. The dot plot shows the distribution of the ages of the viewers in years.

3. There are 25 people attending an event. The ages of the people attending are listed.

3, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10

Ages of Television Show Viewers

a. Create a dot plot showing the distribution of the ages in years of the people in attendance. Ages of People Attending the Event

0 0

Viewer Age (years)

9 10 11 12 13 14 15 a. What shape is the distribution?

Age (years)

The distribution is left-skewed.

b. Based on the dot plot, what is the typical age of a viewer?

b. What is the shape of the distribution?

The typical age of a viewer is around 60 years old.

The distribution is approximately symmetric.

c. Would you say that this television show appeals to a wide age range of viewers? Explain. I would say that it does appeal to a wide age range because the viewers’ ages range from under 10 to over 70.

c. What is the typical age of a person attending the event? The typical age of a person attending the event is 7 years old.

5. When is it not efficient to represent a data set by using a dot plot? Dot plots are not efficient for displaying data for sets with a large number of observations. For instance, it may take a long time to create a dot plot for a sample with over 100 observations. In addition, it might be challenging to create a dot plot for a data set with outliers or for one where the data has a large range.

d. What might the event be? Sample: The event might be a class for children where parents drop off their children and pick them up after the event ends.

P R ACT I C E

263

264

P R ACT I C E

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Math 1 ▸ M1 ▸ TD ▸ Lesson 17

6. A doctor measured the resting heart rates in beats per minute of a large group of middle-aged adults. From the group, the doctor randomly selected 20 adults who are right-handed and 20 adults who are left-handed. The distributions are shown in the dot plots.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 17

Remember For problems 7 and 8, solve the equation. 7. 16 = 7 − 2x − 5x

Resting Heart Rates for Right-Handed Adults

− _9 7

8. − 2_ (x + 20) − 9 = 15 5

−80

9. Which of the following expressions are equivalent to 4x − 2(3x + 1)? Choose all that apply.

44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84

A. 4x − 6x + 2

Heart Rate (beats per minute)

B. 4x − 6x − 2

Resting Heart Rates for Left-Handed Adults

C. − 2x + 2 D. − 2x − 2 E. − 2(x + 1) F. − 2(x − 1)

44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 Heart Rate (beats per minute)

a. In 2 or 3 sentences, compare the distribution for the resting heart rates of right-handed adults with the distribution for the resting heart rates of left-handed adults. The distribution for the right-handed adults looks left-skewed, while the distribution for the left-handed adults is right-skewed. According to the distributions, a typical resting heart rate for these right-handed adults is estimated to be a few beats per minute more than a typical resting heart rate for these left-handed adults. The overall spread of the distribution of right-handed adults is also greater than that of left-handed adults.

10. Danna earns $114 for 8 hours of work. She earns the same amount of money each hour. Write an equation that represents the relationship between Danna’s earnings y in dollars and the number of hours she works x.

y = 14.25x

b. Based on the data, can the fact that an adult is left-handed or right-handed influence that adult’s resting heart rate? Explain your reasoning. No. While the typical resting heart rate for this sample of right-handed adults is greater than that of the left-handed adults, there is not enough evidence to conclude that there is a causal relationship between the two variables. The difference in the typical resting heart rates that we observe could be due to a factor other than being right-handed or left-handed.

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P R ACT I C E

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P R ACT I C E

Teacher Edition: Math 1, Module 1, Topic D, Lesson 18 LESSON 18

Describing the Center of a Distribution Find the mean and median of data shown in a dot plot and estimate the mean and median of a data distribution represented by a histogram. Identify whether the mean, the median, or both appropriately describe a typical value for a given data set.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Name

Date

EXIT TICKET

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Consider the dot plot showing the number of text messages these students sent. Text Messages Sent by Ninth Graders Last Tuesday

A random sample of 10 ninth grade students was asked for the number of text messages they sent last Tuesday.

Student

Number of Text Messages Sent

285

100

150

200

250

300

Number of Text Messages Sent

a. Calculate the mean and median of the data set. The sum of the data values is 560.

560 ___ = 56 10

The mean of the data set is 56 text messages. The data values arranged from least to greatest are 3, 11, 15, 22, 28, 32, 37, 56, 71, and 285. 28 + 32 ______ = 30 2

The median of the data set is 30 text messages.

b. To describe a typical number of text messages these students sent last Tuesday, would you use the mean, median, or both? Explain your answer. I would use the median to describe a typical number of text messages these students sent last Tuesday. Because the distribution is skewed to the right, the median is a better measure of center than the mean.

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EXIT TICKET

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Lesson at a Glance Students work individually or with a partner to calculate the mean and median of the data distributions introduced in the previous lesson. They discuss in groups how to use the measures to compare typical values for the distributions. They engage in a class discussion to formalize their thinking about whether the mean, the median, or both are appropriate measures to describe the center of a data distribution based on its shape. Students analyze a set of data that includes an outlier to explore the effect of outliers on measures of center. They apply their understanding to use measures of center to describe a typical value of a data distribution displayed in a histogram, realizing that the mean and median can only be estimated.

Key Questions • What are some ways to describe the shape of a data distribution? • What are some ways to describe the center of a data distribution? • How are the mean and median of a data set related to the shape of the distribution?

Achievement Descriptors Math1.Mod1.AD13 Represent data with dot plots, histograms, or box plots. (S.ID.A.1) Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers,

and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the context of the

data sets. (S.ID.A.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Test Your Memory Revisited

• None

• Choosing a Measure of Center

Lesson Preparation

• Measures of Center from a Histogram

• None

Land 10 min

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Fluency Find Mean and Median Students find the mean and median of data sets to prepare for determining a typical value for a data set. Directions: Find the mean and median of each data set. 1.

5, 7, 2, 5, 4

6, 1, 0, 3, 8, 6, 46

3.1, 3.3, 2.8, 3.6, 3.5, 2.9

Mean: 4.6

Teacher Note The data sets shown here are also used in the Fluency activities in lessons 19, 20, and 21. Consider having students keep their mean and median calculations easily accessible.

Median: 5 Mean: 10 Median: 6 Mean: 3.2 Median: 3.2

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Launch

Students estimate typical measures from dot plots to elicit and assess their prior knowledge of the mean and median. Display the trial 1 and trial 2 dot plots that show the data collected from lesson 17. Direct students to work with a partner to estimate the typical value for each distribution. Have volunteers share their estimates.

Teacher Note

Trial 1

The goal of the opening activity is to compare a distribution that is approximately symmetric with one that is not symmetric. • If the class-generated dot plots are not suitable for this goal, use the sample dot plots shown, which are also provided in lesson 17.

Number of Letters Remembered

Trial 2

• If using different dot plots from those used in lesson 17, have students work with a partner to estimate a typical value for each distribution. • If using the dot plots from lesson 17, have students recall their estimates for each distribution.

Number of Letters Remembered

Recall that in lesson 17 we talked about typical values for data sets. How did you estimate the typical value for each trial? I estimated the value of the dot that was in the middle of the data. I looked for a dot that had about the same number of dots to the left of it and to the right of it.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Which measures can represent a typical value of a data distribution? The mean and median can represent a typical value of a data distribution. Today, we will look at how the mean and median are used to describe a typical value for a data distribution.

Learn Test Your Memory Revisited Students calculate the mean and median of data sets and determine when it is appropriate to use each measure to represent a typical value of a data distribution. Using the same dot plots from Launch, assign half of the class trial 1 and half of the class trial 2. Have them find the mean and median for their assigned trial. Direct students to the Test Your Memory Revisited section to record their work. Display the answers under each dot plot, have students record the answers, and discuss as a class. Note that the answers that follow correspond to the sample dot plots from lesson 17.

UDL: Representation Activate students’ prior knowledge by reviewing how to find the mean and median for quantitative data by using a sample data set from the Fluency activity.

Trial 1: Sum of the numbers: 257

257 Mean: Approximately 9.9, because ___ ≈ 9.9 8+9 Median: 8.5, because ____ = 8.5

Trial 2:

Sum of the numbers: 685

685 ≈ 26.3 Mean: Approximately 26.3, because ___ 26

21 + 24 Median: 22.5, because ______ = 22.5 2

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Trial 1

Number of Letters Remembered

The mean is approximately 9.9 letters. The median is 8.5 letters.

Trial 2

Number of Letters Remembered

The mean is approximately 26.3 letters. The median is 22.5 letters. What do the mean and median of a data distribution represent? The median represents the middle of a distribution, and the mean represents the balance point. Both are measures of center for a data distribution and can be considered the typical value of a data distribution. The median is the middle of a distribution when the data values are ordered from least to greatest. The mean represents the balance point of the data distribution, where the sum of the distances between each data point and the mean on the left side of the mean is equal to the sum of the distances between each data point and the mean on the right side of the mean.

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For trial 1, how do the values of the mean and median compare to your estimate of a typical value from lesson 17? Both the mean and median are close to what I estimated as a typical value. Describe the shape of the data distribution for trial 1. The data distribution is approximately symmetric and mound-shaped. For trial 2, how do the values of the mean and median compare to your estimate of a typical value from lesson 17? The mean is greater than what I estimated as a typical value. The median is closer to what I estimated as a typical value. Describe the shape of the data distribution for trial 2. The data distribution for trial 2 is skewed to the right. Would you compare typical values for the two distributions by using the mean or by using the median? Why? I would use the median. For trial 1, the mean and median both seem like good estimates of a typical value, so it does not really matter which one I use. For trial 2, the median seems like a better estimate of a typical value because the distribution is skewed. For trial 2, the distribution is skewed to the right, and the mean is greater than the median. In general, when describing skewed distributions, we can say that the mean lies among the values in the direction of the skew. To interpret the difference in the typical values of the distributions in context, we discuss the measures of center in relation to the data being collected. How can this situation be interpreted in context? The typical number of letters memorized in trial 2 is about 14 more than the typical number of letters memorized in trial 1. I am going to read aloud four statements, one at a time. After I read a statement, stand up at your desk if you agree with it. Remain seated if you disagree with it.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Display the first statement. After reading the first statement, ask students to share their thinking. Then clarify the correct answer for the class and have the students sit down. Repeat this process with each of the remaining statements. The mean and median are about the same for distributions that are symmetric or approximately symmetric. Agree—the mean and median are both located near the middle of data distributions that are symmetric or approximately symmetric. The mean and median are about the same for distributions that are skewed, or asymmetric. Disagree—the mean and median will be farther apart for skewed distributions. The mean is a good descriptor of a typical value for skewed distributions. Disagree—the mean will be greater than a typical value if the distribution is right-skewed and the mean will be less than a typical value if the distribution is left-skewed.

Teacher Note Some students may recognize that the median is a good measure of a typical value for both approximately symmetric distributions and skewed distributions, causing them to wonder why it is necessary to ever calculate the mean of a distribution. Explain that, while the median is always a good measure of center, the mean is used to calculate other measures. For example, the mean is used to calculate one of the measures of spread, which students will learn about later in the topic.

The mean is a good descriptor of a typical value for mound-shaped distributions. Agree—the mean is located at or near the middle of a mound-shaped distribution. Students possibly may think that the median is also a good descriptor of a typical value for mound-shaped distributions and decide to remain seated. Once the four statements have been discussed, direct students to the Typical Measures for Data Distributions table. Have them draw a sample dot plot for each case and summarize the information about whether the mean, median, or both measures appropriately describe a typical value for the distributions. In the next section, students will add information to the table about the relative positions of the mean and median for the distributions based on their shape.

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Draw a sample dot plot for each case, and summarize the information about the mean and median. Typical Measures for Data Distributions Distribution Shapes Symmetric or Approximately Symmetric

The mean and median are both appropriate descriptors of a typical value. The mean and median are the same or approximately the same value.

Skewed to the Left

Skewed to the Right

The median is the better descriptor of a typical value.

The mean will be less than the median.

The mean will be greater than the median.

Choosing a Measure of Center Students choose an appropriate measure to describe a typical value for a data distribution, and they describe the effect of an outlier on the mean and median of the distribution. Direct students to the table in problem 1 and have them review the table. Make sure students realize that the populations are measured in thousands. Allow students time to work through problem 1 in pairs. Circulate as students work, and provide support as needed. Listen for pairs who are able to explain their answers, and invite some of them to share their answers during the debrief.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Differentiation: Support

1. The table shows estimates of the population size in thousands of 16 penguin species.

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Species

Population (thousands)

Chinstrap

8000

Adelie

7560

King

4600

Southern rockhopper

2500

Royal

Species

Population (thousands)

Northern rockhopper

530

Erect-crested

150

Snares

African

1700

Humboldt

Magellanic

1300

Fiordland

Gentoo

774

Yellow-eyed

3.4

Emperor

595

Galapagos

1.2

If students need support in recognizing the position of the mean relative to the median in skewed distributions, consider having them mark the estimated center and balance point on the dot plots of the distributions from lesson 17. This may help students to notice that the mean lies among the values in the direction of the skew. Trial 2

8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 balance point middle

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 18

a. Complete the dot plot.

Penguin Species Population Sizes

1000

2000

3000

4000

5000

6000

7000

8000

Population Size (thousands)

b. What is the median population size? The median is halfway between 530 and 595.

530 + 595 ________ = 562.5

2 The median population size is 562.5 thousand penguins, or 562,500 penguins.

c. Without computing the mean, determine whether the mean or median population size is greater. Explain your answer. The mean population size is greater. The distribution is skewed to the right because the data include very large population sizes. So the mean lies among the values toward the upper end of the distribution while the median lies near the center. d. To estimate a typical penguin population size of these species, is it appropriate to use the mean, the median, or both? Why? It is appropriate to use the median because it is a better estimate of a typical penguin population size than the mean because the data distribution is skewed.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

e. The population size of the macaroni penguin species is approximately 18,000,000. If we included the population size for the macaroni penguins in the data set, how would that affect the mean and the median? Which measure would be more affected and why? Including the macaroni penguin data will increase the median population size from

562,500 to 595,000. It will increase the mean population size from approximately 1,741,500 to approximately 2,697,900. The macaroni penguin data will increase the

mean more substantially than the median because the macaroni penguin population size is much larger than the population sizes of the other penguin species. Including the macaroni penguin data will cause the distribution to be skewed to the right, so the mean population size will lie among the greater values. Call the class back together to discuss. Then redirect students to the Typical Measures for Data Distributions table. When a data distribution is right-skewed, how does the mean compare to the median? The mean will be greater than the median. Because the mean represents the balance point of a data set, the mean lies among the values farther to the right, making it greater than the median.

EUREKA MATH2

Promoting the Standards for Mathematical Practice When students consider the effect the data distribution and possible outliers have on the mean and the median, they are making sense of the problem (MP1). Ask the following questions to promote MP1: • What information about the distributions do you need to determine if it is appropriate to use the mean, the median, or both to represent the typical value of the data distribution? • Does it make sense that including 18,000,000 in the data set would cause a large change in the mean value? Why?

When a data distribution is left-skewed, how does the mean compare to the median? The mean will be less than the median. The mean lies among the values farther to the left, making it less than the median. Have students record this information in the table. Then direct students to part (e). The population size of the macaroni penguins is much larger than the population size of the other penguin species. We call this value an outlier. Recall that outliers are values that are either much less than or much greater than the rest of the values in a data set. How did the outlier affect the mean and median of the data set? The outlier increased both the mean and median of the data set, but it increased the mean by much more than it increased the median.

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Measures of Center from a Histogram Students estimate the mean and median from a histogram and decide whether each is appropriate to describe a typical value of a data distribution. Direct students to the histogram shown in problem 2. Ask the following questions to guide students on how to read this type of graph. What is the interval size?

15 Why would a histogram be used for this data set rather than a dot plot? In the histogram, we can group the data into intervals, in this case intervals of 15 days, and display the frequency of cities in each interval rather than displaying the frequency for the individual number of clear days. What information can you get about the average number of clear days from looking at this histogram? It looks like most of the cities have between 80 and 125 clear days each year. There are a few cities that get many more clear days than the typical number of days. What is not evident from the histogram? Exactly how many clear days any one city gets in a year is not evident. Within each interval, I can’t tell where the values fall. Sketch a vertical line on the histogram where the median value is approximately located. In what interval does your estimate of the median lie? My estimate is in the interval of 95 to 110 clear days.

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Is it possible to calculate the exact value of the median of a data set from a histogram? Explain your reasoning. In most cases, it is not possible to calculate the exact value of the median of a data set because it is located within an interval on the histogram, and the exact values in the interval are unknown. How can we estimate the mean from the histogram? We can estimate the balance point for the distribution. Is it possible to calculate the exact value of the mean of a data set from a histogram? Explain your reasoning. No, because the mean is located within an interval on the histogram and the exact values in the interval are unknown. Sketch a vertical line on the histogram where the mean value is approximately located. Describe the location of your vertical line relative to the location of the line representing the median of the data set. Accept all reasonable estimates that are slightly to the right of the median. Example: 100–120. The line I drew to estimate the mean is slightly to the right of the line I drew to estimate the median. The distribution is skewed to the right, so the mean lies to the right of the median.

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Have students complete problem 2 with a partner. 2. The histogram summarizes the average number of clear days in one year for 48 of the largest cities in the United States. A clear day is defined as one in which clouds cover at most 30% of the sky during daylight hours.

Frequency

Clear Days in One Year for US Cities 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

Median

Mean

95 110 125 140 155 170 185 200 215

Average Number of Clear Days in One Year

a. Estimate the median and the mean of the distribution from the histogram. The median is located in the interval of 95 to 110, so the median is between 95 and 110 clear days. The mean is between 100 and 110 clear days.

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b. What is a typical number of clear days in a year for one of these cities? Explain your answer. A typical number of clear days in a year for one of these cities is between 95 and 110 days. The mean is slightly greater than the median. Three of the cities have a much greater number of clear days than the remaining 45 cities, which makes the median a slightly better descriptor of a typical number of clear days for one of these cities. c. Use the histogram for the average number of clear days to construct a relative frequency histogram.

Clear Days in One Year for US Cities

Relative Frequency

0.5 0.4

Differentiation: Support If students need support with recalling how to create a relative frequency histogram, consider asking them how a relative frequency histogram and a frequency histogram look different and how information from a frequency histogram can be used to make a relative frequency histogram.

0.3 0.2 0.1 0

50 65 80 95 110 125 140 155 170 185 200 215 Average Number of Clear Days in One Year

d. Estimate the median and the mean of the distribution from the histogram. The median is located in the interval of 95 to 110, so the median is between 95 and

110 clear days.

The mean is between 100 and 110 clear days.

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Debrief with the following questions. How do your estimates for the median and the mean of the data distribution compare to the estimates from the frequency histogram? The estimates are the same. Would you prefer to estimate the median and the mean from a frequency histogram or from a relative frequency histogram? Explain your reasoning. Estimating the balance point seems to require the same sort of thinking regardless of the type of histogram. The shapes of the graphs are exactly the same, and it is only the scaling of the vertical axis that has changed. I would prefer to estimate the median from the relative frequency histogram because it seems easier because I just need to add up the relative frequencies until I get to 50%.

Land Debrief 5 min Objectives: Find the mean and median of data shown in a dot plot and estimate the mean and median of a data distribution represented by a histogram. Identify whether the mean, the median, or both appropriately describe a typical value for a given data set. Ask each question and invite students to respond. Ensure that students summarize the key takeaways from the lesson in the Typical Measures for Data Distributions table. What are some ways to describe the shape of a data distribution? The shape can be described as symmetric or approximately symmetric or either left-skewed or right-skewed. What are some ways to describe the center of a data distribution? The center can be described by using either the mean or the median.

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How are the mean and the median of a data set related to the shape of the distribution? For approximately symmetric distributions, the mean and the median are approximately equal, and both are located approximately in the center of the distribution. For skewed distributions, the mean lies among the values in the direction of the skew, while the median is not affected by the skew in the same way. If we want to compare the typical value for two different distributions, what measure of center should we use: the mean or the median? Median and mean are both appropriate measures of center for distributions that are symmetric or approximately symmetric. If one or both distributions are skewed, the median is preferred over the mean to compare measures of center.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Recap

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

RECAP Name

Date

2. The histogram shows the distribution of audience ratings for a sample of 100 movies. Audience Ratings for Movies 35

In this lesson, we

determined the mean and median of a distribution from a dot plot.

•

estimated the mean and median of a distribution from a histogram.

•

determined which measures of center are most appropriate to describe a distribution.

Frequency

Describing the Center of a Distribution •

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

The left half of the distribution almost mirrors the right half of the distribution.

20 15 10

Examples

1. The dot plot shows the number of grams of protein per serving in 20 different varieties of breakfast cereal.

0 30

100

Audience Rating (percent)

Protein per Serving of Cereal

a. Describe the shape of the distribution. The distribution is approximately symmetric.

b. Use the histogram to estimate the mean and median of the distribution.

The mean movie rating is about 65.

The median movie rating is between 60 and 70.

Protein (grams)

The balance point is near the center because the distribution is approximately symmetric.

a. Find the mean and median of the data set.

5(1) + 4(2) + 3(3) + 2(4) + 2(5) + 2(6) + 7 + 8 67 __________________________________ = __ = 3.35 20 20

The mean of the data set is 3.35 grams. The median of the data set is 3 grams. b. Determine whether the mean, median, or both would be appropriate to use to describe the typical number of grams of protein per serving of one of these cereals. Explain. The median would be appropriate to use to describe the typical number of grams of protein per serving of one of these cereals because the distribution is right-skewed. © Great Minds PBC

Because there are an even number of data values, the median is the mean of the two middle values.

A histogram does not provide exact data values. The best estimate of the median is within an interval.

c. Determine whether the mean, median, or both would be appropriate to use to describe a typical audience rating for a movie in this sample. Explain.

____

3+3 =3 2

Both the mean and the median are appropriate to use to describe a typical audience rating in this sample because the distribution is approximately symmetric.

In a right-skewed distribution, the mean is usually greater

In an approximately symmetric distribution, the mean and the median are about the same value.

than the median.

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RECAP

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

PRACTICE Name

Date

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For problems 3 and 4, determine whether it is appropriate to use the mean, median, or both to describe a typical value. Explain. 3. A random sample of 70 employees of several different companies was asked how much money they spent on lunch last week.

1. Draw a dot plot of a data distribution representing the ages of 20 people for which the median and the mean are approximately the same value. Explain your reasoning.

Employee Spending on Lunch Last Week (Several Workplaces)

Sample:

Frequency

Age (years)

15 10 5

The mean and median are approximately the same value in an approximately symmetric distribution.

Money (dollars)

Both the mean and median are appropriate to describe a typical value for the data set because the distribution is approximately symmetric.

2. Draw a dot plot of a data distribution representing the ages of 20 people for which the median is less than the mean. Explain your reasoning. Sample:

Age (years)

The median is less than the mean in a right-skewed distribution.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

5. The dot plot shows the movie lengths in minutes of the 8 Best Picture nominees for the 2019 Oscars.

4. Another random sample of 70 employees, this time from the same workplace, was asked how much money they spent on lunch last week. Employee Spending on Lunch Last Week (Same Workplace)

Movie Lengths of Best Picture Nominees for Oscars 2019

Frequency

25 20

120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137

Time (minutes)

The median movie length for the movies is 133.5 minutes. Determine by inspection whether the mean movie length is greater than, less than, or the same as the median movie length. Explain your reasoning.

5 0 0

The mean movie length is less than the median movie length. The value 121 is several minutes less than the other values in the data set, which makes the mean movie length less than the median movie length.

Money (dollars)

The median is appropriate to describe a typical value for the data set because the data distribution is skewed to the right. The mean lies among the greater values, so it is greater than the median and does not describe a typical value in the data set.

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6. An event has 40 people attending it. The ages of the people are as follows:

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c. Using the histogram, estimate the mean and median.

The mean is about 50 years old. The median is between 50 and 60 years old.

d. Would the mean, median, or both appropriately describe the typical age of a person attending this event? Explain. Because the data distribution is approximately symmetric, both the mean and median would appropriately describe the typical age of a person attending this event.

a. Create a histogram of the ages by using the provided axes. Use intervals of 10 years. Ages of People Attending an Event 14 12

Frequency

10 8 6 4 2 0 0

100

Age (years)

b. Would you describe the histogram as symmetric or skewed? Explain your choice. The histogram is approximately symmetric. If I make a vertical line through the center of the histogram and fold the histogram along the line, the two halves almost match up.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

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7. The relative frequency histogram shows the distribution of heights for a sample of 25 players in the Women’s National Basketball Association (WNBA).

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Math 1 ▸ M1 ▸ TD ▸ Lesson 18

Remember For problems 8 and 9, solve the inequality for p.

Heights of 25 WNBA Players

9. 3(p - 5) ≤ -18

8. 5p + 8 > 28

p ≤ -1

p>4

1 0.9

10. The drama club sells bouquets of flowers at the spring musical performances. Each small bouquet sells for $12, and each large bouquet sells for $20. The drama club sells 14 more small bouquets than large bouquets. If the drama club makes $552 in total sales, how many small bouquets do they sell?

Relative Frequency

0.8 0.7 0.6

The drama club sells 26 small bouquets.

0.5 0.4 0.3

11. The soccer coach purchased headbands for the team. The graph shows the relationship between the number of headbands purchased and the total cost. What is the cost per headband?

0.2 0.1

0 65

Height (inches) Total Cost (dollars)

a. Estimate the mean and median of the distribution from the histogram. The median is in the interval of 71 to 74 inches. Because the distribution is skewed to the left, the mean should be less than the median but probably in the interval of 71 to 74 inches.

b. To describe the typical height of these basketball players, is it appropriate to use the mean, median, or both? Explain.

P R ACT I C E

(8, 24)

(4, 12) (2, 6)

Number of Headbands Purchased

Since the distribution is skewed, it is appropriate to use the median because it is a better estimate of the typical height of the basketball players than the mean.

26 24 22 20 18 16 14 12 10 8 6 4 2

The cost per headband is $3.

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Teacher Edition: Math 1, Module 1, Topic D, Lesson 19 LESSON 19

Using Center to Compare Data Distributions Determine the median from data distributions displayed in box plots. Use the median to compare data distributions displayed in box plots.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 19

Name

Date

EXIT TICKET

A forest is divided into equal sections, and 100 of these sections are randomly selected. A scientist asks volunteers to record the number of squirrels and birds seen in each section. The box plots show the distributions of the number of squirrels and birds observed in each section. Animal Sightings in

100 Sections of the Forest

Squirrels Birds

Lesson at a Glance In this lesson, students use the median to compare typical values of two data sets. Facilitated by teacher guidance, students construct a box plot from a given relative frequency histogram and engage in a discussion to interpret the information provided in the box plot. They work in groups to compare data distributions represented in box plots. Students share their key takeaways and critique one another’s reasoning through a class discussion focused on interpreting differences in the shapes and centers of data distributions in terms of the situations described.

Key Questions 0 1 2 3 4 5 6 7 8 9 10 11 12

• Suppose we wanted to compare two data distributions. Should we compare them by using histograms or box plots? Why?

Number of Squirrels and Birds Seen

Is the typical number of squirrels seen in each section of the forest greater than the typical number of birds seen in each section of the forest? Justify your answer. Both box plots show that the median number of each type of animal seen in each section of the forest is 5. So the typical number of squirrels seen in each section of the forest is not greater than the typical number of birds seen in each section of the forest.

• Besides measures of center, which are the mean and median, what are other ways we can compare two data sets?

Achievement Descriptors Math1.Mod1.AD13 Represent data with dot plots, histograms, or box

plots. (S.ID.A.1) Math1.Mod1.AD14 Summarize and compare data distributions by

their shapes, centers, and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the

context of the data sets. (S.ID.A.3) © Great Minds PBC

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Constructing a Box Plot

• None

• Comparing Data Distributions

Lesson Preparation

Land 10 min

• None

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Fluency Describe Distributions Numerically Students describe data sets numerically to prepare for constructing box plots and comparing distributions represented in box plots. Directions: Find the minimum, first quartile, median, third quartile, and maximum for each data set.

Teacher Note These five-number summaries will be useful for calculating range and interquartile range in the lesson 21 Fluency activity. Consider having students keep them easily accessible.

Minimum: 2 First quartile: 3 1.

5, 7, 2, 5, 4

Median: 5 Third quartile: 6 Maximum: 7 Minimum: 0 First quartile: 1

6, 1, 0, 3, 8, 6, 46

Median: 6 Third quartile: 8 Maximum: 46 Minimum: 2.8 First quartile: 2.9

3.1, 3.3, 2.8, 3.6, 3.5, 2.9

Median: 3.2 Third quartile: 3.5 Maximum: 3.6

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Launch

Math 1 ▸ M1 ▸ TD ▸ Lesson 19

Students analyze a relative frequency histogram to determine how the same information can be displayed by using a box plot. Allow students time to work on problem 1, and then have them compare their answers with a partner. Note: The histogram in this problem is specifically designed with clear breaks at 25%, 50%, 75%, and 100% to facilitate the process of representing the distribution by using a box plot. After a few minutes, facilitate a discussion by asking the following questions. What percent of the people surveyed chose prices of $6, $7, or $8?

Language Support If students require a high level of support, have them debrief the meaning of problem 1 with a partner before writing a description of a typical value for the data set.

25% What percent of the people surveyed chose prices of $9 or $10?

25% What percent of the people surveyed chose prices of $11 or $12?

25% What percent of the people surveyed chose prices of $13, $14, or $15?

25% What type of graph can we use to easily see a data distribution divided into quarters? We can use a box plot. Box plots are often a convenient way to display data distributions, particularly when comparing two sets of data. Today, we will construct box plots and compare data distributions that are represented in box plots.

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Sandwich Price

Relative Frequency

1. A restaurant owner invited 100 people to try a new sandwich and to provide feedback by completing a survey. The survey included the question, What price would you be willing to pay for the sandwich, rounded to the nearest dollar? The results are shown in the relative frequency histogram.

0.25 0.2 0.15 0.1 0.05 0

10 11 12 13 14 15 16

Price (dollars)

Write two or three sentences that describe a typical value for the data set and interpret that value context. The median of the data is 10.5. So the people surveyed are typically willing to pay $10 or $11 for the sandwich.

Learn Constructing a Box Plot Students construct a box plot and interpret the information displayed. Direct students to problem 2. Use the relative frequency histogram from problem 1 to construct a box plot. Review the elements of the five-number summary with students: minimum, first quartile, median, third quartile, and maximum. As a class, use the five-number summary to construct the box plot by using the axis provided in problem 2.

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Language Support To support organizing the terms in this topic, consider having students add the term box plot to the Univariate Quantitative Data graphic organizer from lesson 17. Students can sketch a box plot in the section labeled Graphical Displays. They can also label components of the box plot, including the five-number summary.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 19

2. Find each value and create a box plot that shows the distribution from problem 1.

Sandwich Price Minimum: $6 Q1: $8.50 Median: $10.50 Q3: $12.50 Maximum: $15

9 10 11 12 13 14 15 16

Differentiation: Support If students need support with identifying the quartiles, guide them on how to find the first and third quartile values by reminding them that these values (often abbreviated as Q1 and Q3) are the medians of the lower half and upper half of the data set, respectively. Students may also benefit from labeling each component of the five-number summary on their box plot.

Price (dollars)

Once students have constructed the box plot, debrief to ensure they understand how to read and interpret a box plot. Display the relative frequency histogram and the box plot aligned as shown, and then have students draw a copy of their plots from problem 2 under the histogram in problem 1 so they can see that 25% of the data falls within each section of the box plot.

Relative Frequency

Sandwich Price

Students first created box plots in grade 6. Note that in Mathematics I, the focus is less on the construction of the plots and more about the ability to read and interpret them. A common misconception is for students to equate the size of the parts of the box plot with the amount of data in the part. Be sure to emphasize correct understanding of the box plot throughout the lesson.

0.25 0.2 0.15 0.1 0.05 0 5

9 10 11 12 13 14 15 16 Price (dollars)

Teacher Note

Note that when there is an odd number of data points, there is no consensus among mathematicians as to what to do with the median of the data set when calculating Q1 and Q3. In the Eureka Math2 curriculum, the median is excluded from both halves when calculating the values of Q1 and Q3, which is the method students learn in grade 6.

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Then ask the following questions about the data. What percent of the values lie between $12.50 and $15?

25% Explain what this percent means in terms of this survey. This means that 25% of the people surveyed are willing to pay $13, $14, or $15. Is it correct to say that more people are willing to pay somewhere from $13 to $15 than from $11 to $12? No. Each section of the box plot represents 25% of those surveyed, so the same number of people would fall into each section. What are some advantages to displaying data by using a box plot compared with displaying data by using a histogram? It is usually easier to identify values like the minimum, median, first and third quartiles, and maximum from a box plot than from a histogram.

Comparing Data Distributions Students compare data distributions displayed by using box plots. Give students time to work in groups on problem 3. Circulate as students work, and check for understanding about how to read and interpret the information from the box plots.

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UDL: Engagement Structure peer interactions for success by assigning group roles (e.g., facilitator and timekeeper) and defining responsibilities. Review the directions and group norms before groups begin. Suggest a target completion time and project a visual timer.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 19

3. The box plots summarize the life expectancy at birth in countries in Asia and Europe in 2017.

Life Expectancy at Birth

Asia Teacher Note

Europe

68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 Age (years)

a. What is the lowest life expectancy in any country in Europe?

Some students may be more familiar with calling a box plot a box-and-whisker plot. They may interpret the distributions in box plots by referencing the endpoints of whiskers rather than specifically referencing elements from the five-number summary.

71 years b. What is the highest life expectancy in any country in Asia?

85 years c. Is it true that there are more countries in Europe where people have a life expectancy between 76 and 80.5 years than between 80.5 and 82 years? Explain. No. Life expectancy in 25% of the countries in Europe is between 76 and 80.5 years, and the life expectancy in 25% of the countries in Europe is between 80.5 and 82 years. d. Mason says that the number of countries in Asia where people have a life expectancy between 75 and 78 years is the same as the number of countries in Europe where people have a life expectancy between 80.5 and 82 years. Do you agree with Mason? Explain. No. While the same percentage of countries in Europe and Asia have people with life expectancies within the ranges stated, this does not mean that the number of countries in Asia with people who have a life expectancy between 75 and 78 years is the same as the number of countries in Europe with people who have a life expectancy between 80.5 and 82 years because Europe and Asia do not have the same number of countries. © Great Minds PBC

Promoting the Standards for Mathematical Practice Students are constructing viable arguments and critiquing the reasoning of others (MP3) when they discuss their summaries about the life expectancies of people in the two continents and explain why they agree or disagree with the summary statements of others. Ask the following questions to promote MP3: • Is Mason’s reasoning in problem 3(d) valid? How do you know? • How could you change your argument to make it more accurate?

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e. How does a typical life expectancy for people in a country in Asia compare with a typical life expectancy for people in a country in Europe? A typical life expectancy for people in a country in Asia is about 75 years, while a typical life expectancy for people in a country in Europe is greater, at about 80.5 years. Therefore, the typical life expectancy in an Asian country is about 5.5 years less than the typical life expectancy in a European country. f. Write a few sentences comparing the life expectancy for people in an Asian country and the life expectancy for people in a European country. A typical life expectancy for people in a European country is greater than that for people in an Asian country. However, 25% of the Asian countries have people with a life expectancy between 78 and 85 years, so there are Asian countries where the life expectancy is comparable to that of European countries. But 25% of Asian countries have people with a life expectancy between 69 and 71.5 years, which means that the minimum life expectancy of people in Asian countries is less than the minimum life expectancy of people in European countries.

Teacher Note If students ask about why the typical life expectancy in European countries is higher than it is in Asian countries, consider discussing some of the factors that commonly contribute to a higher life expectancy. An example is the further development of health care infrastructure in some European countries, which gives more people access to health care. Because of this development, children born in Europe are more likely to be born in a hospital and receive medical care. Greater access to health care also means that people in Europe are more likely to receive treatment for certain diseases, thus increasing their life expectancy.

Debrief students’ responses as a class by using informal assessment. For instance, pose statements about the distributions, and consider having students give a thumbs-up or thumbs-down to indicate whether they agree or disagree with the statements. The lowest life expectancy for people in any country in Europe is 71 years. (thumbs-up) More countries in Europe have people with a life expectancy between 71 and 82 years than between 82 and 89 years. (thumbs-up) Have different students share one key takeaway based on what they wrote in part (f), and invite the class to respond to each key takeaway by giving a reason why they agree or disagree. Direct students to work on problem 4 either individually or in groups.

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4. The box plots summarize the average points per game for two sets of NBA players. • The first box plot shows the distribution of the average points per game of 21 of the leading scorers in NBA history. • The second box plot shows the distribution of the average points per game of the 21 leading scorers among NBA players who were active through June 2019.

NBA Leading Scorers

Throughout NBA History Active Players Differentiation: Support

Average Points per Game

a. Complete each sentence.

100 % of the leading scorers in NBA history had a higher average than the bottom 75% of the leading scorers who were active through June 2019. Approximately 25 % of the leading scorers in NBA history had a higher average than 100% of the leading scorers who were active through June 2019. b. Describe the shape of each data distribution. Both data distributions are right-skewed.

If students need support with describing the shape of each data distribution, consider advancing their thinking by asking the following questions about each distribution: • What does it mean when the median is not in the center of the box? • Which half of the data is clustered in a smaller range? • What does the long line segment represent?

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c. Compare a typical value for the average points per game for 21 of the leading scorers in NBA history with a typical value for the average points per game for the leading scorers who were active through June 2019. A typical value for 21 of the leading scorers in NBA history is approximately 25 points per game. A typical value for the leading scorers who were active through June 2019 is approximately 21 points per game. The leading players in NBA history typically scored approximately 4 more points per game than the active players. Ask the following question to facilitate a discussion about how students determined the shape of the distribution in problem 4. Use the discussion to further solidify the idea that each section of the box plot represents 25% of the data. How did you know that the data distributions were both right-skewed? For both distributions, the lower 50% of the data is clustered in a smaller range of values than the upper 50% of the data. The median is also located to the left of the center of each box, which means the data between Q1 and the median have a smaller range of values than the data between the median and Q3. This indicates that the distributions are right-skewed.

Land Debrief 5 min Objectives: Determine the median from data distributions displayed in box plots. Use the median to compare data distributions displayed in box plots. Pose the first question, and give a minute of silent time for students to think about the advantages and disadvantages of each type of display. Then ask for students to volunteer answers. Pose the second question. If students do not mention the idea of describing the spread, it can be mentioned as a lead-in to lesson 20 without explicitly naming measures of spread.

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Suppose we wanted to compare two data distributions. Would we prefer to compare them represented in histograms or box plots? Why? We would rather use histograms to compare two distributions because it is easier for us to tell if the data distribution is skewed and, if so, in which direction. We can also determine the number of observations in a distribution that is represented in a frequency histogram, while we cannot determine the number of observations in a distribution that is represented in a box plot or a relative frequency histogram. We would rather use box plots because it is easy to identify the median, which makes comparing the center of two distributions easier than with a histogram. Besides measures of center, mean and median, what are other ways we can compare two data distributions? We could compare their shape. We also might want to describe how spread out the data values are, for instance, by using range.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

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RECAP Name

Date

a. Determine whether each statement is true or false. Most of the swimmers’ maximum speeds when not wearing a wetsuit are less than 1.4 meters per second.

Using Center to Compare Data Distributions In this lesson we

False

•

interpreted information about a distribution from a box plot.

•

compared distributions displayed in box plots.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 19

Example Twelve swimmers each swam 1500 meters two times: once wearing a wetsuit and once not wearing a wetsuit. The box plots show the distributions of the swimmers’ maximum speeds in meters per second.

Only about 25% of the speeds fall into this category.

The range of the swimmers’ maximum speeds when not wearing a wetsuit is less than the range of the swimmers’ maximum speeds when wearing a wetsuit. True

About 50% of the swimmers’ maximum speeds when wearing a wetsuit are between 1.2 meters per second and 1.5 meters per second.

The typical maximum speed for a swimmer when wearing a wetsuit is less than the typical maximum speed for a swimmer when not wearing a wetsuit.

True

False

Swimming Speeds

The median speed for swimmers when wearing a wetsuit is about 1.5 meters per second. The median speed for swimmers when not wearing a wetsuit is about

No Wetsuit

1.45 meters per second.

Wetsuit

b. Did a greater percentage of swimmers reach a maximum speed of at least 1.45 meters per second when wearing a wetsuit or when not wearing a wetsuit? Explain.

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 1.7 1.75 1.8 1.85

A greater percentage, about 75%, of swimmers reached maximum speeds of at least 1.45 meters per second when wearing a wetsuit. This is a greater percentage than the approximately 50% of swimmers who reached this maximum speed when not wearing a wetsuit.

Maximum Speed (meters per second)

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Math 1 ▸ M1 ▸ TD ▸ Lesson 19

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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2. The box plot displays the distribution of heights for a sample of 25 players in the Women’s National Basketball Association (WNBA). Determine whether each statement is true or false and justify your choice.

1. Ana recorded the daily high temperature in degrees Fahrenheit at her house for 10 days.

Heights of 25 WNBA Players

77, 85, 77, 79, 82, 84, 79, 78, 79, 80 Find each value and create a box plot that shows the distribution of the daily high temperatures. Minimum: 77°F

Daily High Temperatures

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 Height (inches)

Q1: 78°F

a. The minimum height of these basketball players is 66 inches.

Median: 79°F Q3: 82°F Maximum: 85°F

True; the lower whisker begins at 66 inches, which means this is the minimum value.

73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 b. Most of these basketball players have a height of 73 inches or less.

Temperature (degrees Fahrenheit)

False; 50% of these basketball players have a height of 73 inches or less.

c. More of these basketball players are between 69 inches and 73 inches tall than between 73 inches and 76 inches tall. False; there are the same number of basketball players, 25% of the total, who have heights in each interval.

d. Approximately 25% of these basketball players are between 76 inches and 78 inches tall. True; a whisker extends from 76 inches to 78 inches, which means 25% of the players have heights in this interval.

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3. A timed quiz was given to 20 students. The same 20 students were given the same quiz the next day, but it was untimed. The distributions of the scores for the quizzes are shown in the box plots.

d. Can the teacher conclude that students had better scores on the second quiz because it was untimed? Explain. The teacher can conclude that the scores were better but not because it was untimed. A variable other than whether the test was timed also might have affected students’ performance. For example, students may have scored better because they have seen the quiz questions before.

Quiz Scores

4. Why are box plots useful to compare two univariate data sets?

Timed

Box plots allow for a direct comparison of the quartiles for each distribution. Because box plots display the distributions of data rather than individual data points, I can directly compare data sets with different sample sizes. The minimum and maximum values are also easy to compare, because they are evident in a box plot.

Untimed

10 11 12 13

5. The box plot displays the distribution of ages for 200 randomly selected residents from Kenya. The relative frequency histogram displays the distribution of ages for 200 randomly selected residents from the United States.

Score

a. Which distribution had a greater percentage of scores that are at least 7 points? Explain.

Ages of US Residents

The untimed score distribution had a greater percentage of scores that are at least 7 points. While 25% of the scores on the timed quiz are at least 7 points, 50% of the scores on the untimed quiz are at least 7 points.

0.08 0.07 Relative Frequency

b. Describe a typical score on the timed quiz and a typical score on the untimed quiz. A typical score on the timed quiz is 5 points. A typical score on the untimed quiz is 7 points. Ages of Kenya Residents

Kenya Residents

c. Based on the box plots, the teacher concludes that these students did significantly better on the untimed quiz than on the timed quiz. Give three pieces of evidence that support this conclusion.

0.06 0.05 0.04 0.03 0.02 0.01 0

The median for the untimed quiz was 2 points higher than the median for the timed quiz. Fifty percent of the scores on the untimed quiz were between 7 and 10 points, and 50% of the scores on the timed quiz were between 5 and 8 points. The minimum score on the timed quiz was 1 point, while the minimum score on the untimed quiz was 3 points.

Age (years)

100

105

Age (years)

a. Which country’s distribution has a greater percentage of people who are at least 20 years old? Explain. The United States distribution has a greater percentage of people who are at least 20 years old. Only approximately 50% of the Kenya residents are at least 20 years old, versus approximately 75% of the United States residents.

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16 9. Suppose Nina and Zara each walk at a constant rate. The equation d = __ t represents

b. Estimate the percentage of the people in each distribution who are at least 50 years old.

Nina’s distance d in miles for t hours of walking. The graph represents Zara’s distance y in miles for x hours of walking.

Kenya: between 0% and 25% United States: between 30% and 35%

Who walks at a faster rate? Explain. y

Remember 3.5

For problems 6 and 7, solve the inequality for c. 7. -4(2c - 8) < -24

Zara’s Distance (miles)

6. -34c + 18c ≤ 8 c ≥ - 1_ 2

c>7

8. Consider the equation |3x - 5| = 4. a. Rewrite the equation as a compound statement.

3x - 5 = 4 or 3x - 5 = -4

3 2.5 2 1.5 1 0.5 0

0.25

0.5

0.75

1.25

1.5

1.75

Time (hours)

16 Because Nina walks at a rate of __ mph, or 3.2 mph, and Zara walks at a rate of 3 mph, 5

b. Use the compound statement you wrote in part (a) to find the solution set of the original equation.

Nina walks at a faster rate than Zara.

_1 {3 , 3}

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Teacher Edition: Math 1, Module 1, Topic D, Lesson 20 LESSON 20

Describing Variability in a Univariate Distribution with Standard Deviation Calculate standard deviation to represent a typical variation from the mean of a data distribution. Use standard deviation to compare two data distributions.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 20

Name

Date

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2. The following data set lists the fuel economy of 5 cars in miles per gallon.

23.4, 24.7, 24.7, 24.9, 27.9 1. The dot plot for data set A shows scores on a math quiz that had 4 questions worth 2 points each. The dot plot for data set B shows scores on a math quiz that had 8 questions worth 1 point each.

The mean of the data set is approximately 25.1 miles per gallon, and the standard deviation is approximately 1.7 miles per gallon.

Which statement correctly compares the mean and the standard deviation of data set A and data set B?

A different data set lists the fuel economy of 6 other cars in miles per gallon.

22.9, 23.7, 23.7, 25.9, 26.5, 50.0

Data Set A

Without calculating the standard deviations, explain why the standard deviation of the second data set is greater than the standard deviation of the first data set. The standard deviation of the second data set is greater than that of the first data set because the second data set has a large outlier of 50.0 miles per gallon.

Quiz Scores

Data Set B

4 Quiz Scores

A. Data set A and data set B have the same mean and standard deviation. B. Data set A has a greater mean and a larger standard deviation than data set B. C. Data set A has a greater mean than data set B, but data set A and data set B have the same standard deviation. D. Data set A and data set B have the same mean, but data set A has a greater standard deviation than data set B.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 20

Lesson at a Glance In this lesson, students engage in a class discussion to compare two data sets with similar measures of center and different amounts of spread, realizing that measures of center alone may be inadequate to compare the data distributions. After being introduced to the standard deviation as a measure of spread, students calculate the standard deviations of the data sets, first by hand and then by using technology. Students use the Always Sometimes Never routine and visual support from a digital interactive to discuss the effect of individual data values on the standard deviation of a data set. This lesson introduces the terms variance and standard deviation.

Key Questions • What does standard deviation reveal about a data distribution? • Why is comparing the mean or median of two data sets not enough to describe differences in their distributions?

Achievement Descriptors Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers,

and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the context of the

data sets. (S.ID.A.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Measuring Spread with Standard Deviation

• Statistics technology

• Using Technology to Calculate Standard Deviation

• None

Lesson Preparation

• Changing Values and Standard Deviation

Land 10 min

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Fluency Find Deviations from the Mean Students find deviations from the mean for individual data values to prepare for calculating standard deviation. Directions: Find the mean of each data set. Then find the deviation from the mean for each value in the data set. For example, if the mean of a data set is 3, a data value of 2 has a deviation of 2 - 3, or -1. 1.

2, 4, 5, 5, 7

0, 1, 3, 6, 6, 8, 46

2.8, 2.9, 3.1, 3.3, 3.5, 3.6

Mean: 4.6 Deviations: -2.6, -0.6, 0.4, 0.4, 2.4 Mean: 10 Deviations: -10, -9, -7, -4, -4, -2, 36 Mean: 3.2 Deviations: -0.4, -0.3, -0.1, 0.1, 0.3, 0.4

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Launch

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Students compare two data distributions to illustrate the need for using a statistical measure of variation. Pose the following questions to students. Do you think the typical number of hours of sleep high school students get differs from the typical number of hours of sleep middle school students get? How could we test your conjecture? Invite students to share their responses. Highlight responses that suggest comparing the typical amounts of sleep for high school students and for middle school students. For example, students could collect data from high school and middle school students selected at random and then compare the shapes of the data distributions. Measures of center and spread for the data set could also be compared. Have students think-pair-share to complete problem 1. Circulate as students work, and listen for students using different characteristics for comparison (e.g., mean and median).

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1. A principal wants to know how much sleep students get each night. She randomly selects 8 students from ninth grade and 8 students from sixth grade and asks them to report the number of hours they slept the night before to the nearest hour. The dot plots display the results.

Hours of Sleep for Ninth Grade Students

10 11 12

Number of Hours of Sleep

Hours of Sleep for Sixth Grade Students

10 11 12

Number of Hours of Sleep

Based on the distributions, compare the typical number of hours of sleep for the ninth grade students with the typical number of hours of sleep for the sixth grade students. The typical number of hours of sleep for the ninth grade students is about the same as the typical number of hours of sleep for the sixth grade students. The mean and median of both distributions is 8 hours. Use the following questions to facilitate a brief discussion about problem 1. Emphasize that a measure of center does not necessarily provide a complete picture of a data distribution. It is okay if students do not recall specific measures of spread, as these measures are addressed in this lesson and in lesson 21. Based on the sample distributions, does the typical number of hours of sleep for the ninth grade students differ from that for sixth grade students? Justify your reasoning. The typical number of hours of sleep for the ninth grade students is about the same as that for the sixth grade students. The mean and median for both groups is 8 hours.

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Can we use these data sets to draw general conclusions about the number of hours of sleep ninth grade students get in relation to the number of hours of sleep sixth grade students get? Why? No, because the principal took random samples of the students but only asked them about how much sleep they got the night before. They could have gotten less sleep just on that night because of a sports event, for example, so it would be better to ask about more nights. No, because the principal only sampled 8 students from each grade, which is not a large sample. In what ways are the distributions different? The distributions have different shapes. The ranges are also different. What other measures besides measures of center can we use to compare the distributions? We can use measures of spread, such as range, interquartile range, or mean absolute deviation. Today, we will learn about new measures of spread and use them to compare distributions of univariate data.

Learn Measuring Spread with Standard Deviation Students measure spread in a small set of data by calculating the standard deviation by hand. Direct students’ attention to problem 2, which shows the data set for ninth grade students from problem 1 in a table. Allow students about 1 minute to determine the mean of the data set if they did not calculate it during Launch.

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Ask the following question to confirm student answers to problem 2. What is the mean of the data set, and what does it represent in this context? The mean of the data set is 8 hours, which represents that a ninth grade student typically slept 8 hours last night. Use the following prompts to introduce the terms variance and standard deviation. Model how to calculate standard deviation by hand by using the data set for the ninth grade students. In previous grades, you calculated spread in a data distribution by using deviations from the mean. What does a deviation from the mean measure? A deviation from the mean measures how far each value is above or below the mean of the data set. Look at the dot plot in problem 1 that displays the distribution of the data for ninth grade students. What is a typical deviation from the mean for these values? Sample: Approximately 1 hour Explain how you estimated a typical deviation from the mean for the data set for ninth grade students. I calculated the mean absolute deviation. I noticed that three values are each equal to the mean, so they are each 0 hours from the mean. Three values are each 2 hours from the mean, and two values are each 1 hour from the mean. So it seems like the typical deviation from the mean is about 1 hour. How do we calculate deviations from the mean? We calculate deviations from the mean by subtracting the mean from the data value. Have students calculate the deviations from the mean for the data set for ninth grade students and record the deviations in the table. Also have students calculate and record the sum of the deviations.

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Note: Students will complete the remaining parts of the table as a class as the lesson progresses. 2.

Ninth Grade Students Number of Hours of Sleep

Deviation from the Mean

Squared Deviation from the Mean

-2

-1

Sum of deviations: 0

Sum of squared deviations: 14

Mean: 8 hours Estimate the typical deviation from the mean of the data set for ninth grade students. The typical deviation from the mean is about 1 hour.

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Ask the following questions to lead a class discussion. What do you notice about the sum of the deviations from the mean of the data set? The sum of the deviations is 0. Why is the sum of the deviations from the mean 0? The mean is the balance point, so the sum of the positive deviations is the opposite of the sum of the negative deviations. In grades 6 and 7, we calculated the mean absolute deviation (MAD) to represent a typical distance from the mean. We can also use a measure called the standard deviation to calculate a typical deviation. This measure is especially useful for distributions that are mound-shaped. Instead of using the absolute value of the deviations for this problem, we use squared deviations to find a typical distance from the mean. Why might we want to find the absolute value of the deviations or square the deviations instead of just finding the average of the deviations? To find an average, we usually add the numbers and then divide by the count, but that will not work with deviations, because they always sum to 0. In addition to ensuring that the sum is not 0, squaring the deviations is preferred in more advanced statistics. Have students calculate the squared deviations from the mean and record them in the table. Also have students calculate and record the sum of the squared deviations. We found the sum of the squared deviations. To find the average squared deviation, we would typically divide the sum by the number of values. However, statisticians have determined that dividing by one less than the sample size provides a more accurate measure of average squared deviation. Because we have 8 values in this distribution, we divide the sum of the squared deviations by 7. The resulting value is called the variance.

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Have students read the definitions of variance and mean absolute deviation included in problem 2, and then facilitate a brief discussion by asking the following questions. How is the calculation for the variance similar to the calculation for the mean absolute deviation? Both calculations involve finding a sum and then dividing the sum by a number related to the size of the data set. How is the calculation for the variance different from the calculation for the mean absolute deviation? The calculation for variance involves finding the sum of the squared deviations rather than finding the sum of the absolute deviations. Also, to find the variance, we divide the sum by one less than the size of the data set, and to find the mean absolute deviation, we divide the sum of the absolute deviations by the size of the data set. Have students read the definition of standard deviation, and then continue with the discussion. What is the relationship between the variance and the standard deviation of a data distribution? The standard deviation is the square root of the variance. Why do you think we use standard deviation instead of the variance as a typical measure of spread for a data distribution? The variance is a squared measure, so it measures a typical squared distance. Standard deviation is measured with linear units, so it represents a typical distance rather than a typical squared distance.

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Variance: a measure of spread for a data distribution. To calculate variance, first sum the squares of all the differences between the data values and the mean of the data set. Then divide the sum by one less than the size of the data set. Mean absolute deviation: a measure of spread for a data distribution. It is the average distance between a data value and the mean of a data distribution. Standard deviation: the square root of the variance. It can be used to measure the typical distance from the mean. 14 Variance: 2, because -=2 7 _ Standard deviation: Approximately 1.4 hours, because √ 2 ≈ 1.4

Math 1 ▸ M1 ▸ TD ▸ Lesson 20

Language Support To support understanding of the terms variance and standard deviation, consider having students add the terms to the Measures of Spread section of the Univariate Quantitative Data graphic organizer from lesson 17. Along with the terms, students can include a brief description of what each value measures, how to calculate the values, and an example of each term.

What does the standard deviation represent in this context?

The ninth grade students’ numbers of hours of sleep typically vary by about 1.4 hours from the mean number of 8 hours of sleep. Ask the following questions. How close is the standard deviation to the typical deviation you estimated? My estimate of 1 hour is pretty close to the standard deviation, which is about 1.4 hours. What does the standard deviation represent in this problem? Record your thoughts in the space at the end of problem 2. In this problem, the standard deviation means that the difference between the typical number of hours of sleep a ninth grade student gets and the mean number of hours of sleep for the sample is about 1.4 hours.

Differentiation: Challenge If students quickly compute the sample standard deviation, encourage them to consider why dividing by n - 1 instead of by n becomes less important when calculating the sample standard deviation as the sample size increases. Challenge students to justify their reasoning in terms of the calculation (the two values will approach one another as n increases) and in terms of the sample size (as sample size increases, it eventually approaches the population size).

Have students turn and talk about what standard deviation measures and how it is calculated. Answer any questions students have about standard deviation. Direct their attention to problem 3. Have students predict whether the standard deviation of the distribution of data for sixth grade students will be greater than, less than, or about the same as that of the data set for ninth grade students. Then have students complete problem 3 with a partner and record the standard deviation to the nearest tenth of an hour.

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Sixth Grade Students

3. Number of Hours of Sleep

Deviation from the Mean

Squared Deviation from the Mean

-1

Sum of deviations: 0

Sum of squared deviations: 2

Mean: 8 hours

2 Variance: 7 Standard deviation: about 0.5 hours, because

√ 2-7 ≈ 0.5

Interpret the standard deviation in this context.

The sixth grade students’ number of hours of sleep typically vary by about 0.5 hours from the mean number of 8 hours of sleep.

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Debrief with the following questions. What does the standard deviation measure? It measures the distance from a typical data value to the mean. Why is the standard deviation for the data set for sixth grade students less than the standard deviation for the data set for ninth grade students? What does that tell us in this situation? The values in the data set for sixth grade students are closer together than the values in the data set for ninth grade students. It tells us that there is less variation in the numbers of hours of sleep in the sixth grade student sample than in the ninth grade student sample. Suppose data are collected from two other sixth grade students. One student got approximately 3 hours of sleep the previous night, and the other student got approximately 13 hours of sleep the previous night. How do these values compare to the other values in the data set? The value of 3 hours is much less than the other values. The value of 13 hours is much greater than the other values. What term do we use to describe a value in a data set that is unusually less than or unusually greater than the other values in the data set? Outlier The values of 3 hours and 13 hours would be considered outliers. How do the outliers affect the mean? They don’t affect the mean because they cancel one another out. Because 3 hours and 13 hours both differ from the mean by 5 hours, but in different directions, their deviations sum to 0. They are exactly balanced on opposite sides of the mean, so they do not change the balance point.

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Have students calculate the standard deviation of the distribution of the data for sixth grade students including the outliers, rounding to the nearest tenth. How do the outliers affect the standard deviation of the distribution of the data for sixth grade students? Explain. The outliers increase the standard deviation from about 0.5 hours to about 2.4 hours because they increase the typical distance from a data value to the mean of the distribution. An outlier increases the standard deviation of a distribution, sometimes greatly.

UDL: Representation Consider pausing after emphasizing that an outlier increases the standard deviation of a distribution. Prompt students to think and generate questions that address how the inclusion of an outlier increases the standard deviation of a data distribution. This prompt supports information processing and signifies the importance of the information.

Using Technology to Calculate Standard Deviation Students calculate standard deviation by using technology to compare variations in two data sets. Direct students’ attention to problem 2. Model how to use technology to calculate standard deviation of the distribution of the data for ninth grade students, and have students record the value. Then have students work in pairs to complete problem 4. Circulate as students work, confirming answers and answering any questions that arise. Then pose the following question. When would it be more efficient to use technology to calculate the standard deviation of a data set? It is more efficient to calculate standard deviation by using technology for data sets with large sample sizes and when the deviations from the mean are not integers.

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Teacher Note An online keyword search of the phrase standard deviation calculator should provide links to applications that allow students to quickly enter data and calculate the standard deviation. Make sure students use a calculator for sample standard deviation. Population standard deviation is found by dividing the sum of the squared deviations by the sample size instead of one less than the sample.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 20

4. The following dot plots show larger random samples of ninth grade students and sixth grade students. Use technology to find the standard deviation of the distribution of each data set. Round each standard deviation to the nearest tenth.

Hours of Sleep for Ninth Grade Students

Hours of Sleep for Sixth Grade Students

Promoting the Standards for Mathematical Practice Students use appropriate tools strategically (MP5) when they calculate the standard deviation for a data set with pencil and paper and by using technology and then reflect on the efficiency of using one tool over the other. Ask the following questions to promote MP5:

10 11 12

Number of Hours of Sleep

10 11 12

Number of Hours of Sleep

a. Calculate the standard deviation of the distribution of the data for ninth grade students.

• Which tool would be more efficient when calculating the standard deviation for a large data set? Why? • Which tool would be more helpful when calculating the standard deviation for a data set with only a few values? Why? • How can you estimate the standard deviation by looking at a dot plot?

The standard deviation of the distribution of the data for ninth grade students is approximately 1.3 hours. b. Calculate the standard deviation of the distribution of the data for sixth grade students. The standard deviation of the distribution of the data for sixth grade students is approximately 1.3 hours. c. Compare the standard deviations of the distributions. The standard deviations of the distributions are nearly the same.

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Changing Values and Standard Deviation Students analyze the effects of changing values on the standard deviation of a distribution. Open and display the Standard Deviation interactive, which shows the data set for sixth grade students from Launch. Direct the students to the statements in problems 5-7. Use the Always Sometimes Never routine to engage students in constructing meaning and discussing their ideas. Give students 2 minutes of silent time to evaluate whether the statements are always, sometimes, or never true. Have students discuss their thinking with a partner. Circulate and listen as students talk. Identify a few students to share their thinking. For problems 5-7, determine whether each statement is always, sometimes, or never true. Explain your reasoning. 5. Decreasing a data value will decrease the standard deviation of a distribution. Sometimes true. Decreasing a data value will decrease the standard deviation only if the value is closer to the mean. 6. Increasing a data value will increase the standard deviation of a distribution. Sometimes true. Increasing a data value will increase the standard deviation only if the data value is farther away from the mean. 7. Inserting an outlier into a distribution will increase the standard deviation. Always true. Inserting an outlier will insert a new, large squared deviation into the calculation of the standard deviation, which will increase its value. Next, facilitate a class discussion. Invite students to share their thinking with the class. Encourage them to provide examples and nonexamples to support their claim. Students can be instructed to use the Standard Deviation interactive to support their reasoning by moving points and examining how this affects the standard deviation of the data set.

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Conclude by coming to the consensus that the statements in problems 5 and 6 are sometimes true and the statement in problem 7 is always true. Elicit the following key takeaways: • A change in a data value that decreases its distance from the mean will decrease the standard deviation of a distribution. • A change in a data value that increases its distance from the mean will increase the standard deviation of a distribution. • Changing a data value may affect the standard deviation of a distribution, but it is not important whether the value is increased or decreased. The standard deviation is affected if the data value’s distance from the mean is changed. • Inserting an outlier will increase the standard deviation of a distribution because it increases the typical distance from a data value to the mean of the data set.

Land Debrief 5 min Objectives: Calculate standard deviation to represent a typical variation from the mean of a data distribution. Use standard deviation to compare two data distributions. Use the following prompts to guide a discussion about standard deviation. What does standard deviation reveal about a data distribution? Standard deviation is the typical distance of a value from the mean of the distribution. Why is comparing the mean or median of two data sets not enough to describe differences in their distributions? The mean and median are measures of center. However, there may be other differences in the distributions that measures of center do not show, such as spread or how much variation there is in each data set.

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All of our examples today were approximately symmetric distributions. Is standard deviation an appropriate measure of the typical variation in a skewed distribution? Explain. I don’t think it would be a good measure of spread. Because standard deviation measures typical distance from the mean and mean is not the most appropriate measure of center to represent typical values in a skewed distribution, standard deviation may not be the most appropriate measure of variation for skewed distributions. What measure of spread might be a more appropriate measure of typical variation for skewed data distributions? Interquartile range

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Recap

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c. Record deviations from the mean and the squared deviations from the mean in the table. Subtract the mean from each data value.

Describing Variability in a Univariate Distribution with Standard Deviation In this lesson, we •

calculated the variance and standard deviation of a distribution, by hand and with technology.

•

used standard deviation to compare two distributions.

•

explored the effects of changing data values on the standard deviation of a distribution.

Terminology The variance is a measure of spread for a data distribution. To calculate variance, first sum the squares of all of the differences between the data values and the mean of the data set. Then divide the sum by one less than the size of the data set.

Example The list shows the ages in years of 5 children on the swings at a playground.

6, 8, 8, 11, 12 a. What is the range of the distribution?

12 - 6 = 6 The range of the distribution is 6 years.

The standard deviation is a measure of variability appropriate for data distributions that are approximately symmetric. Standard deviation is often used to describe the typical distance from the mean. It is calculated by taking the

Age (years)

Deviation from the Mean

Squared Deviation from the Mean

-3

-1

The deviation from the mean is negative when the data value is less than the mean. The squared deviation is always positive.

d. What is the variance of the distribution?

_____________

9+1+1+4+9 =6 4

The variance of the distribution is 6.

square root of the variance of a data set.

To calculate variance, divide by 1 less than the sample size. The sample size is 5, so divide by 4.

e. What is the standard deviation of the distribution? Round to the nearest tenth.

√6 ≈ 2.449

b. What is the mean of the distribution?

6 + 8 + 8 + 11 + 12 _______________ =9

The standard deviation is the square root of the variance.

The standard deviation of the distribution is approximately 2.4 years.

The mean of the distribution is 9 years. f.

Interpret the standard deviation in this context. The difference between the typical age and the mean age of a child at the playground is about 2.4 years.

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Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 20

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3. The following list shows the quiz scores for 5 students selected at random.

6, 7, 9, 9, 9 a. What is the range of the data distribution?

1. At a high school, ninth grade students are required to run a timed mile as part of a yearly physical fitness test. Each data set shows the times it took 5 randomly selected ninth grade students to run one mile, rounded to the nearest minute.

3 points

Data Set A: 9, 9, 9, 9, 9

b. What is the mean of the data distribution?

8 points

Data Set B: 7, 8, 9, 10, 11 a. Write a sentence that compares the centers and spreads of the data sets.

c. Record the deviations from the mean and the squared deviations from the mean in the table.

The centers of the data sets are the same, but data set B is more spread out than data set A.

Quiz Score

Deviation from the Mean

Squared Deviation from the Mean

-2

-1

b. Interpret what the centers and spreads represent in this situation. The typical time it took the students to run one mile is the same for both data sets. There is no variation in the students’ times for data set A. For data set B, the students’ times to run one mile differ from the typical time by about 1.5 minutes.

2. A manager of a trampoline park recorded attendance at the park each day over several months. The mean of the data set is 85 people, and the standard deviation is 6 people. a. Explain what the mean of this data set tells you about the number of people who attended the trampoline park during this time. Typically, 85 people attended the trampoline park each day during this time.

d. What is the variance for the data distribution?

2 b. Explain what the standard deviation of this data set tells you about the number of people who attended the trampoline park during this time.

e. What is the standard deviation for the data distribution? Round to the nearest tenth. Approximately 1.4 points

The typical deviation between the attendance at the trampoline park each day and the mean attendance during this time is 6 people.

Interpret the standard deviation in this context. The typical difference between the quiz scores and the typical quiz score for these students is approximately 1.4 points.

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4. At a track meet, there are three men’s 100-meter races. The times in which the runners completed each race are recorded to the nearest 0.1 second and are shown in the dot plots.

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Race 1

5. A grocery store manager wondered whether there was a consistent number of apples in the bags of apples sold at the store. The manager randomly selected 20 bags of apples and recorded the number of apples in each bag. The mean number of apples in each bag was 8, and the standard deviation was approximately 2 apples.

Race 2

Random1y Se1ected Bags of App1es

Race 3

10.5

11.0

11.5

12.0

12.5

13.0

Time (seconds)

a. Without calculating, rank the races in order from least standard deviation to greatest standard deviation.

Number of App1es

Race 3, race 1, race 2

a. How could the manager redistribute the apples into bags so that the mean of the distribution is unchanged but the standard deviation is 0?

b. Explain how you were able to rank the standard deviations of the distributions without calculating.

All the bags of apples would need to have exactly 8 apples in them.

I noticed that the values for the race 3 data set are clustered more closely to the mean than the other data sets, so it would have the least standard deviation. I also noticed that the typical distance of the values from the mean seemed greatest for race 2.

b. How could the manager redistribute the apples into bags so that the mean and range of the distribution are unchanged but the standard deviation is made as great as possible?

c. Which estimate is closest to the standard deviation for the data from race 1? A. 0.1 second

The manager would need to place 5 apples each in 10 bags and then 11 apples each in the other 10 bags.

B. 0.75 seconds C. 1.15 seconds D. 1.30 seconds

c. Suppose another bag of apples is added to the original sample. Describe what would happen to the standard deviation if the bag contains the following numbers of apples:

d. Use your calculator to find the mean and standard deviation for race 1. Round each value to the nearest hundredth. How close was your estimate of standard deviation to the actual value?

i. 9 apples Standard deviation would decrease.

Race 1 mean:

ii. 11 apples

11.73 seconds

Standard deviation would increase.

Race 1 standard deviation:

iii. 3 apples

0.77 seconds

Standard deviation would increase.

My estimate was within a few hundredths of a second of the actual standard deviation.

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Remember For problems 6 and 7, solve the inequality. 4 6. 5 - _ m ≤ -31 5

7. -0.5(-4m - 6) > 4

m ≥ 45

m > 0.5

8. Which equations or inequalities have no solution? Choose all that apply. A. │x + 2│ = -5

B. -│x + 2│ = 5 C. -│x + 2│ = -5 D. │x + 2│ < -5 E. │x + 2│ ≥ -5 9. Find the slope of the line. y 10 9 8 7 6 5 4 3 2 1 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 -1

9 10

-2 -3 -4 -5 -6 -7 -8 -9 -10

-1_ 5

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Teacher Edition: Math 1, Module 1, Topic D, Lesson 21 LESSON 21

Estimating Variability in Data Distributions Estimate and compare variation in data distributions represented by histograms. Use the interquartile range to compare the variation in data distributions represented by box plots and dot plots.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Name

EXIT TICKET

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

2. Two different locations have a speed limit of 70 kilometers per hour. Ana uses a speed detector to collect random samples of the speeds in kilometers per hour of cars traveling past the two locations in one day. The box plots display the distributions of those random samples.

1. The histogram shows the distribution of the weights in pounds of 50 dogs at a dog show.

Speeds of Cars at Two Locations

Weights of Dogs Location 1

18 16

Location 2

Frequency

14 12 10

8 6

100

Speed (kilometers per hour)

Suppose Ana claims that the amount of variation in the driving speeds at location 1 is greater than the amount of variation in the driving speeds at location 2.

2 0 0

a. Provide one form of evidence from the data that supports Ana’s claim.

Weight (pounds)

The range is greater for the distribution of the driving speeds at location 1 than for the distribution of the driving speeds at location 2.

Which value is the best estimate of the standard deviation of the distribution? A. 15 pounds B. 30 pounds C. 37.5 pounds

b. Provide one form of evidence from the data that refutes Ana’s claim.

D. 75 pounds

The interquartile range is greater for the distribution of the driving speeds at location 2 than for the distribution of driving speeds at location 1.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Lesson at a Glance In this lesson, students work in pairs and engage in discussions to determine appropriate measures of spread to represent and compare data distributions. Students compare the standard deviation of a mound-shaped distribution with that of a distribution that is skewed, leading to a class discussion of whether standard deviation is an appropriate measure of spread for skewed data distributions. Students estimate and compare variations of distributions represented with histograms, in which standard deviation can only be estimated. Students use the interquartile range to compare the variations in distributions displayed with box plots, including distributions that contain outliers. As with previous lessons, discussions focus on the importance of interpreting comparisons of the shape, center, and spread of data distributions within real-world contexts.

Key Questions • How does the shape of a distribution influence your decision about which measures of center and variation to use to compare data sets? • How can the presence of outliers affect measures of center and variation?

Achievement Descriptors Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers,

and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the context of the

data sets. (S.ID.A.3)

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Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Estimating Spread from Histograms

• None

• Measuring Variation from Box Plots

Lesson Preparation

Land 10 min

• None

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Fluency Find Range and Interquartile Range Students find the range and interquartile range of data sets to prepare for comparing variation in distributions represented by box plots and dot plots. Directions: Find the range and the interquartile range of each data set. 1.

5, 7, 2, 5, 4

6, 1, 0, 3, 8, 6, 46

3.1, 3.3, 2.8, 3.6, 3.5, 2.9

Range: 5 Interquartile range: 3 Range: 46 Interquartile range: 7 Range: 0.8 Interquartile range: 0.6

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Launch

Students compare data sets with mound-shaped distributions to skewed distributions to illustrate that the range and standard deviation are not always appropriate measures of spread. Direct students’ attention to the dot plots in problem 1. Allow students 2 minutes to think-pair-share to compare the spreads of the distributions without calculating. Circulate as students work, and listen for varying responses and reasoning, including student responses that apply the definition of standard deviation as a typical distance from the mean of a data set and those that compare the spread by using range. 1. As part of a project, Mrs. Allen’s Mathematics I class collected data to determine whether the amount of homework increased as grade level increased. One group surveyed a random sample of 50 ninth grade students and 50 eleventh grade students and asked them how many hours each week they spent on homework. The dot plots show the results of the survey.

Weekly Hours of Homework for Ninth Grade Students

Weekly Hours of Homework for Eleventh Grade Students

0 1 2 3 4 5 6 7 8 9 1011121314

Time (hours)

Compare the spreads of the distributions. Explain your reasoning. The ninth grade distribution has a greater spread. The range of the ninth grade distribution is 11 hours, while the range of the eleventh grade distribution is 10 hours. 410

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Ask the following questions to facilitate a brief discussion. Some of you said that the ninth grade data distribution had a greater spread. What evidence supports this conclusion? The range is greater for the ninth grade data distribution. If students do not mention range, help them recall that the range is the difference between the maximum and minimum data values and can be used to describe the overall spread in a data set. Some of you said that the eleventh grade data distribution had a greater spread. What evidence supports this conclusion? The average distance from a typical value to the mean looks like it is greater for the eleventh grade data distribution than for the ninth grade data distribution. What measure of spread represents the distance between a typical value in the data distribution and the mean of the data distribution? Standard deviation Would it be appropriate to compare the spreads of the data distributions for these ninth grade and eleventh grade students by using standard deviation? Explain. No. Since standard deviation is a measure of spread related to the mean, it seems appropriate to use standard deviation to compare the spreads of approximately symmetric data distributions. The eleventh grade distribution is skewed, which means standard deviation would not be an appropriate measure to compare the spreads of the distributions. In previous grades, we also measured the range of the middle 50% of the data to represent the spread for a data distribution. What do we call that measure? Interquartile range For some distributions, range and standard deviation are not the most representative measures of spread, either because the distributions are skewed or because the data are summarized in a way where standard deviation cannot be calculated. Today, we will continue to compare the spreads in distributions by using estimations of standard deviation and the interquartile range (IQR).

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Learn Estimating Spread from Histograms Students estimate the standard deviations of distributions displayed in histograms and use the estimates to compare distributions of univariate data. Direct students’ attention to problem 2. Read the scenario aloud to the students and discuss it with them. Have the students analyze the graph for brand D for about 1 minute, and answer any questions that arise. It may be beneficial to point out that the histogram displays relative frequencies. Then allow students to work on problem 2(a) and 2(b) with a partner. Circulate as students work, and listen for students’ reasoning as they estimate the mean and standard deviation. 2. A product review team is conducting a study of various brands of batteries. The team measures the battery life of a random sample of 100 batteries from brand D and the battery life of a random sample of 100 batteries from brand E.

Brand D Battery Life

Relative Frequency

0.40 0.30 0.20 0.10 0 85

105 115 125 135 145 155 165 Battery Life (hours)

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a. Estimate the mean of the battery life distribution for brand D. The mean battery life appears to be about 110 hours. b. Circle the best estimate of the standard deviation of the battery life distribution for brand D.

3 hours 10 hours 30 hours Once most students have finished, invite students to share their responses and reasonings, highlighting the following points: • Because the distribution for brand D battery life is approximately symmetric, the mean should have a value similar to that of the median, near the center of the distribution. • We can estimate the mean to be about 110 hours, but we can’t find its exact value because histograms do not display exact values. • The typical deviation is likely greater than 3 hours from the estimated mean because the interval size is 10 hours. • Almost all of the data is at a distance of 30 hours or less from the mean, so 30 is too large to represent a typical distance from the mean. • While 10 hours is the best estimate of the standard deviation given the three options provided, this value is an approximation.

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Ask the following questions. Approximately what fraction of the values in the distribution are at a distance of less than 3 hours from the center?

Differentiation: Support

Approximately one-third Approximately what fraction of the values in the distribution are at a distance greater than 3 hours from the center? Approximately two-thirds Because standard deviation measures the typical distance of a value from the mean of a distribution, is 3 hours a good estimate of the standard deviation for the distribution of brand D battery life? Explain your thinking. No, it is too small. There are clearly more values that are greater than 3 hours from the mean than there are values less than 3 hours from the mean. Repeat the same process for the estimates of 10 hours and 30 hours. Students should recognize that 10 hours is the best estimate of the three choices. The goals of this discussion are the following: • Help students identify a rough estimate for the standard deviation and identify values that are clearly too small or too large. • Lead students to recognize that the interval between one standard deviation less than the mean and one standard deviation greater than the mean should incorporate approximately two-thirds of the data for a mound-shaped distribution. • Emphasize to students that the values for the mean and standard deviation discussed are both estimates. • Highlight that because these histograms do not display exact data values, the exact mean or standard deviation cannot be calculated, but estimations can be made and used to compare the distributions. Allow students to work on parts (c) through (e) independently. Then have them briefly share their responses with a partner.

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If students need support estimating the standard deviation from the distribution, help them visualize the estimates by encouraging them to sketch a vertical line segment at the center of the distribution. Then have them sketch a vertical line segment 3 units to the left of the estimated mean and sketch another segment 3 units to the right of the estimated mean.

Promoting the Standards for Mathematical Practice When students interpret the measures of center and spread for battery life data in context, they are reasoning abstractly and quantitatively (MP2). Ask the following questions to promote MP2: • Do the estimated standard deviations show reasonable amounts of variation in battery life? • What does comparing estimated standard deviations tell you about the variation in battery life for brands D and E?

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Brand E Battery Life

Relative Frequency

0.40 0.30 0.20 0.10 0 85

105 115 125 135 145 155 165 Battery Life (hours)

c. Estimate the mean of the battery life distribution for brand E. The mean battery life appears to be about 130 hours. d. Is the standard deviation for this distribution greater than, about the same as, or less than the standard deviation for the brand D battery life distribution? Explain your reasoning. The spreads for both distributions look about the same, so the standard deviations for the distributions are probably similar. e. Compare the distribution of brand D battery life with the distribution of brand E battery life. Interpret the comparison in context. The typical battery life of brand E is greater than that of brand D. The overall variation in battery life is about the same for both battery brands.

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Invite a few students to share their answers with the class. Focus on the comparisons of the mean and standard deviation and their meaning in context by having students describe what differences in center or spread indicate about battery lives. The goal of problem 2 is for students to recognize that the typical battery life of brand E is greater than that of brand D and that the overall variation in battery life is about the same for both battery brands. Suppose the product review team tested one additional battery for brand E and found that its life was 300 hours. How would this added value affect the mean and standard deviation of the distribution? Explain your reasoning. This added value would increase both the mean and the standard deviation of the distribution because it is an outlier that is much greater than the other values in the distribution. Display the relative frequency histogram for brand F batteries. Then conclude the activity by asking the following questions to facilitate a brief discussion reviewing when standard deviation is most appropriate to use as a measure of variation for a distribution.

Brand F Battery Life

Relative Frequency

0.40 0.30 0.20 0.10 0 85

105 115 125 135 145 155 165 Battery Life (hours)

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Look at the relative frequency histogram for the battery life of brand F batteries. Would standard deviation be an appropriate measure of variation for this distribution? Explain your thinking. Standard deviation would not be an appropriate measure of variation because the distribution is skewed. Measures of center for skewed distributions are more appropriately estimated by using the median, so it would not make sense to use standard deviation, which measures variation from the mean, as a measure of variation for a skewed distribution. What measures of center and variation might be more appropriate for this distribution? Median and interquartile range

Measuring Variation from Box Plots Students use the interquartile range to compare variation in distributions displayed in box plots and dot plots. To transition to measuring variation for data distributions represented in box plots, ask students to think–pair–share about the following question.

Language Support To support recall with terms revisited from grades 6 and 7, have students add the terms range and interquartile range to the Measures of Spread section of the Univariate Quantitative Data graphic organizer from lesson 17. Students could include a brief description of what each measure means, how to calculate the values, and an example of each term.

What other graphical display can be used to represent a distribution if we are only provided with a histogram? Explain. I cannot represent the data by using dot plots because histograms do not record exact values. However, the distributions could be represented by using box plots if the quartiles are easy to identify from the histogram.

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Direct students’ attention to problem 3. What do you notice? What do you wonder? I notice the range of maximum speeds of the wooden roller coasters is less than the range of maximum speeds of the steel roller coasters. I notice the median of maximum speeds of the wooden roller coasters is greater than the median of maximum speeds of the steel roller coasters. I wonder what the mean for each distribution is. I wonder about the points to the right of the box plot for the distribution of maximum speeds of the steel roller coasters. If students do not notice or wonder about the points to the right of the box plot for the distribution of maximum speeds of the steel roller coasters, suggest it as your own wondering. The values represented by the points to the right of the box plot for the distribution of maximum speeds of the steel roller coasters are outliers. Box plots may display outliers as points that are separated from the rest of the plot. This means that the rightmost edge of the box plot represents the maximum value, excluding outliers.

Differentiation: Challenge If students seem especially interested in knowing how to determine whether values in a data set are outliers, consider presenting them with one common definition of an outlier. Outlier: any data value that is more than 1.5 times the IQR above the third quartile, or less than 1.5 times the IQR below the first quartile. Have students apply the definition to verify that the speeds of the two fastest steel roller coasters are outliers.

Teacher Note

What do you think the leftmost edge of the box plot represents? The leftmost edge of the box plot represents the minimum value, excluding outliers. Have students label the outliers on the plot in problem 3. Give students time to complete problem 3 in pairs. Circulate while students work, listening for students’ reasoning for selecting a particular measure of variation to compare the distributions. If students have difficulty identifying the interquartile ranges from the box plots, remind them that the interquartile range (IQR) is the difference between the third quartile and the first quartile.

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Technology can be used to identify outliers. Some calculators or apps may use different rules to determine which data points qualify as outliers.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

3. A company that owns amusement parks was studying wooden and steel roller coasters of their competitors. A researcher selected a random sample of 75 wooden roller coasters and 75 steel roller coasters and recorded their maximum speeds. The distribution of the data is shown in the box plots.

Maximum Speeds of Roller Coasters

Wooden Steel

100

110

120

130

Speed (miles per hour)

a. How many outliers does each distribution have? Estimate their values. The distribution of the maximum speeds of the wooden roller coasters has no outliers. The distribution of the maximum speeds of the steel roller coasters has outliers of about 118 miles per hour and about 127 miles per hour. b. What is the approximate difference in the medians of the distributions? Median for the maximum speeds of the wooden roller coasters: approximately 50 miles per hour Median for the maximum speeds of the steel roller coasters: approximately 47 miles per hour The approximate difference in the medians of the distributions is 3 miles per hour. c. What is the approximate difference in the ranges of the distributions? Range of the distribution for wooden roller coasters: approximately 36 miles per hour Range of the distribution for steel roller coasters: approximately 122 miles per hour The approximate difference in the ranges of the distributions is 86 miles per hour.

UDL: Action & Expression To support the completion of problems 3(b) through 3(f), consider providing a reference sheet that includes the following: • A box plot with labels that identify the minimum, first quartile, median, third quartile, and maximum • Directions for finding the interquartile range, IQR = Q3 - Q1 Additionally, encourage students to label the first and third quartiles on the box plots in their books.

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d. What is the approximate difference in the interquartile ranges of the distributions? IQR of the distribution for wooden roller coasters: approximately 10 miles per hour IQR of the distribution for steel roller coasters: approximately 35 miles per hour The approximate difference in the IQRs of the distributions is 25 miles per hour. e. Which measure more accurately compares the variation in the distributions? Justify your reasoning. The interquartile range more accurately compares the variation in the distributions. The outliers in the distribution of the maximum speeds for the steel roller coasters greatly increase the range of the distribution, but they do not affect the interquartile range. Therefore, the interquartile range is a better estimate of the spread in the distribution of the maximum speeds of the steel roller coasters. f. Compare the distributions of the maximum speeds of the wooden and steel roller coasters. Interpret your findings in context. The typical maximum speed for the wooden roller coasters is about the same as the typical maximum speed for the steel roller coasters. However, the maximum speeds of the steel roller coasters vary more than the maximum speeds of the wooden roller coasters. Facilitate a brief discussion comparing the distributions. How does the difference in the medians of the distributions compare to the differences in the interquartile ranges or ranges of the distributions? The difference in the medians of the distributions is much smaller than the differences in ranges or interquartile ranges. Which is a better measure of variation to compare the distributions: the range or the IQR? Why? The IQR is a better measure for comparing the variations of the distributions because outliers in the distribution of the maximum speeds of the steel roller coasters greatly increase the range, making it a less appropriate measure of variation.

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Suppose the distribution of the maximum speeds of the steel roller coasters was skewed to the right. Would this affect your choice of measure to describe the variation in the distribution? No, the IQR still represents the range of the middle 50% of the data in the distribution, regardless of the shape of the distribution. Ask for a few volunteers to share their responses to part (f) and discuss the comparison of the distributions as a class, making sure that students compared both the typical maximum speeds and the variation of the maximum speeds. Direct students’ attention back to the dot plots from problem 1. Have students work in pairs to calculate the medians and interquartile ranges of the distributions. Then facilitate a brief discussion comparing the distributions. What are the median and IQR for the ninth grade data distribution? The median is 5 hours, and the IQR is 4 hours. What are the median and IQR for the eleventh grade data distribution? The median is 11 hours, and the IQR is 4 hours.

Teacher Note Consider assigning half of the pairs to calculate the measures for the ninth grade data distribution and half of the pairs to calculate the measures for the eleventh grade data distribution.

Compare the distributions. Explain any differences in context. The ninth grade data distribution is approximately symmetric and mound-shaped, while the eleventh grade data distribution is skewed to the left. The typical number of hours spent completing math homework weekly is about 6 hours greater for eleventh grade students than for ninth grade students, and the variability in the number of hours spent completing math homework weekly is the same for eleventh grade students and ninth grade students.

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Land Debrief 5 min Objectives: Estimate and compare variation in data distributions represented by histograms. Use the interquartile range to compare the variation in data distributions represented by box plots and dot plots. Use the following prompts to guide a discussion about estimating variation in distributions represented by histograms or box plots. From which graphs can we estimate the mean and standard deviation? What about the median, range, and interquartile range? Dot plots and histograms can be used to estimate the mean and standard deviation. Dot plots, histograms, and box plots can be used to estimate the median, range, and interquartile range. How does the shape of a distribution influence your decision about which measures of center and variation to use to compare data sets? For skewed distributions, the median and interquartile range are the most appropriate measures to use to estimate the center and variation in distributions. For approximately symmetric distributions, any of the measures we have learned about are appropriate. How can the presence of outliers affect measures of center and variation? Outliers can increase or decrease the mean drastically so that the mean is not an appropriate measure of a typical value in a distribution. Outliers also increase the range and standard deviation so those are not appropriate measures of spread in a distribution. Outliers have little to no effect on the median and interquartile range.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Recap

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

RECAP Name

Date

b. Circle the best estimate of the standard deviation of the age distribution.

8 years 16 years

Estimating Variability in Data Distributions

24 years

In this lesson, we •

estimated the mean and standard deviation of a distribution from a histogram.

•

compared spreads of distributions in box plots and dot plots.

c. On the histogram, mark the locations that are approximately one standard deviation greater than the mean and one standard deviation less than the mean.

Examples

2. The box plot summarizes the numbers of videos that the random sample of video streaming service subscribers streamed over the last month.

1. The histogram shows the ages of a random sample of 100 subscribers of one streaming video service.

A data point separated from the rest of the plot in a box plot is an outlier.

The distribution is mound-shaped. About two-thirds of the data values lie within one standard deviation of the mean of the distribution.

Ages of Subscribers of One Streaming Video Service

The mean is located at approximately

10 12 14 16 18 20 22 24 26 28 30 32

a. Identify any outliers in the distribution.

25 Frequency

Videos Streamed Last Month

Number of Videos

the center of the distribution.

0 videos

b. What is the range of the distribution?

30 videos

c. What is the interquartile range (IQR) of the distribution?

5 0

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

8 videos 10

Include outliers when calculating measures of center and spread. The minimum value in this distribution is 0 videos, not 13 videos.

30 _ 0 = 30

IQR is the difference between Q3 and Q1.

24 _ 16 = 8

d. Is the range or IQR a better measure of the spread for this distribution? Explain.

Age (years)

The IQR is a better measure of spread for this distribution. The outlier increases the range of the distribution significantly, but it has little to no effect on the IQR. So the IQR better represents the overall spread of the data.

a. Estimate the mean age of a subscriber to the streaming video service. Mark it on the histogram with a vertical line. About 40 years © Great Minds PBC

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

PRACTICE Name

Date

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 21

2. Suppose all the members of a different high school swim team were asked how many hours they studied last week, estimated to the nearest half hour. The box plot shows the data distribution. Weekly Time Spent Studying Swim Team B

1. All the members of a high school swim team were asked how many hours they studied last week, estimated to the nearest half hour. The histogram summarizes the results. Weekly Time Spent Studying Swim Team A 12 0

Frequency

Study Time (hours)

a. What is the range of the distribution?

33 hours

4 2

b. Interpret the range in this context.

0 2.5

7.5

12.5

17.5

22.5

27.5

32.5

37.5

The student who studied the longest amount of time studied 33 hours more than the student who studied for the least amount of time.

42.5

Study Time (hours)

a. Which value best estimates the mean study time?

c. Identify any outliers in the distribution.

A. 5 hours

There are no outliers in this distribution.

B. 10 hours C. 20 hours

d. What is the interquartile range (IQR) of the distribution?

14 hours b. Which value best estimates the standard deviation of the study time? A. 2.5 hours

e. Interpret the IQR in this context.

B. 5 hours

The middle 50% of the study times lies within a range of 14 hours.

C. 10 hours D. 20 hours

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

3. Now, suppose all the members of a high school track team were asked how many hours they studied last week. The box plot shows the data distribution.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 21

4. City A hosts a marathon. The ages of the runners in the marathon are summarized in the histogram.

Weekly Time Spent Studying Track Team

Ages of Participants City A Marathon 16,000 14,000

Frequency

12,000 0

Study Time (hours)

a. Identify any outliers in the distribution.

10,000 8,000 6,000 4,000

40 hours is an outlier.

2,000 0

Age (years)

b. What is the range of the distribution?

a. Estimate the mean age of runners in the city A marathon. Mark this value on the histogram with a vertical line. Interpret the mean in this context.

37 hours

The mean age of runners in the city A marathon is about 40 years old. c. What is the IQR of the distribution?

b. Circle the best estimate of the standard deviation of the ages of runners in the city A marathon.

11 hours

5 years 10 years 15 years

d. Is the range or IQR a better measure of the spread in the distribution? Explain your reasoning. The IQR is a better measure of spread. The outlier increases the range of the data set by a lot while it has either no effect or a smaller effect on the IQR. Because the IQR is less affected by the outlier, it better represents the overall spread of the distribution.

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c. On the histogram, mark with vertical lines the locations that are approximately one standard deviation greater than the mean and less than the mean. The standard deviation of the ages of the runners in the city A marathon is about 10 years.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

5. Consider the box plots of the distributions of delay times in minutes of 60 randomly selected Big Air flights in November 2019 and December 2019.

d. Approximately what fraction of the runners’ ages lies between one standard deviation less than the mean and one standard deviation greater than the mean? About two-thirds of the runners’ ages lies within one standard deviation of the mean age.

Delay Times for Big Air Flights

e. City B also hosted a marathon. The ages of the runners in the marathon are summarized in the histogram.

November

Ages of Participants City B Marathon

December

Frequency

100 80

0 10 20 30 40 50 60 70 80 90 100 110 120

Delay Time (minutes)

Compare the distributions of delay times in context by using measures of shape, center, and spread.

Both distributions are right-skewed. The typical delay times as measured by the medians of the distributions are about the same at 30 minutes. The variation in delay times is greater for the November flights, as both the IQR and range are greater for the November distribution than for the December distribution.

20 0 18

Age (years)

Estimate the mean of the age of the runners in the city B marathon. The mean age of runners in the city B marathon is about 48 years old. f.

Circle the best estimate for the standard deviation of the runners’ ages in the city B marathon.

6. Nina selected 10 students at random from her English class and 10 students at random from her math class. She asked each student how many movies they had watched in the previous month. The students’ responses are included in the lists shown.

5 years 10 years 15 years g. For which distribution is the standard deviation of the ages of the runners greater? Why is it greater?

Class

Number of Movies Watched by Students Last Month

English

0, 0, 1, 1, 2, 2, 3, 4, 4, 6

Math

0, 0, 0, 1, 2, 2, 2, 4, 4, 16

The standard deviation of the ages of runners is greater for the city B distribution than for the city A distribution. The ages of runners in the city B marathon are typically farther from their mean age compared to the city A runners’ ages. The ages of runners in the city A marathon are more clustered around the mean age. © Great Minds PBC

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

a. Calculate the mean and median for the data from each class.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

7. Three data sets are shown in the dot plots.

Class

Mean

Median

English

2.3 movies

2 movies

Math

3.1 movies

2 movies

Data Set A Data Set B Data Set C

b. Calculate the standard deviation and IQR for the data from each class. Round to the nearest tenth. Class

Standard Deviation

Interquartile Range (IQR)

English

1.9 movies

3 movies

Math

4.8 movies

4 movies

20 21 22 23 24 25 26 27 28 29 30 a. Compare the distributions by using measures of shape, center, and variation. Data sets A and B are approximately symmetric, while data set C is left-skewed. The medians of data sets A and B are both 25, and the median of data set C is 24.5, which is only 0.5 units less than those of A and B. The IQR of data set A is 2, the IQR of data set B is 5, and the IQR of data set C is 3. Using interquartile range, data set A has the least spread, followed by data set C, and then data set B.

c. Identify the outliers in either data set.

b. Which measures of center and variation did you choose to compare the data sets? Why?

The math class data set has an outlier of 16 movies.

I chose the median and IQR to compare the data sets because data set C is skewed, so the median and IQR may be more appropriate measures of center and spread than the mean and standard deviation.

d. Which measure of spread is more affected by the outlier? Explain your reasoning. The standard deviation is affected more than the IQR because the outlier increases the average distance between the values in the data set and the mean, while the outlier does not affect the range of the middle 50% of the data as much.

c. How do these data sets illustrate the importance of measuring variation to describe a data distribution?

e. Is the typical number of movies watched last month by students in Nina’s English class different from the typical number of movies watched last month by students in Nina’s math class? Explain your reasoning.

If I only used the centers to compare the distributions, all I can conclude is that they have about the same typical value. Analyzing the measures of variation tells us that these distributions differ from one another in how spread out they are.

The typical number of movies watched by students in Nina’s English class is about the same as the typical number of movies watched by students in her math class. The median values for the data sets are identical.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 21

11. Write the equation of the line in slope-intercept form.

Remember

For problems 8 and 9, solve the inequality. 8. _25 > 8(_t _ 3) 3 _ 3_ > t 8

2 t _ 4 t ≤ 30 3 9

t ≥ _27

8 6 4 2

10. The dot plot shows the distribution of the time spent at the gym on Tuesday for a random sample of 40 gym members. Times are rounded to the nearest 5 minutes.

-10

-8

-6

-4

-2

Time Spent at Gym

-4 -6 -8 -10

y = _3x + 6 0

5 10 15 20 25 30 35 40 45 50 55 60 65 Time (minutes)

a. What is the shape of the distribution of time spent at the gym for this sample of gym members? The distribution of time spent at the gym for these gym members is left-skewed.

b. Is 30 minutes a typical amount of time spent at the gym for this sample of gym members? Justify your answer. No, 30 minutes is not a typical amount of time spent at the gym for these gym members. Three-fourths of the sample spent at least 40 minutes at the gym.

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Teacher Edition: Math 1, Module 1, Topic D, Lesson 22 LESSON 22

Comparing Distributions of Univariate Data Compare two or more data sets by using shape, center, and variability. Interpret differences in data distributions in context.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Name

EXIT TICKET

Date

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

EUREKA MATH2

b. Compare the centers of the data distributions in this context. Because the distribution represented by the dot plot is mound-shaped, the mean and the median are both appropriate measures of center. The mean of this sample is approximately 10 push-ups, and the median is 10 push-ups.

The dot plot displays the distribution of the number of push-ups done in a row by a random sample of 20 people.

Because the distribution represented by the box plot is skewed to the right, the median is the most appropriate measure of center of the distribution. The median of the sample is 10 push-ups. The typical number of push-ups completed in a row by one person is about the same for both samples.

Number of Push-ups

The box plot displays the distribution of the number of push-ups done in a row by a random sample of 200 people.

c. Compare the variation of the data distributions in this context. The variation of the second sample is greater than the variation of the first sample. The interquartile range of the second sample is 10, which is greater than the range of the first sample, which is 8. This means that variation in the middle 50% of the second sample is greater than the variation in the entire first sample. Therefore, there is greater variation in the number of push-ups completed in a row in the second sample than in the first sample.

Number of Push-ups

a. Compare the shapes of the data distributions. The distribution of the number of push-ups displayed by the dot plot is mound-shaped. The distribution of the number of push-ups displayed by the box plot appears right-skewed.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Lesson at a Glance In this lesson, students work in groups to compare two or more sets of univariate data by using tables, graphs, and summary statistics provided to them. They work in their groups to informally compare the sets of data they are assigned, and they select measures of center and spread appropriate to the shapes of the distributions to compare the data sets. Students interpret differences in the data sets in terms of the real-world context and present their findings to the rest of the class. Students conclude the topic by debriefing the process of comparing two or more sets of univariate data in a class discussion. Note: To save time, students are provided with graphical representations and the measures for the data sets they compare. If time permits, consider providing only the raw data, and have students display the data in a graph and calculate appropriate measures of center and spread to compare the data sets.

Key Questions • What do the shape, center, and spread reveal about distributions of univariate data? • Why is it important to always interpret data in context?

Achievement Descriptors Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers,

and variabilities. (S.ID.A.2) Math1.Mod1.AD15 Interpret differences in summary measures in the context of the

data sets. (S.ID.A.3)

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Agenda

Materials

Fluency

Teacher

Launch 5 min

• None

Learn 30 min

Students

• Comparing Sets of Univariate Data

• Statistics technology

• Presenting the Findings

• Copies of Data Sets (1 problem per student group and 1 copy per student)

Land 10 min

Lesson Preparation • Make enough copies of each problem in the Data Sets pages for each student to have a copy of their group’s assigned problem. This should be approximately 7-12 copies of each problem.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Fluency Analyze Distributions with Appropriate Vocabulary Students describe shapes of distributions to prepare for comparing two or more data sets by using shape, center, and variability. Directions: Use the appropriate data display to answer each question. 9 8 7 6 5 4 3 2 1 0 0

What is the shape of the distribution in the histogram?

What measure of center is appropriate to describe a typical data value for the data in the histogram: mean, median, or both?

What is the shape of the distribution in the dot plot?

What measure of center is appropriate to describe a typical data value for the data in the dot plot: mean, median, or both?

Mound-shaped

Both

Left-skewed

Median

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Launch

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Students informally compare univariate data sets. There are three data sets provided for this activity in the Data Sets pages. Divide students into groups of four, and assign each group to one of the three sets: Home Runs, Rainfall, or Gasoline. Distribute to each group the information associated with their data set. Direct students to read the information provided. Invite students to write the name of their assigned data set and one sentence describing the context for their group’s assigned data set. Data sets you were assigned: In one sentence, describe the context for the data sets you will compare. Direct students to the displays of their data sets. Have them analyze and match each data display to the provided measures of center and spread. Once all groups have finished, confirm that the matches are correct. Today, we will analyze the distributions of the data sets provided and choose appropriate measures of center and spread to compare them. We will interpret differences between the data sets in context and present our findings.

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Learn Comparing Sets of Univariate Data Students use data displays and measures of center and spread to compare sets of univariate data. Direct the students’ attention to the bulleted list of tasks to complete for this activity. Answer any initial questions students have about the expectations. Then allow students to work in their assigned groups, only intervening if a group has difficulty progressing. Students should have access to statistics technology. However, students should consider when and how to use technology to analyze the data.

Promoting the Standards for Mathematical Practice When students are given data sets from a real-world scenario and prepare to analyze and compare them, they are making sense of the problem (MP1). Ask the following questions to promote MP1:

Compare the data sets assigned to your group by completing the following tasks:

• How can you describe this data set in your own words?

• Use the given data displays to determine which measures of center and spread are appropriate to use to compare the distributions.

• What information do you need to compare the data sets?

• Compare the shapes, centers, and spreads of the data distributions. Interpret any differences in context.

• What is your plan to compare the data sets?

Circulate as students work, and listen as groups compare the data sets for the problem they are assigned. A primary focus should be on constructing viable arguments about which plots and measures of center and spread to use to represent the data. Another area of focus should be on explaining what the comparisons of shape, center, and spread mean in context. There is not one correct set of measures of center and spread that students must use for their comparisons. For example, the centers of the distributions of the home run data could be compared by using either the mean or median, as both distributions are roughly symmetric and do not contain outliers. It is more important for students to be able to defend their choice of data displays and measures than for them to make a particular selection.

UDL: Representation Consider having students use the Univariate Quantitative Data graphic organizer from lesson 17 to determine which measures of center and spread are appropriate to use to compare the distributions.

If groups have difficulty determining which statistical measures to use to compare the data sets, lead a brief discussion by asking the following questions about how the shapes of data distributions provide insight into which measures of center and spread best represent the distributions:

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

• If your distribution is skewed, how does the skewness affect the mean and median? How does the skewness affect the range, standard deviation, and interquartile range? • Is there always one correct measure of center and one correct measure of spread to use to compare data distributions? If not, when might more than one measure of center or spread provide an accurate description of a data distribution? If groups are comparing the shapes and measures of center and spread for the distributions but are not interpreting them in context, ask students to restate their comparisons in terms of the situation. Example: • I heard you say that the median for the Pittsburgh data is 27.3 inches greater than the median for the Los Angeles data. What does this difference tell you about the rainfall in these cities? As groups finalize their comparisons, have them create a visual to share with the class. The visual should be a graphic organizer that includes any measures of center and spread they used to compare the distributions, and it should briefly compare the data sets in context. Groups can create the visual on a sheet of paper to be displayed under a document camera, or time permitting, they could create a more elaborate visual on chart paper or by using technology.

Presenting the Findings Students use a visual to explain how data sets compare. Have students present their visuals to the class. They should briefly describe the context of the data, explain their reasoning for choosing plots and statistical measures for comparison, and interpret the results in context. Consider having groups that analyzed the same data present in succession to facilitate a coherent class discussion. Invite students to ask questions about the groups’ choices for plots and statistics or the conclusions they made.

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Teacher Note As time permits, encourage students to use their experiences to form conjectures about the conclusions drawn from comparisons of the data sets. For instance, some students might suggest that the American League teams typically hit more home runs in a baseball season because the American League uses a designated hitter to bat for the pitchers, who are not generally considered powerful hitters.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Home Runs Sample: Both distributions are approximately symmetric with no outliers. Using the mean, American League teams averaged 193.3 home runs, compared to National League teams averaging only 179 home runs. So these data do support the claim that the typical number of home runs hit by American League teams is greater than the typical number of home runs hit by National League teams. The variation in the American League distribution is somewhat greater than the distribution for the National League, as the standard deviation is 4.4 home runs greater and the interquartile range is 7 home runs greater than the respective values for the National League.

Rainfall Sample: The annual rainfall distributions for both Seattle and Pittsburgh appear approximately symmetric, while the annual rainfall distribution for Los Angeles appears right-skewed. The distributions for Pittsburgh and Los Angeles both contain outlier values that are much greater than most of the other values in these distributions, so we chose the median to compare the centers of the distributions and the IQRs to compare the spreads of the distributions. Seattle’s typical annual rainfall is similar to that of Pittsburgh, and both cities typically get about 28 inches more annual rainfall than Los Angeles. The interquartile range of the distribution for Seattle is 11.1 inches, which is greater than the interquartile ranges of the distributions for Los Angeles and Pittsburgh. The annual rainfall in Seattle varies more from year to year than it does in Pittsburgh and Los Angeles.

Gasoline Sample: According to the data, the typical price of gasoline does not differ based on whether a state is one of the first 25 states or last 25 states when listed alphabetically. Because the shape of the distribution of average prices for group A is right-skewed, I chose to compare the medians. The median average gas price for group B is only 2 cents more than the median average gas price for group A. However, group A has average gas prices that are more variable than group B, as measured by both the ranges and the IQRs.

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Land Debrief 5 min Objectives: Compare two or more data sets by using shape, center, and variability. Interpret differences in data distributions in context. Use the following questions to facilitate a brief discussion with the students about comparing sets of univariate data. What do the shapes, centers, and spreads reveal about distributions of univariate data? The shapes of distributions can reveal information about whether data values are clumped together or more evenly spread out. The shape of a distribution can also help determine which measures of center and spread are appropriate to use to characterize a set of data. Measures of center provide information about typical values in a data set, while measures of spread provide information about the overall variation in a data set. Why is it important to always interpret data in context? The different data sets aren’t just different sets of numbers. They can each be used to answer a different statistical question. The meanings of the similarities and differences we found change depending on what data set we analyzed.

Exit Ticket 5 min Provide up to 5 minutes for students to complete the Exit Ticket. It is possible to gather formative data even if some students do not complete every problem.

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Teacher Note Assign the Practice problems for completion outside of class or use them in class if time remains after the lesson. Refer students to the Recap for support.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Recap

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

RECAP Name

Date

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Center: The median flight time from Boston to San Francisco is about 356 minutes. The median flight time from San Francisco to Boston is between 295 minutes and 300 minutes. The typical flight time for a flight from Boston to San Francisco is about 55 to 60 minutes longer than the typical flight time from San Francisco to Boston.

Comparing Distributions of Univariate Data In this lesson, we •

compared distributions by using shape, center, and variability.

•

interpreted differences in distributions in context.

EUREKA MATH2

Spread: The IQR for the distribution of flight times from Boston to San Francisco is about 28 minutes. The IQR for the distribution of flight times from San Francisco to Boston is between 10 and 20 minutes. The typical variation in flight times from Boston to San Francisco is about 10 to 20 minutes more than the typical variation in flight times from San Francisco to Boston.

Example The data displays show the distributions of flight times for a random sample of flights from Boston to San Francisco and a random sample of flights from San Francisco to Boston. Flights from San Francisco to Boston

Frequency

Flights from Boston to San Francisco

325 330 335 340 345 350 355 360 365 370 375 380 385 390 Flight Time (minutes)

7 6 5 4 3 2 1 0 285

290

295

300

305

310

315

320

325

Flight Time (minutes)

Compare the distributions of flight times in context by using shape and measures of center and spread. Shape: The distribution of flight times from Boston to San Francisco appears approximately symmetric, while the distribution of the flight times from San Francisco to Boston appears right-skewed.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

PRACTICE Name

Date

2. The relative frequency histogram summarizes the ages of a random sample of people in Bangladesh in 2018. Ages of People in Bangladesh

1. The following dot plot represents the ages of 60 randomly selected people from Japan in 2018. Ages of People from Japan

0.12

Relative Frequency

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

100

Age (years)

0.10 0.08 0.06 0.04 0.02 0 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

a. What is the shape of the distribution?

Age (years)

The distribution is approximately symmetric.

a. Compare the shapes of the distributions of the ages in the sample from Japan in problem 1 and the sample from Bangladesh.

b. How does the shape of a distribution help us decide which measures of center and spread to use?

The distribution of ages in the sample from Japan is approximately symmetric, while the distribution of the ages in the sample from Bangladesh is skewed to the right.

The shape of the distribution provides information about whether the mean, range, and standard deviation are appropriate measures of center and spread for a data set. When the shape is approximately symmetric, all the measures can be used; when it’s skewed, we should only use the median and IQR.

b. Which measures of center and spread would you use to compare the distributions? Justify your reasoning.

c. What is the typical age of a person in the sample?

I would use the median and interquartile range because the distribution of ages in Bangladesh is skewed.

The typical age of a person in the sample is 51 years old. d. What measure of center did you use to calculate a typical age in part (c)? Justify your choice.

c. Compare the distributions by using appropriate measures of center and spread.

I chose the median because it was easier to calculate than the mean. I can easily find the locations of the 30th and 31st ordered values, rather than adding all the values.

The median age in the sample from Bangladesh is between 25 and 30 years. This is between 21 to 26 years less than the median age of the sample from Japan. The spread of ages is similar in the samples for Bangladesh and Japan. The interquartile ranges of the two samples are both between 30 and 40.

e. What is an estimate of the range for the distribution? An estimate of the range for the distribution is 100 years.

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

3. A principal selected 100 eighth grade students at random to participate in a reading fluency test. Teachers recorded the number of words students read accurately in one minute. Students were tested in the fall and then again in the spring. The results of each test are summarized in the box plots.

Compare the distributions in context by using shape and measures of center and spread. The distribution of runs scored by the Blue Socks team is symmetric, while the distribution of runs scored by the Black Socks team is right-skewed. Both distributions have a median of 5 runs, which means the typical number of runs scored is the same for both teams. The interquartile range of the distribution for the Blue Socks is 4 runs, while the interquartile range of the distribution for the Black Socks is 9 runs. Based on the IQR, the Black Socks’ distribution is more variable.

Reading Fluency Test Scores

Spring

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Math 1 ▸ M1 ▸ TD ▸ Lesson 22

5. The box plots summarize the distributions of weekly sales of chocolate chip cookies from three bakeries.

Fall

Weekly Chocolate Chip Cookie Sales

90 100 110 120 130 140 150 160 170 180 190 200 210 220 Bakery A

Fluency (words per minute)

Write two or three sentences about the eighth grade reading fluency test results. Bakery B

The numbers of words read fluently in the spring and fall tests both have approximately symmetric distributions. The typical number of words read fluently in the spring test is about 20 more than the typical number of words read fluently in the fall test. The overall spread in the number of words read fluently in the fall and spring tests is similar, with a slightly smaller spread in the spring.

Bakery C

4. The following graphs show the distributions of the number of runs scored per game for a random sample of 30 games for the Blue Socks baseball team and a random sample of 31 games for the Black Socks baseball team.

0 1 2 3 4 5 6 7 8 9 10

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Number of Runs Scored

110

130

150

170

190

210

230

250

Compare the distributions in this context by using measures of shape, center, and spread.

Runs Scored by Black Socks Baseball Team

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Runs Scored by Blue Socks Baseball Team

345

The distribution for bakery A is approximately symmetric with an outlier. The distribution for bakery B is left-skewed. The distribution for bakery C is approximately symmetric. The median for bakery A is about 120 cookies, while the medians for bakery B and bakery C are about 145 cookies and 150 cookies, respectively. This means that bakeries B and C typically sold about the same number of chocolate chip cookies each week, which was about 25 more than the typical number of chocolate chip cookies sold each week by bakery A. The IQR for bakery A is 60 cookies, the IQR of bakery B is 120 cookies, and the IQR of bakery C is about 95 cookies. This means that the variation in weekly cookie sales from least to greatest is bakery A, bakery C, and bakery B.

346

P R ACT I C E

441

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

EUREKA MATH2

Math 1 ▸ M1 ▸ TD ▸ Lesson 22

9. Graph y = _ x - 5. 2 3

Remember For problems 6 and 7, solve the equation.

6. 6a - 4a - 3(4a + 1) = 47

7. -8a + 2(-3a + 3) - a = 1

10 9

_ 1 3

-5

8 7 6 5 4 3 2 1

8. A random sample of ninth grade students and tenth grade students at one school was asked how many hours they volunteered last month. The box plots show the distributions of time spent volunteering.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 -1

9 10

-2 -3 -4

Volunteer Hours

-5 -6 -7 -8 -9

Tenth Grade

-10

Ninth Grade

Time (hours)

Is the typical number of volunteer hours for ninth grade students in the sample greater than, less than, or equal to the typical number of volunteer hours for tenth grade students in the survey? Justify your answer. Sample: The typical number of volunteer hours for ninth grade students is greater than the typical number of volunteer hours for tenth grade students. The box plots show that the median number of volunteer hours for ninth grade students is 4, and the median number of volunteer hours for tenth grade students is 3.

442

P R ACT I C E

347

348

P R ACT I C E

188 182 216 135 205 155 214 166 267 227 176 150 194 217 208

176 175 167 172 210 235 128 218 170 186 157 162 133 205 191

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Link to American League data: https://www.baseball-reference.com/leagues/AL/

Teacher Edition: Math 1, Module 1, Topic D, Lesson 22

Link to National League data: https://www.baseball-reference.com/leagues/NL/

American League

National League

Home Runs by Team

The following table shows the total number of home runs hit by each team in the National League and each team in the American League for the 2018 baseball season.

Major League Baseball is a professional baseball league in the United States and Canada. The league has two subdivisions: the National League and the American League. Some fans feel as though the American League is the hitter’s division, which means that there are more powerful batters in the American League. One way to measure powerful batting is by counting the number of home runs hit by each team in a baseball season.

Home Runs

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

443

160

200

220

Number of Home Runs per Team

180

260

280

Interquartile range: 50 home runs Range: 132 home runs Range: 107 home runs

Median: 194 home runs Median: 175 home runs Interquartile range: 43 home runs

Standard deviation: 33.0 home runs Standard deviation: 28.6 home runs

Statistics Mean: 193.3 home runs

American League

Number of Home Runs

120 140 160 180 200 220 240 260 280

Mean: 179 home runs

Statistics

Number of Home Runs

0.1

0.2

0.3

0.4

0.5

240

American League

Home Runs Hit in Major League Baseball in 2018

140

National League

120

120 140 160 180 200 220 240 260 280

National League

0.1

0.2

0.3

0.4

0.5

National League

American League

Relative Frequency

This page may be reproduced for classroom use only. Relative Frequency

444

Home Runs Hit in Major League Baseball in 2018

Do these data support the claim that the typical number of home runs hit by American League teams is greater than the typical number of home runs hit by National League teams?

Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets EUREKA MATH2

36.7 38.2 41.3

32.8 28.8 34.8

1992 1993 1994

27.1

1988

39.2

29.9

1987

35.4

37.4

38.3

1986

1991

38.5

25.1

1985

52.2 (outlier)

35.3

1984

44.8

41.4

40.9

1983

1990

39.3

1982

42.5

37.5

35.4

1981

34.7

39.5

35.6

1980

1989

Pittsburgh

Seattle

Year

Annual Rainfall (inches)

20.7

16.6

5.2

4.2

7.2

15.9

9.3

7.8

29.5 (outlier)

13.7

11.4

21.6

Los Angeles

Seattle is known as a rainy city because of its large amount of annual rainfall. Is this reputation supported by data? The following table shows annual rainfall amounts in inches for the cities of Seattle, Pittsburgh, and Los Angeles from 1980 to 2018.

Rainfall

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

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445

446

Pittsburgh

28.9 45.5 34.5 34.2 36.2 40.1 35.7 32.3 41 57.4 (outlier) 41.2 34.9 40.7 39.7 32.8 37.9 43.5

Seattle

42.6 50.7 43.3 44.1 42.1 28.7 37.6 31.4 41.8 31.1 35.4 48.8 39 30.7 38.4 47 36.4

Year 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011

Annual Rainfall (inches)

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9.9

7.5

4.9

9.2

18.8

16.3

9.5

6.9

27.1

16.1

23.3

Los Angeles

Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets EUREKA MATH2

41.7 36.7 36.8 40.6 35 42.2 57.8 (outlier)

48.3 32.6 48.5 44.8 45.2 47.9 35.8

2012 2013 2014 2015 2016 2017 2018

7.8

12.3

10.3

8.3

3.7

8.9

Los Angeles

Link to Los Angeles data: http://www.laalmanac.com/weather/we09aa.php (LAX)

Link to Pittsburgh data: https://xmacis.rcc-acis.org/ (Pittsburgh)

Link to Seattle data: http://www.seattleweatherblog.com/rain-stats/

Pittsburgh

Seattle

Year

Annual Rainfall (inches)

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

This page may be reproduced for classroom use only.

447

Relative Frequency Frequency Relative

0.1 0

0.3 0.2

0.5 0.4

Annual Rainfall (inches)

0.1 0

0.3 0.2

0.5 0.4

Los Angeles

Pittsburgh

Annual Rainfall (inches)

0 5 10 15 20 25 30 35 40 45 50 55 60

Annual Rainfall (inches)

0 5 10 15 20 25 30 35 40 45 50 55 60

Annual Rainfall (inches)

0.1 0

0.3 0.2

0.5 0.4

Annual Rainfall from 1980 to 2018 Seattle

0 5 10 15 20 25 30 35 40 45 50 55 60

Seattle

Pittsburgh

Los Angeles

Relative Frequency Frequency Relative

This page may be reproduced for classroom use only. Relative Frequency Frequency Relative

448

Annual Rainfall from 1980 to 2018

Does the distribution of annual rainfall in Seattle differ from the distributions of annual rainfall in Pittsburgh and Los Angeles?

Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets EUREKA MATH2

Statistics

Range: 27.5 inches

Interquartile range: 8.6 inches

Median: 9.9 inches

Standard deviation: 6.4 inches

Mean: 11.7 inches

Los Angeles

Range: 30.7 inches

Interquartile range: 6.3 inches

Median: 38.2 inches

Standard deviation: 6.4 inches

Mean: 38.9 inches

Pittsburgh

Range: 25.6 inches

Interquartile range: 11.1 inches

Median: 37.6 inches

Standard deviation: 6.5 inches

Mean: 38.4 inches

Seattle

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

This page may be reproduced for classroom use only.

449

450

Group B

2.84 2.51 3.51 (outlier) 3.46 (outlier) 3.40 (outlier) 3.07 3.05 2.95 2.92 2.87 2.85 2.85 2.83 2.81

Group A

2.51 3.36 3.14 2.55 4.08 (outlier) 2.79 2.98 2.77 2.72 3.58 (outlier) 2.72 2.76 2.90 3.03

Average Gasoline Price per Gallon (dollars)

Do you think average gas prices would vary based on the name of a state? The data shows the average gasoline price per gallon in dollars on a particular day in 2019 for each state in the United States. Each state is grouped by whether it is one of the first twenty-five states (group A) or the last twenty-five states (group B) when listed alphabetically.

Gasoline

Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets EUREKA MATH2

2.79 2.77 2.76 2.75 2.74 2.71 2.70 2.65 2.6 2.55 2.52

3.08 2.71 2.63 2.50 2.51 2.62 2.76 2.81 2.83 2.84 2.94 Link to gas price data: https://www.gasbuddy.com/USA

Group B

Group A

Average Gasoline Price per Gallon (dollars)

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

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451

This page may be reproduced for classroom use only. Relative Frequency

452 Relative Frequency 0

0.1

0.2

0.3

0.4

0.5

0.1

0.2

0.3

0.4

0.5

3.5

Group A

Average Gasoline Price per Gallon (dollars)

2.5

4.5

Average Gasoline Price per Gallon (dollars)

2.50 2.60 2.70 2.80 2.90 3.00 3.10 3.20 3.30 3.40 3.50 3.60 3.70 3.80 3.90 4.00 4.10

Group B

Average Gasoline Price per Gallon (dollars)

2.50 2.60 2.70 2.80 2.90 3.00 3.10 3.20 3.30 3.40 3.50 3.60 3.70 3.80 3.90 4.00 4.10

Group A

Group B

State Average Gasoline Prices in 2019

Does the typical price of gasoline differ based on whether a state is in group A or group B?

Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets EUREKA MATH2

Statistics

Range: $1.01

Interquartile range: $0.23

Median: $2.81

Standard deviation: $0.26

Mean: $2.86

Group B

Range: $1.58

Interquartile range: $0.34

Median: $2.79

Standard deviation: $0.36

Mean: $2.88

Group A

EUREKA MATH2 Math 1 ▸ M1 ▸ TD ▸ Lesson 22 ▸ Data Sets

This page may be reproduced for classroom use only.

453

Teacher Edition: Math 1, Module 1, Standards Resource

Standards Module Content Standards Create equations that describe numbers or relationships. A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. A.CED.A.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. A.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V = IR to highlight resistance R. Understand solving equations as a process of reasoning and explain the reasoning. A.REI.A.1

Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.

Solve equations and inequalities in one variable. A.REI.B.3

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

Interpret the structure of expressions. A.SSE.A.1

Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients.

454

EUREKA MATH2

Math 1 ▸ M1

Build a function that models a relationship between two quantities. F.BF.A.1 Write a function that describes a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context. Reason quantitatively and use units to solve problems. N.Q.A.1

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

N.Q.A.2

Define appropriate quantities for the purpose of descriptive modeling.

Summarize, represent, and interpret data on a single count or measurement variable. S.ID.A.1 Represent data with plots on the real number line (dot plots, histograms, and box plots). S.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. S.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).

Standards for Mathematical Practice MP1

Make sense of problems and persevere in solving them.

MP2

Reason abstractly and quantitatively.

MP3

Construct viable arguments and critique the reasoning of others.

MP4

Model with mathematics.

MP5

Use appropriate tools strategically.

MP6

Attend to precision.

MP7

Look for and make use of structure.

MP8

Look for and express regularity in repeated reasoning.

455

Teacher Edition: Math 1, Module 1, Achievement Descriptors: Proficiency Indicators

Achievement Descriptors: Proficiency Indicators Math1.Mod1.AD1 Use units to understand and guide the solution path of multi-step problems. RELATED CCSSM

N.Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

Partially Proficient

Proficient

Highly Proficient

Use units to understand and guide the solution path of multi-step problems. Huan drives to a beach that is 225 miles away. He drives at an average speed of 65 miles per hour. Before he began driving, he purchased 15 gallons of gas at $3.00 per gallon. Huan’s car has a gas mileage of 25 miles per gallon. What is the cost of the amount of gas Huan used for the entire drive to the beach?

Math1.Mod1.AD2 Define appropriate quantities for the purpose of descriptive modeling. RELATED CCSSM

N.Q.A.2 Define appropriate quantities for the purpose of descriptive modeling.

Partially Proficient

Proficient

Highly Proficient

Define appropriate quantities for the purpose of descriptive modeling. Students and teachers plan a car wash fundraiser. They have $300 available to buy supplies. Their fundraising goal is $2500. What quantities should be defined during their planning? Explain your thinking.

456

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD3 Interpret parts of linear expressions, such as terms, factors, and coefficients, in context. RELATED CCSSM

A.SSE.A.1.a Interpret parts of the expression, such as terms, factors, and coefficients.

Partially Proficient

Proficient

Highly Proficient

Interpret parts of linear expressions, such as terms, factors, and coefficients, in context. Evan purchases one pair of shorts and one tank top. The original price of the pair of shorts is s dollars. The original price of the tank top is t dollars. He receives a discount on the price of each item. The total price of both items with the discounts is given by (s − 5) + 0.75t. Interpret the meaning of the expressions s − 5 and 0.75t.

457

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD4 Create equations and inequalities in one variable and use them to solve problems. RELATED CCSSM

A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

Partially Proficient

Proficient

Identify equations and inequalities that model real-world contexts. Match each situation to the equation or inequality that represents it. Write one equation or inequality in each1 box Math ▸ M1from the given answer choices. Teacher Edition: Math 1, Module 1, Resource Table Situation

Equation or Inequality

Huan has at most $95 to order n baskets. Each basket costs $5, and there is a shipping fee of $10 per order. Determine the maximum number of baskets Huan can purchase in one order.

Ana spends $50 at the art supply store. She purchases n rolls of paper for $5 each and an easel for $15. How many rolls of paper does Ana purchase?

Create equations and inequalities in one variable and use them to solve problems, including those in a real-world context. Tiah has $50 in her savings account and saves $10 per week from her babysitting job wages.

5n + 15 = 50

EUREKA MATH2

Part A Write an inequality that can be used to find the Answer Choices minimum number of weeks Tiah must work to save at least $375. Part B Solve the inequality you 5n +wrote 10 ≤ 95 in part A.

Emma pays $15 per month and a $50 annual fee for a gym membership. Determine the number of months n that Emma has been a member of the gym if she pays a total of $95.

Highly Proficient

5n + 10 ≥ 95

15n + 50 = 95

50n + 15 = 95

15n + 5 = 50

Expression

Property or Operation

6(3 + x) + 1 − 7x 18 + 6x + 1 − 7x

Answer Choices

5n + 15 = 50

5n + 10 ≥ 95

5n + 10 ≤ 95

15n + 50 = 95

50n + 15 = 95

15n + 5 = 50

Expression

Property or Operation

18 + 1 + 6x − 7x 19 − x

6(3 + x) + 1 − 7x

458

18 + 6x + 1 − 7x

18 + 1 + 6x − 7x

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD5 Represent constraints by using equations and inequalities. RELATED CCSSM

Partially Proficient Interpret constraints of equations or inequalities in one variable.

Proficient

Highly Proficient

B. $75.80 C. $89.75 D. $99.30 E. $125.60

459

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD6 Interpret solutions to equations and inequalities in one variable as viable or nonviable options in

a modeling context. RELATED CCSSM

Partially Proficient

Proficient

Highly Proficient

Interpret solutions to equations and inequalities in one variable as viable or nonviable options in a modeling context by writing equations or inequalities and finding their solution sets. Riku sells pins for $3 each. With his earnings, he plans to pay his sister the $25 that he owes her, but he wants to have at least $100 left. Part A Write an inequality to represent the given situation. Part B How many pins must Riku sell? Explain your answer.

460

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD7 Rearrange formulas to highlight a quantity of interest. RELATED CCSSM

A.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V = IR to highlight resistance R.

Partially Proficient

Proficient

Rearrange formulas to highlight a quantity of interest in one step by using the same reasoning as in solving equations.

Rearrange formulas to highlight a quantity of interest in two or more steps by using the same reasoning as in solving equations.

The formula for finding the volume V of a rectangular prism is V = lwh, where l represents its length, w represents its width, and h represents its height. Which equation represents the width of a rectangular prism in terms of its volume, length, and height?

The formula for finding the surface area SA of a prism is SA = 2B + ph, where B represents the area of the base, p represents the perimeter of the base, and h represents the height. Which equation represents the height of a prism in terms of its surface area, the area of its base, and the perimeter of its base?

A. w = V lh

___

B. w = l Vh

C. w = lh V

D. w = h lV

A. h =

Highly Proficient

p _______ SA − 2B

SA − 2B B. h = _______ p

C. h = p(SA − 2B) D. h = SA − 2B − p

461

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD8 Explain each step in demonstrating that algebraic expressions are equivalent. EUREKA MATH2

Math 1 ▸ M1 RELATED CCSSM

A.REI.A.1 Explain each step Teacher Edition: Math 1, Module 1, Resource Tablein solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.

Situation

Huan has at most $95 to order n baskets. Each basket costs $5, and there is a shipping fee of $10 per order. Determine the maximum number of baskets Huan can purchase in one order. Emma pays $15 per month and a $50 annual fee for a gym membership. Determine the number of months n that Emma has been a member of the gym if she pays a total of $95. Ana spends $50 at the art supply store. She purchases n rolls of paper for $5 each and an easel for $15. How many rolls of paper does Ana purchase?

Equation or Inequality

Partially Proficient

Answer Choices

5n + 15 = 50

Proficient

5n + 10 ≥ 95

5n + 10 ≤ 95

15n + 50 = 95

50n + 15 = 95

15n + 5 = 50

Expression

Property or Operation

Highly Proficient

Explain each step in demonstrating that algebraic expressions are equivalent. Show that 6(3 + x) + 1 − 7x is equivalent to 19 − x. Complete the two-column table. Write the property or operation used in each step.

6(3 + x) + 1 − 7x 18 + 6x + 1 − 7x 18 + 1 + 6x − 7x 19 − x

462

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD9 Explain steps required to solve linear equations in one variable and justify solution paths. RELATED CCSSM

A.REI.A.1 Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.

Partially Proficient

Proficient

Identify steps required to solve linear equations in one variable by filling in missing steps that justify a given solution path. The steps to solve the equation −3(x + 2) + 15 = −5x − 7 are shown. Circle an answer choice from each list to describe each step.

Solve linear equations in one variable and explain the property used to justify each step.

Construct viable arguments to justify or critique solution paths to linear equations in one variable.

Solve the equation −10x + 3 = 4(x − 6) + 1. Write the property or operation that justifies each step.

Kadir solved the equation 13 − 3(x + 1) + 5x = 9 − 2x. His work is shown.

13 − 3(x + 1) + 5x = 9 − 2x 13 − 3x − 3 + 5x = 9 − 2x

− 3(x + 2) + 15 = −5x − 7 − 3x − 6 + 15 = −5x − 7 −3x + 9 = −5x − 7 2x + 9 = −7 2x = −16 x = −8

(A)

13 − 3 − 3x + 5x = 9 − 2x

(B)

10 − 3x = 9 − 2x

(C)

1=x

(D)

Is Kadir’s solution path correct? Why?

(E)

Commutative property of addition

Addition of like terms

Distributive property

Addition property of equality

Associative property of addition

Multiplication property of equality

Commutative property of addition

Highly Proficient

463

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD9 Explain steps required to solve linear equations in one variable and justify solution paths. RELATED CCSSM

Partially Proficient

Proficient

Addition property of equality

Multiplication property of equality

Addition of like terms

Addition property of equality

Associative property of addition

Highly Proficient

Associative property of addition

Commutative property of addition

E Distributive property Associative property of addition Multiplication property of equality Commutative property of addition

464

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD10 Solve linear equations in one variable. RELATED CCSSM

A.REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

Partially Proficient

Proficient

Solve linear equations in one variable with numerical coefficients.

Solve linear equations in one variable with coefficients represented as letters.

Solve the equation −10(x − 3) = 4x + 16.

Solve the equation x(c − 12) = −3x + d for x.

Highly Proficient

Math1.Mod1.AD11 Solve linear inequalities in one variable. RELATED CCSSM

A.REI.B.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

Partially Proficient

Proficient

Solve linear inequalities in one variable by using the addition property of inequality.

Solve linear inequalities in one variable by using the multiplication property of inequality.

Solve the inequality x + 7 ≤ 8.

Highly Proficient

Solve the inequality 3 − 6x < − 15.

Solve compound linear inequalities in one variable. Solve the inequality −5 < 3x − 4 < 11.

465

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD12 Write equivalent expressions for contexts that can be modeled by arithmetic sequences. RELATED CCSSM

F.BF.A.1.a Determine an explicit expression, a recursive process, or steps for calculation from a context.

Partially Proficient

Proficient

Identify equivalent expressions for contexts that can be modeled by arithmetic sequences.

Write equivalent expressions for contexts that can be modeled by arithmetic sequences.

Consider the pattern of circles.

Figure 1

Figure 2

Figure 3

Suppose n represents the figure number. Which expressions can represent the number of circles in any figure given the figure number? Choose all that apply.

Highly Proficient

Suppose n represents the figure number. Write two expressions that each represent the number of circles in any figure given the figure number.

A. n + 4 B. 2n + 1 C. 2n + 4 D. 4n + 2 E. 2(n + 2) F. 2(2n + 1)

466

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD13 Represent data with dot plots, histograms, or box plots. RELATED CCSSM

S.ID.A.1 Represent data with plots on the real number line (dot plots, histograms, and box plots).

Partially Proficient

Proficient

Highly Proficient

Represent data with dot plots, histograms, or box plots. A random sample of 30 people was asked about the number of text messages they sent last month.

274, 609, 415, 239, 823, 548, 113, 226, 342, 465, 329, 453, 748, 176, 408, 624, 378, 637, 506, 461, 317, 527, 293, 302, 481, 396, 521, 464, 525, 329 Represent the given data distribution with a box plot.

467

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD14 Summarize and compare data distributions by their shapes, centers, and variabilities. RELATED CCSSM

S.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets.

Partially Proficient

Proficient

Describe the shape, center, and variability of data sets.

Compare and identify differences in shape, center, and variability of two or more data sets, including effects of outliers, if any.

Which histogram shows a distribution that is left-skewed and has a median between 40 and 50? A. Frequency

Time Spent at Restaurant A 12 10 8 6 4 2 0

Highly Proficient

Consider the following dot plots. Veterinary Clinic A

Time (minutes)

2 4 6 8 10 12 14 16 18 20 22 24 26

B. Frequency

Time Spent at Restaurant B 12 10 8 6 4 2 0

Number of Animals

Veterinary Clinic B

Time (minutes)

C. Frequency

Time Spent at Restaurant C 12 10 8 6 4 2 0

2 4 6 8 10 12 14 16 18 20 22 24 26 Number of Animals

Circle an answer choice from each list to make each statement true. 0

Time (minutes)

Frequency

40 Time (minutes)

468

the

(A) (B)

The variability of the first distribution is variability of the second distribution.

Time Spent at Restaurant D 12 10 8 6 4 2 0

The center of the first distribution is center of the second distribution.

the

less than

greater than

equal to

EUREKA MATH2

Math 1 ▸ M1

Math1.Mod1.AD15 Interpret differences in summary measures in the context of the data sets. RELATED CCSSM

S.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).

Partially Proficient

Proficient

Highly Proficient

Interpret differences in data distributions in the context of the data.

Interpret differences in data distributions in the context of the data, including the effects of outliers.

Interpret measures of variation and use them to support or refute a claim.

The box plot and dot plot display the data distributions of customers’ shoe sizes at two different shoe stores in one afternoon. Compare the distribution of the customers’ shoe sizes at Store A with the distribution of the customers’ shoe sizes at Store B.

Employees at two veterinary clinics created dot plots to show the number of animals they have treated each day for one month.

The box plot displays the distribution of screen time in minutes for 200 students at Northside High School. Northside High School

Veterinary Clinic A

Customer Shoe Sizes at Store A

200

250

2 4 6 8 10 12 14 16 18 20 22 24 26 5

Number of Animals

Veterinary Clinic B

Sizes

300

350

400

450

500

Screen Time (minutes)

The dot plot displays the screen time in minutes for 25 students at West Dale High School. West Dale High School

Customer Shoe Sizes at Store B

2 4 6 8 10 12 14 16 18 20 22 24 26 Number of Animals

11 Sizes

Part A Compare the distribution of the number of animals treated at Veterinary Clinic A with the distribution of the number of animals treated at Veterinary Clinic B during this month. Use appropriate summary measures to justify your reasoning. Part B Suppose Veterinary Clinic B hosted an event one day this month where all animals received a free check-up. The clinic treated 100 animals on this day. How does this value affect your comparison of the distributions of the number of animals treated at these veterinary clinics during this month?

200

250

300

350

400

450

500

Screen Time (minutes)

Levi says there is more variation in screen time at West Dale High School compared to Northside High School. He reasons that this is true because the range of screen time at West Dale High School is greater than the range of screen time at Northside High School. Do you agree with Levi? Explain your answer.

469

Teacher Edition: Math 1, Module 1, Terminology Resource

Terminology The following terms are critical to the work of Mathematics I module 1. This resource groups terms into categories called New, Familiar, and Academic Verbs. The lessons in this module incorporate terminology with the expectation that students work toward applying it during discussions and in writing. Items in the New category are discipline-specific words that are introduced to students in this module. These items include the definition, description, or illustration as it is presented to students. At times, this resource also includes italicized language for teachers that expands on the wording used with students. Items in the Familiar category are discipline-specific words introduced in prior modules or in previous grade levels. Items in the Academic Verbs category are high-utility terms that are used across disciplines. These terms come from a list of academic verbs that the curriculum strategically introduces at this grade level.

New absolute value equation An equation that can be written in the form |bx − c| = d, where b, c, and d are real numbers. For values of x in its solution set, |bx − c| = d indicates bx – c is a distance of d units from 0. This equation has the same solution set as the compound statement bx − c = d or −(bx − c) = d. (Lesson 14) absolute value inequality A statement that compares the expressions |bx – c| and d, where b, c, and d are real numbers, using one of the four inequality symbols (Lesson 15)

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algebraic expression A number, a variable, or the result of placing previously generated algebraic expressions into the blanks of one of the four operators ( ) + ( ), ( ) − ( ), ( ) ⋅ ( ), ( ) ÷ ( ), or into the base blank of an exponentiation with an exponent that is a rational number, ( )( ) (Lesson 3) compound statement Two or more statements connected by logical modifiers, like and or or (Lesson 12) element of a set An item in the set (Lesson 6) Example: 3 is an element of the set {1, 3}. empty set The set that contains no elements, denoted { } (Lesson 6) mound-shaped distribution A symmetric distribution that looks like a smooth, round hill. A mound-shaped distribution has exactly one peak. (Lesson 17) set-builder notation A representation of a set that builds the set by describing what is in it. The set is denoted by curly braces containing a variable on the left followed by a vertical bar and constraints listed on the right of the bar. (Lesson 6)

x x >5

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standard deviation A measure of variability appropriate for data distributions that are approximately symmetric, used to describe the typical distance from the mean. It is calculated by taking the square root of the variance of a data set. (Lesson 20) statement A sentence that is either true or false, but not both (Lesson 12) uniform distribution A quantitative data distribution where the values have the same frequencies (Lesson 17)

Frequency

Lyla’s Neighborhood

absolute value of a number box plot center of a data distribution coefficient data distribution data set dot plot endpoint (of the graph of the solution set of a single-variable inequality)

equivalent expressions

formula

histogram

inequality

interquartile range (IQR) 0

Number of Pets per Household

univariate quantitative data Observations on one numerical variable (Lesson 17) variance A measure of spread. To calculate variance, first sum the squares of all the differences between the data values and the mean of the data set. Then divide the sum by one less than the size of the data set. (Lesson 20)

Familiar

mean median observation observational unit outlier

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properties: addition property of equality addition property of inequality associative property of addition associative property of multiplication commutative property of addition commutative property of multiplication multiplication property of equality multiplication property of inequality properties of exponents range relative frequency histogram

472

set skewed data distribution solution set spread of a data distribution statistical question symmetric data distribution variable

Academic Verbs generalize

Teacher Edition: Math 1, Module 1, Math Past

Math Past Algebra Begins When did algebra begin? Who first wrote about it? What is the origin of the word algebra? Invite students to examine this algebra problem.

because he was the first mathematician to establish systematic procedures for solving problems that we represent with equations. Al-Khwārizmī was so respected for his work that his likeness was depicted in sculpture and on a postage stamp. The sculpture in the photo is in modern-day Khiva, Uzbekistan—near where al-Khwārizmī was born.

You have [separated] ten into two parts, and you have divided one by the other; the obtained quotient is four. [Find the two parts.]1 The language is a bit old-fashioned, so parsing this problem could benefit from a few hints. First, clarify to students that the phrase separated ten into two parts means that two numbers add to 10 and that the words divided and quotient have their usual meanings. Next, encourage students to test some pairs of numbers—to take guesses, in other words. For example, students might test 5 and 5, 6 and 4, or some other combinations, including those with fractions. If they test 8 and 2, students will realize that they have found the correct answer. But algebra is not about guessing. Algebra uses general rules that lead directly to solutions to problems like this one. Students may be intrigued to learn that this problem appeared about 1200 years ago in a book that represents the beginning of algebra. The author of that book was Persian scholar Muh.ammad ibn Mūsā al-Khwārizmī (780–850). Al-Khwārizmī lived in the city of Baghdad as a member of the House of Wisdom, a center of learning. The year 820 saw the publication of al-Khwārizmī’s book Kitāb al-jabr wa al-muqābala. The title means “Book of Restoration and Balancing.” The word al-jabr in the title gives us the modern word algebra. Historians call al-Khwārizmī the “father of algebra” 1

Roshdi Rashed, Al-Khwārizmī: The Beginnings of Algebra, 148.

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Let’s go back to al-Khwārizmī’s algebra problem and see how he solved it. We thus infer that you posit one of the two parts as one thing and the other as ten minus one thing.2 More old-fashioned language. The terms one thing and ten minus one thing are al-Khwārizmī’s version of our expressions x and 10 − x. The use of symbols to represent numbers was unknown in al-Khwārizmī’s time, and every problem was a word problem. Then you divide ten minus one thing by one thing, in order to get four.3 2

Rashed, The Beginnings of Algebra, 148.

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10 − x This would translate into the modern equation _____ x = 4.

Next, al-Khwārizmī jumps out of the flow of the problem to make a general observation. You know that when you multiply the quotient by the divisor, you find the amount you had divided once again.4 Al-Khwārizmī tells us here that multiplying a fraction (quotient) by the denominator (divisor) gives us the numerator (the amount you had divided).

Math 1 ▸ M1

Al-Khwārizmī used al-jabr (restoration) to move a negative quantity away from one side to become a positive quantity on the other side. There is another part of al-Khwārizmī’s procedure, al-muqābala (balancing), which means to remove the same quantity from both sides by adding or subtracting it. Al-Khwārizmī described the procedures of al-jabr and al-muqābala to interpret any problem leading to a linear or quadratic equation as one of the following six types.8

In this problem, the quotient is four and the divisor is one thing. Multiply four by one thing; the result is four things equal to the amount you have divided; that is ten minus one thing.5

Now al-Khwārizmī uses an unusual word: restore.

The word restore comes from the Arabic word al-jabr in the title of al-Khwārizmī’s book and connotes the reuniting of broken parts, such as when a doctor realigns a broken bone. Al-Khwārizmī regards minus one thing as a broken part that must be restored to where it belongs, together with the four things. He is saying that we should add x to both sides of 10 − x = 4x and obtain 10 = 5x … the result is five things equal to ten, and one thing is two, which is one of the two parts.7 To complete the problem, al-Khwārizmī found that one of the two parts (one thing) is 2. He did not state that the other part is 8 and apparently regarded this detail as obvious enough that readers could fill it in for themselves. 4 5 6 7

Rashed, The Beginnings of Algebra, 148. Rashed, The Beginnings of Algebra, 148. Rashed, The Beginnings of Algebra, 148. Rashed, The Beginnings of Algebra, 148.

(squares equal to roots)

(2) ax 2 = c

(squares equal to a number)

(3) bx = c

(roots equal to a number)

(4) ax + bx = c

(squares plus roots equal to a number)

(5) ax 2 + c = bx

(squares plus a number equal to roots)

(6) bx + c = ax 2

(roots plus a number equal to squares)

Al-Khwārizmī says that multiplying both sides of the 10 − x equation _____ x = 4 by x results in 10 − x = 4x. Restore ten by the thing, and add it to the four things …6

(1) ax 2 = bx

Squares means “multiples of the second power of the unknown number.” Roots means “multiples of the first power of the unknown number.” A number simply refers to a known constant. Al-Khwārizmī required the coefficients a, b, and c to be positive. Type (3) on al-Khwārizmī’s list is a linear equation. It is the only combination of squares, roots, and numbers where no squares actually occur. The problem that al-Khwārizmī just explained falls into type (3), as do many of the problems throughout Mathematics I. Al-Khwārizmī’s work shows that a wide variety of problems can be reduced to a type (3) problem and then solved. Interestingly, while al-Khwārizmī’s work is foundational to one of the most well-studied and important areas of pure mathematics, he did not see it this way. In Book of Restoration and Balancing, al-Khwārizmī states that the book contains only “what is easiest

Rashed, The Beginnings of Algebra, 23–24.

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Math 1 ▸ M1

and most useful in arithmetic,” such as the mathematics required “in cases of inheritance, legacies, partition, law-suits, and trade.”9 You and your students have probably noticed that al-Khwārizmī followed pretty much the same steps that we follow in solving

EUREKA MATH2

algebra problems today. After all, we are using his procedures! If he were alive today, we could show him how to use x instead of referring to the variable as one thing. Then al-Khwārizmī would be ready to learn Eureka Math2!

Jeff Suzuki, Mathematics in Historical Context, 87.

476

Teacher Edition: Math 1, Module 1, Materials

Materials The following materials are needed to implement this module. The suggested quantities are based on a class of 24 students and one teacher. 72

Colored pencils

Personal whiteboards

Teacher computer or device

Personal whiteboard erasers

Dry-erase markers

Projection device

Index cards

Scientific calculators

Learn books

Statistics technology

Painter's tape

Computers or devices

Paper or notebook

Teach book

Pencils

478

Teacher Edition: Math 1, Module 1, Fluency

Fluency Fluency activities allow students to develop and practice automaticity with fundamental skills so they can devote cognitive power to solving more challenging problems. Skills are incorporated into fluency activities only after they are introduced conceptually within the module. Each lesson in A Story of Functions begins with a Fluency segment designed to activate students’ readiness for the day’s lesson. This daily segment provides sequenced practice problems on which students can work independently, usually in the first few minutes of a class. Students can use their personal whiteboards to complete the activity, or you may distribute a printed version, available digitally. Each fluency routine is designed to take 3–5 minutes and is not part of the 45-minute lesson structure. Administer the activity as a bell ringer or adapt the activity as a teacher-led Whiteboard Exchange or choral response.

Whiteboard Exchange This routine builds fluency through repeated practice and immediate feedback. A Whiteboard Exchange maximizes participation by having every student record solutions or strategies for a sequence of problems. These written recordings allow for differentiation: Based on the answers you observe, you can make responsive, in-the-moment adjustments to the sequence of problems. Each student requires a personal whiteboard and a whiteboard marker with an eraser for this routine. 1. Display one problem in the sequence. 2. Give students time to work. Wait until nearly all students are ready.

Bell Ringer

3. Signal for students to show their whiteboards. Provide immediate and specific feedback to students one at a time. If revisions are needed, briefly return to validate the work after students make corrections.

This routine provides students with independent work time to determine the answers to a set of problems.

4. Advance to the next problem in the sequence and repeat the process.

1. Display all the problems at once. 2. Encourage students to work independently and at their own pace. 3. Read or reveal the answers.

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Math 1 ▸ M1

Choral Response

Sprints

This routine actively engages students in building familiarity with previously learned skills, strengthening the foundational knowledge essential for extending and applying math concepts. The choral response invites all students to participate while lowering the risk for students who may respond incorrectly.

Sprints are activities that develop mathematical fluency with a variety of facts and skills. A major goal of each Sprint is for students to witness their own improvement within a very short time frame. The Sprint routine is a fun, fast-paced, adrenaline-rich experience that intentionally builds energy and excitement. This rousing routine fuels students’ motivation to achieve their personal best and provides time to celebrate their successes.

1. Establish a signal for students to respond to in unison. 2. Display a problem. Ask students to raise their hands when they know the answer. 3. When nearly all hands are raised, signal for the students’ response. 4. Reveal the answer and advance to the next problem.

Count By This routine actively engages students in committing counting sequences to memory, strengthening the foundational knowledge essential for extending and applying math concepts.

Each Sprint includes two parts, A and B, that feature closely related problems. Students complete Sprint A, followed by two count by routines—one fast-paced and one slow-paced—that include a stretch or other physical movement. Then students complete Sprint B, aiming to improve their score from Sprint A. Each part is scored but not graded. Sprints can be given at any time after the content of the Sprint has been conceptually developed and practiced. The same Sprint may be administered more than once throughout a year or across grade levels. With practice, the Sprint routine takes about 10 minutes.

1. Establish one signal for counting up and counting down and another signal for stopping the count. 2. Tell students the unit to count by. Establish the starting and ending numbers between which they should count. 3. Begin the count by providing the signals. Be careful not to mouth the words as students count

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Directions 1.

Have students read the instructions and sample problems. Frame the task by encouraging students to complete as many problems as they can—to do their personal best.

Time students for 1 minute on Sprint A. Do not expect them to finish. When time is up, have students underline the last problem they completed.

Read the answers to Sprint A quickly and energetically. Have students call out “Yes!” if they answered correctly; have them circle the answer if they answered incorrectly.

Have students count their correct answers and record that number at the top of the page. This is their personal goal for Sprint B.

Celebrate students’ effort and success on Sprint A.

To increase success with Sprint B, offer students additional time to complete more problems on Sprint A or ask sequencing questions to analyze and discuss the patterns in Sprint A.

Lead students in the fast-paced and slow-paced count by routines. Include a stretch or other physical movement during the count.

Remind students of their personal goal from Sprint A.

Direct students to Sprint B.

10.

Time students for 1 minute on Sprint B. When time is up, have students underline the last problem they completed.

11.

12. 13.

14.

Sample Vignette Have students read the instructions and complete the sample problems. Frame the task: • You may not finish, and that’s okay. Complete as many problems as you can—do your personal best. • On your mark, get set, think! Time students for 1 minute on Sprint A. • Stop! Underline the last problem you did. • I’m going to read the answers quickly. As I read the answers, call out “Yes!” if you got it right. If you made a mistake, circle the answer. Read the answers to Sprint A quickly and energetically. • Count the number of answers you got correct and record that number at the top of the page. This is your personal goal for Sprint B. Celebrate students’ effort and success. Provide 2 minutes to allow students to complete more problems or to analyze and discuss patterns in Sprint A using sequencing questions. Lead students in the fast-paced and slow-paced count by routines. Include a stretch or other physical movement during the count. • Point to the number of answers you got correct on Sprint A. Remember, this is your personal goal for Sprint B. Direct students to Sprint B.

Read the answers to Sprint B quickly and energetically. Have students call out “Yes!” if they answered correctly; have them circle the answer if they answered incorrectly.

• On your mark, get set, improve!

Have students count their correct answers and record that number at the top of the page.

• Stop! Underline the last problem you did.

Have students calculate their improvement score by finding the difference between the number of correct answers in Sprint A and in Sprint B. Tell them to record the number at the top of the page. Celebrate students’ improvement from Sprint A to Sprint B

Time students for 1 minute on Sprint B. • I’m going to read the answers quickly. As I read the answers, call out “Yes!” if you got it right. If you made a mistake, circle the answer. Read the answers to Sprint B quickly and energetically. • Count the number of answers you got correct and record that number at the top of the page. • Calculate your improvement score and record it at the top of the page. Celebrate students’ improvement.

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Math 1 ▸ M1

The table below provides implementation guidance for the Sprints recommended in this module.

Sprint Name

Administration Guidelines

Sequence Questions

Count By Routines

Add and Subtract Fractions

Students add and subtract fractions to build and maintain procedural fluency for module 1.

How are problems 10–18 related?

Fast-paced: Count by tens from 0 to 120.

Administer anytime in module 1.

Evaluate Expressions with Absolute Values

Multiply and Divide Fractions

Operations with Integers

How do problems 10–18 relate to problems 19–27?

Students evaluate numerical expressions with absolute values.

What do you notice about problems 6–9?

Fast-paced: Count by fours from 0 to −36.

Administer before module 1 lesson 14 or in place of the module 1 lesson 14 Fluency.

How are problems 18–20 related to problems 21 and 22?

Slow-paced: Count by threes from −30 to 30.

Students multiply and divide fractions to build and maintain procedural fluency for module 1.

What do you notice about problems 2 and 3? About problems 10 and 11?

Fast-paced: Count by fives from 0 to 100.

Administer anytime in module 1.

Do you think problem 13 or problem 18 was easier? Why?

Students add, subtract, multiply, and divide integers to build and maintain procedural fluency for module 1.

How are problems 12–14 related? How can you use problem 18 to answer problem 19?

Administer anytime in module 1.

Solve One- and Two-Step Equations

Students solve one- and two-step equations to prepare for solving multi-step equations in module 1. Administer before module 1 lesson 7 or in place of the module 1 lesson 7 Fluency.

Solve One- and Two-Step Inequalities

Students solve one- and two-step inequalities to prepare for solving multi-step inequalities in module 1. Administer before module 1 lesson 11 or in place of the module 1 lesson 11 Fluency.

Slow-paced: Count by fours from 0 to 44.

Slow-paced: Count by threes from 0 to 48. Fast-paced: Count by nines from 0 to 99. Slow-paced: Count by eights from 72 to 8.

How do problems 7–9 compare to problems 10 and 11?

Fast-paced: Count by fives from 0 to −50.

How can you use problem 20 to answer problem 21?

Slow-paced: Count by twos from −20 to 20.

How is the strategy you used for problems 1–3 the same as or different from the strategy you used for problems 4–6?

Fast-paced: Count by powers of 3 from 3 to 243.

What do you notice about problems 9–12?

Slow-paced: Count by powers of from

__1 to _1_ . 81

_1_ 3

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Math 1 ▸ M1

Sprint Name

Administration Guidelines

Sequence Questions

Count By Routines

Solve Two-Step Equations

Students solve two-step equations to prepare for solving multi-step equations in module 1.

Do you think problem 13 or problem 14 was easier? Why?

Fast-paced: Count by powers of 2 from 2 to 128.

Administer before module 1 lesson 6.

Write Equivalent Expressions

Students rewrite simple numeric and algebraic expressions in equivalent forms to prepare for using properties to demonstrate equivalence in module 1. Administer before module 1 lesson 4 or in place of the module 1 lesson 4 Fluency.

484

What pattern do you notice in problems 14–16? How are problems 23 and 24 related?

Slow-paced: Count by powers of from

__1 to _1_ . 64

_1_ 2

How are problems 1–5 and 6–11 related?

Fast-paced: Count by sevens from 0 to 77.

How is the strategy you used for problems 1–5 the same as or different from the strategy you used for problems 6–11?

Slow-paced: Count by sixes from 48 to 6.

4 7

_ _6 7

__7 11

__4 11

3 11

__ __1 0 3 10

__ _1 2

__7 __1 13 __ 18

5 18

__ 11 __ 20

19 20

__ 17 __

1 3 + 7 7

_ _ _1 + 5_ 7

__8 − __1 11

__8 − __4 11

__8 − __5 11 7 8 − 11 11

__ __

__8 − __8 1 + 1 5 10

_ __ _1 + __3 5

3 1 + 10 5

__ _ 1 − 1 9 18

_ __ _7 − __1 9

__7 − 1_ 18

__9 + __1 20

__1 + __9 20

__9 − __1

3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

366

3 7

_1 + _2

_1 + _1

Add or subtract.

Math 1 ▸ M1 ▸ Sprint ▸ Add and Subtract Fractions

Add and Subtract Fractions

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

__7 − __5

__4 + __9

_5 − __4

__5 + _3

__7 + _7

_ _ 5 2 − 8 9

_ _

5 2 + 8 9

_1 − __1

_1 + __1

_4 − __9

_5 − __9

__9 + 2_

_ __

7 3 − 4 24

_2 + __3

_3 + __9

_ __

5 − 4 9 27

_3 + __4

_5 + _2

Number Correct:

__ 11 36

__ 53 50

__ 37 54

19 __

49 __

29 __

61 __

__3

__ 17 70

19 __

11 __

25 __

__ 11 24

13 __

21 __

__ 11 27

13 __

11 __

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EUREKA MATH2 Math 1 ▸ M1

485

486

_5 7 9

__8 13

__5 13 4 13

__ __2 0

_3 _5 8

7 8

_ __1 14

__9 3 14

__ __9 16

15 16

__ __

1 4 + 9 9

_ _ 1 6 + 9 9

_ _

__9 – __1 13

__9 – __4 13

__9 – __5 13 9 – 7 13 13

__ __

9 9 – 13 13

__ __ 1 1 + 4 8

_ _ _1 + _3 4

1 3 + 8 4

_ _ _1 – __1 7

5 – 1 7 14

_ __ __5 – 1_ 14

__7 + 1_ 16

__1 + _7 16

_7 – __1

3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

368

4 9

_1 + 3_

13 16

2 9

_1 + _1

Add or subtract.

Math 1 ▸ M1 ▸ Sprint ▸ Add and Subtract Fractions

Add and Subtract Fractions

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

__7 – __4

__9 + __3

_ __

5 7 – 8 12

__7 + _5

__4 + 4_

_ _ 5 2 – 6 7

_ _ 5 2 + 6 7

_ __ 1 – 1 9 10

_ __ 1 + 1 9 10

_ __ 2 – 4 3 21

_ __

3 2 – 7 28

__7 + _5

_ __ 7 3 – 5 30

_ __ 3 2 + 7 35

_ __ 5 2 + 3 12

_ __

8 4 – 5 15

__5 + 4_

_ _ 2 7 + 3 9

Improvement:

Number Correct:

__ 19 60

__ 39 70

__ 11 24

__ 61 48

__ 32 45

__ 23 42

__ 47 42

__1

__ 19 90

__ 10 21

__5

__ 47 85

__ 11 30

__ 13 35

__ 13 12

__4

__ 17 21

__ 13 9

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

|a4|

34. 35.

20 20

40 a40

2|a4(a5)| a2|a4(a5)|

21. 22.

370

a40

44.

43.

42.

41.

40 a|a2(4)(a5)|

|2(4)(a5)|

a|2(4)(5)|

20.

19.

18.

|a4(5)|

40.

39.

20 17.

a40

38.

|a4(a5)|

20 16.

37.

a20

a|a4(5)| 15.

36.

a20

a|a4(a5)|

|4(a5)|

32.

14.

13.

|4(5)|

a20

a|20|

11. 12.

33.

a20

a|a20|

31.

10.

29.

28.

a|a6|

27.

26.

30.

25.

24.

a|6|

|a6|

|6|

|5|

|a5|

23.

|4|

|3|

Evaluate the expression.

Math 1 ▸ M1 ▸ Sprint ▸ Evaluate Expressions with Absolute Values

Evaluate Expressions with Absolute Values

a|3(2) + 5| + 5

|a3(a2) a 5|

|a3(2) + 5|

a|a3(2)| a 5

a|a3(a2)| + 5

|3(a2)| + 5

|a3( a 2)| a 5

|a3 a 2| a 5

|a3 + 2| + 5

a5 a |3 a 2|

a5 + |2 + 3|

|a6| a 5

|6| + 5

5 a |6|

5 + |a6|

a|a5 a 1|

a|a5 + 1|

a|5 + 1|

|a5 a 1|

|a5 + 1|

|5 a 1|

|5 + 1|

Number Correct:

a11

EUREKA MATH2

EUREKA MATH2 Math 1 ▸ M1

487

488 |–8|

34. 35.

18 18

36 –36

2|–3(–6)| –2|–3(–6)|

21. 22.

372

–36

44.

43.

42.

41.

36 –|–2(3)(– 6)|

|2(3)(–6)|

–|2(3)(6)|

20.

19.

18.

|–3(6)|

40.

39.

18 17.

–36

38.

|–3(–6)|

18 16.

37.

–18

–|–3(6)|

15.

36.

–18

–|–3(–6)|

|3(–6)|

32.

14.

13.

|3(6)|

–18

–|18|

11. 12.

33.

–18

–|–18|

31.

10.

29.

–9

28.

–|–9|

27.

26.

30.

25.

–9

24.

–|9|

|–9|

|9|

|7|

|–7|

23.

|8|

|2|

Evaluate the expression.

Evaluate Expressions with Absolute Values

–|3(9) + 6| + 6

|–3(–9) – 6|

|–3(9) + 6|

–|–3(9)| – 6

–|–3(–9)| + 6

|3(–9)| + 6

|–3(–9)| – 6

|–3 – 9| – 6

|–3 + 9| + 6

–6 – |3 – 9|

–6 + |9 + 3|

|–9| – 3

|9| + 3

3 – |9|

3 + |–6|

–|–6 – 3|

–|–6 + 3|

–|6 + 3|

|–6 – 3|

|–6 + 3|

|6 – 3|

|6 + 3|

Improvement:

Number Correct:

–27

–33

–21

–12

–6

–9

–3

–9

Math 1 ▸ M1 EUREKA MATH2

22. 23.

__1 _

1 1

_5 _3 __2

1 1 ˜ 5 3

_ _

3 ˜ 1_ 5

1 1 ÷ 3 3

_ _

1 1 ÷ 5 5

_ _

1 1 ÷ 3 5

_ _

1 1 ÷ 5 3

_ _ 1 2 ˜ 3 5

_ _ 2 1 ˜ 3 5

3. 4. 5. 6. 7. 8. 9. 10.

__3

3 ÷ 1_ 5

_ _ 1 3 ˜ 4 7

13. 14.

_1 ÷ 3 18.

374

__1

7 ÷ 1_ 17.

_ _

21 __ 3 1 ÷ 4 7

16.

__3

_ _

15.

3 1 ˜ 4 7

_ _

5 6

1 2 ÷ 3 5

12.

__2

_ _

5 3

3 5

11.

5 ˜ 1_

21.

_ _ 15

1 15

1 1 ˜ 3 5

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

20.

_ 19.

1 9

_ _

1 1 ˜ 3 3

Multiply or divide.

Multiply and Divide Fractions

5 7

3 ÷ _4 ˜ 1_

_ _

2 4 ˜ ÷3 3 5

_ _ _ 5 7

2 3 1 ÷ ˜ 3 5 7

__8

15 28

10 63

__ 16 15

__ 70 27

175

24 ___

_ _ _

2 4 1 ˜ ÷ 3 5 2

5 9 3

7 ˜ _ ˜ 2_

_ _ _

2 3 4 ˜ ˜ 5 7 5

80 21

5 2 ˜8˜ 7 3

21 ___ 320

__ 15 64

__5

__ 35 27

__ 36 35

__ 20 63

21 __

12 __

__ 28 15

__ 21 20

__ 12 35

_ _ __

1 3 7 ˜ ˜ 4 8 10

_ _ _

1 3 5 ˜ ˜ 2 4 8

5 ÷7 9

_ _ 7 3 ÷ 9 5

_ _ 4 5 ÷ 7 9

_ _

5 4 ˜ 9 7

3÷_

3 ˜4 5

_ _ 4 3 ÷ 5 7

_ _ 3 4 ÷ 5 7

_ _

4 3 ˜ 7 5

Number Correct:

EUREKA MATH2 Math 1 ▸ M1

489

490 22. 23.

28. 29.

_5 _

1 1

_7 5

5 7

_ __2 __2

_1 u _1

7 5

1 1 ÷ 5 5

_ _ 1 1 ÷ 7 7

_ _

_1 ÷ _1 5 1 1 ÷ 7 5

_ _ 1 2 u 5 7

_ _

_2 u _1

3. 4. 5. 6. 7. 8. 9. 10.

32. 33.

__5 __5

_ _ _1 u _5

14.

36.

__1

_ 18.

376

35.

36 1 ÷5 9

9 ÷ _1 17.

34.

_ _

45 4

5 1 ÷ 4 9

16.

4 9

31.

15.

5 1 u 4 9

5 ÷ _1 13.

30.

_ _

7 10

1 2 ÷ 5 7

12.

27.

26.

25.

5 7

7 5

11.

7 u 1_

24.

21.

__1

_ _ 5 u _1

20.

__1

1 1 u 5 7

2. 35

__ 19.

1 25

_ _

1 1 u 5 5

Multiply or divide.

Math 1 ▸ M1 ▸ Sprint ▸ Multiply and Divide Fractions

Multiply and Divide Fractions

3 7

3 ÷ _2 u _1

_ _

4 2 u ÷3 7 5

_ _ _ 7 5

__9

3 2 1 ÷ u 7 3 5

105

8 ___

21 10

__ 24 35

___ 162 35

175

48 ___

_ _ _

4 2 1 u ÷ 5 7 3

6 3 7 5

9u_u_

_ _ _

3 4 4 u u 5 7 5

135 28

___ 7

_3 u 9 u _5

21 ___ 200

__ 35 72

__4

__ 20 21

__ 15 28

__ 12 35

__ 35 3

14 __

__ 21 10

__ 14 15

__6

_ _ __

1 3 7 u u 4 5 10

_ _ _

1 5 7 u u 2 4 9

4 ÷5 9

_5 ÷ _3

_3 ÷ 4_

5 7

_4 u _3

5÷_

_2 u 7

_3 ÷ 2_

_2 ÷ _3

7 5

_3 u _2

Improvement:

Number Correct:

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

30. 31.

0 ˜1 ˜1 1

˜1 ˜ (˜1)

41. 42.

˜10 10 2 ˜50 ˜50 50

˜4 ˜ 6 4 ˜ (˜6) ˜4 ˜ (˜6)

15. 16. 17.

˜14 ÷ (˜2) 22.

378

˜14 ÷ 2

˜10 ˜ (˜5)

10 ˜ (˜5) 21.

20.

19.

˜7

40.

˜5

˜2 + (˜3) 14.

˜10 ˜ 5

˜1

2 + (˜3)

13.

18.

39.

˜2 + 3

12.

44.

43.

38.

37.

36.

35.

34.

33.

˜1 ÷ (˜1)

11.

32.

˜1 ÷ 1

˜1 ˜ (˜1)

1 ˜ (˜1)

27.

26.

25.

10.

˜1

29.

1 ˜ (˜1)

˜1 ˜ 1

˜2

˜1 ˜ 1

28.

˜2

˜1 + (˜1)

24.

1 + (˜1)

23.

˜1 + 1

Evaluate.

Operations with Integers

(8 ˜ (˜2)) ˜ (˜3 ˜ 4)

(˜8 ˜ 2) + (˜4 ˜ 3)

6 + 5 ˜ (˜2)

(˜5 + 6) ˜ 2

˜5 ˜ 2 ˜ 6

˜5(2 ˜ 6)

˜1 ˜ 1 ˜ 1

˜1(1 ˜ 1)

˜2 ˜ 7 ˜ (˜3)

˜2 + 7 + (˜3)

˜4 + (˜13)

˜13 ˜ 4

4 ˜ 13

˜12 ˜ (˜6)

˜6 + (˜12)

˜12 ÷ 6

6 ˜ (˜12)

˜9 ˜ (˜3)

˜9 ÷ (˜3)

˜3 + (˜9)

˜9 ˜ (˜3)

Number Correct:

˜4

˜28

˜4

˜16

˜2

˜6

˜17

˜52

˜9

˜18

˜2

˜6

˜12

EUREKA MATH2 Math 1 ▸ M1

491

492 30. 31.

0 a4 a4 4

a2 a (a2)

41. 42.

a15 15 3 a80 a80 80

a6 a 9 6 a (a9) a6 a (a9)

15. 16. 17.

a18 a (a3) 22.

380

a18 a 3

a20 a (a4)

20 a (a4) 21.

20.

19.

40.

a4 a (a5)

14.

a20 a 4

4 a (a5)

13.

18.

39.

a4 a 5

12.

44.

43.

38.

37.

36.

35.

34.

33.

a2 a (a2)

11.

32.

a2 a 2

a2 a (a2)

2 a (a2)

27.

26.

25.

10.

29.

2 a (a2)

a2 a 2

28.

a2 a (a2)

24.

2 a (a2)

23.

a2 a 2

Evaluate.

Math 1 ▸ M1 ▸ Sprint ▸ Operations with Integers

Operations with Integers

(5 a (a4)) a (a2 a 7)

(a5 a 4) a (a7 a 2)

6 a 5 a (a3)

(a5 a 6) a 3

a6 a 3 a 5

a6(3 a 5)

a2 a 2 a 2

a2(2 a 2)

a1 a 9 a (a4)

a6 a (a15)

a15 a 6

6 a 15

a21 a (a7)

a7 a (a21)

a21 a 7

7 a (a21)

a12 a (a4)

a4 a (a12)

a12 a (a4)

Improvement:

Number Correct:

a34

a23

a21

a90

147

a28

a16

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

31. 32.

0 15 17 19 2 4 8

x+6=6 x − 3 = 12 x − 5 = 12 x − 7 = 12 8x = 16 4x = 16 2x = 16

3. 4. 5. 6. 7. 8. 9.

43. 44.

4 4 −4 −2 −1− 2

−3 −1− 3

−12 12 −12

8 + x = −4 −8 + x = −4 x − 8 = −4 x − (−8) = 4 −5x = 10 10x = −5 2x = −6 −6x = 2 1 x = −6 2

−1− x = −6 2

−−1 x = 2 6

12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

382

42.

−12

1 x=4 3

11.

−

41.

40.

39.

38.

37.

36.

35.

34.

33.

−

−1 x = 4

10.

30.

29.

28.

27.

26.

25.

24.

x+4=6

23.

x+2=6

Solve each equation.

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Equations

Solve One- and Two-Step Equations

−4 −4 2 2

−1− x + 1 = 2 −2 = 1− x − 1 3 x+6=9 2 3 (x + 4) = 9 2

−2− x + 1 = 1

−1− 2

−

1 = −2 x + 2 3 3

−2− (x − 1) = 1

3 4

−1−

−− (x + 8) = −9 3

−20

−9 = − (x + 8) 3 4

3 4

9 = − (x + 8)

−

−4

−1− (x − 4) = 2 4

−5

−4

12 = −3(x + 1)

−3x − 3 = 12

−x − 3 − x − x = 12

12 = −x − x − x

12 = −3x

18 = 3(x + 2)

3x + 6 = 18

2x + x + 6 = 18

2x + x + 3 = 18

2x + x = 18

2x = 18

Number Correct:

EUREKA MATH2

EUREKA MATH2 Math 1 ▸ M1

493

494 31. 32. 33.

0 12 14 16 2 3 6 15

x+7=7 x − 2 = 10 x − 4 = 10 x − 6 = 10 9x = 18 6x = 18 3x = 18

_1 x = 5

3. 4. 5. 6. 7. 8. 9.

41. 42.

3 3 −3 −3 −1_ 3

−2 −_1

6 + x = −3 −6 + x = −3 x − 6 = −3 x − (−6) = 3 8x = −24 −24x = 8 −6x = 12 12x = −6

12. 13. 14. 15. 16. 17. 18. 19.

21.

384

−_1 x = 3

−12

−1 x = −4 3 22.

−12

_1 x = −4 20.

−9

_1 x = 5

11.

44.

43.

40.

39.

38.

37.

36.

35.

34.

10.

30.

29.

28.

27.

26.

25.

24.

x+5=7

23.

x+3=7

Solve each equation.

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Equations

Solve One- and Two-Step Equations

2 2 8 −32 8

_5 x + 10 = 15 _5 (x + 4) = 15 16 = _4 (x + 12) −16 = 4_ (x + 12) −_4 (x + 12) = −16

−_2

_ 3 6 2=− x+ 4 4

−_2

−_ (x − 2) = 2 3 4

3 4

−_ x + 2 = 2

−12

−3 = _1 x − 1 2

−12

−_1 x + 1 = 3 6

−12

−_1 (x − 6) = 3 6

−5

−4

16 = −4(x + 1)

−4x − 4 = 16

−x − 4 − x − x − x = 16

16 = −x − x − x − x

16 = −4x

20 = 5(x + 2)

5x + 10 = 20

4x + x + 10 = 20

4x + x + 5 = 20

4x + x = 20

4x = 20

Improvement:

Number Correct:

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

x>6 x>4 x < 12 x < 14 x < 16 x<9 x>6 x>5 x > -25 x > -5 x > 25 x > -1 x>2 x > -4 x>8

x + 4 > 10 x + 6 > 10 x-3<9 x-5<9 x-7<9 2x < 18 3x > 18 10 + x > 15 10 + x > -15 -10 + x > -15 -10 + x > 15 4x > -4 4x > 8 4x > -16 4x > 32

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

386

x>8

x + 2 > 10

Solve each inequality for x.

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Inequalities

Solve One- and Two-Step Inequalities

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

18.

17.

x<6

-_1 x > -2

x > -10 x < -10

-12 - _ x < -6 3 (x + 20) < 6 5

3 5

x < -27

_2 (x + 9) < -12 3

x > -27

-6 - 2_ x < 12 3

x < 27

_2 x - 6 < 12 3

x < -9

x < -9 -4x - 12 > 24

-4(x + 3) > 24

3(x + 3) > 18

x>3

x > -6

-1_ x < 3 2

x < -4

1 x < -2 2

x > 20

x > 10

x < -10

x>5

x < -4

-x < -20

-2x < -20

2x < -20

-4x < -20

5x < -20

Number Correct:

EUREKA MATH2

EUREKA MATH2 Math 1 ▸ M1

495

496 x>7 x>5 x < 10 x < 12 x < 14 x<4 x>2 x>5 x > -35 x > -5 x > 35 x > -1 x>2 x > -3 x>6

x + 5 > 12 x + 7 > 12 x-2<8 x-4<8 x-6<8 4x < 16 8x > 16 15 + x > 20 15 + x > -20 -15 + x > -20 -15 + x > 20 6x > -6 6x > 12 6x > -18 6x > 36

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

388

x>9

x + 3 > 12

Solve each inequality for x.

Math 1 ▸ M1 ▸ Sprint ▸ Solve One- and Two-Step Inequalities

Solve One- and Two-Step Inequalities

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

18.

17.

x < 20

-1_ x > -4

x < -50 x > -21 x < -21

2 (x + 15) < -14 5

-18 - _ x < -9 3 (x + 42) < 9 7

3 7

x > -50

-6 - 2_ x < 14 5

x < 50

2 x - 6 < 14 5

x < -7

x < -7 -4x - 8 > 20

-4(x + 2) > 20

4(x + 2) > 16

x>2

x > -20

-1_ x < 5 4

x < -12

1 x < -3 4

x > 28

x > 14

x < -14

x>4

x < -7

-x < -28

-2x < -28

2x < -28

-7x < -28

4x < -28

Improvement:

Number Correct:

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

5 5 5 5 4 5 6 7 10 9 7 6 6 7 8 12 11

2x c 5 d 15 2x c 6 d 16 2x c 7 d 17 2x c 8 d 18 2x a 2 d 6 2x a 4 d 6 2x a 6 d 6 2x a 8 d 6 2x a 6 d 14 2x a 6 d 12 2x c 4 d 18 2x c 4 d 16 2bx c 2c d 16 2bx c 2c d 18 2bx c 2c d 20 2bx a 2c d 20 2bx a 2c d 18

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

390

2x c 4 d 14

Solve each equation.

Math 1 ▸ M1 ▸ Sprint ▸ Solve Two-Step Equations

Solve Two-Step Equations

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

a3 15 3 15 0 0 18

1 x a 2 d a3 3

f1 x a 3 d 2 af1 x c 3 d 2 a2 d af1 bx a 9c af1 bx a 9c d 3 3 d af1 x c 3 a1f x c 3 d a3 3

3 d f1 x c 2 3

f1 bx c 6c d 3 3

3bx c 2c d a6

6 d a3bx a 2c

a3bx a 2c d a12

a3x a 6 d a12

3x a 12 d a6

a6 d 3x c 12

3x c 12 d 6

3x c 6 d 12

12 d 3bx c 2c

Number Correct:

EUREKA MATH2

EUREKA MATH2 Math 1 ▸ M1

497

498 2 2 2 2 3 4 5 6 8 7 4 3 3 4 5 11 10

3x + 5 = 11 3x + 6 = 12 3x + 7 = 13 3x + 8 = 14 3x a 3 = 6 3x a 6 = 6 3x a 9 = 6 3x a 12 = 6 3x a 6 = 18 3x a 6 = 15 3x + 9 = 21 3x + 9 = 18 3(x + 3) = 18 3(x + 3) = 21 3(x + 3) = 24 3(x a 3) = 24 3(x a 3) = 21

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

392

3x + 4 = 10

Solve each equation.

Math 1 ▸ M1 ▸ Sprint ▸ Solve TwoaStep Equations

Solve TwoaStep Equations

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

22.

21.

20.

19.

a8 24 8 24 0 0 32

1 x a 2 = a4 4 1 xa4=2 4

a1a x + 4 = 2 a2 = a1a (x a 16) a1a (x a 16) = 4 4 = a1a x + 4 a1a x + 4 = a4 4

4 = 1a x + 2 4

1 (x + 8) = 4 4

4(x + 3) = a12

12 = a4(x a 3)

a4(x a 3) = a20

a4x a 12 = a20

4x a 20 = a12

a12 = 4x + 20

4x + 20 = 12

4x + 12 = 20

20 = 4(x + 3)

Improvement:

Number Correct:

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

3 3 4 5 2x 3x 3x 4x 5x 6x 3x 4x 5x 8x 9x 10x 9x 5x 0 ax a2x

1a1a1 (1 a 1) a 1 (1 a 1) a (1 a 1) (1 a 1) a (1 a 1 a 1) xax xaxax (x a x) a x (x a x) a (x a x) (x a x) a (x a x a x) (x a x) a (x a x a x a x) 2x a x 3x a x x a 4x x a 7x 2x a 7x 7x a 3x 10x a x 10x a 5x 10x a 10x 10x a 11x 10x a 12x

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

394

1a1

44.

43.

42.

41.

40.

39.

38.

37.

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

Number Correct:

(2x a 3y) a (2x a 3y)

(2x a 3y) a (3y a 4x)

(2x a 3y) a (4x a y)

(11 a 5x) a (2 a 4x)

(5x a 11) a (2 a 4x)

(11 a 5x) a (4x a 2)

(3x a 9) a (3x a 9)

2x a (4 a 3x)

4x a (4 a 3x)

6x a (4 a 3x)

6x a (4x a 3)

3x a (4x a 3)

(4x a 3) a 2x

(4x a 3) a x

4x a 6x a 4

4x a 6x a 4x

4x a 6x a 12x

Rewrite each expression as an equivalent expression in standard form.

Math 1 ▸ M1 ▸ Sprint ▸ Write Equivalent Expressions

Write Equivalent Expressions

a2x

6x a 4y

ax a 9

x a 13

ax a 13

a18

6x a 18

ax a 4

xa4

3x a 4

9x a 4

10x a 3

7x a 3

6x a 3

5x a 3

10x a 4

a2x

EUREKA MATH2

EUREKA MATH2 Math 1 ▸ M1

499

500 4 4 5 6 3x 4x 4x 5x 6x 6x 5x 6x 9x 12x 14x 16x 8x 4x 0 nx n2x

1+1+1+1 (1 + 1) + (1 + 1) (1 + 1 + 1) + (1 + 1) (1 + 1 + 1) + (1 + 1 + 1) x+x+x x+x+x+x (x + x) + (x + x) (x + x + x) + (x + x) (x + x + x) + (x + x + x) (x + x) + (x + x) + (x + x) 4x + x 5x + x x + 8x x + 11x 3x + 11x 11x + 5x 9x n x 9x n 5x 9x n 9x 9x n 10x 9x n 11x

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

396

1+1+1

44.

43.

42.

41.

40.

39.

38.

37.

36.

35.

34.

33.

32.

31.

30.

29.

28.

27.

26.

25.

24.

23.

Improvement:

Number Correct:

(4x + 9y) n (4x n 9y)

(4x n 9y) + (9y n 8x)

(4x n 9y) + (8x n 3y)

(4x + 9y) + (8x + 3y)

(6 n 13x) n (3 n 8x)

(13x + 6) + (3 n 8x)

(6 n 13x) + (8x + 3)

(5x n 7) n (5x + 7)

(5x + 7) + (5x n 7)

(5x n 7) + (5x n 7)

(5x + 7) + (5x + 7)

3x + (3 n 4x)

5x + (3 n 4x)

7x + (3 n 4x)

7x + (3 + 4x)

7x + (3x + 4)

3x + (3x + 4)

(3x + 4) + 2x

(3x + 4) + x

3x + 5x + 3

3x + 5x n 3x

3x + 5x n 10x

Rewrite each expression as an equivalent expression in standard form.

Math 1 ▸ M1 ▸ Sprint ▸ Write Equivalent Expressions

Write Equivalent Expressions

18y

n4x

12x n 12y

12x + 12y

n5x + 3

5x + 9

n5x + 9

n14

10x

10x n 14

10x + 14

nx + 3

x+3

3x + 3

11x + 3

10x + 4

6x + 4

5x + 4

4x + 4

8x + 3

n2x

EUREKA MATH2

Math 1 ▸ M1 EUREKA MATH2

Teacher Edition: Math 1, Module 1, Sample Solution Resource

Sample Solutions Expect to see varied solution paths. Accept accurate responses, reasonable explanations, and equivalent answers for all student work.

EUREKA MATH2

Math 1 ▸ M1

Mixed Practice

Name

EUREKA MATH2

Math 1 ▸ M1 ▸ Mixed Practice 1

4. Which point represents the solution to the system of linear equations shown in the graph?

Date

1. Evaluate 4(_1 x − 1) for x = −_1 . 2 2

y 3

A. Point F B. Point G

−5

C. Point J

D. Points F, G, and J

F −3

−2

−1

−1 −2 −3

2. Solve 6 + _7 x = 4 + _3 x + 8 for x.

5. Solve the system of linear equations algebraically.

y = 3_ x + 16

{y = −_15 x 5

3. Write the number that completes the equation so it has infinitely many solutions.

4(2x +

(−20, 4)

) = 8x + 12

502

349

350

EUREKA MATH2

Math 1 ▸ M1

Math 1 ▸ M1 ▸ Mixed Practice 1

6. Match each scatter plot to its description.

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Math 1 ▸ M1 ▸ Mixed Practice 1

7. Each student at Sparks Middle School takes one elective class. A random sample of students was asked which elective class they currently take. The two-way table shows the number of students in the sample by grade level who take each elective class. Use the data in the table to calculate each proportion in parts (a) and (b). The scatter plot shows an outlier.

Elective Classes

Grade Level The scatter plot shows a negative, nonlinear association.

Band

Art

Yearbook

Total

Grade 6

Grade 7

Grade 8

Total

125

a. What proportion of students from the sample take art class? If necessary, round the answer to the nearest hundredth. A. 0.27

The scatter plot shows clustering.

B. 0.31 C. 0.36 D. 0.42 b. What proportion of seventh grade students from the sample take yearbook class? If necessary, round the answer to the nearest hundredth.

The scatter plot shows a positive, linear association.

A. 0.21 B. 0.27 C. 0.30 D. 0.38

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Math 1 ▸ M1 ▸ Mixed Practice 1

8. Over a period of several years, Tiah recorded the number of fish she caught each time she went fishing. She plans to go fishing again today. The table shows the number of fish Tiah caught on previous trips and the probability that she catches that number of fish today.

Number of Fish

Probability

0.1

0.05

0.09

0.25

0.3

0.2

or more

0.01

a. What is the probability that Tiah will catch exactly 4 fish today? The probability that Tiah will catch exactly 4 fish today is 0.3, or 30%.

b. What is the probability that Tiah will catch at least 3 fish today? The probability that Tiah will catch at least 3 fish today is 0.76, or 76%.

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Mixed Practice

Math 1 ▸ M1

Name

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Math 1 ▸ M1 ▸ Mixed Practice 2

3. Match each statement to the inequality that models it. An inequality may be matched to more than one statement.

Date

1. Which expressions are equivalent to 43 ⋅ 4−8? Choose all that apply.

Five less than 7 times a number is at least 21.

−24

A. 4

7x − 5 < 21

7x − 5 ≤ 21

Five less than 7 times a number is less than 21.

B. 4−5

Five less than 7 times a number is at most 21.

C. 411 43 D. ___ 4−8

7x − 5 > 21

7x − 5 ≥ 21

Five less than 7 times a number is no more than 21.

4 E. __ 48 1 F. __ 45 3

Five less than 7 times a number is greater than 21.

4. Water flows from a hose at a constant rate of 14 gallons per minute. a. Write an equation to represent the number of gallons g that flows from the hose in t minutes.

2. Which expressions are equivalent to 1.2 × 10 ? Choose all that apply.

g = 14t

A. 120,000 6

B. (1.0 × 10 ) + (2.0 × 10 ) C. (2.0 × 103) (6.0 × 102) 4 . 8 × 107 D. ________2 4 . 0 × 10

b. Complete the table.

E. (2.0 × 105) − (8.0 × 104)

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356

Time, t (minutes)

Number of Gallons, g

140

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c. Graph the relationship between time and gallons of water.

6. The table represents a function. Which ordered pairs can be input–ouput pairs of the function? Choose all that apply.

g 160 Number of Gallons

EUREKA MATH2

Math 1 ▸ M1 ▸ Mixed Practice 2

140 120 100 80 60 40

n10

20 0

9 10 11

A. (4, n13)

B. (1, n1)

Time (minutes)

C. (n4, n1) D. (0, n1)

5. Calculate the slope of the line twice by using two different pairs of points.

E. (6, n10)

F. (3, n12)

(0, 6)

7. A right triangle has one leg with a length of 12 cm and a hypotenuse with a length of 20 cm. What is the length of the other leg?

(1, 3) (2, 0) n10

16 cm

(3, n3) n5

n10

6 n 3 ___ Using the points (0, 6) and (1, 3), m = ______ = 3 = n3. 0n1

3 n 0 ___ Using the points (1, 3) and (2, 0), m = ______ = 3 = n3. 1n2

The slope of the line is n3.

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Math 1 ▸ M1 ▸ Mixed Practice 2

¯ that will result in an image with 8. Ji-won says that a translation is the only transformation of AB ¯. Do you agree with Ji-won? Explain. the same length as AB

A No, I do not agree with Ji-won. While Ji-won is correct that a translation will result in a segment ¯, it is not the only transformation that will do so. A reflection or a that is the same length as AB ¯. Translations, reflections, and rotation will also result in a segment with the same length as AB rotations all preserve length.

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Teacher Edition: Grade A1, Module 4, Works Cited

Works Cited National Governors Association Center for Best Practices, Council of Chief State School Officers (NGA Center and CCSSO). Common Core State Standards for Mathematics. Washington, DC: National Governors Association Center for Best Practices, Council of Chief State School Officers, 2010.

O’Connor, John J. and Edmund F. Robertson. “Arabic Mathematics: Forgotten Brilliance?” St. Andrews, Scotland: School of Mathematics and Statistics, University of St. Andrews. 2019. http://mathshistory.st-andrews.ac.uk/HistTopics /Arabic_mathematics.html#13.

O’Connor, John J. and Edmund F. Robertson. “Abu Ja’far Muhammad ibn Musa Al-Khwarizmi.” St. Andrews, Scotland: School of Mathematics and Statistics, University of St. Andrews. 2019. http://mathshistory.st-andrews.ac.uk /Biographies/Al-Khwarizmi.html.

Rashed, Rashdi. Al-Khwarizmi: The Beginnings of Algebra. London: Saqi Books, 2010.

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Teacher Edition: Grade A1, Module 4, Credits

Credits Great Minds® has made every effort to obtain permission for the reprinting of all copyrighted material. If any owner of copyrighted material is not acknowledged herein, please contact Great Minds for proper acknowledgment in all future editions and reprints of this presentation. Common Core State Standards for Mathematics © Copyright 2010 National Governors Association Center for Best Practices and Council of Chief State School Officers. All Rights Reserved.

Cover, Crockett Johnson, Logarithms, 1966, Masonite and wood, 56 cm x 66.3 cm x 3.8 cm; 22 1/16 in x 26 1/8 in x 1 1/2 in. Courtesy of Ruth Krauss in memory of Crockett Johnson, Smithsonian National Museum of American History; page 97 Courtesy Gutenberg Museum, Mainz. Photograph by D. Bachert; page 171 and 474 Getty Images/iStockPhoto; page 352 bakabuka/Shutterstock

For a complete list of credits, visit http://eurmath.link /media-credits.

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Teacher Edition: Grade A1, Module 4, Acknowledgments

Acknowledgments Adriana Akers, Tiah Alphonso, Lisa Babcock, Chris Black, Joseph Phillip Brennan, Beth Brown, Amanda H. Carter, Leah Childers, Mary Christensen-Cooper, Monique Colbert, Jill Diniz, Karen Eckberg, Dane Ehlert, Samantha Falkner, David Gertler, Krysta Gibbs, Winnie Gilbert, Ryan Grover, Kimberly Hager, Robert Hollister, Alanna Jackson, Travis Jones, Emily Koesters, Robin Kubasiak, Sara Lack, Alonso Llerena, Gabrielle Mathiesen, Maureen McNamara Jones, Lauren Nelson, Ben Orlin, Selena Oswalt, Brian Petras, Lora Podgorny, Rebecca A. Ratti, Meri Robie-Craven, Bridget Soumeillan, Ashley Spencer, Danielle Stantoznik, Tara Stewart, Mike Street, Cathy Terwilliger, Carla Van Winkle, Cody Waters Ana Alvarez, Lynne Askin-Roush, Trevor Barnes, Brianna Bemel, Carolyn Buck, Lisa Buckley, Adam Cardais, Christina Cooper,

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Kim Cotter, Lisa Crowe, Brandon Dawley, Cherry dela Victoria, Delsena Draper, Sandy Engelman, Tamara Estrada, Ubaldo Feliciano-Hernández, Soudea Forbes, Jen Forbus, Liz Gabbard, Diana Ghazzawi, Lisa Giddens-White, Laurie Gonsoulin, Adam Green, Dennis Hamel, Cassie Hart, Sagal Hassan, Kristen Hayes, Marcela Hernandez, Abbi Hoerst, Libby Howard, Elizabeth Jacobsen, Ashley Kelley, Lisa King, Sarah Kopec, Drew Krepp, Cindy Medici, Ivonne Mercado, Sandra Mercado, Brian Methe, Patricia Mickelberry, Mary-Lise Nazaire, Corinne Newbegin, Tara O’Hare, Max Oosterbaan, Tamara Otto, Christine Palmtag, Laura Parker, Katie Prince, Gilbert Rodriguez, Todd Rogers, Karen Rollhauser, Neela Roy, Gina Schenck, Amy Schoon, Aaron Shields, Leigh Sterten, Mary Sudul, Lisa Sweeney, Tracy Vigliotti, Dave White, Charmaine Whitman, Glenda Wisenburn-Burke, Howard Yaffe

Exponentially Better Knowledge2 In our tradition of supporting teachers with everything they need to build student knowledge of mathematics deeply and coherently, Eureka Math2 provides tailored collections of videos and recommendations to serve new and experienced teachers alike. Digital2 With a seamlessly integrated digital experience, Eureka Math2 includes hundreds of clever illustrations, compelling videos, and digital interactives to spark discourse and wonder in your classroom. Accessible2 Created with all readers in mind, Eureka Math2 has been carefully designed to ensure struggling readers can access lessons, word problems, and more. Joy2 Together with your students, you will fall in love with math all over again—or for the first time—with Eureka Math2.

What does this painting have to do with math? Although he is best known as a cartoonist for his comic strip Barnaby and the book Harold and the Purple Crayon, the American artist Crockett Johnson also took a keen interest in mathematics, painting more than 100 pieces relating to mathematics and mathematical physics during his life. His painting Logarithms shows two sequences of rectangles. How do the lengths of the rectangles in the top row relate to those in the bottom row? On the cover Logarithms, 1966 Crockett Johnson, American, 1906–1975 Masonite and wood Smithsonian National Museum of American History, Washington, DC, USA Crockett Johnson (1906–1975), Logarithms, 1966. Masonite and wood, 56 cm x 66.3 cm x 3.8 cm; 22 1/16 in x 26 1/8 in x 1 1/2 in. Courtesy of Ruth Krauss in memory of Crockett Johnson, Smithsonian National Museum of American History

ISBN 979-8-88588-174-6

798885 881746

Module 1 Equations, Inequalities, and Data in One Variable Module 2 Equations, Inequalities, and Data in Two Variables Module 3 Functions and Their Representations Module 4 Transformations of the Plane and Constructions Module 5 Linear and Exponential Functions Module 6 Modeling with Data and for Contexts