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  RUSH HOU UR   The strrategies of “R Rush Hour” caan be used to address num merous NY Staate Math Stan ndards in the  hich the Deteective  Integraated Algebra curriculum.  TThis document cites speci fic examples of ways in wh Metho od and Divide and Conquer relate to staandards undeer the Algebraa strand.    The De etective Method    ethod, the pro oblem is defined, questionns are asked b based on cau use and effectt, and the  In the Detective Me em is exposed d and solved in a back to frront way.  Thiis supports th he strategy off equation solving  proble which uses inverse operations to o undo steps,, starting withh the last onee first.  This m method can bee used to  most type of e equations and inequalitiess, including o nes addresseed in the follo owing standarrds.  An  solve m examp ple is illustrate ed below.    A.A.6 - Analyze an nd solve verbaal problems whose w solutionn requires sollving a linear equation in oone variablle or linear in nequality in on ne variable A.A.7 - Analyze an nd solve verbaal problems whose w solutionn requires sollving systemss of linear equuations in two vaariables A.A.8 - Analyze an nd solve verbaal problems th hat involve quuadratic equattions A.A.9 - Analyze an nd solve verbaal problems th hat involve exxponential groowth and decay 0 - Solve systtems of two liinear equation ns in two variiables algebraaically A.A.10 A.A.11 1 ‐ Solve a system of one linear and one e quadratic eqquation in two variables, w where only factoring  is requ uired  Note: TThe quadratic equation sho ould represen t a parabola and the soluttion(s) should d be  integers.  A.A.22 2 - Solve all types t of linearr equations in n one variablee A.A.23 3 - Solve literral equations for a given vaariable A.A.24 4 - Solve lineear inequalitiees in one variaable A.A.25 5 - Solve equations involving fractionall expressions Note: Expreessions whichh result in lineear equatio ons in one variable. A.A.26 6 - Solve algeebraic proporttions in one variable v whichh result in linear or quaddratic equationns       

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The De etective Meth hod    1) Defiine the objecttive.   

  d from  2) Ask:  “How am I being blocked achievving my objective?”      3) Und do the problem in a back to o front way.

    

Appplication of SStrategy to Eq quation Solvin ng:    Solve: 7xx + 15 = 36    olate x.   The objeective is to solve for, or iso     a) x is beeing multiplieed by 7.  b) 15 is bbeing added to the 7x.      a) To unndo “15 is beinng added to 7 7x”, use inverrse  operatioons to subtracct 15 from bo oth sides of th he  equationn.    7x + 15 == 36        ‐ 15    ‐15  7x         == 21    b) To unndo “x is beingg multiplied b by 7”, use inveerse  operatioons to divide bboth sides of the equation n by 7.    7x         == 21  7               7             x == 3   

 

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e and Conque er  Divide   nquer strategyy states that w when a probllem is too diffficult to solvee in one or tw wo steps, it  The Divide and Con e dismantled iinto its primaary componen nts and each ccomponent ccan be solved separately.  This is  can be particu ularly applicable to problems that comb bine standardds from the polynomial (beelow) and equ uation  solving g (above) secttions of the In ntegrated Alggebra Standarrds.  An exam mple is illustraated below.    2 - Multiply and a divide mo onomial expreessions with a common baase, using the properties off A.A.12 exponeents Note: Use Us integral exxponents onlyy. A.A.13 3 - Add, subtrract, and multtiply monomiials and polynnomials A.A.14 4 - Divide a polynomial p by y a monomiall or binomial,, where the quuotient has noo remainder A.A.15 5 - Find valuees of a variable for which an a algebraic ffraction is unddefined A.A.16 6 - Simplify fractions f with h polynomialss in the numerrator and denominator by ffactoring bothh and renamiing them to lo owest terms A.A.17 7 - Add or sub btract fraction nal expression ns with monoomial or like bbinomial denoominators A.A.18 8 - Multiply and a divide alg gebraic fractio ons and expreess the producct or quotientt in simplest fform A.A.19 9 - Identify an nd factor the difference off two perfect ssquares A.A.20 0 - Factor alg gebraic expresssions compleetely, includinng trinomials with a lead ccoefficient of one (after factoriing a GCF)      

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Divide e and Conque er    1) Deconstruct the problem.   

  2) Solvve each comp ponent.   

  blem.  3) Reconstruct prob

Applicattion of Strateegy to Equatiion Solving In nvolving  the Simpliffication of Po olynomials:    Solve forr r:  4(2r + 1)) + 5r = 41 – ((6r + 21)/3    n down into tw wo  The aboove problem ccan be broken equivaleent expressionns:    6r + 21)/ 3  a) 4(2r ++ 1) + 5r     and     b) 41 – (6     Follow tthe order of ooperations to ssimplify each h  expressiion:    a) 4(2r ++ 1) + 5r         8r +  4   + 5r      (m multiplication))       13r +  4                (addition)    b) 41 – ((6r + 21)/ 3      41 ‐   22r + 7          (ddivision)      34 –  22r                 (ssubtraction)      he two expresssions  To reconnstruct the prroblem, set th equal too one anotherr (as they werre in the original  problem m) and solve.     13r + 44 = 34 – 2r  + 2r                 + 2r    15r + 44 = 34            ‐ 44    ‐ 4   15r          = 30  15                15         r = 2  

      

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QUORIDO OR The strrategies of “Q Quoridor” can n be used to aaddress numeerous NY Statte Math Stand dards in the In ntegrated  Algebrra curriculum.  This docum ment cites specific examplees of ways in w which they reelate to stand dards  under the Algebra sstrand.  The SStoplight Method, Self‐Blo cking, and Paaving the Route can be inteegrated  ne strategy ussed to solve vvirtually any m math problem m, while the EEffective Alloccation of Reso ources can  into on be use ed to generate e real life exaamples which demonstratee the necessitty of math in the successfu ul  planning of resourcces.    The Sttoplight Meth hod, Self‐Bloccking, and Paving the Rou ute    The Stoplight Method, Self‐Blockking, and Pavving the Routee, can be insttrumental in ccompleting viirtually all  math p problems.  Students often solve problems using matth that they aare comfortab ble with, whicch is not  alwayss the correct math for thatt problem.  Fo or example, ggiven the prob blem (7x + 3)(x2 – 3x + 12), students  sometimes approacch it as if it we ere (7x + 3) + + (x2 – 3x + 12)), without sto opping to reco ognize that th he  em is a multip plication, rather than an ad ddition probleem.   proble   The strrategy illustraated below, w which is a com mbination of tthe Stoplight Method, Selff‐Blocking, an nd Paving  the Ro oute, can be u used to solve virtually any math problem m, such as these from the Variables and  Expresssions section n of the Integrrated Algebraa curriculum.   2 - Multiply and a divide mo onomial expreessions with a common baase, using the properties off A.A.12 exponeents Note: Use Us integral exxponents onlyy. A.A.13 3 - Add, subtrract, and multtiply monomiials and polynnomials A.A.14 4 - Divide a polynomial p by y a monomiall or binomial,, where the quuotient has noo remainder A.A.15 5 - Find valuees of a variable for which an a algebraic ffraction is unddefined A.A.16 6 - Simplify fractions f with h polynomialss in the numerrator and denominator by ffactoring bothh and renamiing them to lo owest terms A.A.17 7 - Add or sub btract fraction nal expression ns with monoomial or like bbinomial denoominators A.A.18 8 - Multiply and a divide alg gebraic fractio ons and expreess the producct or quotientt in simplest fform A.A.19 9 - Identify an nd factor the difference off two perfect ssquares A.A.20 0 - Factor alg gebraic expresssions compleetely, includinng trinomials with a lead ccoefficient of one (after factoriing a GCF)    

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The Sttoplight Meth hod for Math h, which integgrates  Self‐Blockin ng and Pavingg the Route:   d analyze the e problem.  Red Ligght:  Stop and   What ttype of problem is it?  How do you know?   

  w Light:  Determine the strrategy require ed to  Yellow solve tthe problem ccorrectly.  By writing out the  type of problem and appropriate e strategy,  nts block them mselves from using an inco orrect  studen metho od and pave the route for successfully  answe ering the quesstion.    quired to solvve this proble em  What sstrategy is req correctly?      Green Light:  Carefully employ the strategy.

Application of Strategy tto Example M Math  Problem m:    2)  Probblem:  (7x + 33)(x2 – 3x + 12   Thiss is a multipliccation problem m.  I know beecause  therre are two polynomials sep parated by  pareentheses and there is no op peration sign n  betw ween the pareentheses.      Mulltiply every teerm in the firsst set of paren ntheses  by eevery term in the second seet of parentheeses.   Thenn combine likke terms. 

  Mulltiply:    7x(xx2) = 7x3  7x(‐33x) = ‐21x2  7x(112) = 84x    3(x2)) = 3x2  3(‐33x) = ‐9x  3(122) = 36    mbine like term ms:  Com   7x3 –– 21x2 + 84x         +  3x2  ‐   9x ++ 36    7x3 –– 18x2 + 75x ++ 36   

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n of Resource es  Effectiive Allocation   d to plan for tthe use of lim mited resourcees, such as treees and crudee oil, to ensurre that we  In real life, we need have aa sufficient supply until we e are no longe er dependentt on them.  Th he example below illustrattes the  necesssity of math to plan for the e effective allocation of re sources.    (Example 1)    here are t full grown trees on earth.  Ea ch year, f fulll grown trees are destroyeed in  Situation:  Today th on, paper, annd other uses.  Based on cu urrent analysis and  forest fires and c arre cut down for constructio onal a trees reach full grow wth each yea r.  planning, an additio   omework, havve students re esearch the actual values oof t, f, c, and aa.  For ho   Standard S Examplee Question R Related to “Quoridor” A.A.1 ‐‐ Translate a quantitative vverbal phrase e  a) Use th e informationn above to set up an expreession in  terms of  t, f, c, and a tthat represen into an n algebraic exxpression  nts the numbeer of full  grown treees that will eexist after y yyears.    tt – (f+c)y + ayy    A.A.4 - Translate verbal sentencces into b) Use th e variables deefined above to translate tthe  mathem matical equattions or inequ ualities followingg into an equaation:    mber of full grrown trees, n,, after y yearss, is the  “The num original nnumber of fulll grown minu us the numberr of trees  that havee been destrooyed over y yeears plus the n number  of trees tthat have reacched full grow wth over y years.”    n == t – (f+c)y + a ay    c) Re‐writte the equatioon in part b sso that it reprresents a  situation  in which the total numberr of full grown n trees is  less than  500.    5000 > t – (f+c)y ++ ay  or  t – ((f+c)y + ay < 5 500        330 East 85th h Street, Suite C C • New York,  NY 10028 • Teel: (212) 717‐0 0265 


A.A.5 - Write algeb braic equation ns or present a situaation inequaalities that rep

A.A.3 - Distinguish h the differencce between an n algebraic expression n and an algeebraic equation   A.A.6 - Analyze an nd solve verbaal problems uires solving a linear whose solution requ on in one variiable or linearr inequality in n equatio one vaariable A.A.22 2 - Solve all types t of linearr equations in n one vaariable A.A.24 4 - Solve lineear inequalitiees in one variablle A.A.25 5 - Solve equations involving fractionall expresssions Note: Expressions E which w result in linear equations in one variable..    

d) Set up  an equation in terms of t, f, c, and a th hat could  be used tto solve for thhe number of years we havve  remainin g before all trrees are depleted.    n == t – (f+c)y + a ay  becomes  0 == t – (f+c)y + a ay,  w which can be re‐arranged to solve for yy:  y == ‐t / (‐f – c + a)    hat  e) Set up  an inequalityy in terms of tt, f, c, and a th ber of new fu ull grown  could be  used to solvee for the numb we have  trees we  would need eeach year to eensure that w trees to last A AT LEAST 300 years.  enough tr   n == t – (f+c)y + a ay  becomes  0 ≤ t –– (f+c)300 + a a(300)  becausse the total number of fulll grown trees in 300  years neeeds to be greeater than orr equal to 0.  TThis can  be re‐arrranged to sollve for a:  a ≥≥ (t/‐300) + f ++ c    f) What m makes the staatement in pa art a an expreession?   What maakes the stateement in part b an equatio on?    g) Replacce t, f, c, and aa in the equation in part d d with the  values yoou researchedd for homewo ork.  Solve for the  actual nuumber of yearrs we have remaining befo ore all  trees are  depleted.    with the  h) Replacce t, f, and c, iin the inequality in part e w values yoou researchedd for homewo ork.  Solve for the  number oof new full groown trees wee would need each  year to ennsure that wee have enoug gh trees to lasst AT  LEAST 3000 years.     

 

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A.A.21 1 - Determinee whether a giiven value is a solutio on to a given linear l equatio on in one variablle or linear in nequality in on ne variable

   

i) Your friiend determinnes that an ad dditional 2 biillion  new full ggrown trees eeach year wou uld last exacttly 400  years.  Dooes this agreee with your da ata?    a ≥≥ (t/‐300) + f ++ c  from m part e becom mes  2,000,0000,000 = (t/‐40 00) + f + c    for t, f, and c ccan b plugged d in and the eequation  Values fo can be  worked out tto determine if it comes ou ut to be  true.    j) Your friiend determinnes that we n need AT LEASTT 2  billion neew full grown trees each to o last 400 yea ars.  Does  this agreee with your ddata?    In this ssituation, y=4 400 and  00,000.  So, (t/‐300) + f + c ≥ 2,000,00   a ≥≥ (t/‐300) + f ++ c  becomes  2,000,0000,000 ≤ (t/‐40 00) + f + c    d in and the in nequality  Values foor t, f, and c ccan b plugged can be  worked out tto determine if it comes ou ut to be  true.     

 

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The ab bove example e can be slighttly modified tto address th e following sttandards.    bal expression n that matche es a given maathematical expression  A.A.2 ‐‐ Write a verb A.A.7 - Analyze an nd solve verbaal problems whose w solutionn requires sollving systemss of linear equuations in two vaariables A.A.8 - Analyze an nd solve verbaal problems th hat involve quuadratic equattions nd solve verbaal problems th hat involve exxponential groowth and decay A.A.9 - Analyze an 0 - Solve systtems of two liinear equation ns in two variiables algebraaically A.A.10 A.A.11 1 ‐ Solve a system of one linear and one e quadratic eqquation in two variables, w where only factoring  is requ uired  Note: TThe quadratic equation sho ould represen t a parabola and the soluttion(s) should d be  integers.  A.A.23 3 - Solve literral equations for a given vaariable A.A.26 6 - Solve algeebraic proporttions in one variable v whichh result in linnear or quadraatic equations   For example: 

  

students caan be provid ded an equatiion and key and asked foor the situatiion that the eequation represents uations by sttudying the availability of two or moore students caan create a system of equ resources over o a speciffic time periiod students caan be asked to examine the t depletionn and/or production of reesources thaat increase or decrease at an exponential rate or quadraticc rate

Studen nts can also b be asked to an nalyze their findings to thee above situattions on a coo ordinate plan ne.  In this  situation, questionss can be devissed which address the foll owing standaards.    A.A.32 2 - Explain sllope as a rate of change bettween dependdent and indeependent variaables A.A.33 3 - Determinee the slope off a line, given the coordinattes of two points on the linne A.A.34 4 - Write the equation of a line, given itts slope and thhe coordinatees of a point oon the line A.A.35 5 - Write the equation of a line, given th he coordinatees of two poinnts on the linee A.A.36 6 - Write the equation of a line parallel to the x- or yy-axis A.A.37 7 - Determinee the slope off a line, given its equation iin any form A.A.38 8 - Determinee if two lines are parallel, given g their eqquations in anny form A.A.39 9 - Determinee whether a giiven point is on o a line, giv en the equatioon of the linee A.A.40 0 - Determinee whether a giiven point is in i the solutionn set of a systtem of linear inequalities A.A.41 - Determinee the vertex an nd axis of sym mmetry of a pparabola, giveen its equation     

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OOPS    Strategiess taught in the game Oops can be applied to nearly aany math top pic, including tthose in num mber  sense, alggebra, geometry, statistics,, and measurrement.  The eexamples bellow illustrate the use of “M Magic  Rules”, “TThe Movie Camera Method d”, and “One Step at a Tim me” to a varieety of problem ms in numberr  sense, alggebra, and geo ometry.    Magic Rules    ere are “Maggic Rules” in th he game Oop ps, there are ““Magic Rules”” in every areea of math.  A An  Just as the example o of a “Magic Rule” in numb ber sense is th he order of opperations.  An ny time theree are multiple  operation ns in a problem m (any combination of mu ultiplication, ddivision, addittion, and/or ssubtraction),  there is a certain orderr in which the e operations n need to be doone:     1) Parentheses  2) Exxponents  3) Multiplication/ M / Division  raction  4) Addition/Subt A   es of “Magic Rules” in geometry are forrmulas and anngle relationsships.  For exaample, the arrea of  2 example a triangle can always b be defined as Areatriangle = ½ ½ (base)(heig ht) and when n 2 parallel lin nes are cut byy a  transversaal, corresponding angles w will always be equal.       Problem SSolving One SStep at a Time    (Algebra EExample – Co ombining Like e Terms)    problems mayy seem so com mplicated thaat it appears tthere is no waay of getting tto a correct  Initially, p solution.  For example,, when asked to simplify:   6a + 7ab2 – 8b2 + 2b2 – – ab2 + 4a2b – 3a + 9a2 – 6aab2 – 2a2b + 1 6b2 – 19a2b,   the proble em may seem m daunting at first.  Howevver, if we breaak it down intto its constitu uents, it beco omes  much simpler.  First, id dentify all of tthe different ttypes of term ms in the prob blem:    Types of  a  ab2  b2  a2b  a2  terms:    Next, list o out all of the terms from tthe problem tthat fall into eeach categoryy.  (Magic Rulle:  Be sure to o  keep the ssymbol that ccomes before e each term w with that term m.)  As you fill  each term in nto the  appropriaate column, crross out the tterm in the orriginal probleem.     

 


Types of  terms:  Terms fro om  problem: 

ab2 

b2 

a2b 

a2 

6a  – 3a   

– ab2  + 7ab2  – 6ab2 

– 8b2  + 2b2  + 16b2

+ 44a2b  – 22a2b  – 119a2b 

+ 9a2     

  Finally, co ombine all of the terms witthin each colu umn:    Types of  a  b2  ab2  a2b  a2  terms:  Terms fro om  6a  – ab2  – 8b2  + 44a2b  + 9a2  2 2 2 problem:  – 3a  + 7ab   + 2b   – 22a b    2 2 2   – 6ab   + 16b – 119a b    Sum of ea ach  3a  0  10b2  ‐17a2b  9a2  column:    The final aanswer is  3a + 10b2 ‐17a2b + 9a  b     (Geometrry Example – Dissecting a Word Proble em involving A Area of a Circcle and Area of a Square)   Many stud dents react to o a complex w word problem m with frustraation.  They th hink it’s too ccomplicated aand  give up be efore they even begin.  Keep in mind th hat all problem ms can be takken apart into o smaller, sim mpler  problems.   Rememberr the Magic R Rules with word problems:     1) Determine what the proble em is asking. 2) Draw a picture e.   3) Write down im W mportant information from m the problem m.  4) Write down an W ny relevant fo ormulas.  5) Answer the qu uestion.    Example p problem:  The e area of a square is x2.  W What is the circcumference o of the largest circle, in term ms of  x, that can n be inscribed d in the squarre?    At first, th his problem m might seem vaague and com mplex.  When  broken down n into parts, h however, the  problem b becomes morre clear.    1) Th he problem iss asking for a circumferencce.  2) (P Picture is belo ow.)  2 3) Areasquare = x   4) Areasquare = (sid de)2  Circumference ecircle = (π)(diameter)  5) (A Answer is belo ow.)       


on is asking fo or the circum mference of th he circle.  Sincce  The questio Circumfere encecircle = (π)((diameter), thhe only unkno own is the diaameter of thee  circle.  Draw w the diametter of the circcle.  Notice th hat the diameeter is the sam me  length as th he side of the e square.   

 

 

 

 

 

Since we arre given:          Areasquare = ((x)2  And we kno ow:                   Areasquare = ((side)2,    we can con nclude that eaach side of thhe square (and d the diameteer of the circlle) is  equal to x ((remember th hat (x)(x) = (x 2)). 

 

 

 

now:          Cirrcumferenceccircle = (π)(diam meter)  Since we kn   And we dettermined that the diameteer is equal to  x,                                          Circcumferenceci rcle = (π)(x). 

x      x 

 

 

   

     

      

 


The Movvie Camera M Method    (Algebra EExample – So olving Equatio ons)    nished movie e.   Imagine th he equation iss the last scen ne in a moviee, and you neeed to  An equatiion is like a fin figure outt what the pre evious sceness were that le ed to the equ ation.  For exxample, if thee equation 2x + 15  = 29 is the e last scene in n a movie, traace the steps that led from m “x” to “2x + 15”:    Scene 2:  Sceene 3:  Scene 1:        2x  x  2x + 15        Add  15. h x.                                 Multiply by 2 2.                               Start with   Now, worrk backwards to go from th he last scene to the first sccene:    Scene 2:  Sccene 3:  Scene 1:        2x  x  2xx + 15        Su ubtract 15.    Divide by 2.                                   End with xx.                                   Remembe er, there are 2 2 Magic Ruless with solvingg 2‐step equa tions:    1) Use inverse op perations und do steps, startting with the  last one first (The Movie C Camera Meth hod  illlustrated abo ove).  2) Whatever you  W de of the equation, you muust do to the other (illustrrated below). do to one sid   Sccene 3:  Scene 2:  Scene 1:        2x  x  2xx + 15        Su ubtract 15.  End with xx.                                   Divide by 2.                                         299  7  14        Su ubtract 15.  7.                                   Divide by 2.                                   End with 7   We’re don ne!  The value e of x is 7.  Byy employing tthe Movie Cam mera Method d and following the Magic   Rules, we have successsfully solved tthe equation!     

 


DRAG GON TREASUR RE    Dragon Trreasure can b be related to p probability an nd improving  the probability that studeents will be  successful in life.  For instance, if a sspider web re epresents a ppoor decision,, is it possiblee to improve tthe  of not makingg a poor decision by payingg close attenttion to the deecisions that yyou and the  chances o people around you maake and the co onsequences of these chooices?  Each caard in the gam me of Dragon n  Treasure can represent something iin real life.  Fo or example, tthe treasure cchest can rep present money and  the doll caan represent family.    (Probability Example)    1) It's you ur turn and yo ou are trying tto find 4 treassure chests.  You have alreeady revealed d 3, as well ass 2  diamond rings and a doll.  There are e 14 facedow wn cards to chhoose from.  YYou know thee location of 1 1 out  of the 3 sp pider webs.  YYou do not kn now what anyy of the otherr 14 cards aree.    a) What's the probability of selectin ng a 4th treassure chest verrsus selectingg a spider web b?    b) Previou usly, your frie end had reveaaled one of th he other spideer webs.  If yo ou had paid m more attentio on  and remembered the location of this spider web b, would yourr chances of sselecting the 4 4th treasure cchest  improve?  If so, by how w much? 


The game SET can c be modiffied to addreess numerouss NY State M Math Standaards, includinng those from m the mber Sense, Algebra, A Geo ometry, Meaasurement, and a Statisticss and Probabbility strandss of the Integgrated Num Algeebra curricullum. Specifiic examples of an introdu uctory lessoon and new vversion of SE ET are outlinned in this docuument, which h address staandards A.A A.12 and A.A A.13 below. This knowleedge is criticcal to learninng standards A.A.14 through A.A.20, whiich can be ad ddressed in later l versionns of SET. Select standard ds from the Algebra A strand of the NY N State In ntegrated Allgebra Standards: ply and divid de monomia al expressio ons with a coommon basse, using thee properties of A.A..12 - Multip expoonents Notee: Use integrral exponents only. A.A..13 - Add, subtract, and d multiply monomials m and polynom mials A.A.14 - Divide a polynomial by a mono omial or bin nomial, wherre the quotiennt has no rem mainder A.A.15 - Find vaalues of a vaariable for wh hich an algeebraic fractioon is undefinned A.A.16 - Simpliffy fractions with w polynomials in the numerator aand denominnator by facttoring both aand renaming g them m to lowest teerms A.A.17 - Add orr subtract fraactional exprressions with h monomial or like binom mial denomiinators A.A.18 - Multiply and divide algebraic fractions f and d express thee product or quotient in ssimplest form m A.A.19 - Identify fy and factor the differen nce of two peerfect squarees A.A.20 - Factor algebraic ex xpressions co ompletely, in ncluding trinnomials withh a lead coeff fficient of onne (after factooring a GCF))


ke Terms” Introductory Leesson: “Lik Matterials: 12 of the 81 card ds used in the new versio on of SET (illustrated beelow) 1 sett of 12 cards for each gro oup of 2-3 sttudents blackkboard/ whitteboard and chalk/ whiteeboard mark kers Aim m: The objective off the lesson is for studentts to identify y, define, andd add like teerms as well as multiply non-like term ms. Warrm-up: Divide students into groups of 2-3. Givee each group p the set of 112 cards illusstrated below w. Ask studeents to categgorize the fo ollowing card ds into 3 gro oups of 4 carrds each. Leet them know w that they w will be expeccted to suppport their reaasoning for how h they div vided the card ds. Howeveer, do not givve them any guidance ass to how the cards should be divided. d

x

3x2

-3y y

4x2

6yy

-x

-2xx

5y

9x x2

-y

4xx

-2x2

Whille students are a dividing the t terms intto groups, write w “Group 1 Group 2 Group 3” on the boardd as follows. Be suure to leave room below w and to the right r of the columns. c Affter studentss have divideed the cards into 3 grouups, ask indiv viduals from m 3 different groups who have correcctly divided tthe cards intto groups of “like terms”” to wrrite their ressponses in 3 columns on the board, as a follows. T The vertical order of the terms does nnot matter. Group p1 x -x -2x 4x

Group 2 3x2 4x2 9x2 -2x2

Group 3 -3y 6y 5y -y

Lessson: Ask students thee difference between b the cards in gro oups 1 and 3.. (The termss have differrent variablees – the term ms in group 1 have an “x” while the terms t in group 3 have a “y”.) Ask students if the terms in groups g 1 and d 2 have the same variabble. (Yes, booth groups 1 and 2 contaain an “x” ass theirr variables.) Ask studen nts the differeence betweeen the cards iin groups 1 aand 2. (The terms have different expoonents – the terms t in gro oup 1 have an n exponent of o 1 while the terms in grroup 2 have an exponent of 2.) Define “like term ms” with students: (likee terms - term ms that have the same va ariables raiseed to the sam me exponentt. Like termss can be addded or subtrracted. Non n-like terms can c NOT be added or su ubtracted). a each collumn of like terms in theeir notebook s, while 3 sttudents add tthe columns on the Ask students to add boarrd:


(SU UM of eacch column)

Group p1 x -x -2x 4x 2x

Group 2 3x2 4x2 9x2 -2x2 14x2

Group 3 -3y 6y 5y -y 7y

Reviiew with students. Note that althoug gh non-like terms t can NO OT be addedd or subtractted, they cann be multtiplied or div vided. (It is suggested th hat students have h previouusly learnedd to multiply terms with a common base using the laaws of expon nents.) Ask students to multiply m eac h row of term ms in their nnotebooks, w while 4 studeents multiply y the rows on the board: (PRO ODUCT of Group Group 3 p1 eachh row) Group 2 x 3x2 -3y -9x3y -x 4x2 6y -24x3y 2 -2x 9x 5y -90x3y 4x -2x2 -y 8x3y 2 (SU UM of 2x 14x 7y eacch column) Ask students if the new colu umn consists of like term ms. Since it ddoes, you maay have studdents add thee terms in thee new column.


w Version off SET: “Opeerations witth Monomia als” New Aim m: The objective off the game iss to recognize and collect sets of 3 caards from thee 12 cards laaid out on thhe table. on: Starrting Positio Studdents are diviided into gro oups of 2-4 students. s Eaach group recceives 81 caards. One off the players,, (the dealeer), lays 12 cards c face up p in a rectangular formattion on the taable. The reest of the carrds are stackked face downn in a pile in n the center. The Rules of the Game: The rules of the game are ideentical to thee original veersion of set with the folllowing exceptions: 11.

Each card d has a term written w on it. Students are to identify 2 terms that caan be added, ssubtracted, muultiplied, or r divided to o result in a 3rd term in the set. For exaample, the caards x, 2x, an nd 2x2 are a set because (x)(2x) =

2x2. Stud dents may allso support this t selection n by noting tthat (2x2)/(2xx) = x or (2xx2)/(x) = 2x. 22. Some of the terms co ontain numbeers with no variables. v E Each set of 3 cards must contain at leeast one variable. For example, the cards 1, 2, and ½ are NOT a sset because nnone of thesse cards conttains a ( though h (1)/(2) = ½). ½ variable (even


mples of Setts (Note, as evidenced in n rule 1 abov ve, that theree are multiplle valid reasoons for eachh set. Also, Exam thesee are just som me of the setts created ou ut of the 81 cards; c there aare many othhers.):

C C1

C2 2

C3 3

x x x x x x x x x x x x 22x 22x 22x 22x 22x 22x 22x 22x 3x 3x 3x 3x x y 22x 22y 3x 3y

x x x x 2x x 2x x 2x x 2x x 3x x 3x x 3x x 3x x 2x x 2x x 2x x 2x x 3x x 3x x 3x x 3x x 3x x 3x x 3x x 3x x y x 2y y 2x x 3y y 3x x

2x x2 0 1 3x 2x2 -x 1/2 2 4x 3x2 -2x x 1/3 3 4x 4x2 0 1 5x -x 6x2 2/3 3 6x 9x2 0 1 x/y y y/x x x/y y y/x x x/y y y/x x

Reason: C1 (op peration) C2 = C3 Addition Multip plication Subtraaction Divisio on Addition Multip plication Subtraaction Divisio on Addition Multip plication Subtraaction Divisio on Addition Multip plication Subtraaction Divisio on Addition Multip plication Subtraaction Divisio on Addition Multip plication Subtraaction Divisio on Divisio on Divisio on Divisio on Divisio on Divisio on Divisio on

C1

C2

C3

y y y y y y y y y y y y 2y 2y 2y 2y 2y 2y 2y 2y 3y 3y 3y 3y x y x x y y

y y y y 2y 2y 2y 2y 3y 3y 3y 3y 2y 2y 2y 2y 3y 3y 3y 3y 3y 3y 3y 3y -1 -1 -2x -2x -2y -2y

2y y2 0 1 3y 2y2 -y 1/2 4y 3y2 -2y 1/3 4y 4y2 0 1 5y -y 6y2 2/3 6y 9y2 0 1 -x -y -2x2 -1/2 -2y2 -1/2

Reason: C1 (operaation) C2 = C3 Addition Multiplicaation Subtractioon Division Addition Multiplicaation Subtractioon Division Addition Multiplicaation Subtractioon Division Addition Multiplicaation Subtractioon Division Addition Multiplicaation Subtractioon Division Addition Multiplicaation Subtractioon Division Multiplicaation Multiplicaation Multiplicaation Division Multiplicaation Division


Mind Lab Sample Math Connections for High Schools