Lesson 25: Adding and Subtracting Rational Expressions


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1 Lesson 2: Adding nd Subtrcting Rtionl Expressions Student Outcomes Students perform ddition nd subtrction of rtionl expressions. Lesson Notes This lesson reviews ddition nd subtrction of frctions using the fmilir number line technique tht students hve seen in erlier grdes. This leds to n lgebric explntion of how to dd nd subtrct frctions nd n opportunity to prctice MP.7. The lesson then moves to the process for dding nd subtrcting rtionl expressions by converting to equivlent rtionl expressions with common denomintor. As in the pst three lessons, prllels re drwn between rithmetic of rtionl numbers nd rithmetic of rtionl expressions. Clsswork The four bsic rithmetic opertions re ddition, subtrction, multipliction, nd division. The previous lesson showed how to multiply nd divide rtionl expressions. This lesson tckles the remining opertions of ddition nd subtrction of rtionl expressions, which re skills needed to ddress AAPR.C.6. As discussed in the previous lesson, rtionl expressions re worked with in the sme wy s rtionl numbers expressed s frctions. First, the lesson reviews the theory behind ddition nd subtrction of rtionl numbers. Exercise (8 minutes) First, remind students how to dd frctions with the sme denomintor. Allow them to work through the following sum individully. The solution should be presented to the clss either by the techer or by student becuse the process of dding frctions will be extended to the new process of dding rtionl expressions. Exercises. Clculte the following sum: One pproch to this clcultion is to fctor out from ech term Scffolding: If students need prctice dding nd subtrcting frctions with common denomintor, hve them compute the following. * + +, / , *0,* *0
2 Ask students for help in stting the rule for dding nd subtrcting rtionl numbers with the sme denomintor. If, b, nd c re integers with b 0, then b + c b + c b nd b c b c b. The result in the box bove is lso vlid for rel numbers, b, nd c. But wht if the frctions hve different denomintors? Let s exmine technique to dd the frctions * + nd, /. Recll tht when we first lerned to dd frctions, we represented them on number line. Let s first look t * +. And we wnt to dd to this the frction, /. If we try plcing these two segments next to ech other, the exct loction of the endpoint is difficult to identify. The units on the two originl grphs do not mtch. We need to identify common unit in order to identify the endpoint of the combined segments. We need to identify number into which both denomintors divide without reminder nd write ech frction s n equivlent frction with tht number s the denomintor; such number is known s common denomintor. Since is common denomintor of * + nd,, we divide the intervl [0, ] into prts of equl length. Now / when we look t the segments of length * + nd, plced next to ech other on the number line, we cn see tht / the combined segment hs length,,,+. How cn we do this without using the number line every time? The frction * : is equivlent to, nd the +,+ frction, + is equivlent to. We then hve /, Thus, when dding rtionl numbers, we hve to find common multiple for the two denomintors nd write
3 ech rtionl number s n equivlent rtionl number with the new common denomintor. Then we cn dd the numertors together. Hve students discuss how to rewrite the originl frction s n equivlent frction with the chosen common denomintor. Discuss how the identity property of multipliction llows one to multiply the top nd the bottom by the sme number so tht the product of the originl denomintor nd the number gives the chosen common denomintor. Generlizing, let s dd together two rtionl nd A. The first step is to rewrite both frctions s B equivlent frctions with the sme denomintor. A simple common denomintor tht could be used is the product of the originl two denomintors: b + c d d bd + bc bd. Once we hve common denomintor, we cn dd the two expressions together, using our previous rule for dding two expressions with the sme denomintor: b + c d + bc. d bd We could use the sme pproch to develop process for subtrcting rtionl numbers: b c d bc. d bd Now tht we know to find common denomintor before dding or subtrcting, we cn stte the generl rule for dding nd subtrcting rtionl numbers. Notice tht one common denomintor tht lwys works is the product of the two originl denomintors. If, b, c, nd d re integers with b 0 nd d 0, then b + c d d + bc bd nd b c d d bc. bd As with the other rules developed in this nd the previous lesson, the rule summrized in the box bove is lso vlid for rel numbers. Exercises 2 ( minutes) Ask students to work in groups to write wht they hve lerned in their notebooks or journls. Check in to ssess their understnding. Then, hve students work in pirs to quickly work through the following review exercises. Allow them to think bout how to pproch Exercise, which involves dding three rtionl expressions. There re multiple wys to pproch this problem. They could generlize the process for two rtionl expressions, rerrnge terms using the commuttive property to combine the terms with the sme denomintor, nd then dd using the bove process, or they could group the ddends using the ssocitive property nd perform ddition twice π + 2
4 π + 2 π 20 2 π m + b 2m c m m + b 2m c m 2 2m + b 2m 2c 2 + b 2c 2m 2m Discussion (2 minutes) Before we cn dd rtionl numbers or rtionl expressions, we need to convert to equivlent rtionl expressions with the sme denomintors. Finding such denomintor involves finding common multiple of the originl denomintors. For exmple, 60 is common multiple of 20 nd. There re other common multiples, such s 20, 80, nd 00, but smller numbers re esier to work with. To dd nd subtrct rtionl expressions, we follow the sme procedure s when dding nd subtrcting rtionl numbers. First, we find denomintor tht is common multiple of the other denomintors, nd then we rewrite ech expression s n equivlent rtionl expression with this new common denomintor. We then pply the rule for dding or subtrcting with the sme denomintor. If, b, nd c re rtionl expressions with b 0, then b + c b + c b nd b c b c b. Exmple (0 minutes) Work through these exmples s clss, getting input from students t ech step. Exmple Perform the indicted opertions below nd simplify.. Nb + 2Ob A common multiple of nd is 20, so we cn write ech expression s n equivlent rtionl expression with denomintor 20. We hve Nb Nb nd 2Ob 8Ob, so tht Nb + 2Ob Nb + 8Ob Nb b. c. x x 2 A common multiple of x nd x 2 is x 2, so we cn write ech expression s n equivlent rtionl expression with denomintor x 2. We hve x 20x x2 x xO9 x2 x 2. 2x 2 N2x + x 2 OxO Since 2x 2 + 2x 2x(x + ) nd x 2 x (x )(x + ), common multiple of 2x 2 + 2x nd x 2 x is 2x(x + )(x ). Then we hve 2x 2 N2x + x 2 OxO (xo) 2x(xN)(xO) + 2x 2x(xN)(xO)
5 xo2 2x(xN)(xO). Exercises 8 (8 minutes) Hve students work on these exercises in pirs or smll groups. Exercises 8 Perform the indicted opertions for ech problem below.. xo2 + x xo8 A common multiple is (x 2). x 2 + x x 8 20 (x 2) + x (x 2) x + 20 (x 2) 6. 7m mo + m Om Notice tht ( m) (m ). A common multiple is (m ). 7m m + m m 7m m + m m 7m m m m 2m m 7. b 2 b 2 O2bcNc 2 b boc A common multiple is b c b c. b 2 b 2 2bc + c b 2 b c b 2 b c b c b 2 bc b c b c bc (b c) 2 8. x x 2 O 2x x 2 NxO2 A common multiple is x x + x + 2. x x 2 2x x 2 + x 2 x x x + 2x x x + 2 x x + 2 x x + (x + 2) 2x x + x x + x + 2 x 2 (x )(x + )(x + 2)
6 Exmple 2 ( minutes) Complex frctions were introduced in the previous lesson with multipliction nd division of rtionl expressions, but these exmples require performing ddition nd subtrction opertions prior to doing the division. Remind students tht when rewriting complex frction s division of rtionl expressions, they should dd prentheses to the expressions both in the numertor nd denomintor. Then they should work inside the prentheses first following the stndrd order of opertions. Exmple 2 Simplify the following expression. b 2 + b 2b 8 (b + ) First, we cn rewrite the complex frction s division problem, remembering to dd prentheses. b 2 + b 2b 8 (b + ) b2 + b 2b 8 (b + ) Remember tht to divide rtionl expressions, we multiply by the reciprocl of the quotient. However, we first need to write ech expression s rtionl expression in lowest terms. For this, we need to find common denomintors. MP.7 b 2 + b 2b 8 b + b2 + b 2b b2 b 2b 2b 2b b + b + 8 b + b b + b b + Now, we cn substitute these equivlent expressions into our clcultion bove nd continue to perform the division s we did in Lesson 2. b 2 + b 2b 8 (b + ) b2 + b 2b 8 (b + ) b2 b 2b b b + b(b ) b + 2b (b ) b(b + ) (2b )
7 Closing (2 minutes) Ask students to summrize the importnt prts of the lesson in writing, to prtner, or s clss. Use this opportunity to informlly ssess their understnding of the lesson. In prticulr, sk students to verblly or symboliclly rticulte the processes for dding nd subtrcting rtionl expressions. Lesson Summry In this lesson, we extended ddition nd subtrction of rtionl numbers to ddition nd subtrction of rtionl expressions. The process for dding or subtrcting rtionl expressions cn be summrized s follows: Find common multiple of the denomintors to use s common denomintor. Find equivlent rtionl expressions for ech expression using the common denomintor. Add or subtrct the numertors s indicted nd simplify if needed. Exit Ticket ( minutes)
8 Nme Dte Lesson 2: Adding nd Subtrcting Rtionl Expressions Exit Ticket Perform the indicted opertion.. /?N* + 0?O+ 2. 0U UN/ + U
9 Exit Ticket Smple Solutions Perform the indicted opertion.. N2 + O ( + 2)( ) 2. r rn r r r + r r2 r r + r + r r + r2 r r r +
10 Problem Set. Write ech sum or difference s single rtionl expression b c. x + 2x 2. Write s single rtionl expression.. x xo b. x 2y x 6y + x y c. Ob 2 + d. po2 pn2 e. po2 + 2Op f. bn b Nb g. +p h. pnq r poq 2 i. sor + s rns j. xo + 2 Ox k. n no2 + 2On l. 8x yo2x + 2y 2xOy m. 2mOn 2mNn m m 2 On 2 n. (2Ob)(Oc) + (boc)(bo2) o. b 2 N b 2 O + bn2 + bo2. Write ech rtionl expression s n equivlent rtionl expression in lowest terms.. O 2 b. x 2 N x O x c. x \ N x 2 \ x \ 7 O x 2 \
11 . Suppose tht x 0 nd y 0. We know from our work in this section tht x y tht x + y is equivlent to? Provide evidence to support your nswer. xny is equivlent to. Is it lso true xy. Suppose tht x 2t t2 2 nd y +t +t 2. Show tht the vlue of x2 + y 2 does not depend on the vlue of t. 6. Show tht for ny rel numbers nd b, nd ny integers x nd y so tht x 0, y 0, x y, nd x y, y x x y x+by x+y x by x y 2 b. 7. Suppose tht n is positive integer.. Rewrite the product in the form P Q for polynomils P nd Q: + n + n+. b. Rewrite the product in the form P Q for polynomils P nd Q: + n + n+ + n+2. c. Rewrite the product in the form P Q for polynomils P nd Q: + n + n+ + n+2 + n+. d. If this pttern continues, wht is the product of n of these fctors?
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