Oxford Maths Standard Standard 2 Year 12 Sample Chapter

1

Module

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Lessons •

Lesson 1.1 Graphs of the form y = mx + c

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Lesson 1.2 Linear models

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Lesson 1.3 Identifying solutions to simultaneous equations

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Lesson 1.4 Solving simultaneous equations graphically

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Lesson 1.5 Solving simultaneous equations by substitution

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Lesson 1.6 Solving simultaneous equations by elimination

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Lesson 1.7 Break-even analysis

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Using spreadsheets

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Lesson 1.8 Review: Simultaneous linear equations

Online resources Take the diagnostic pre-test to assess your knowledge of the prerequisite skills listed.

Diagnostic pre-test: Simultaneous linear equations

After completing the diagnostic pre-test, brush up on your knowledge of the prerequisite skills by using the downloadable support sheets.

Support sheet: Substitution Support sheet: Solving linear equations Support sheet: Changing the subject of a formula Support sheet: Plotting linear relationships Support sheet: Reading values from graphs Support sheet: Identifying features of a linear graph

Syllabus links •

Algebraic relationships (MST-12-S2-01) ◦ A student represents the relationships between quantities algebraically and graphically to solve problems and make predictions in practical contexts.

▪ Simultaneous linear equations

•

Working mathematically (MAO-WM-01) ◦ A student develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently and clearly. © NESA

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Lesson 1.1

Graphs of the form y​ = mx + c​ Learning intentions By the end of this lesson, you will be able to ... → graph linear relationships.

Straight-line graphs •

This lesson reviews content in Module 2 of Oxford Maths Standard Year 11.

•

The graph of a linear relationship between two variables is a straight line.

•

If the equation of a linear relationship is written in the form y​ = mx + c​, then: → the gradient of the line is m ​ ​and the y​ ​-intercept is c​ ​ → x​ ​ is the independent variable, as any value may be used, and y​ ​ is the dependent variable, as it depends on the value of x​ ​.

•

There are many ways to draw a linear graph. These include: → completing a table of values and plotting points on the Cartesian plane → identifying the y-intercept ​(0, c)​, and a second point on the graph. The second point can be found using the gradient, m ​ ​, or by substitution → graphing applications and digital tools.

•

A point ​(x, y)​lies on a linear graph if, when x​ ​is substituted into the equation of the graph, the result is y​ ​.

Worked example 1.1A Completing a table of values to graph a linear relationship Use a table of values to draw the graph of y​ = 5 − 2x​.

Think

Write

Prepare a table with values of x​ ​ranging from ​ − 2​ to ​2​.

Substitute each ​x​-value into the equation ​y = 5 − 2x​to complete the table.

x​ = − 2​: ​y = 5 − 2 × (− 2) = 9​ ​x = − 1​: ​y = 5 − 2 × (− 1) = 7​ ​x = 0​: ​ y = 5 − 2 × 0 = 5​ ​x = 1​: ​ y = 5 − 2 × 1 = 3​ ​x = 2​: ​ y = 5 − 2 × 2 = 1​ ​​ x ​y​

Write the corresponding points from the table as a list.

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​− 2​ ​9​

​− 1​ ​7​

​0​ ​ ​ 5

​1​ ​ ​ 3

​2​ ​ ​ 1

​(− 2, 9)​, ​(− 1, 7)​, ​(0, 5)​, ​(1, 3)​, ​(2, 1)​

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Think

Write

Plot the points on the Cartesian plane and draw a straight line through them, extending beyond the plotted points.

y 10 8 6 4 2

−4

−2

0

4 x

2

−2

Worked example video: Completing a table of values to graph a linear relationship

Understanding, fluency and communicating Answers: pXXX

Helpful hints •

Substitute each x​ ​-value into the linear relationship to find the corresonding y​ ​-value.

•

The graph of a linear relationship is always a straight line. If you plot the points from a table of values of a linear relationship and the line is not straight, check you have substituted the x​ ​-values correctly.

1 Draw each linear relationship by completing the given table of values and plotting the corresponding points on the Cartesian plane. a ​y = 2x + 1​ ​​ x ​y​

​− 2​

b ​y = 2x + 3​ ​− 1​

​0​

​1​

c ​y = − 3x − 4​ ​​ x ​y​

​− 4​

​− 2​

​− 4​

​− 2​

​0​

​2​

​4​

​− 1​

​0​

​1​

​2​

​− 1​

​0​

​1​

​2​

d ​y = 4 − 5x​ ​− 2​

​0​

​2​

​​ x ​y​

​4​

e ​y = 2x − 7​ ​​ x ​y​

​​ x ​y​

​2​

f ​− 1​

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​0​

​1​

​2​

​− 2​

​y = 2 − 2x​ ​​ x ​y​

​− 2​

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2 Use a table of values to draw the graphs of the following linear relationships.

Hint: Choose the x​ ​-values in the table carefully to make the points easier to plot. For example, choose multiples of ​2​for part e. a ​y = x − 3​ b ​y = x + 2​ c ​y = 2x − 1​ d ​y = 1 − 3x​ 3 e ​y = _ ​ 1 ​x​ f ​y = _ ​ 2x + ​​ g ​y = _ ​ 2x ​ − 6​ h ​y = 4 − 2x​ 2 2 3 x 3 Use a table of values to draw the graph of y​ = _ ​ ​ + 5​. Explain your choice of ​x​-values used in the table and 10 the scale used on the axes of the Cartesian plane.

Worked example 1.1B Writing the gradient and y​ ​-intercept of a line from its equation Write the gradient and y​ ​-intercept of the linear relationships with the following equations.

2x ​​ a ​y = − 3x + 2​ b ​y = − 5 + ​ _ c ​4x + 2y = 1​ 3 Part

Think

Write

a

The equation is in the form y​ = mx + c​.

Identify the value of ​m​as the gradient and c​ ​as the ​y​-intercept. The y​ ​-intercept is located at (​ 0, c)​.

y​ = − 3x + 2​ ​gradient = m = − 3​ ​c = 2​, y​ ​-intercept: ​(0, 2)​

b

Rearrange the equation to be in the form y​ = mx + c​.

Identify the value of ​m​as the gradient and c​ ​as the ​y​-intercept. The y​ ​-intercept is located at (​ 0, c)​.

2x ​ y​= − 5 + ​ _ 3 ​​​ ​​ ​ 2x _ y​= ​ ​ − 5 3 ​gradient = m = _ ​ 2 ​​ 3 ​c = − 5​, y​ ​-intercept: ​(0, − 5)​

c

Rearrange the equation to be in the form y​ = mx + c​.

Identify the value of ​m​as the gradient and c​ ​as the ​y​-intercept. The y​ ​-intercept is located at (​ 0, c)​.

4x + 2y​= 1 2y ​​ ​​= −​4x + 1​​​ 1 ​ y​= − 2x + ​ _ 2 ​gradient = m = − 2​ ​c = _ ​ 1 ​​, y​ ​-intercept: ​​​(​​0, _ ​ 1 ​​)​​​​ 2 2

Worked example video: Writing the gradient and y-intercept of a line from its equation

Helpful hints •

When a linear relationship is in the form y​ = mx + c​, the gradient is m ​ ​and the y​ ​-intercept is located at ​(0, c)​.

4 Write the gradient and y​ ​-intercept of the straight lines with the following equations. a ​y = 4x + 6​ b ​y = 9 − 2x​ c ​y = _ ​ 3 ​x − 8​ 4 d ​y = − 2 + x​ e ​y = − x + 4​ f ​y = − 1.2x − 2.8​ 3x ​ + ​ _ 9 ​​ x ​​ g ​y = 6.5 + 5.4x​ h ​y = − ​ _ i ​y = 7 + ​ _ 8 2 2 5 ​​ x ​ − 2​ x ​= 2​ j ​y = _ ​ x ​ + ​ _ k ​y = − ​ _ l ​y − ​ _ 6 3 3 2 m ​4x + y = 8​ n ​3x + 2y = − 4​ o ​2x − y = 5​

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Worked example 1.1C Using the gradient and y​ ​-intercept to write the equation of a straight line Write the equation of the straight line with: a gradient ​4​and y​ ​-intercept ​− 3​ b gradient − ​ 6​and y​ ​-intercept ​1​ c gradient 1 ​ ​and y​ ​-intercept ​0.9​ Part

Think

Write

a

Substitute the values for the gradient, m ​ ​, and y​ ​-intercept, ​ c​, into y​ = mx + c​. ​m = 4​, c​ = − 3​

​y = 4x − 3​

b

​m = − 6​, c​ = 1​

​y = − 6x + 1​

c

​m = 1​, c​ = 0.9​

​y = x + 0.9​

Worked example video: Using the gradient and y-intercept to write the equation of a straight line

5 Write the equation of the straight line with: a gradient ​7​and y​ ​-intercept ​− 1​ c gradient _ ​​ 3 ​​ and y​ ​-intercept ​− 2​ 7 e gradient ​− 1​and y​ ​-intercept ​2.7​

b gradient ​− 2​and y​ ​-intercept ​5​ d gradient ​1​and y​ ​-intercept ​4​ f

gradient ​0.2​and y​ ​-intercept ​3​ h gradient _ ​​ 5 ​​ and y​ ​-intercept _ ​​ 7 ​​ 6 12 j gradient _ ​​ 7 ​​ and y​ ​-intercept ​24​.

5

g gradient ​− 1.4​and y​ ​-intercept ​− 0.3​ i

gradient ​− 0.56​and y​ ​-intercept ​− 1.24​

Worked example 1.1D Finding the gradient, y​ ​-intercept and equation from a linear graph For the straight lines shown, find the i gradient, ii y​ ​-intercept and iii equation of the line. y 4

y 4

3

3

2

2

1

1

a

−4 −3 −2 −1 0 −1

b

1

2

3

4 x

−4 −3 −2 −1 0 −1

−2

−2

−3

−3

−4

−4

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1

2

3

4 x

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Part

Think

Write

ai

The line is sloping up from left-to-right, so the gradient is positive.

Choose any two points on the line, say ​(− 1, − 3)​ and (​ 2, 3)​, and draw a right-angled triangle as shown, to find the gradient of the line.

The vertical rise is ​6​and the horizontal run is 3 ​ .​ Substitute the rise and run into the formula: rise ​​. ​gradient = ​ _ run

y 4 3 2 1

−4

Identify the y​ ​-intercept as the point where the graph crosses the y​ ​-axis.

a iii

In the equation y​ = mx + c​, m ​ ​is the gradient and ​ c​is the ​y​- coordinate of the y​ ​-intercept.

Substitute the values for m ​ ​and ​c​into y​ = mx + c​ to find the equation of the line.

bi

The line is sloping down from left-to-right, so the gradient is negative.

Choose any two convenient points on the line, say ​(0, 2)​and (​ 2, 1)​, and substitute the values x1 = 0, y1 = 2, x2 = 2 and y2 = 1 into the formula ​y2​ ​​ − ​y​ 1​​ ​gradient = ​ _ ​x​ ​​ − ​x​ ​​​​ .

2

1

Simplify to find the gradient

1

2

4 x

3

−2 −3

a ii

+6

−4 −3 −2 −1 0 −1

+3

rise ​ gradient​= ​ _ run ​​ ​​= _ ​ 6 ​​ ​​ ​ 3 ​= 2 ​y-intercept = (0, − 1)​ ​ = 2,​c​ = − 1​ m The equation of the line is ​y = 2x − 1​.

y 4 3 2

(0, 2) (2, 1)

1 −4 −3 −2 −1 0 −1

1

2

3

4 x

−2 −3 −4

​(​x1​ ​, ​y1​ ​) = (0, 2)​, ​(​x2​ ​, ​y2​ ​) = (2, 1)​ ​y2​ ​​ − ​y​ 1​​ gradient​= ​ _ ​x2​ ​​ − ​x​ 1 ​​​ 2 ​​ ​ ​​ ​= _ ​ 1 − ​​ 2−0 1 ​ ​= − ​ _ 2 b ii

Identify the y​ ​-intercept as the point where the graph crosses the y​ ​-axis.

​y-intercept = (0, 2)​

b iii

Substitute the values for m ​ ​and ​c​into y​ = mx + c​ to find the equation of the line.

1 ​,​ ​c = 2​ ​ = − ​ _ m 2 1 ​x + 2​.

The equation of the line is ​y = − ​ _ 2

Worked example video: Finding the gradient, y-intercept and equation from a linear graph 8

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Helpful hints •

The working is easier if you can find points whose coordinates are whole numbers. Often using the points where the graph intersects an axis makes calculations easier.

•

Recall that there are two formulas for the gradient. You can use either formula as they will always give the same answer. rise ​​ → ​gradient = ​ _ run ​_ y2​ ​​ − ​y​ 1​​ → ​gradient = ​ ​x​ ​​ − ​x​ ​​​​ 2 1

6 For the straight lines shown, find the: i

ii ​y​-intercept

gradient

a

b

y 4

3

2

2

1

1 1

2

3

−4 −3 −2 −1 0 −1

4 x

−2

−2

−3

−3

−4

−4

y 4

d

5

2

4

1

3 1

2

3

−3

−4 −3 −2 −1 0 −1

−4

−2

f

3

4

2

3

1

−4 −3 −2 −1 0 −1

−4 −3 −2 −1 0 −1

−2

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1

2

3

4 x

4 x

1

2

3

4 x

1

2

3

4 x

y 4

5

1

3

1

−2

2

2

2

4 x

y 6

1

y 6

3

−4 −3 −2 −1 0 −1

e

y 4

3

−4 −3 −2 −1 0 −1

c

iii equation of the line.

−2 −3 −4

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Worked example 1.1E Sketching a linear graph Sketch the following straight lines using the method described in brackets. a ​y = 3x − 2​(Identify rise and run from the gradient) b ​y = − 2x + 1​(Use substitution to identify a second point on the graph) Part

Think

Write

a

Identify the values of ​m​and ​c​from the equation as the gradient and the y​ ​-intercept.

Plot the ​y​-intercept on the Cartesian plane.

Because the gradient is ​3 = _ ​ 3 ​​, move up ​3​ units 1 then right 1 ​ ​unit. This gives a second point ​(1, 1)​ on the line.

Rule a line through these two points.

​ radient = 3​ g ​y​-intercept: ​(0, − 2)​ y 2

(1, 1)

1 0 −1

1

2

3

4

x

−2 (0, −2) −3

b

Identify the value c​ ​from the equation as the ​y​-intercept.

Plot the ​y​-intercept on the Cartesian plane.

Substitute another value of ​x​into the equation of the line. Choose a value of ​x​that is easy to work with. Plot the resulting point on the Cartesian plane.

Rule a line through these two points.

y​ ​-intercept: ​(0, 1)​ ​x = 1​: y​ = − 2 × 1 + 1 = − 1​ ​(1, − 1)​lies on the line. y 2 1 0 −1 −2

(0, 1) 1

2 3 4 x (1, −1)

−3

Worked example video: Sketching a linear graph

Helpful hints •

When identifying rise and run from the gradient, you can write the gradient as any equivalent fraction.

In Worked example 1.1E part a, you could write ​3 = _ ​ 6 ​​and then move six units up and two units 2 across to find a second point. This will always result in the same line being sketched.

•

When using substitution to find a second point, you can substitute any other value of x​ ​. Often it is simplest to substitute x​ = 1​, but you can substitute any non-zero value to sketch the graph.

7 Sketch the lines in question 4 a–e using each of the following methods. i

Identify rise and run from the gradient.

ii Use substitution to identify a second point on the graph.

8 Compare each pair of graphs sketched in question 7. What do you notice?

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Problem solving and reasoning 9 More than one graph can be sketched on the same Cartesian plane. a Sketch, or use a graphing application to draw, the graphs of these linear relationships on the same set of axes. i

ii ​y = 2x + 3​

​y = 2x​

iii ​y = 2x − 2​

iv ​y = 2x + 1​

b What do you notice about the graphs in part a? c What is the effect of the constant term? d Sketch, or use a graphing application to draw, the graphs of these linear relationships on the same set of axes. i ​y = x + 1​ ii ​y = 2x + 1​ iii ​y = 3x + 1​ iv ​y = _ ​ 1 ​x + 1​ 2 e What do you notice about the graphs in part d? f

What is the effect of changing the coefficient of x​ ​?

10 Use a graphing application to draw the graphs with the linear relationships in questions 1 to 3. Compare the accuracy of your hand-drawn graphs with those obtained using technology.

11 Use any method to sketch the graphs of the following linear relationships. y a ​y = 4 − 3x​ b ​y = _ ​ 2 ​x − 3​ c ​x − y = 2​ d _ ​​ x ​ + ​ _ ​= 1​ 3 2 4 12 The equation of a horizontal line passing through ​(0, c)​can be written in the form y​ = c​. a Determine the equations of the following horizontal lines. y

i

y

ii

y 4

iii

2 x

x

x

−3

b Sketch the following horizontal lines on the Cartesian plane. i

ii ​y = 3​

​y = 1​

iii ​y = − 2​

iv ​y = 0​

c Are horizontal lines written in the form y​ = mx + c​? Explain. d Using a gradient formula, show that the gradient of a horizontal line is zero.

13 The equation of a vertical line passing through ​(a, 0)​can be written in the form x​ = a​. a Determine the equations of the following vertical lines. i

2

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y

ii

y

x

−2

y

iii

x

−4

x

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b Sketch the following vertical lines on the Cartesian plane. i

​x = 1​

ii ​x = 3​

iii ​x = − 1​

iv ​x = 0​

c Using a gradient formula, show that the gradient of a vertical line is undefined.

14 The x​ ​- and y​ ​-axes can be thought of as a horizontal line and a vertical line. Write the equation which represents the: i

​x​-axis

ii ​y​-axis.

Challenge 15 An equation such as 5 ​ x + 4y = 60​is not written in gradient-intercept form. a Explain how this equation can be rearranged into gradient-intercept form to be graphed. b Explain how the same equation can be graphed by finding the x​ ​-intercept and y​ ​-intercept, instead of the gradient. c Which method is easier when graphing an equation of this form?

Related resources 1.1 Quick quiz

Lesson 1.2

Linear models Learning intentions By the end of this lesson, you will be able to ... → model practical situations using linear relationships.

Mathematical models

12

•

This lesson reviews content in Module 2 of Oxford Maths Standard Year 11.

•

A mathematical model is a representation of the structure, workings or relationships within a situation. Most models simplify the situation in some way, so care should be taken to consider any limitations. Extending beyond any limitations means any tests or predictions become less reliable.

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Linear models •

A linear model uses a linear relationship to model a practical situation.

•

Linear models can be represented by a straight-line graph on the Cartesian plane, a rule or relationship describing the relationship and a list of ordered pairs, often represented as a table of values.

•

Most linear models can be written in the form ​y = mx + c​, where the gradient (​m​) and y​ ​-intercept (​c​) of the corresponding linear graph have practical meanings in each case. → The constant c​ ​is a fixed amount in the linear relationship. → The gradient m ​ ​describes the rate of change between the two variables in the relationship. → Usually, the pronumerals x​ ​and y​ ​are replaced by ones which better match the situation being modelled. For example, C ​ ​may represent the “cost” and n ​ ​the “number of items”.

Modelling cost and revenue •

Businesses can use mathematical models to compare revenue and cost. We will consider examples where these models are linear.

•

The (production) cost is the cost (or expense) of acquiring, producing or maintaining an item for sale.

This can involve both a fixed cost and a variable cost that depends on the number of items produced. → Cost can be modelled by a linear relationship which considers the cost to produce each item and any fixed cost(s).

•

Revenue is the total income generated by a business through the sale of items or services. → Revenue can be modelled by a linear relationship which considers the number of items sold and the income generated by those sales.

Worked example 1.2A Modelling cost The Magnificent Muffin Shop has a fixed cost of ​$100​per day and a variable production cost of ​$1.50​per muffin. The daily cost can be modelled using the equation C ​ = 1.5n + 100​. a Explain each term in the equation. b Draw the graph of C ​ = 1.5n + 100​for 0 ​ ≤ n ≤ 200​using the vertical intercept and substitution to identify a second point on the graph. c Use the graph to find the number of muffins produced for $​ 220​.

Part

Think

Write

a

There are three terms in the equation.

The first term is the subject, ​C​, which represents the total production cost.

The variable cost relates to the gradient, which is the coefficient of n ​ ​.

The fixed cost is the intercept on the vertical axis, which is the constant term in the equation.

Terms: ​C​, ​1.5n​, ​100​ ​C​represents the total production cost in dollars.

The coefficient 1 ​ .5​gives the cost per muffin as ​ $1.50.​There are n ​ ​muffins, so ​1.5n​represents the variable cost in dollars of producing n ​ ​ muffins. ​100​is the fixed daily cost in dollars.

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Part

Think

Write

b

Identify the value c​ = 100​from the equation as the C ​ ​-intercept.

Plot the ​C​-intercept on the Cartesian plane.

Substitute another value of ​n​into the equation of the line. Choose a value of ​ n​that is easy to work with. Here the maximum value of n ​ ​that needs to be included in the graph is chosen. Plot the resulting point on the Cartesian plane.

Rule a line through these two points, making sure the graph shows all values of n ​ ​ from ​0​ to ​200​

​ ​-intercept: ​(0, 100)​ C ​n = 200​: ​C = 1.5 × 200 + 100 = 400​ ​(200, 400)​lies on the line.

Cost 400 360

C = 1.5n + 100

Cost in $ (C )

320 280 240 200 160 120 80 40 0

c

Draw a line across from the vertical axis at ​220​until it meets the graph. Draw down to the horizontal axis and read the value of ​n​.

20 40 60 80 100 120 140 160 180 200 Number of muffins (n)

​80​muffins are produced for $​ 220​.

Worked example video: Modelling cost

Understanding, fluency and communicating Answers: pXXX

Helpful hints •

Graphs of linear models can be created using any method that produces a linear graph. These include: → plotting points from a table of values → using the vertical intercept and identifying rise and run from the gradient → using the vertical intercept and substitution to identify a second point on the graph → using a digital tool or graphing application.

•

14

In Worked example 1.2A part c, the graph was used to find the number of muffins produced for ​$220.​This is equivalent to substituting C ​ = 220​into the equation and solving algebraically for n ​ ​.

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1 The Supertight Tie Company has a fixed cost of $​ 1200​per day and a variable production cost of $​ 3​ per tie.

The daily cost can be modelled using the equation C ​ = 3n + 1200​. a Explain each term in the equation C ​ = 3n + 1200​. b Draw the graph of C ​ = 3n + 1200​for ​0 ≤ n ≤ 250​. c Use your graph from part b to find the number of ties produced when the daily cost is $​ 1500​. d Verify your answer to part c by substituting $​ 1500​into the equation and solving for n ​ ​.

2 The Munchy Sandwich Shop has a monthly fixed expense of ​$2000​and the cost of making each sandwich is $​ 1.20​. a Write an equation modelling this information. Use C ​ ​for monthly cost in dollars and n ​ ​for the number of sandwiches made. b Draw the straight-line graph for the linear model from part a for ​0 ≤ n ≤ 2500​. c What is the vertical intercept and what does it represent? d What is the value of the gradient and what does it represent? e Use the graph to find the number of sandwiches made in a month if the cost is $​ 4400​.

Verify your answer to part e by substituting $​ 4400​into the equation and solving for n ​ ​.

f

3 The Tight Squeeze Fruit Juice Company has a weekly fixed cost of $​ 300​. The cost to produce a cup of fruit juice is ​$1.50​. a Write a linear equation modelling this information. Use C ​ ​for the weekly cost in dollars and n ​ ​ for the number of cups of fruit juice. b Draw the straight-line graph for the linear model from part a for ​0 ≤ n ≤ 1000.​ c What is the vertical intercept and what does it represent? d What is the value of the gradient and what does it represent? e Use the graph to find the number of cups of juice made in a week if the cost is $​ 1050​.

Verify your answer to part e by substituting $​ 1050​into the equation and solving for n ​ ​.

f

Worked example 1.2B Modelling revenue The Magnificent Muffin Shop sells its muffins for $​ 4​each. The sales revenue can be modelled using the equation ​R = 4n​. a Explain each term in the equation ​R = 4n​. b Graph the straight line with equation ​R = 4n​for 0 ​ ≤ n ≤ 100​by using the gradient to identify rise and run. c Use the graph to find the number of muffins that are sold if the revenue is $​ 240​.

Part

Think

Write

a

There are two terms in the equation.

The first term is the subject, ​R​, which represents the revenue.

The variable cost relates to the gradient, which is the coefficient of n ​ ​.

Terms: ​R​, ​4n​ ​R​represents the revenue in dollars.

The coefficient 4 ​ ​gives the price per muffin as ​$4.​ There are n ​ ​muffins, so ​4n​represents the revenue (total income) in dollars from selling the muffins.

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Think

Write

b

Identify the values of ​m​and ​c​from the ​ radient = 4​ g equation as the gradient and the ​R-​ intercept: ​(0, 0)​ ​R​-intercept.

Plot the ​R​-intercept on the Cartesian 400 plane.

400 _ 360 Because the gradient is ​4 = ​ ​​, move up ​ 100 320 400​units then right 1 ​ 00​units. This gives +400 a second point ​(100, 400)​on the line.

280 Rule a line through these two points and 240 extend until the point where n ​ = 100​.

Revenue in $ (R)

Part

Revenue +100 R = 4n

200 160 120 80 40

0

c

Draw a line across from the vertical axis at ​240​to the graph and then down to the ​ n​-axis to read the value of n ​ ​.

10 20 30 40 50 60 70 80 90 100 Number of muffins (n)

​ 0​muffins must be sold for the revenue to be ​ 6 $240​.

Worked example video: Modelling revenue

4 The Supertight Tie Company sells ties for $​ 15​each. The sales revenue may be modelled using the equation ​ R = 15n​. a Explain each term in the equation R ​ = 15n​. b Graph the straight-line graph of the equation R ​ = 15n​for ​0 ≤ n ≤ 300​by using the gradient to identify rise and run.

Hint: Use ​15 = _ ​ 4500 ​​.

300 c Use the graph from part b to find the number of ties sold when the revenue is $​ 1800​. d Verify your answer to part c by substituting $​ 1800​into the equation and solving for n ​ ​.

5 The Munchy Sandwich Shop sells sandwiches for $​ 3.50​ each. a Write an equation modelling this information. Use R ​ ​for revenue in dollars and n ​ ​for the number of sandwiches sold. b Draw a straight-line graph for the linear model from part a for ​0 ≤ n ≤ 3000​. c What is the gradient of this straight line and what does it represent? d What is the vertical intercept and what does it represent? e Use the graph from part b to find the number of sandwiches sold when the revenue is $​ 2800​. f

16

Verify your answer to part e by substituting $​ 2800​into the equation and solving for n ​ ​.

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6 The Tight Squeeze Fruit Juice Company sells juice for $​ 4.00​per cup. a Write an equation modelling this information. Use R ​ ​for revenue in dollars and n ​ ​for the number of cups of juice sold. b Draw the straight-line graph for the linear model from part a for ​0 ≤ n ≤ 1000​. c What is the gradient of this straight line and what does it represent? d What is the vertical intercept and what does it represent? e Use the graph from part b to find the number of cups of juice sold when the revenue is $​ 3200​. f

Verify your answer to part e by substituting $​ 3200​into the equation and solving for n ​ ​.

Problem solving and reasoning 7 The distance, d ​ ​kilometres, travelled by a train in a time of t​ ​hours is ​d = 80t​. a Use any method to draw the graph of d ​ = 80t​for ​0 ≤ t ≤ 5​. b Identify the d ​ ​-intercept. What does this represent in the linear model? c What is the gradient of this graph and what does it represent in the linear model? d Use the graph to find how far away the train is after ​3__ ​ 1 ​​ hours.

2 e Use the graph to find when the train is ​200 km​away. f

Describe one limitation of this linear model.

8 Tina travels ​200 km​to her friend’s home at an average speed of ​80 km/h​. The distance Tina is from her friend’s home after travelling for t​ ​hours can be modelled using the linear equation ​D = 200 − 80t​. a Explain each term in the equation D ​ = 200 − 80t​. b Use any method to draw the graph of D ​ = 200 − 80t​for ​0 ≤ t ≤ 4​. c Use the graph to find how far Tina is from her friend’s home after 1 ​ .5​hours. d Use the graph to find how long it takes for Tina to reach her friend’s home. e What is the value of D ​ ​ when t​ = 4​? f

What is the maximum value of t​ ​that this model is suitable for?

9 The cost of hiring a taxi is $​ 6​flagfall and ​$3.60​per kilometre travelled. a Write a linear relationship modelling this information to relate cost and distance travelled. b Use any method to draw the graph of this linear model for distances between 0 ​ km​and ​20 km​. c Use the graph to find the cost of travelling ​10 km​. d Use the graph to estimate how far you could travel for $​ 30​. e Is this model suitable for distances greater than ​20 km​? Discuss.

10 On a building site, columns with a fixed cross-sectional area of 0 ​ .78 ​m​​ 2​​are to be constructed by pouring concrete into moulds of different heights. a Write a linear relationship modelling this information to relate volume of concrete (in cubic metres) and height of a column (in metres). b Use any method to draw the graph of this linear model for height values up to 7 ​ m​. c Use the graph to estimate the volume of concrete needed to construct a column that is 4 ​ m​high. d Use the graph to estimate the height of a column that can be constructed using 2 ​ .4 ​m​​ 3​​ of concrete. e Is this model suitable for heights over ​7 m​? Discuss.

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Challenge 11 The conversion from temperature in degrees Celsius ​(C)​to temperature in degrees Fahrenheit ​(F )​is given by the equation F ​ =_ ​ 9 ​ C + 32​.

5 a Use a graphing application or digital tool to draw the graph of this relationship for values of C ​ ​ between ​0​ and ​100​. b A “rule of thumb” conversion is to double the temperature in degrees Celsius and add 3 ​ 0​. Add the linear graph for this model to the same graph as in part a for values of C ​ ​ between ​0​ and ​100​. c Compare the two graphs. When is the “rule of thumb” more accurate?

12 An inflatable pool containing ​1000 L​of water starts leaking at a rate of ​5 L​per minute. a Write a linear model to represent this situation, using V ​ ​for the volume (in litres) of water in the tank after ​ t​ minutes. b How long will it take for half the water to leak out of the tank? c The leak is fixed when there is ​500 L​remaining and a hose is turned on to refill the pool. The hose is running water at a rate of ​10 L​per minute. Find how many minutes it will take for the pool to reach ​ 1000 L​. d Find the equation of the line that represents the volume of water in the pool when it is being refilled, where ​ t​still represents the number of minutes after the leaking began. e Using a graphing application, graph both equations and explain why only parts of these lines make sense in the situation.

Related resources 1.2 Quick quiz

Investigation: Water tank

Worksheet: Further linear models

Investigation: The freezing icy pole

Lesson 1.3

Identifying solutions to simultaneous equations Learning intentions By the end of this lesson, you will be able to ... → identify the point of intersection between two linear graphs.

Points of intersection •

When two distinct linear graphs are plotted on the same Cartesian plane, either: → the two lines are parallel and therefore they never cross, or → the two lines cross each other at a single point, called the point of intersection.

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•

This graph shows three lines labelled A ​ ​, B ​ ​and C ​ ​

y

→ ​A​and C ​ ​have one point of intersection

A

→ ​B​and C ​ ​have one point of intersection → ​A​and B ​ ​are parallel, so they have no point of intersection.

•

B C

The point of intersection between two linear graphs can be found by identifying the unique point that lies on both lines. x

Simultaneous solutions •

The coordinates of any point on a line give a solution to the corresponding linear equation. → For example, the point ​(− 3, 1)​is on the line y​ = 2x + 7​, so the values x​ = − 3​and y​ = 1​form a solution to the linear equation. To confirm this, substitute x​ = − 3​and y​ = 1​into the equation: ​ 1 = 2 × (− 3) + 7​.

•

If the same point lies on two different lines, we obtain a solution to both linear equations simultaneously.

•

Hence to “solve a pair of simultaneous linear equations” means to find values for the variables that make both equations true. → This is the same as finding the point of intersection between the two lines.

Worked example 1.3A Finding points of intersection from tables of values Consider the following table of values. It has one row for each of the linear equations ​ y = 2x + 7 ​ and ​ y = − 4x − 5​. ​​ x ​y = 2x + 7​

​− 3​ ​1​

​− 2​ ​3​

​− 1​ ​5​

​0​ ​ ​ 7

​1​ ​ ​ 9

​ ​ 2 ​11​

​ ​ 3 ​13​

​y = −4x − 5​

​7​

​3​

​− 1​

​− 5​

​− 9​

​− 13​

​− 17​

Use this table of values to find the coordinates of the point of intersection between the lines ​ y = 2x + 7 ​ and ​ y = − 4x − 5​.

Think

Write

In the column where ​x = − 2​, both linear equations have y​ ​values of ​y = 3​.

This means that both lines pass through the point ​(− 2, 3)​.

x​ = − 2​and y​ = 3​is a solution to both equations, so the point ​(− 2, 3)​lies on both lines.

The point of intersection is ​(− 2, 3)​.

Worked example video: Finding points of intersection from tables of values

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Understanding, fluency and communicating Answers: pXXX

1 For each pair of linear equations in the tables below, find the coordinates of the point of intersection between the two lines. a

b

c

d

​​ x ​y = 2x​

​− 3​ ​− 6​

​− 2​ ​− 4​

​− 1​ ​− 2​

​0​ ​ ​ 0

​1​ ​ ​ 2

​2​ ​ ​ 4

​3​ ​ ​ 6

​y = 6 − x​

​9​

​8​

​7​

​6​

​5​

​4​

​3​

​​ x ​y = 2x − 1​

​ ​ 0 ​− 1​

​1​ ​ ​ 1

​2​ ​ ​ 3

​3​ ​ ​ 5

​4​ ​ ​ 7

​y = 8 − x​

​8​

​7​

​6​

​5​

​4​

​​ x ​y = 3x − 5​

​5​ ​10​

​6​ ​13​

​7​ ​16​

​8​ ​19​

​9​ ​22​

​y = 23 − x​

​18​

​17​

​16​

​15​

​14​

​x​

​− 4​

​− 2​

​0​

​2​

​4​

​y = _ ​ 1 ​x + 5​

​3​

​4​

​5​

​6​

​7​

​10​

​4​

​− 2​

​− 8​

​− 14​

2 ​y = −3x − 2​

Worked example 1.3B Checking the solution to simultaneous linear equations Check whether each pair of x​ ​and ​y​values is the solution to the pair of simultaneous linear equations. a ​x = 2​, y​ = 4​ b ​x = 5​, y​ = − 3​ y​= 2x − 11 ​1​ y​= 3x − 2 ​1​ ​​ ​ ​ ​ ​​​ ​ ​ ​ ​​​ ​​ y​= − 2x + 7 ​2​ y​= − 2x + 8 ​2​

20

Part

Think

Write

a

Substitute ​x = 2​and y​ = 4​into the left-hand side (LHS) and right-hand side (RHS) of equation ​​1​​ .

Since ​LHS = RHS​, x​ = 2​and y​ = 4​is a solution to equation ​​1​​ .

Substitute ​x = 2​and y​ = 4​into the LHS and RHS of equation 2​​ ​​ .

Since ​LHS = RHS​, x​ = 2​and y​ = 4​is a solution to equation ​​2​​ .

​​1​ y = 3x − 2​ When x​ = 2​and y​ = 4​, LHS​= 4 ​ ​ RHS​ ​× 2 − 2​ = 3 ​​ ​= 4 So L ​ HS = RHS​.

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​​2​ y = –2x + 8​ When x​ = 2​and y​ = 4​, LHS​= 4 ​ ​ RHS​ ​2 × 2 + 8​ = − ​​ ​= 4 So L ​ HS = RHS​.

As x​ = 2​, y​ = 4​satisfies both equations, this is the solution to the pair of simultaneous equations.

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Part

Think

Write

b

Substitute ​x = 5​and y​ = − 3​into the LHS and RHS of equation 1​​ ​​ .

Since ​LHS ≠ RHS​, x​ = 5​and y​ = − 3​is not a solution to equation ​​1​​.

Since ​x = 5​and y​ = − 3​is not a solution to one of the equations, it cannot be simultaneously a solution to both.

​​1​ y = 2x − 11​ When x​ = 5​and y​ = –3​, LHS​= − 3 ​​ RHS​ ​ 5 − 11​ = 2 × ​​ ​= − 1 So ​LHS ≠ RHS​.

As x​ = 5​, y​ = − 3​does not satisfy equation 1​​ ​​ , it is not a solution to the pair of simultaneous equations.

Worked example video: Checking the solution to simultaneous linear equations

Helpful hints •

Substitute the given ​x​- and y​ ​-values into each equation and check if they make both equations true statements.

•

If the given values are not a solution to one of the equations, it does not matter whether or not they are a solution to the other equation.

2 Check if each given x​ ​-value and ​y​-value is the solution to the pair of simultaneous linear equations. a ​x = 1​, y​ = 5​ y​= x + 4 ​​ ​ ​ ​​ y​= − 2x + 7

b ​x = − 2​, y​ = 1​ y​= x + 3 ​​ ​ ​ ​​ y​= − 3x − 5

c ​x = 5​, y​ = − 2​ y​= x − 9 ​​ ​ ​ ​​ y​= − 2x + 7

d ​x = 3​, y​ = − 7​ y​= − 2x − 1 ​​ ​ ​​​ y​= − x − 4

e ​x = 4​, y​ = 3​ y​= 2x − 5 ​​ ​ ​​​ y​= − x + 1

f

​x = − 6​, y​ = 0​ y​= x − 6 ​​ ​ ​ ​​ y​= − x − 6

Worked example 1.3C Identifying the solution to simultaneous linear equations from graphs y Consider the following pair of simultaneous linear equations. y​= x − 7 ​ ​ ​ ​ ​​ y​= − 3x + 9 Use the diagram of the corresponding graphs drawn on the same Cartesian plane. a Write the coordinates of the point of intersection of the two lines. b Use the coordinates of this point to write the solution to the simultaneous linear equations.

10 8 6 4

y=x–7

2 −4 −2 0 −2 −4 −6

2

4

6

8 10 x

y = –3x + 9

−8

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Part

Think

Write

a

Identify the point where the two lines cross and write the x​ ​-value and the y​ ​-value of the point as coordinates.

Point of intersection at (​ 4, − 3)​.

b

Use the coordinates of the point to write the solution.

The solution to the simultaneous equations ​ y = x − 7 ​ and ​ y = − 3x + 9​ is x​ = 4​, y​ = − 3​.

Worked example video: Identifying the solution to simultaneous linear equations from graphs

Helpful hints •

Identify the point of intersection of the two lines and list the coordinates.

•

Write the solution as the x​ ​-value and the y​ ​-value that makes both equations true.

3 Consider each pair of simultaneous linear equations and their graphs. i

Write the coordinates of the point of intersection of the two lines.

ii Use the coordinates of this point to write the solution to the simultaneous linear equations. a y​​ ​​= 2x − 5​​​ y​= − x + 7

​= − 4x + 4 b y​​ ​ ​​​ y​= x − 6

y

y

8

y = 2x – 5

6 4

y=x–6

4 2

2 −2 0 −2

6

2

6

4

−4

8 10 12 14 x

2

4

8 10 12 x

6

y = –4x + 4

−8

c y​​ ​​= 2x + 8 ​​​ y​= x + 3

y = 2x + 8

−4 −6

y=x+7

−6

−4 −2 0 −2

+5 d y​​ ​​= x ​​ ​ y​= 3 y

y

8

6

6

5

4 2

−8 −6 −4 −2 0 −2 −4

4

y=x+3 2

4

6 x

y=3

3 y=x+5

2 1

−6 −5 −4 −3 −2 −1 0 −1

1

2

3 x

4 Use substitution to check your solutions to the simultaneous linear equations in question 3.

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Problem solving and reasoning

a How many cards does Suzy need to sell to recover all her costs?

Cost and revenue 240 210 Amount ($)

5 Suzy makes hand-painted cards. She invests $​ 60​in a set of paintbrushes and enough paint for 4 ​ 0​cards. The cost of each plain white card is $​ 2​and she sells the painted cards for ​$6​ each. The production cost and the revenue generated by the sale of the cards can be modelled with two linear equations and shown on a graph.

Revenue

180 150 120

Cost

90

b How much profit does Suzy make if she sells ​30​cards?

60

c What is the maximum profit Suzy can make?

30 0

5 10 15 20 25 30 35 40 Number of cards

6 Daniel is deciding which plan to use for his mobile phone data. He is comparing the ​10 GB​Plan and the ​ 100 GB​Plan. The ​10 GB​Plan costs $​ 20​but any data use beyond the included 1 ​ 0 GB​is charged at a rate of ​ 50​cents per gigabyte. The 1 ​ 00 GB​Plan costs ​$60​and any data use beyond the included ​100 GB​is charged at a rate of ​20​cents per gigabyte. This graph shows the costs of these plans per month.

Mobile data plan rates 140

10 GB Plan

120 Cost ($)

100

100 GB Plan

80 60 40 20 0

20 40 60 80 100 120 140 160 180 200 220 240 Data usage (GB)

a If Daniel expects to use ​40 GB​of mobile data per month, which plan should he use? Why? b If Daniel expects to use ​120 GB​of mobile data per month, which plan should he use? Why? c For how many gigabytes of data per month are the costs of the two plans equal? What is this cost? d Explain why these two graphs are not simultaneous linear equations.

Challenge y

7 Consider the straight lines ​A​, B ​ ​and ​C​shown on this graph. a Write the coordinates of the point of intersection of lines B ​ ​and C ​ ​ and hence give the solution to the simultaneous equations for these lines. b Use the graph to solve the equations for lines A ​ ​and C ​ ​ simultaneously. c Can the equations for lines A ​ ​and B ​ ​be solved simultaneously? Explain why or why not.

A

4

B

3 2 1 −3 −2 −1 0 −1

1

2

3

C 4 x

−2 −3

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25 Cost ($)

8 Maria is comparing the cost of using two taxi services.

One company charges ​$2.40​per kilometre, while a second one charges a flagfall of ​$3​, and $​ 2.20​ per kilometre. She creates a linear model for both services and creates the graph shown. From this, Maria concludes that one of the services is always better than the other.

Explain why Maria is incorrect.

20 15 10 5 0

2 4 6 8 Distance travelled (km)

10

Related resources 1.3 Quick quiz

Lesson 1.4

Solving simultaneous equations graphically Learning intentions By the end of this lesson, you will be able to ... → solve pairs of simultaneous linear equations graphically.

Solving simultaneous linear equations graphically •

When asked to solve a pair of simultaneous linear equations, one approach is to find the point of intersection of the two lines on a Cartesian plane. To do this, we must sketch both graphs on the same set of axes.

•

Sketching two graphs on the same set of axes is no different to sketching just one graph, except that we need to choose the range of values on the x​ ​- and y​ ​-axes to accommodate both graphs.

•

It is important to sketch both graphs accurately to ensure the point of intersection in your sketch is correct. → A simultaneous solution found from a graph should always be substituted back into both equations to confirm that it is indeed a solution to both equations.

•

Graphing applications and digital tools such as Desmos can be used to find points of intersection graphically. → Note: The coordinates of a point of intersection found using digital tools are typically given as ­decimal approximations.

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Worked example 1.4A Sketching graphs to solve simultaneous linear equations Consider the linear equations 1 ​x​. ​ y = x − 2 ​ and ​ y = 4 − ​ _ 2 a Sketch both graphs on the same Cartesian plane. b Identify the point of intersection of the two lines. c Verify that the point of intersection gives a solution to both equations. d Write the solution to the pair of simultaneous linear equations 1 ​x​. ​ y = x − 2 ​ and ​ y = 4 − ​ _ 2 Part

Think

a

The line ​y = x − 2​has a y​ ​-intercept at ​(0, − 2)​.

Find a second point on this line by letting ​x = 2​: When x​ = 2​, y​ = 2 − 2 = 0​, so the point ​(2, 0)​is also on the line.

Plot the points ​(0, − 2)​and (​ 2, 0)​, then use a ruler to connect them with a straight line, making sure to extend the line in both directions.

1 ​x​has a y​ ​-intercept at ​(0, 4)​.

The line ​y = 4 − ​ _ 2 1 ​​to find a second point on this line: Use the gradient ​− ​ _ 2 The rise is − ​ 1​and the run is 2 ​ ​, so move down one unit and right two units from the point (​ 0, 4)​to the point ​(2, 3)​.

Use a ruler to connect the two points with a straight line, making sure to extend the line in both directions.

Write y 5 4 –1 3

(0, 4)

2 1

(2, 3) +2

(2, 0)

−1 0 −1

1

−2

(0, –2)

2

3

4

5

6 x

b

Identify the point where the two lines cross and write the coordinates of this point.

The point of intersection is ​(4, 2)​.

c

Substitute ​x = 4​and y​ = 2​into both sides of each equation and check that the left-hand side (LHS) and right-hand side (RHS) are equal.

y​ = x − 2​: LHS = 2 RHS​​= 4 ​​ − 2​ ​​ ​= 2 1 ​x​: ​y = 4 − ​ _ 2 1 ​ × 4 LHS = 2 RHS​= 4 − ​ _ 2​ ​​ ​ ​​ ​ ​= 4 − 2 ​= 2

d

As x​ = 4​and y​ = 2​is a solution to each of the two equations, it is the solution for the pair of simultaneous equations.

The solution is x​ = 4​, y​ = 2​.

Worked example video: Sketching graphs to solve simultaneous linear equations

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Understanding, fluency and communicating Answers: pXXX

Helpful hints •

A linear graph can be plotted by connecting the ​y​-intercept and any other point satisfying the equation with a straight line. Review Worked example 1.1E, if necessary.

•

Always confirm the point of intersection in your sketch is accurate by substituting the coordinates back into both equations and checking that both sides of each equation are equal.

1 For each of the following pairs of linear equations, sketch both graphs on the same Cartesian plane.

3 _ ​1​ a y​= x + 2 b y​= 2x + 2 ​1​ c y​= 2x − 2 ​1​ d y​= ​ 2 ​ x − 2 ​1​ ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​​​ ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​​​ 1 ​ x ​2​ y​= 2x − 1 ​2​ y​= 3x + 1 ​2​ y​= 4 − x ​2​ y​= 2 − ​ _ 2 2 Identify the point of intersection for each pair of lines in question 1 and hence state the solution to each pair of simultaneous equations.

3 Solve each of the following pairs of simultaneous linear equations by finding the point of intersection of their graphs. ​1​ a y​= 6 − x b y​= x − 3 ​1​ c y​= x − 1 ​1​ ​ ​ ​ ​​​ ​ ​ ​​​ ​ ​ ​ ​​​ ​​ ​​ ​​ y​= 10 − 2x ​2​ y​= 1 − x ​2​ y​= 2x ​2​ ​1​ d y​= 4 − x ​​ ​ ​ ​ ​​​ y​= 5 − 2x ​2​

1 ​ x ​1​ e y​= 3 − ​ _ 2 ​ ​ ​​ ​ ​​​ y​= 2x − 2 ​2​

y​= x + 7 ​1​ ​​ ​ ​ ​ 2 ​ x + 2 ​2​​​​ y​= − ​ _ 3

f

Worked example 1.4B Solving simultaneous linear equations graphically Use a graphical method to solve the pair of simultaneous linear equations ​ 3x + y = − 4 ​and ​ x = 5y + 2​.

Think

Write y

Rearrange both equations to be in the form y​ = mx + c​: First equation: 3x + y​= − 4 ​ ​ ​ ​ ​​ y​= − 3x − 4 Second equation: x​= 5y + 2 x − 2 ​ ​ ​​= 5y​ ​ ​​ 2 ​ y​= _ ​ 1 ​ x − ​ _ 5 5 Sketch both graphs on the same Cartesian plane.

Locate the coordinates of the point where the lines cross.

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1 −3 –2 –1 0 −1 (–1.125, –0.625)

1 2 y = —x – — 5 5 1

2

3 x

−2 −3 −4

y = –3x – 4

The point of intersection is at ​(− 1.125, − 0.625)​.

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Think

Write

Substitute ​x = − 1.125​and y​ = − 0.625​into both of the original equations to check ​LHS = RHS​for both equations.

First equation: LHS​= 3(− 1.125) + (− 0.625) ​

​​ ​ = − 3.375 − 0.625​ ​= − 4

​ ​

​ RHS = − 4​ So L ​ HS = RHS​.

Second equation: ​ LHS = − 1.125​ RHS​= 5(− 0.625) + 2 ​

= − 3.125 ​​ + 2​ ​ ​= − 1.125

​ ​

So L ​ HS = RHS​.

The solution to this pair of simultaneous linear equations is ​x = − 1.125​, y​ = − 0.625​ Worked example video: Solving simultaneous linear equations graphically

Helpful hints •

Find the point of intersection of the two lines and use the coordinates to write the solution to the simultaneous equations.

•

Many graphing applications allow you to enter the equation of a graph in any form. → For example, you can enter the equation ​2y + 4x = 3​directly into Desmos, without first rearranging to make y​ ​the subject (which would be y​ = _ ​ 3 ​ − 2x​).

2

4 Consider the linear equations ​y = 2x + 3​and ​y = 8 − 3x​. a Use a graphing application to plot the graphs of both linear equations on the same set of axes. b Find the point of intersection of these lines. c Verify that the point of intersection gives a solution to both equations. d Hence write the solution to the simultaneous linear equations ​

y = 2x + 3 ​ and ​ y = 8 − 3x​.

5 Use a graphing application to find the solution to each of the following pairs of simultaneous linear equations. a y​= x − 3 ​1​ ​​ ​ ​ ​​​ y​= 1 − x ​2​

b y​= x − 1 ​1​ ​​ ​ ​ ​ ​​​ y​= 2x ​2​

c 4x + 3y​= 10 ​1​ ​​ ​ ​ ​ ​​​ x − 2y​= − 3 ​2​

6 Solve the following pairs of simultaneous linear equations using their graphs.

5 _ 1 _ c y​= ​ 4 ​ − ​ 2 ​ x ​1​ ​1​ a y​= 5x − 8 b 4x​+ 3y = 9.5 ​1​ ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​ ​​​ 2x​+ 3y = 10 ​2​ y​= 2.5 ​2​ x​+ 3y = _ ​ 7 ​ ​2​ 2 1 ​ ​1​ _ x ​ − 2y = − ​ f e y​= 27.5 − 5x ​1​ g 2x​+ 5y = 17 ​1​ 2 ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​​​ 5 3x​+ 4y = 25 ​2​ x​+ 3y = 10 ​2​ x​+ 4y = _ ​ ​ ​2​ 2 Oxford University Press

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d

3x + y​= 5 ​1​ ​​ ​ ​ ​ ​​​ 2x − 3y​= 18 ​2​

​1​ d x​+ y = 5 ​​ ​ ​ ​ ​​​ y​= 2x − 2 ​2​ h 2x​− y = − 3.5 ​1​ ​​ ​ ​ ​ ​​​ 4x​+ 4y = 8 ​2​

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Problem solving and reasoning 7 The cost (​$C​) of hiring a car from company A is ​$50​per day; that is, C ​ = 50d​, where ​d​is the number of days for which the car is hired. Company B charges a flat fee of $​ 60​plus $​ 40​per day; that is, C ​ = 60 + 40d​. a Use any method to plot both equations on the same set of axes. b Use your graph to solve this pair of simultaneous equations. c Explain the meaning of the solution in the context of the two car-hire companies. d For what lengths of car hire is it cheaper to hire from company B? 8 Joanna is a salesperson who earns her income from commissions. Her manager offers her a chance to change the way her income is calculated. Currently, she receives a straight commission of ​10%​on her sales: ​I = 0.1S​, where ​I​is income and S ​ ​is sales in dollars. The proposed alternative would be to receive a retainer of $​ 300​plus 6 ​ %​on her sales: ​I = 300 + 0.06S​. a Use any method to plot both equations on the same set of axes. b Use your graph to solve this pair of simultaneous equations. c Explain the meaning of the solution in the context of the two payment options. d How much would Joanna need to make in sales for the proposed alternative to be a better option for her?

9 Consider two cars that are identical apart from the type of fuel they use. One car has a petrol engine and uses 1 ​ 0.0 L/100 km​. If the cost of petrol is ​$1.80/L​then the cost (​$C​) of fuel to drive the car is given by ​C = 0.18d​, where ​d​is the distance travelled. The other car has a diesel engine and costs ​$2000​more than the car with a petrol engine, but it only uses ​6.0 L/100 km​. If the cost of diesel fuel is $​ 2.00/L​, the relative cost of choosing the diesel model of the car is given by C ​ = 2000 + 0.12d​. a Find the solution to this pair of simultaneous linear equations. b How far would you need to travel for the cost of driving each model to be the same?

10 The perimeter of a printed rectangular photo is 5 ​ 0 cm​. a Use l​​for length and b​ ​for breadth to write a linear equation for the perimeter of the photo. b Use l​​and w ​ ​to write a linear equation relating the length and breadth of the photo, given that the length is ​ 5 cm​more than the breadth. c Solve the pair of simultaneous equations using their graphs and hence state the dimensions of the photo.

11 At the cinema, a large popcorn costs ​$2​more than an ice-cream. Liam buys four boxes of popcorn and six ice-creams for $​ 58​. a Write two linear equations to model this information. b Solve the pair of simultaneous equations using their graphs to find the price of each item.

12 Tom rides his bicycle from Town A to Town B at an average speed of 2 ​ 0 km/h​. Julia walks from Town B to Town A along the same road at an average speed of 5 ​ km/h​. The distance along the road between the two towns is ​15 km​and they each start off at the same time. a Write two linear equations to model the two journeys. In each case, relate the distance in kilometres from Town A to the time travelled in hours. b Solve the pair of equations simultaneously using a graphical method to find the time at which Tom and Julia meet along the road between the two towns. c How far has each person travelled before they meet?

13 Solve the following pairs of simultaneous linear equations graphically and explain your results. a ​y = 2x + 1​and y​ = 2x − 3​ b ​x − 2y = 4​and ​2x − 4y = 8​

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Challenge 14 A confectionery company produces packets of lollies containing lollipops and gummies. A packet with ​6​ lollipops and ​8​gummies weighs 1 ​ 32 g​. A packet containing ​7​lollipops and ​6​gummies weighs 1 ​ 29 g​. By setting up two equations, and using graphing software, find how much a lollipop and a gummy weigh each.

Related resources 1.4 Quick quiz Worksheet: Solving simultaneous linear equations graphically Investigation: Train travel

Lesson 1.5

Solving simultaneous equations by substitution Learning intentions By the end of this lesson, you will be able to ... → solve pairs of simultaneous linear equations algebraically using substitution.

Solving simultaneous equations by substitution •

In the previous lesson, we found the solution to a pair of simultaneous linear equations graphically by finding the point of intersection of the two graphs.

•

In this lesson, we use the substitution method to find simultaneous solutions algebraically.

•

At the point of intersection, corresponding x​ ​-values and ​y​-values are equal for both equations.

•

When x​ ​or y​ ​is the subject of the equation, we can substitute into the other equation and solve for the unknown variable.

•

Once we know the value of one variable, we can substitute again to find the other.

•

Write the solution to both equations as a coordinate pair or in the form x​ = a​, y​ = b​, depending on the context.

Oxford University Press

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Worked example 1.5A Solving by substitution Find the solution to the pair of equations by substitution. y = 2​x − 1 ​ ​ ​ ​​​ y = −​x + 2 Think

Write

Label your equations 1​​ ​​ and 2​​ ​​ .

y = 2​x − 1 ​1​ ​​ ​ ​ ​​​ y = −​x + 2 ​2​

Substitute equation 2​​ ​​ into equation 1​​ ​​ .

​(− x + 2) = 2x − 1​

Solve to find x​ ​.

− x + 2​= 2x − 1 2​​= 3x ​​ ​​ ​− 1​ 3​= 3x x​= 1

Now we know that x​ = 1​, substitute this into either equation and solve to find y​ ​.

y​= 2(1) − 1 ​​ ​ ​​​ y​= 1

Write the solution to both equations as a coordinate pair.

The simultaneous solution to both equations is (​ 1, 1).​

Check the solution by substituting ​x = 1​and ​y = 1​into both sides of each equation.

Alternatively, use Desmos or a similar tool to find the point of intersection between the two graphs.

Equation ​​1​​ : − 1​ ​ ​LHS = 1 RHS​​= 2(1) ​​ ​= 1 Equation ​​2​​ : (1) + 2​ ​ ​LHS = 1 RHS​​= − ​​ ​= 1

Worked example video: Solving by substitution

Understanding, fluency and communicating Answers: pXXX

Helpful hints •

Check that x​ ​or y​ ​is the subject of the equation before substituting.

•

You can always verify your answer by graphing on software like Desmos.

1 Solve the following pairs of equations simultaneously by substitution. Write the solution as a coordinate pair.

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a y​​ ​​= 2x​ ​​ y​= x + 1

b y​​ ​​= 2x​ ​​ y​= − x + 1

c y​​ ​​= 2x​ ​​ y​= x − 2

d y​= 2x ​​y​​= _ ​ 1​​ ​ x ​ + 2

e y​= x + 1 ​​ ​ _ ​ ​​ y​= ​ 1 ​ x + 2 2

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y​= 2x − 7 ​​ ​ ​ ​​ y​= − 10x + 5

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Worked example 1.5B Changing the subject then solving by substitution Find the solution to the pair of equations 3x​− y = 5 ​ ​ ​​ 5x​+ 3y = − 1

​ ​

by changing the subject and then substituting.

Think

Write

Label the equations ​​1​​ and 2​ ​​ .​

3x − y​= 5 ​1​ ​​ ​ ​ ​ ​​​ 5x + 3y​= − 1 ​2​

Rearrange equation 1​​ ​​ to make ​y​the subject.

3x − y​= 5 3x​​= y ​+ 5​​​ ​​ 3x − 5​= y ​y = 3x − 5 ​1​​

Substitute equation 1​​ ​​ into equation 2​​ ​​ .

​5x + 3(3x − 5) = − 1​

Solve for x​ ​.

5x + 3(3x − 5)​= − 1 5x + 9x − 15​= − 1 ​​

14x − 15​= − ​​ 1​​ 14x​= 14 x​= 1

Substitute the ​x​-value you found into equation 1​​ ​​ and solve for y​ ​.

y​= 3(1) − 5 ​​ ​ ​​​ y​= − 2

Write the solution as a coordinate pair.

The solution is (​ 1, − 2) .​

Worked example video: Changing the subject then solving by substitution

Helpful hints •

Some equations are easier to substitute than others. Inspect each equation and choose the one that will be easiest to substitute in.

2 Solve the following pairs of equations simultaneously by substitution, you may wish to change the subject of an equation first. x​+ 2y − 3 = 0 ​+ x + 1 = 0 a x​​ ​​+ y = 6​​​ b ​​ c y​​ d ​​ y​​− 2 = − x​ ​ ​​​ ​ ​​​ ​​ x​− y = 2 2x​− 2y − 6 = 0 y​= 2x + 1 6​− 2y + 3x = 0 ​+ 2y = 10 e 3x ​​ ​ ​​​ 4x​+ 3y = 13

Oxford University Press

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f ​​

y​− 2x = 1 ​ ​ ​​ 3x​+ 2y = 16

x​− 3 = 4y − 6 g ​​ ​ ​​​ 2x​− 5y = 3

x​+ 2 + y = 9 h ​​ ​ ​​​ 2x​− 3y = 4

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Worked example 1.5C Forming equations to solve practical problems Yuri is hosting a party and has to pick between two ice-cream vendors. Vendor A charges ​$500​for initial hire and charges $​ 6​per scoop. Vendor B charges $​ 200​for initial hire and charges $​ 8​per scoop. a Form equations to model each vendor offer. b Use your equations from part a and solve simultaneously to find the number of scoops which must be sold for the total cost from both vendors to be equal. c Under what circumstances should Yuri pick Vendor A over Vendor B? Part

Think

Write

a

Each vendor offer consists of an initial charge and a charge per ice-cream scoop served. These charges can be combined into a total charge, which depends on the number of ice-cream scoops served. So, assign one pronumeral for the variable “total charge” and one for the variable “number of scoops”.

The initial hire fee is a fixed amount that is charged before any scoops of ice-cream are served, that is, when n ​ = 0​. This value will be the height of the vertical-axis intercept.

The charge per scoop is a rate and so tells us the gradient of the line.

Let C ​ ​be the total cost and n ​ ​be the number of scoops.

Vendor A: ​C = 6n + 500​ Vendor B: ​C = 8n + 200​

b

Solve simultaneously by substitution to find when the total cost from both vendors is equal.

​6n + 500​= 8n + 200 300​= 2n n = 150​ After selling ​150​scoops the vendors cost the same.

c

Check which vendor is cheaper for ​n < 150​by substituting n ​ = 100​into both equations.

Vendor A: C = 6(100) + 500​ ​​ ​​​ ​​ ​= 1100 Vendor B: C = 8(100) + 200​ ​​ ​​​ ​​ ​= 1000 If Yuri expects there will be fewer than​ 150​scoops of ice-cream sold, they should pick Vendor B. Otherwise, they should pick Vendor A.

Worked example video: Forming equations to solve practical problems

Helpful hints •

Think of these problems in terms of the structure ​ variable rate × measurement + fixed cost​.

This is the standard form of a linear equation, ​y = mx + c​.

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3 Kelvin is choosing between two gym memberships.

Big Muscles Gym: charges a $​ 150​sign-up fee plus $​ 25​a month.

Plus-Ultra-Mega Fit: Charges a ​$50​sign-up fee plus $​ 35​a month. a Form equations to model each gym offer. b Use your equations from part a and solve simultaneously to find the number of months Kelvin must keep his membership for the total cost from both gyms to be equal. c Under what circumstances should Kelvin pick one gym over the other?

4 Georgia needs to rent a car for a one-day trip. She is comparing two rental companies.

Gogo car hire: $​ 2.60​per kilometre travelled, no other costs.

Speedy Car: $​ 50​per day, $​ 2​per kilometre travelled. a Under what circumstances is the cost to hire a car the same for both companies? b If Georgia is planning to travel ​120​kilometres, which car company should she choose? Justify your answer with appropriate calculations.

5 Clare is hiring a jumping castle for a party. She obtained quotes from two companies as shown below: Jumpy McJumpface: $​ 1000​set-up plus $​ 150​per hour.

BigFunCastle: $​ 250​an hour. No set-up fee.

Clare wants to hire the jumping castle for 1 ​ 0​hours. Show that the cost is the same for both companies by solving simultaneously.

Problem solving and reasoning 6 Martyn’s mother is now four times as old as Martyn. Five years ago, she was five times his age. What are their ages now?

7 A bus company has ​8​buses. Some buses carry ​100​passengers and the other buses carry ​50​passengers.

When all ​8​buses are full, they can carry a total of 7 ​ 00​passengers. How many buses of each size does the company own?

8 Terrence and Jerimiah go shopping and buy some watermelons and hats. They both spend $​ 142​. Terrence buys five watermelons and eight hats. Jerimiah buys 1 ​ 2​watermelons and five hats. Form a pair of simultaneous equations to find the cost of a watermelon and a hat.

Challenge 9 Find the solution to the pairs of linear equations. y 2y + 1 _ 6 ​= ​ _ ​ 2x − ​ ​ a 2x − ​ 4 ​​= 5 b _ 4 3 ​ ​​ ​​ ​ ​ ​​ ​​ ​ 6y 2y + 3 1 ​= ​ _ _ 2x + ​ _ ​​= − 2 ​ 4x − ​ ​ + 1 4 3 2

c 2​​(​​3x + 4​)​​​​ = 2​​(​​5 − y​)​​​ ​​ ​ ​ ​​ y−4 ​ _ ​ − 1​= 2​​(​​3 − 2x​)​​​ 3

10 A jar contains 3 ​ 8​coins, made up of ​$1​and $​ 2​coins. The total value of the $​ 1​coins is ​$11​more than the value of the $​ 2​ coins. a Set up a pair of simultaneous equations to represent the situation. b Solve the simultaneous equations to find the number $​ 1​and $​ 2​ coins.

Related resources 1.5 Quick quiz Oxford University Press

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Worksheet: Using the substitution method Module 1 Simultaneous linear equations

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Lesson 1.6

Solving simultaneous equations by elimination Learning intentions By the end of this lesson, you will be able to ... → solve pairs of simultaneous linear equations algebraically using elimination.

Combining whole equations •

Consider two equations ​​1​ x + 2​= 2y ​ ​​ ​ ​ ​ ​​ ​​2​ 3x + 1​= y − 2.

•

We can “combine” these equations by adding or subtracting their left-hand sides, then performing the same operation on their right-hand sides. → For example, adding equation ​​1​​ to equation ​​2​​ would result in ​

(x + 2) + (3x + 1) = (2y) + (y − 2)​.

→ After collecting like terms and simplifying, this equation becomes 4 ​ x + 3 = 3y − 2​. → Similarly, subtracting equation ​​2​​ from equation ​​1​​ would result in the equation ​

(x + 2) − (3x + 1) = (2y) − (y − 2)​.

→ Again, this can be simplified to − ​ 2x + 1 = y + 2​.

•

We can also perform more than one operation when combining whole equations. → For example, we could create a new equation from ​​2​ − 2 × ​1​​, which would be ​

(3x + 1) − 2 × (x + 2) = (y − 2) − 2 × (2y)​.

→ By expanding, collecting like terms and simplifying, this becomes 3x + 1 − 2x − 4​= y − 2 − 4y − 2x + 1 − 4​​ = y − 4y − ​​ 2​​

​ ​ 3x x − 3​= − 3y − 2.

Solving simultaneous equations by elimination

34

•

In addition to the substitution method, another algebraic strategy for solving simultaneous linear equations is the elimination method.

•

This method combines the pair of simultaneous equations by adding or subtracting in such a way that one of the variables is completely eliminated.

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•

For example, when the equation x​ + y = 2​is added to the equation ​2x − y = 4​, the “​+ y​” in the first equation is cancelled out by the “​− y​” in the second equation: x + y​= 2 ​ ​ +

​= 4]​ ​​​ [2x − y 3x + 0​= 6 → This leaves the simpler equation ​3x = 6​, which involves only one variable.

•

In general, elimination is used to find a new equation containing only one of the original variables, which can then be solved to find the value of that variable. → In the previous example, ​3x = 6​can be simplified to find x​ = 2​.

•

The second variable can always be found by substituting the known variable back into one of the original equations. → For example, we can substitute the value x​ = 2​in the example above into the first equation x​ + y = 2​, to find ​(2) + y = 2​, and therefore y​ = 0​.

•

The other original equation can be used to check your answer. → In our example, substituting x​ = 2​and y​ = 0​into the equation ​2x − y = 4​gives ​ LHS = 2x − y = 2(2)− (0) = 4 = RHS​.

Steps for solving by elimination 1 Label your equations before you start working to assist in communication and clarity.

2 Look for terms that have matching coefficients and if none exist, adjust one or both equations by multiplying.

3 Write out the operation(s) that you will use to eliminate one of the variables.

4 Perform the elimination and solve for the variable that was not eliminated.

5 Substitute the value of the known variable back into one of the original equations and solve to find the remaining variable.

6 Write the solution to both equations as a coordinate pair or in the form x = a, y = b, depending on the context.

7 Verify the solution by substituting back into the unused equation.

Worked example 1.6A Solving by elimination Use elimination to find the simultaneous solution to the pair of equations. y​+ 2x = 5 ​ ​ ​ ​​​ 2y​− 2x = 7 Think

Write

Label the equations ​​1​​ and 2​​ ​​ and stack them so that their ​x​and ​y​ terms are aligned.

y + 2x​= 5 ​1​ ​​ ​ ​ ​​​ 2y − 2x​= 7 ​2​

Since the coefficient of x​ ​in equation 1​​ ​​ is the negative of the coefficient of x​ ​in equation 2​​ ​​ , we can eliminate ​x​by adding the two equations together. That is, find 1​​ ​​ + ​​2​​.

y + 2x​= 5 ​1​ 2y − 2x​​= 7​ ​ ​ ​2​​​​ ​​ +

3y + 0x​= 12 ​1​ + ​2​

Oxford University Press

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Think

Write

Simplify and collect like terms, then solve for y​ ​.

3y​= 12 ​​ ​ ​​​ y​= 4

Substitute the ​y​-value back into either of the original equations and solve for x​ ​.

Equation ​​1​​ : (4) + 2x​= 5 2x​​= 1​​​ ​ ​ x​= _ ​ 1 ​ 2 Solution is x​ = _ ​ 1 ​​, y​ = 4​.

2

Write the solution.

Verify your solution by checking that it satisfies the equation you didn’t use to find x​ ​.

Alternatively, use Desmos or a similar tool to find the point of intersection between the two graphs.

Equation ​​2​​ :

1 ​​ ​​​ LHS​= 2(4) − 2​​(​​ ​ _ 2) ​ ​ ​​ ​ ​ = 8​− 1 ​​ ​= 7 ​= RHS

Worked example video: Solving by elimination

Understanding, fluency and communicating Answers: pXXX

Helpful hints •

Label your equations.

•

Identify what operation to do (addition or subtraction).

•

Choose whichever variable is easier to eliminate first.

•

Don’t forget to find the value of both variables.

1 Consider the following pair of simultaneous equations: y​= 2x + 5 ​1​ ​ ​ ​ ​ ​​​ y​= − 2x + 5 ​2​ a Use substitution to find the solution to the pair of simultaneous equations. b Find and simplify the equation ​​1​ + ​2​.​ c Find and simplify the equation ​​1​ − ​2​.​ d Using your answer to part b, write out the remaining steps to solve the pair of simultaneous equations using elimination.

2 Solve the following pairs of equations simultaneously by elimination.

36

a y​​ ​​= 3x + 6​ ​​ y​= − 3x + 6

b y​​ ​​= 4x + 1​ ​​ y​= − 2x + 1

​+ y = 5 e 2x ​​ ​ ​ ​​ y​= 10 − x

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3x​− 2y = 8 ​​ ​ ​ ​​ 2y​+ 8x = − 19

c y​​ ​​= 4x + 5​ ​​ y​= − 5x − 4

d x​​ ​​+ 4y = 7​​​ x​− 4y = 3

9​+ 6y − 2x = 0 g ​​ ​ ​​​ 3x​+ 6y = 36

h

2 ​ x = − 1 _ ​ 1 ​ y​− ​ _ 2 ​ 3 ​​ ​ ​​ 1 ​ y = 0 − 10​+ 3x + ​ _ 2 Oxford University Press

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Worked example 1.6B Manipulating the equations before solving via elimination Use elimination to find the solution to the pair of simultaneous equations.

4x​− 3y = 12 ​ ​ ​ ​​ ​ 8x​+ y = 24 Think

Write

Label the equations ​​1​​ and 2​​ ​​ and stack them so that their ​x​ and ​y​terms are aligned.

​ x − 3y = 12 ​1​ 4 ​​​ ​ ​ ​​​ ​8x + y = 24 ​2​

Neither the x​ ​- nor the y​ ​-terms have a matching coefficient; however, the coefficient of x​ ​in the equation ​​2​​ is twice the value of the x​ ​- coefficient in equation ​​1​​ .

Hence, multiply the whole of equation 1​​ ​​ by 2 ​ ​. Label the new equation ​​3​​ .

​2 × ​1​ ​:

Subtract equation 3​​ ​​ from equation 2​​ ​​ .

8x + y​= 24 ​2​ [8x − 6y​​= 24]​ ​3​​​​ ​​ −

​ ​ 0 + 7y​= 0 ​2​ − ​3​

Collect like terms, simplify and solve for the remaining variable.

7y​= 0 ​​ ​ ​​​ y​= 0

Substitute ​y = 0​into one of the original equations and solve for x​ ​.

When y​ = 0​, equation 1​​ ​​ becomes:

Write the solution.

Solution is x​ = 3​, y​ = 0​.

Verify your solution by checking that it satisfies the equation you didn’t use to find x​ ​.

Alternatively, use Desmos or a similar tool to find the point of intersection between the two graphs.

Equation ​​2​​ : LHS​= 8(3) + (0) ​​ ​ = 24​ ​​​ ​= RHS

2 × [4x − 3y​= 12] ​1​ ​​

​ ​ ​ ​​​ 8x − 6y​= 24 ​3​

4x − 3(0)​= 12 ​​ ​​​ 4x​= 12 x​= 3

Worked example video: Manipulating the equations before solving via elimination

Helpful hints •

Inspect both equations and identify what operations you need to perform to be able to eliminate a variable by addition or subtraction.

Oxford University Press

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3 For each of the following pairs of equations, write out a sequence of operations that could be used to form a new equation with only one of the variables. a 2x​+ 5y = 12 ​1​ ​​ ​ ​ ​ ​​​ x​− 3y = − 5 ​2​

b 6x​− 5y = 27 ​1​ ​​ ​ ​ ​​​ 3x​− 10y = 9 ​2​

​1​ c 3x​+ y = 7 ​​ ​ ​ ​ ​​​ 5x​− 4y = 6 ​2​

d 2x​+ 5y = 4 ​1​ ​​ ​ ​ ​ ​​​ 4x​+ y = − 1 ​2​

​1​ e 6x​+ 5y = 2 ​​ ​ ​ ​ ​​​ 4x​− 10y = 30 ​2​

f

​1​ g 7x​− 6y = 9 ​​ ​ ​ ​ ​​​ x​+ 2y = 13 ​2​

​1​ h 2x​= 9 − 3y ​​ ​ ​ ​ ​​​ 5x​= 2y + 13 ​2​

x​− 4y = 13 ​1​

​​ ​​ ​ ​ ​​​ 2y​− 3x + 19 = 0 ​2​

4 Perform the operations you suggested in each part of question 3 and hence solve each pair of simultaneous equations.

5 Manipulate both equations to solve simultaneously by elimination.

1 ​ y = 3 a _ ​ 1 ​ x​+ ​ _ ​1​ ​1​ b 0.3x​+ 0.4y = 2.6 2 3 ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​ ​​​ 2 ​ y + 1 = 0 ​2​ 0.5x​− 0.3y + 0.5 = 0 ​2​ _ ​ x ​​− ​ _ 4 3 2 1 ​ y = _ d _ ​ ​ x​+ ​ _ ​ 3 ​ ​1​ e 0.2x​+ 0.3y = 1.7 ​1​ 3 2 2 ​​ ​ ​ ​ ​​​ ​​ ​ ​ ​​​ 5 3 0.5x​− 0.2y = 1.4 ​2​ _ _ ​ ​ x​− ​ ​ y = 15 ​2​ 4 2

1 ​ y = 27 ​1​ c _ ​ 1 ​ x​+ ​ _ 4 2 ​​ ​ ​ ​ ​​​ 2 _ _ ​ ​ x​− ​ 2 ​ y = _ ​ 8 ​ ​2​ 3 3 5 y f 0.05x + ​ _ ​​= 2.5 ​1​ 10 ​​ ​ ​​​ ​ ​ x _ ​ ​ − 0.04y​= _ ​ 6 ​ ​2​ 5 5

Problem solving and reasoning 6 Kenny and Fatima are at a farmer’s market. Kenny buys ​3​apples and ​4​oranges for $​ 7.80​. Fatima purchases ​ 5​apples and ​2​oranges for $​ 8.10​.

Create a pair of equations to model the situation and solve simultaneously by elimination.

7 An architect is designing a rectangular garden. The design has two specific constraints based on the garden’s length (​L​) and breadth (​B​).

Constraint 1: One-third of the length plus one-half of the breadth must total ​10​metres.

Constraint 2: Two-thirds of the length minus one-quarter of the breadth must total ​5​metres.

Create a pair of equations to model the situation and solve simultaneously by elimination to determine the exact length and breadth of the garden.

8 An archaeologist discovers a small leather pouch containing ​26​ancient coins. The collection is a mix of gold and silver coins and weighs a total of 2 ​ 90​grams.

From historical texts, it is known that each gold coin should weigh 1 ​ 2.5​grams, and each silver coin should weigh ​8​grams. How many gold and silver coins do you expect are in the pouch?

9 Show that ​y = _ ​ x ​ + 1​ and ​x − 2y = 4​has no solution by attempting to solve the pair of equations 2 simultaneously using elimination.

10 The total weekly wage of three chefs and five waiters is $​ 11 900​, and the total fortnightly wage of two chefs and eight waiters is $​ 29 400​. Calculate the annual income for these chefs and waiters.

Challenge 11 The lines ​ax + by = 100​and b​ x + ay = 82​intersect at the point (​ 10, 4)​. Find the values of ​a​and ​b​.

Related resources 1.6 Quick quiz

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Worksheet: Using the elimination method

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Lesson 1.7

Break-even analysis Learning intentions By the end of this lesson, you will be able to ... → model and analyse break-even points using simultaneous linear equations.

Profit and loss •

Revenue is the total income generated by a business through the sale of items or services. → If the revenue is greater than the cost, a profit is made. ​ profit = revenue − cost​ → If the revenue is less than the cost, a loss is made. ​ loss = cost − revenue​

Break-even analysis •

Linear models for cost and revenue were introduced in Lesson 1.2 Linear models. → A cost model can be written in the form: ​ C = m ​ C​ ​​ n + c​, where C ​ ​is the total cost in dollars, n ​ ​is the number of items produced, ​​m​ C​​​(the gradient) is the cost to produce one item and c​ ​is the sum of any fixed costs. → A revenue model can be written in the form: ​ R = m ​ R​ ​​ n​, where R ​ ​is the total revenue, ​n​is the number of items sold and m​ ​​ R​​​(the gradient) is the selling price of one item.

•

The linear models for cost and revenue for the same business can be graphed on the same Cartesian plane. If there is a point of intersection, this is called the break-even point. This is the point where the revenue is equal to the cost. → Sales above the break-even point result in a profit for the business.

•

•

The coordinates of the break-even point may not be integers. As a business is usually unable to sell part of an item, round up to find the number of items which must be sold to cover all costs.

Regions can be shaded on the combined graph of both models to show the profit zone and the loss zone for the business. These regions are a visual representation of when the business is making a profit or a loss.

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Profit zone

Amount ($)

→ Sales below the break-even point result in a loss for the business.

Cost Revenue

Break-even point Loss zone

Number of items (n) Module 1 Simultaneous linear equations

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Worked example 1.7A Determining the break-even point from a given graph Holly’s Hot Dogs makes and sells hot dogs. There is a fixed cost per day of $​ 150​and each hot dog costs ​ $1.60​to produce. The hot dogs are sold for $​ 4​each. The graphs of the cost and revenue made from selling hot dogs are drawn on the same set of axes.

Cost and revenue Amount ($)

400 320 240

Cost Revenue

160 80 0

10 20 30 40 50 60 70 80 90 100 Number of items (n)

a How many hot dogs must be sold to break even? b What is the revenue at the break-even point? Give your answer to the nearest ​$10​. c Find the profit made when 1 ​ 00​hot dogs are sold.

Part

Think

Write

a

The break-even point is the point of intersection, which is where the revenue is equal to the cost.

Identify the n ​ ​- coordinate of the point of intersection. The minor gridlines on the given graph divide the horizontal axis into multiples of ​2.5​. The break-even point sits on the gridline for n ​ = 62.5​. As this value is not a whole number, round up to find the number of hot dogs that need to be sold.

At least 6 ​ 3​hot dogs must be sold to break even.

b

Estimate the value for the revenue from the vertical axis at the break-even point.

At the break-even point, revenue is approximately $​ 250​.

c

Read the values from the vertical axis for the revenue line and the cost line when ​n = 100​.

Subtract the cost from the revenue to find the profit.

When n ​ = 100​: ​revenue = $400​, ​cost = $310​ profit​= revenue − cost ​​ ​

​ = 400 ​​ − 310​ ​= $90

Worked example video: Determining the break-even point from a given graph

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Understanding, fluency and communicating Answers: pXXX

Helpful hints •

The break-even point occurs where the lines for revenue and cost intersect.

•

Read the value for the revenue at the point of intersection of the two lines.

•

A business makes a profit when the revenue line is above the cost line. It makes a loss when the cost line is above the revenue line.

1 Suki’s Sushi Station makes and sells sushi rolls. There is a fixed cost per day of $​ 200​and each sushi roll costs $​ 1.50​ to produce. The sushi rolls are sold for $​ 4​each. The graphs of production cost and revenue are drawn on the same set of axes.

Cost and revenue 450

350

b What is the revenue at the break-even point?

300

2 Peter’s Pie Palace makes and sells gourmet pies. There is a fixed cost per day of $​ 220​and each pie costs $​ 2.60​to produce. The pies are sold for $​ 6​ each.

Amount ($)

a How many sushi rolls must be sold to break even? c Subtract the cost from the revenue to find the profit made when ​ 100​sushi rolls are sold.

50 0

600 500

Cost Revenue

400 300 200 100 20 40 60 80 100 Number of pies (n)

b What is the revenue at the break-even point to the nearest $​ 10​? c Find the profit made by Jenny’s Juice Joint when ​100​cups of juice are sold. d Find the difference between the revenue and cost when 2 ​ 0​cups of juice are sold. Does Jenny make a profit? Explain your answer.

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Cost and revenue 400 Amount ($)

0

a How many cups of juice must be sold to break even? Give your answer to the nearest ​10​.

Oxford University Press

20 40 60 80 100 120 Number of sushi rolls (n)

Cost and revenue

Amount ($)

3 Jenny’s Juice Joint makes and sells fresh juices. There is a fixed cost per day of ​$100​ and each cup of juice costs ​$1.80​ to produce. The juices are sold for $​ 3.50​each. The graphs of production cost and revenue are drawn on the same set of axes.

200

100

a How many pies must be sold to break even?

c Find the profit made when ​ 80​pies are sold to the nearest $​ 10​.

250

150

The graphs of the production cost and the revenue made from selling the pies are drawn on the same set of axes. b What is the revenue at the break-even point to the nearest $​ 10​?

Cost Revenue

400

300

Cost Revenue

200 100 0

20 40 60 80 100 120 Number of cups of juice (n)

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4 The daily production costs and revenue from the sale of items is shown on the graph.

Daily cost and revenue 5500

a Use the graph to determine the number of items that must be sold to break even.

5000 4500

b To the nearest $​ 100​, how much profit or loss is made when: ii ​900​items are sold? c How many items need to be sold for the company to make a: i

4000

​100​items are sold?

​$500​ profit?

Amount ($)

i

Cost Revenue

ii ​$500​ loss?

3500 3000 2500 2000 1500 1000 500 0

100 200 300 400 500 600 700 800 900 1000 Number of items (n)

Worked example 1.7B Analysing the break-even point by drawing a graph The Merry Muffin Shop has a fixed cost of $​ 320​per day and a variable production cost of $​ 2​ per muffin. The selling price of each muffin is $​ 6​. The maximum daily production is 1 ​ 20​muffins. a Write an equation for the cost, ​$C​, of producing n ​ ​ muffins. b Graph the linear equation for the production cost. c Write an equation for the revenue, $​ R​, for the number of muffins sold, n ​ ​. Draw the straight line for the revenue equation on the same set of axes as the cost equation in part b. d Find the point of intersection of the two lines. e How many muffins must be sold to break even? f

What is the break-even revenue?

g On the graph drawn for parts b and c, shade the loss zone and the profit zone.

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Part

Think

Write

a

The cost, ​C​, can be modelled using a linear relationship in the form ​C = m ​ C​ ​​ n + c​, where ​​mC​ ​​​is the cost per item and ​c​is the sum of any fixed daily costs.

The cost per muffin, ​​m​ C​​​ is ​$2​.

The fixed daily cost, ​c​, is $​ 320​.

The cost equation is ​C = 2n + 320​.

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Think

b

The maximum number of muffins is 1 ​ 20​ so the graph needs to show the values from ​ n = 0​to n ​ = 120​.

Use any method to draw the graph.

For example, identify the vertical intercept as ​(0, 320)​and use substitution to find the point ​(40, 400)​.

Write Cost and revenue 1000 R = 6n

800 Amount ($)

Part

600 C = 2n + 320 400 200

0

20

40 60 80 100 120 140 Number of muffins (n)

c

The revenue, R ​ ​, can be modelled using a linear relationship in the form ​R = m ​ R​ ​​ n​, where ​​mR​ ​​​(the gradient) is the selling price of one item. There is no fixed revenue, so the equation has no constant value.

Both ​C​and ​R​are values in dollars and so can both be represented on the vertical axis, labelled as “Amount (​$ ​)”.

Use any method to draw the graph. For example, identify the vertical intercept as ​ (0, 0)​and use the gradient, 6 ​ ​, to find that ​ (100, 600)​is a second point on the graph.

The revenue equation is ​R = 6n​.

The graph is drawn on the same set of axes as ​C = 2n + 320​in part b.

d

Find the coordinates of the point where the lines intersect.

The point of intersection has the coordinates ​(80, 480)​.

e

Read the value of the n ​ ​-coordinate for the point of intersection of the two lines.

​80​muffins must be sold to break even.

f

Read the amount in dollars for the second coordinate of the point of intersection.

The break-even revenue is ​$480​.

g

For the loss zone, shade the zone between the two lines to the left of the break-even point to indicate where the revenue is less than the cost for making n ​ ​ items.

For the profit zone, shade the zone between the two lines to the right of the break-even point to indicate where the revenue is more than the cost for making n ​ ​ items.

Cost and revenue

Amount ($)

1000 R = 6n

800 600 400 200 0

Profit zone

C = 2n + 320 Loss zone

Break-even point

20 40 60 80 100 120 140 Number of muffins (n)

Worked example video: Analysing the break-even point by drawing a graph Oxford University Press

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Helpful hints •

The profit zone is the region on the combined graph where the revenue is greater (above) than the cost.

•

The loss zone is the region on the combined graph where the cost is greater (above) than the revenue.

5 A plastics company has a maximum daily production of 7 ​ 00​items. a There is an initial cost of $​ 3000​per day plus $​ 8​per item produced. This can be represented by the linear equation C ​ = 8n + 3000​, where C ​ ​is the production cost in dollars and n ​ ​is the number of items. Complete this table of values for the production cost.

Number of items ​(n)​

​0​

​100​

​300​

​500​

​700​

Production cost ​(C​, $​ )​ b The selling price of each item is $​ 15​. This can be represented by the linear equation R ​ = 15n​, where R ​ ​ is the revenue in dollars on the sale of n ​ ​items. Complete this table of values for the revenue.

Number of items ​(n)​

​0​

​100​

​300​

​500​

​700​

Revenue ​(R​, $​ )​ c Choose a suitable scale and plot these lines on the same set of axes: i

production cost versus number of items

ii revenue versus number of items. d Find the point of intersection of the two graphs. This is the break-even point. e How many items, to the nearest ​25​, must be sold each day to break even? f

What is the break-even revenue? Give your answer to the nearest $​ 500​.

6 The Tin Lid factory produces sports caps. There is a fixed monthly cost of $​ 1800​and it costs $​ 3​to produce a cap. The caps are sold for $​ 8​each. The maximum monthly production of caps is 8 ​ 00​. a Use the equation C ​ = 3n + 1800​, where C ​ ​is the production cost in dollars to produce n ​ ​sports caps, to graph the cost of production. b Use the equation R ​ = 8n​, where R ​ ​is the revenue in dollars on the sale of n ​ ​sports caps, to graph the revenue on the same set of axes as the cost equation in part a. c Find the point of intersection of the two lines. d To the nearest ​10​, how many caps must be sold in a month to break even? e To the nearest $​ 100​, what is the break-even revenue?

7 The Supertight Tie Company has a fixed cost of $​ 1200​per day and a variable cost of $​ 3​per tie. This can be represented by the equation C ​ = 3n + 1200​, where C ​ ​is the cost of producing n ​ ​ties. The revenue of $​ 15​ per tie sold can be represented by the equation R ​ = 15n​, where R ​ ​is the revenue on the sale of n ​ ​ ties. a Graph these two equations on the same set of axes for ​0 ≤ n ≤ 250​. b Find the break-even point. c How many ties must be sold to break even? d What is the break-even revenue?

8 The Munchy Sandwich Shop has a fixed expense of ​$2000​per month and a cost of $​ 1.20​per sandwich.

The sandwiches are sold for $​ 3.50​ each. a Write the linear equations that can be used to model the production cost and revenue for n ​ ​ sandwiches. b Graph straight lines modelling the cost and revenue for ​0 ≤ n ≤ 3000​. c Find the break-even point.

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d How many sandwiches must be sold to break even? e What is the break-even revenue? f

The maximum number of sandwiches that can be made in a month is 8 ​ 00​. Explain the significance of this quantity.

g On the graph drawn for part b, shade the loss zone. Explain the significance of this region.

9 The Tight Squeeze Juice Company has a weekly fixed cost of $​ 300​and a cost of $​ 1.50​per cup of juice. The cups of juice are sold for $​ 4.00​ each. a Graph straight lines modelling the cost and revenue for up to ​1000​cups of juice. b Find the break-even point. c How many cups of juice must be sold to break even? d What is the break-even revenue? e On the graph drawn for part a, shade the loss zone. f

The shop becomes short-staffed and the maximum number of cups of juice per week is reduced to 1 ​ 00​.

Is the business profitable? Explain.

10 Consider the graphs you have drawn to model the cost and revenue for each company in questions 5, 6 and 7. a Shade the loss zone for the graph drawn in question 5. b Shade the profit zone for the graph drawn in question 6. c Shade the loss zone for the graph drawn in question 7.

Worked example 1.7C Finding the break-even point algebraically Cotton the Act sells t-shirts for $​ 30​each. It costs $​ 18​to produce each t-shirt and the company has a fixed daily cost of $​ 600​related to t-shirt production. a Write equations to represent the cost and revenue of producing and selling the t-shirts. Use the variable ​ A​to represent the revenue and cost amounts in dollars, instead of R ​ ​for revenue and C ​ ​for cost. b Solve the pair of simultaneous linear equations from part a. c What does the answer to part b represent?

Part

Think

Write

a

Write the linear relationships for the cost and revenue models using the variable A ​ ​instead of C ​ ​and ​R​.

Cost: A ​ = 18n + 600​ Revenue: A ​ = 30n​

b

As the revenue equation has A ​ ​ as the subject of the equation, use the substitution method to solve the pair of simultaneous linear equations.

A​= 18n + 600 ​1​

​​ ​ ​ ​ ​​ ​ A​= 30n ​2​ Substitute 2​​ ​​ into ​​1​​ and solve using inverse operations: 30n​= 18n + 600 ​​12n

​= 600 ​​ ​ ​ n​= 50 Substitute ​n = 50​into ​​2​​: A​= 30n ​​ ​= 30 ×​ ​​ 50​ ​= 1500 The solution is n ​ = 50​, A ​ = 1500​, or ​(50, 1500)​.

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Part

Think

Write

c

The solution to the pair of simultaneous equations is the point of intersection of the two lines, which is the break-even point.

The break-even point is found by solving the pair of simultaneous equations. This means that 5 ​ 0​t-shirts must be sold for the cost to be equal to the revenue, which would be $​ 1500​.

Worked example video: Finding the break-even point algebraically

Helpful hints •

To solve a pair of simultaneous linear equations, the equations must include the same two variables.

If you are working with cost and revenue models, use the variable ​A​in both equations, instead of C ​ ​ for cost and R ​ ​for revenue.

11 For each situation in questions 5 to 9: a Write a related pair of simultaneous linear equations. b Solve the pair of simultaneous linear equations algebraically and compare the solution with the graph already drawn. c Interpret the meaning of the answer to part b.

12 Recall Holly’s Hot Dogs from Worked example 1.7A. a Write a pair of simultaneous linear equations to represent the cost and revenue from producing and selling hot dogs. b Solve the pair of simultaneous linear equations and find the exact solution. c Explain why you need to round up to find the number of hot dogs that need to be sold to break even.

Problem solving and reasoning 13 The cost of producing cotton bags is initially $​ 2000​plus $​ 2.50​per item. Each bag is sold for $​ 5​. a Model the production cost and the revenue with linear equations. b Use a graphical method to find the break-even point. c How many bags need to be sold to break even? d How much profit or loss is made when: i

​100​bags are sold?

ii ​1000​bags are sold?

e How many bags need to be sold for the company to make: i

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​$1000​ profit?

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ii ​$1000​ loss?

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14 The production cost of printing books is initially $​ 8000​plus $​ 2500​per 1 ​ 000​books printed. The books are sold for $​ 15​ each. a Model the production cost and the revenue with linear equations. b Use any method to find the break-even point. c How many books need to be sold to break even? d How much profit or loss is made when: i

​500​books are sold?

ii ​1000​books are sold?

iii ​2000​books are sold?

e How many books need to be sold for the company to make: i

a loss of $​ 3000​?

ii a profit of $​ 2500​?

iii a profit of $​ 10 000​?

15 Recall question 4 part c. a Find the average number of items sold each day if one day makes a $​ 500​profit and the next day makes a ​ $500​ loss. b Describe how a profit and loss of the same amount relate to the break-even point. c If ​350​items are sold, the loss is $​ 1000​. How many items must be sold to make a profit of $​ 1000​?

Challenge

​

C=m ​ C​ ​​ n + c ​ and ​ R = m ​ R​ ​​ n​ Find the number of items which must be sold to breakeven in terms of m​ ​​ C​​, m ​ R​ ​​​ and ​c​.

Revenue Cost

Amount ($)

16 A company has the following cost and revenue models, where C ​ ​is the cost in dollars, ​R​is the revenue in dollars, ​ n​is the number of items sold, m​ ​​ C​​​is the production cost of one item, ​c​is the sum of any fixed costs and ​​m​ R​​​ is the selling price of one item.

Cost and revenue

(20, 620)

17 A business is selling toy sets and the graph below represents the income and costs associated with this business. The business has fixed running costs each day n Number of toy sets sold (n) and there are certain costs that are incurred for each toy set produced. The graph shows that ​20​toy sets need to be sold each day for the business to break even.

Suppose the business makes a profit of ​$40​when it sells ​25​toy sets. a Show that the sale price of the toy set is $​ 8​more than what it costs to produce the toy set. b Find the value of the vertical intercept for the cost equation. c If the fixed costs of the business were to increase by ​$40​a day, how much would the business need to increase the sale price of each toy set to ensure they still only need to sell 2 ​ 0​items to break even?

Related resources 1.7 Quick quiz Investigation: Cost and revenue models

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Using spreadsheets Syllabus links Identify the break-even point and solve problems involving profit and loss using a spreadsheet (MST-12-S2-01)

Simultaneous linear equations Answers: pXXX

See Oxford Digital for sample spreadsheets for questions in this lesson.

1 Shelby is a potter who creates ceramic mugs to sell at a local market. She models her costs and revenue in the following spreadsheet.

A

B

C

1 2

Number of mugs

3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

D E Break-even analysis

Equipment Cost of Total cost Revenue cost making mugs $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360 $360

$0 $120 $240 $360 $480 $600 $720 $840 $960 $1,080 $1,200 $1,320 $1,440 $1,560 $1,680 $1,800

$360 $480 $600 $720 $840 $960 $1,080 $1,200 $1,320 $1,440 $1,560 $1,680 $1,800 $1,920 $2,040 $2,160

$0 $160 $320 $480 $640 $800 $960 $1,120 $1,280 $1,440 $1,600 $1,760 $1,920 $2,080 $2,240 $2,400

F

G

H

Cost of making one mug

$12

Selling price Equipment cost

$16 $360

Break-even point

=H4/(H3–H2)

a Write the equations of the cost and revenue models Shelby is using. b How many mugs must Shelby sell to “break even”? c How much profit does Shelby make if she sells ​120​mugs? d What formula could Shelby have entered in: i cell D5? ii cell E17? e Shelby decides to add a “Profit” column in column F of her spreadsheet.

f

i What formula could she enter into cell F3 to give the “Profit” when she sells ​0​mugs? ii What could Shelby do to complete the “Profit” column? iii Shelby notices that some of the values in the “Profit” column are negative. What do these values represent? iv What formula would Shelby enter into cell F3 if she decided to have a “Loss” column instead? What would negative values represent in this case?

Explain why the formula given in cell H6 calculates the break-even point.

2 Create your own version of Shelby’s spreadsheet, including the “Profit” column. Use formulas so the table entries columns B to F depend on column A and the values entered in cells ​H2​, ​H3​ and ​H4​.

3 Adapt your spreadsheet to show the cost and revenue models described in Lesson 1.7 Break-even analysis questions 5 to 9.

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Lesson 1.8

Review: Simultaneous linear equations Review questions 1.8A Mathematical literacy review Answers: pXXX

Key terms The following key terms are used in this module: • break-even point • linear model • cost • linear relationship • dependent variable • loss • elimination method • mathematical model • gradient • point of intersection • independent variable • profit

• • • • • •

revenue simultaneous equations substitution substitution method variables ​y​-intercept

Check your understanding 1 Explain what a linear relationship is in your own words, using the equations y​ = 2​x​​ 2​ + 1​ and ​3x + 2y = 5​as an example and a non-example.

2 Label each pronumeral in the equation y​ = mx + c​using the terms gradient, y ​ ​-intercept, dependent variable and independent variable.

3 Write one to three sentences explaining and relating the key terms mathematical model, linear model, and variables.

a break-even point b revenue c cost d loss zone

Amount ($)

4 Match each of the following key terms with one of the marked features on the cost vs. revenue graph below and write a short description for each term in your own words.

e profit zone 5 Which key term describes what you should use to check that the point ​(1, 1)​lies on the line with equation y​ = 4x − 3​? Explain how you would do this.

Number of items (n)

6 Suppose you are given two linear equations ​​1​​ and ​​2​​. a What does it mean to find the point of intersection between ​​1​​ and ​​2​​ ? b What does it mean if we call equations ​​1​​ and ​​2​​ a pair of simultaneous equations? c How is the point of intersection related to a pair of simultaneous equations?

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7 Consider the following pairs of linear equations.

2x − y​= 2 ​y = 2x + 1 ​ A: ​ ​ ​​ ​ and ​ B: ​​ ​ ​​ x + y​= 1 ​y + 2 = − 3x a To solve one of these pairs of simultaneous equations, the substitution method is more suitable. Which is it, and why? b To solve the other pair of simultaneous equations, the elimination method is more suitable. Which is it, and why?

Review questions 1.8B Multiple-choice questions 1 What is the equation of the straight line with gradient of _ ​​ 1 ​​ and y​ ​-intercept ​− 1​?

2 1 ​x​ A ​y = 1 − ​ _ 2 1 ​​ B ​y = x + ​ _ 2 1 _ C ​y = ​ ​x − 1​ 2 _ D ​y = ​ 1 ​ − x​ 2 2 What is the equation of the line shown? y

4 The equation C ​ = 3n + 150​models the costs for the sandwich shop Tasty Toasties. What might the ​ 150​ represent?

A number of sandwiches sold B the cost per sandwich C fixed daily cost D number of sandwiches made 5 The linear relationships y​ = _ ​ 3 ​x − 3​and y​ = _ ​ 1 ​x + 2​ 4 2 are graphed on the same Cartesian plane, as shown.

What is the point of intersection between the two lines?

4 3 2

y

1

4

−4 −3 −2 −1 0 −1

1

2

3

3

4 x

−3

1 y=—x+2 4

2

−2 −4

3 y=—x–3 2

1 −1

0 −1

1

2

3

4

x

−2

A ​y = _ ​ 2 ​x + 1​ 3 _ B ​y = ​ 3 ​x + 1​ 2 2 ​​ C ​y = x + ​ _ 3 _ D ​y = x + ​ 3 ​​ 2 3 The Cactus Water Company purifies water. The weekly fixed cost is $​ 450​and the cost per bottle is $​ 2​. If C ​ ​represents the total cost and n ​ ​ is the number of bottles, which of the following equations models the given information?

A ​C = 2n + 450​ B ​C = 450n + 2​ C ​C = 450n − 2​ D ​C + 450 = n​

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−3 −4

A ​(− 8, 0)​ B ​(4, 1.5)​ C ​(3, 4)​ D ​(4, 3)​ 6 Which line does the line y​ = x + 8​intersect with at the point ​(− 5, 3)​?

A ​4x − 3y = 29​ B ​3x − 2y = − 21​ C ​5x − y = − 22​ D none of the above

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7 Use a graphical method to find the solution to the simultaneous equations y​ = 2x − 1​and y​ = 14 − 3x​.

y

D (–1, 4)

A ​x = 5​, y​ = 3​

(1, 3)

B ​x = 0.5​, y​ = 0​ C ​x = 7​, y​ = 13​ D ​x = 3​, y​ = 5​

x

8 Which of the following diagrams shows a pair of linear graphs with a simultaneous solution of 1 ​​, y​ = _ ​x = − ​ _ ​ 4 ​​?

3 3 y A

(–1, 2)

(2, –2)

(–1, –2)

9 Select the equation that would arise when solving the following pair of simultaneous equations by substitution.

(2, 2)

x

​

y = 2x + 8 ​ and ​

2y + 5x = 7​

A ​(2y + 5x) − 2y = 7 − 2(2x + 8)​ B ​4x + 8 + 5x = 7​ C ​2(2x + 8) + 5x = 7​

(1, –3) (–1, –4)

B

D ​2x + 8 = 2y + 5x = 7​ 10 Charlie has ​200​shares of Darkrock and she purchases five shares a week. Ruby has ​50​shares of Darkrock and she purchases ten shares a week.

y (1, 4)

After how many weeks will Charlie and Ruby have the same number of shares?

(–1, 3)

A ​30​ x

C ​300​ D ​350​

(1, –2)

(–2, –2)

B ​35​

11 For which pair of equations would you begin solving via the elimination method by first finding ​​ 2​ + 3 × ​1​​ ? y

C

A

3y​= 2x ​1​ ​​ ​ _ ​ ​ ​​​ x y − 4​= ​ ​ + 4 ​2​ 2

B

x + y​= 2 ​1​ ​​ ​ ​ ​​​ 3x − 3y​= 7 ​2​

(–1, 4) (1, 3)

x (–1, –2)

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(2, –2)

C y + x − 2​= 0 ​1​ ​​ ​ ​ ​​​ 3x − 2y​= 6 ​2​ 3 ​ ​1​ D y=_ ​ 2 ​ x​+ ​ _ 3 ​ 2​ ​​ ​​​ − x − y​= 9 ​2​

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12 Which pair of equations has a solution of ​x = 2​, y​ = 0​?

3y​= 2x ​1​ A ​​ ​ _ ​ ​ ​​​ x y − 4​= ​ ​ + 4 ​2​ 2

Cost and revenue 6000

x + y​= 2 ​1​ ​​ ​ ​ ​​​ 3x − 3y​= 7 ​2​

Amount ($)

B

14 From the graph, determine the number of items that need to be sold for the business to break even.

C y + x − 2​= 0 ​1​ ​​ ​ ​ ​​​ 3x − 2y​= 6 ​2​

13 Caly’s business has costs modelled by the equation ​ C = 550 + 33n​and revenue modelled by the equation R ​ = 55n​. What is her business’s revenue when she sells enough of her product to break even?

Costs

4000 2000 0

3 ​ ​1​ D y=_ ​ 2 ​ x​+ ​ _ 3 ​ 2​ ​​ ​​​ − x − y​= 9 ​2​

Revenue

200

400 600 800 1000 1200 Number of items (n)

A ​600​ B ​4000​ C ​650​ D ​700​

A ​$550​ B ​$1125​ C ​$25​ D ​$1375​

Review questions 1.8C Set 1 1 A linear relationship can be written in the form ​y = mx + c​. a State the gradient and y​ ​-intercept of the line y​ = _ ​ 3 ​x − 2​.

2 b Write the equation of the line with gradient of ​4​and y​ ​-intercept ​3​. c Complete this table of values for the linear relationships in parts a and b. ​​ x ​ya​

​− 3​

​− 2​

​− 1​

​0​

​1​

​yb​ d Use your table of values to find the point of intersection of these two lines. e Use a graphing application or digital tool to confirm your answer to part d.

2 The distance, d ​ ​kilometres, travelled by a train over t​ ​hours is ​d = 70t​. a Determine the value of d ​ ​ when t​ = 0​and when t​ = 5​. b Plot the points found in part a on a Cartesian plane, then use them to sketch the graph of d ​ = 70t​.

1 __ c How far will the train have travelled after ​3​ ​​ hours?

2 d When is the train ​200 km​away?

3 Use a graphical method to solve the simultaneous equations y​ = _ ​ 1 ​x + 1​and y​ = _ ​ 3 ​x​.

4 2 4 Solve the equations y​ = − 3x − 4​and x​ + 2y = 2​simultaneously using a graphical method.

5 Solve each pair of equations simultaneously by elimination.

2x + 3y​= 24 a ​​ ​ ​​​ 3y − 4x​= − 3

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8x − 12y​= 9 b ​​ ​ ​ ​​ 2(6 − x)​= 6y

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6 The Happy Feet Shoe Company has a weekly fixed cost of $​ 3725​and a cost of $​ 7.50​per pair of shoes. The shoes are sold for $​ 45​per pair. a Graph straight lines modelling the cost and revenue for up to ​300​pairs of shoes on the same set of axes. b Find the break-even point. c How many pairs of shoes must be sold to break even? d What is the break-even revenue? e Shade the loss zone on the graph you produced in part a.

7 Jeffrey needs to hire a Ute for the day and has two quotes as shown below.

Company A: $​ 500​plus $​ 1.40​per kilometre Company B: $​ 200​plus $​ 1.90​per kilometre a For what distance driven is the cost of hire from each company equal? b In what situation is it better to hire from Company A than Company B?

Review questions 1.8D Set 2 1 Find the gradient of the line passing through the points ​(1, 3)​and ​(5, 8)​.

2 Consider the linear relationship y​ = 3x + 2​. a Complete this table of values for the given linear relationship. ​​ x ​y​

​− 3​

​− 2​

​− 1​

​0​

​1​

​2​

​3​

b Use the table of values to graph y​ = 3x + 2​on a Cartesian plane. c Use a graphing application or digital tool to produce a graph of y​ = 3x + 2​and compare it with your answer to part b.

3 The Smart Tie Company has a fixed cost of $​ 1400​per day and a variable cost of $​ 4​per tie. The daily cost can be modelled using the straight-line equation C ​ = 4n + 1400​. a Explain each term in the equation C ​ = 4n + 1400​. b Use any method to draw the straight-line graph of C ​ = 4n + 1400​for ​0 ≤ n ≤ 250​. c From the graph, determine the number of ties produced when the cost is $​ 1500​.

Fuel costs

Cost ($)

4 The fuel cost of driving a large 4WD when running on unleaded petrol (ULP) is shown in the graph. When the vehicle is converted to run on liquid petroleum gas (LPG or autogas), the cost for travelling the same distance was calculated and drawn on the same set of axes. From the graph, estimate the distance travelled before the costs are the same.

14 000 12 000 10 000 8000 6000 4000 2000 0

ULP LPG

10 20 30 40 50 Distance travelled (1000 km)

5 Solve the following pairs of simultaneous linear equations using a graphical method. y​= 2x − 3 2x​+ 3y = 0 a ​​ ​ ​​​ b ​​ ​ ​ ​​ y​= 6 − x 3x​− y = − 11

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6 The Supersheer Stocking Company has a fixed cost of $​ 1500​per day and a variable cost of $​ 2​per pair of stockings. This can be modelled by the equation C ​ = 2n + 1500​, where C ​ ​is the cost in dollars of producing ​n​pairs of stockings. The revenue of $​ 12​per pair of stockings sold can be represented by the equation ​R = 12n​, where R ​ ​is the revenue in dollars on the sale of n ​ ​pairs of stockings. a Graph these two equations on the same set of axes for ​0 ≤ n ≤ 300​. b Find the break-even point. c How many pairs of stockings must be sold to break even? d What is the break-even revenue?

Review questions 1.8E Set 3 1 Consider the linear relationship y​ = 5x − 2​. a Complete this table of values for the given linear relationship. ​​ x ​y​

​− 3​

​− 2​

​− 1​

​0​

​1​

​2​

​3​

b Use the table of values to graph y​ = 5x − 2​on a Cartesian plane. c Use technology to produce a graph of y​ = 5x − 2​and compare it with your answer to part b.

2 The Silky Scarf Company sells scarves for $​ 18​each. Sales revenue may be modelled using the equation R ​ = 18n​. a Explain each term in the equation R ​ = 18n​. b Use any method to draw the straight-line graph of R ​ = 18n​. c From the graph, find the number of scarves sold when the revenue is $​ 2700​.

3 This graph models the cost for a company to cater for different numbers of people.

Catering charges

a How much would it cost to cater for ​35​people?

300

c Find the gradient. What is its meaning? d Find the intercept on the vertical axis. What is its meaning? e The model for another catering company is represented by the equation C ​ = 8n​, where C ​ ​ is the cost in dollars to cater for n ​ ​people. Copy the given graph and add the line for C ​ = 8n​ on the same set of axes. f

Cost ($)

b How many people could be catered for $​ 300​?

200

100

0

10 20 40 30 Number of people (n)

50

For how many people is the cost of catering the same for each company? What is this cost?

4 Solve the equations x​ − 3y = 11​and ​5x − 2y = 16​simultaneously using a graphical method.

5 Show that ​(0, − 5)​is a solution to both y 5 ​ y = x − 5 ​ and ​ ​ _ ​ + ​ _ ​ + x = 0​.

2 2 6 The cost (​$C​) of hiring a car from company A is given by C ​ = 60d​, where d ​ ​is the number of days the car is hired. For company B, the cost of hire is C ​ = 100 + 40d​. a Draw the graphs of cost versus number of days for each company on the same set of axes, for up to 8 ​ ​days. b Find the solution to the simultaneous equations C ​ = 60d​and C ​ = 100 + 40d​. c For how many days is the cost the same for both companies?

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7 The Stay Dry Company produces umbrellas. There is a fixed monthly cost of $​ 2800​and it costs $​ 7​ to produce an umbrella. The umbrellas are sold for $​ 22​each. The maximum monthly production of umbrellas is ​300​. a Write two linear equations that can be used to model the production cost and revenue. b Graph straight lines modelling the cost and revenue for ​0 ≤ n ≤ 300​. c Find the point of intersection of the two lines. d How many umbrellas must be sold in a month to break even? e What is the break-even revenue?

8 Plane A took off from Airport Z, at ​1300​, and flew directly west at a speed of ​600 km/h​.

Plane B took off from Airport Y, at ​1300​, ​1800 km​west of Airport Z and also flew west at a speed of ​500 km/h​.

After how many hours will the planes be at the same location?

Review questions 1.8F Set 4 1 Consider the linear relationship y​ = 7 − 2x​. a Complete this table of values for the given linear relationship. ​​ x ​y​

​− 3​

​− 2​

​− 1​

​0​

​1​

​2​

​3​

b Use the table of values to graph y​ = 7 − 2x​on a Cartesian plane. c Use a graphing application to produce a graph of y​ = 7 − 2x​and compare it with your answer to part b.

2 The Tropical Delight Fruit Juice Company has a weekly fixed cost of $​ 400​. The cost per cup of juice is $​ 1.75​. a Write an equation modelling this information. Use C ​ ​for cost and n ​ ​for the number of cups of juice. b Use any method to draw the straight-line graph of your cost equation. c Use the graph to estimate the number of cups of juice made when the cost is $​ 1000​.

a How many items need to be sold for the company to break even?

Cost and revenue

Amount ($)

3 A plastics company has an initial cost of $​ 3000​ per day plus $​ 8​per item produced. The selling price of the items is $​ 15​each, and the maximum daily production is ​1000​items. The revenue and cost models are represented in the graph.

12 000 10 000 8000 6000 4000 2000

b What is the break-even revenue?

Revenue Cost

0

100

200 300 400 500 Number of items

600

700

4 Consider the linear relationships y​ = 8 − x​and y​ = 2x + 3y = 23​. a Draw both linear relationships on the same Cartesian plane. b Find the point of intersection of the two lines. c Hence, write the solution to the simultaneous equations ​

y = 8 − x ​ and ​

2x + 3y = 23​.

5 Millie has two water tanks on her property. Water tank A has just been damaged by a goat and is leaking water at a rate of ​2 L/h​. Water tank B is currently being filled at a rate of ​118 L/h​. After ​125​hours the tanks have the same amount of water in them.

How much water did tank A start with?

6 Tickets to a festival cost $​ 200​for adults and $​ 150​for children. If ​20 000​people paid to go to the festival and the total money paid was ​$3 750 000​. How many adults and children attended?

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7 The cost (​$C​) of electricity to run a home is $​ 200​per month. This may be written as C ​ = 200m​, where m ​ ​ is the number of months. For the same house, the cost of installing a solar power system is $​ 4000​, but the monthly cost of electricity decreases to $​ 150​. The total cost of electricity using a solar system is then ​C = 4000 + 150m​. a Use any method to graph the following lines on the same Cartesian plane ​ C = 200m ​ and ​ C = 4000 + 150m.​ b Find the simultaneous solution of the linear equations. c After installing the solar system, how long would it take to start saving money?

Review questions 1.8G Practice examination questions 1 Consider the linear relationship y​ = 3 − 2x​. a Complete this table of values and use it to graph the given linear relationship. (2 marks) ​​ x ​y​

​− 3​

​− 2​

​− 1​

​0​

​1​

​2​

​3​

b What is the gradient of the line? (1 mark)

Cost and revenue

c What is the y​ ​-intercept of the line? (1 mark) 90 000

2 The graph shows the total costs and revenue for a manufacturer of calculators.

b Will the manufacturer make a profit when ​1000​ calculators are produced? Give reasons for your answers. (1 mark)

70 000 Amount ($)

a Estimate the total cost to the manufacturer of producing ​1000​calculators, to the nearest ​ $5000​. (1 mark)

80 000 Costs

60 000

Revenue

50 000 40 000 30 000 20 000

c How many calculators must be produced for the manufacturer to break even? (1 mark)

10 000

d How many calculators must be produced to make a profit of $​ 10 000​? (1 mark)

0

1000 2500 500 1500 2000 Number of calculators (n)

e What is the initial set-up cost to the manufacturer? (1 mark) f

By identifying the coordinates of two points in the graph, determine the cost to produce a single calculator.

(2 marks)

3 The Creative Cake Company makes cupcakes. The cost of production is $​ 200​per day and the variable production cost is $​ 2.50​per cupcake. Cupcakes are sold for $​ 7.50​each. Maximum daily production is ​100​cupcakes. a Write an equation for the cost $​ C​of producing n ​ ​ cupcakes. (1 mark) b Write an equation for the revenue $​ R​of selling n ​ ​ cupcakes. (1 mark) c Graph both equations for ​0 ≤ n ≤ 100​on the same set of axes. (1 mark) d How many cupcakes must be sold to break even? (1 mark) e If all ​100​cupcakes are sold, how much profit is made? (1 mark) TOTAL: 16 marks

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Checklist Now that you have completed this module, reflect on your ability to do the following.

Module checklist: Simultaneous linear equations

I can do this ⃞ plot linear relationships using a table of values ⃞ sketch linear relationships of the form ​y = mx + c​by hand ⃞ draw graphs of linear relationships using graphing applications ⃞ model cost and revenue using linear relationships ⃞ model practical situations using linear relationships ⃞ identify the solution from tables or graphs when solving linear equations simultaneously ⃞ find the point of intersection between two straight-line graphs using technology ⃞ solve simultaneous linear equations graphically and interpret the solution

I need to review this Go back to Lesson 1.1 Graphs of the form y = mx + c

Go back to Lesson 1.2 Linear models Go back to Lesson 1.3 Identifying solutions to simultaneous equations

⃞ solve simultaneous equations using the substitution method ⃞ change the subject of an equation before using the substitution method to solve simultaneous equations ⃞ solve practical problems which involve simultaneous equations using the substitution method ⃞ solve simultaneous equations using the elimination method ⃞ manipulate equations before using the elimination method to solve simultaneous equations ⃞ solve practical problems which involve simultaneous equations using the elimination method ⃞ solve practical problems using simultaneous linear equations ⃞ determine and interpret the break-even point of a simple business problem where cost and revenue are represented by linear equations ⃞ identify the break-even point and solve problems involving profit and loss using a spreadsheet

Go back to Lesson 1.4 Solving simultaneous equations graphically Go back to Lesson 1.5 Solving simultaneous equations by substitution

Go back to Lesson 1.6 Solving simultaneous equations by elimination

Go back to Lesson 1.7 Break-even analysis

Go back to Using spreadsheets

Related resources Module review quiz: Simultaneous linear equations

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Worksheet: Challenge 1

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