7h1 Simplifying Rational Expressions. Goals:

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1 h Simplifying Rtionl Epressions Gols Fctoring epressions (common fctor, & -, no fctoring qudrtics) Stting restrictions Epnding rtionl epressions Simplifying (reducin rtionl epressions (Kürzen) Adding nd subtrcting rtionl epressions Solving equtions involving rtionl epressions Vocbulry Rtionl epression Common Fctoring Restrictions Epnding Simplifying (reducin/kürzen Steps. Fctor. Restrict. Cncel Nottion \ { }

2 h Prctice Eercises ID A. Fctor the common Fctor out of the following epressions yr mth ctoring nd restrictions Common fctoring nd restrictions nrgey XLnLvCC.9 RAlPl0 dr6icgehxtksy OrBe9swergvehdf.t ) ) n 0n + 0 mon fctor out u 90pRg of kkkudtp ech LSUoRfntGwnrgeY epression. XLnLvCC.9 RAlPl0 dr6icgehxtksy OrBe9swergvehdf.t Fctor the common fctor out of ech epression. 6 ) n 6n ) n 0n + 0 ) ) n ) 0n ID n ) b b ) + 6 b ) n + 6 6n b + 6 b ) 6 6) 6) + 6 y y0 y y 0y 0y + 6 b 6) y y 0y ) ) b 0 + y 6 + 0y b b 6) ) 9n ) y + 9n 9mn + y 9mn + 6m 0y + 6m y 0 ) 9n + 9mn + 6m ) 0 y 9) + z 0y 0 + y ) 9n + 9mn + + y 0) 0) p + q p 6m qm m y + y 0) p q m 9) z ) 0z + y + y 0) p q m + 0y 0y ) ) 6 6 pq pq + p + q p + q + pqr pqr y + 9 ) 6 pq + p q + pqr ) 0z + 0y + 9 ) 6 pq + p q + pqr Stte the restriction on the vrible for ech of the following rtionl epressions (frctionl terms) B. Stte restrictions on ech of the following rtionl epressions iction on the vrible for ech of the following rtionl epressions (frctionl n + terms) Stte the restriction on the vrible for ech of the following rtionl epressions (frctionl terms) ) ) 6 n + + 6n + 6 ) ) n + 9 ) 9 6) r + r ) n + 6n + 6 ) 9r 6 n + 6n + 6 6) ) ) ) 6) v b 6 + v b 6 v + 6 6) v + v v + 6 6) v ) b 6 + v b 6 v + 6 ) 0n 0 ) 0 ) + ) + v + 6 9) ) 60v ) v + 6 ) 9) 60v 0) 6 0) 9) 9n 90n 9) 9n n 90n 0 n 0 n + ) 0k ) n 0k + 0k ) n + 0k ) + 0) 0k 90 n + 0k ) 6 ) 6 ) + 0 ) ) p p + 0 p 0 ) p p + 0 ) p p p 0 n + 0 ) n + ) n 0 + 0n n + 0n ) m m m m ) m ) m m m m Y rifggh6t0ss qr qevsetrvdeedr.k p pe09q I omoduex jkvuktk wi9tghh psoafttnwjkrew EI pnpfoiznoi0the dlflscm.n AlHgeybrdr O BArlJlY rifggh6t0ss A9. qr qevsetrvdeedr.k I omoduex wi9tghh EI pnpfoiznoi0the AlHgeybrdr A9. Worksheet by Kut Softwre LLC Worksheet by Kut Softwre LLC

3 h C. Reducing/ Simplifying (Kürzen) Rtionl Epressions Simplify s fr s possible. Simplify s fr s possible. Indicte wht Stte restrictions. equivlent originl nd cncelled terms the substitutions stnd for. 9 s u ( u) ( ) u t t b b c b b b bc t ( t) ( y ) ( y ) t t y z k) Simplify s fr s possible. ( ) 6 6 y z ( ) 6 6 u u u k) Decide if the rtionl epression cn be simplifiy, nd if so, simplify. b b b b Epnd the rtionl epression so tht 6 ( + ) becomes the denomintor. 6 the terms Clculte the missing numbers Clculte the missing numbers 6 6 y y y y 6 y 6 y Write s frction 9 with the denomintor, Write with the denomintor s frction, with the denomintor + with the denomintor, with the denomintor, 0 with the denomintor + Write ech pir with common denomintor. s 6 Clculte s the 9 missing terms 6 Mtch ech pir of equivlent rtionl epressions nd their domin. 9 0 Write ech pir with common denomintor. Mtch ech pir of equivlent rtionl epressions n their domin. Mtch ech pir of equivlent rtionl epressions n their domin. (Klett p) 6 The rtionl epression is given. Clculte f() if you substitute ech of the following numbers for 0. Wht substitution of gives you the following numbers? The rtionl epression is given. Clculte f() if you substitute ech of the following numbers for 0.

4 h D. Opertions with Rtionl Epressions Solution sld for Find equivlent epressions tht use only whole numbers s coefficients (by epndin 0, 0, , Indicte two rtionl epressions with the vrible for which the vlue of the epression is equivlent for ll Q \ { 0 }, butnotfor=0,=. Indicte frction term which for ll Q \ { } is equivlent to, but not for = ½ nd =. (Klett p) Clculting with rtionl epressions Simplify s fr s possible. Substitute the number for in the simplified term in order to find the vlues in the solution sld k) In the following coordinte system the grphs of three functions re drwn in different colours. Clculte nd simplify s fr s possible. 6. 0, t t Insert suitble term. k) t t t t t 0 k) 6 Clculte nd simplify s fr s possible. z z ( ) z z ( ) Find the grph which belongs to ech of the following si functions nd eplin the connections between the grphs f ( ) f ( ) s f ( ) f ( ) f ( ) f 6 ( ) ( ) 9 Clculte nd simplify s fr s possible u uv v 6 6 y y y y y 6 y y y y y y ( y )

5 h Opertions (continue (Klett p) Klus did the following clcultion Crin did it this wy Simplify these terms s fr s possible % k) 9 ( ) Wht is the quickest wy to check if there nswers re the sme? E. Solving Equtions with Rtionl epressions (Klett p) Solve the frctionl equtions Solve the equtions in your hed

6 h Solving equtions with rtionl epressions (continue (Klett p) Consider the function f Define the the function -intercepts vlue is of the function Define the symptotes of the function grph f nd sketch the grph Wht re the substitutions for, for which the function vlues is? For which -vlues is -vles is true? Define the intersections of the grphs of f with tht of the function 0. g. Add the function grph of g to tht of f 6 Jürgen is trying to solve the eqution He clcultes 0 And And writes writes The numbers the solution to the The eqution. number zero is the solution to the eqution. Wht did he do wrong? Wht did he do wrong? 9 Cross-multiplying Frnks sys To solve equtions in the form c ll I hve to do is ^cross-multiply_. b d He clcultes c b d d cb Eplin why his eplntion is correct. Use the pproch for the following frctionl equtions 6 Cn you lso solve this frctionl eqution in the sme wy? 0 Find the errors which hve been mde in solving the frctionl equtions Solve the frctionl equtions k) Solve the frctions equtions

7 h F. Applictions of Working with Rtionl Epressions (klett p 6) If you dd the sme number of the numertor nd denomintor of the frction you get the frction. Wht is the number? A frction when cncelled gives. If you dd the number to the uncncelled numertor nd denomintor you get the frction.. Wht is the uncncelled frction? Is there frction with the sme vlue s the number 0. whose denomintor is greter thn the numertor? For which frctionl eqution for 0. do the numertor nd denomintor dd up to 00? Solve the following formuls for the vribles in the brckets. m ( V ) A ( c ) h ( c ) V c F t B ( ( t ) h F t b F gt ( t ( b The Wittke fmily is going on holidy by cr. Rmon hs clculted tht they first need to drive 6 km on the Autobhn nd then 90 km on the regionl rods. Mr. Wittke sys tht they cn drive twice s fst on the Autobhn s on the regionl rods. Mrs. Wittke then sys YSo we will need hours for the journey.z Wht speeds does she bse her clcultion on? p 0 It is lwys possible to find the reciprocl frction of s frction by dding n ppropite number to the numertor nd the denomintor. Try this first with few emples nd then write out n pproprite frctionl eqution for one emple nd solve ist. it. Try to find, in generl terms, the number to dd to frction. b If you divide by nturl number nd by the following number, the difference of this quotient is the sme s the quotient of 0 nd the product of the two consecutive numbers. Wht re the numbers?

8 h Review Clculting with rtionl epressions To reduce (kürzen) rtionl epressions, the numertor nd denomintor re divided by the sme term. To epnd, the numertor nd denomintor re multiplied by the sme term. Rtionl epressions with the sme denomintor cn be dded (or subtrcte by dding (or subtrctin the numertors nd keeping the sme denomintor. Rtionl epressions cn be multiplied with ech other by multiplying the numertor with the numertor nd the denomintor with the denomintor. Identicl fctors on the top nd bottom cncel ech other out. (reduce/kürzen) You cn divide rtionl epressions by multiplying the first epression by the reciprocl of the second rtionl epression. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) Frctionl equtions Frctionl equtions need to be trnsformed into denomintorfree equtions by multiplying by the common denomintor. ( ) ( ) ( )( ) Answers (Solutions with steps re vilble online see h solutions) A ) 9(9 + 0 ) ) (n n + ) ) ( 6n n ) ) ( + ) ) b( + 9b + 9 6) y (9 + y + 0) ) 0( y + y ) ) 9(n + mn + m) 9) 9(9z + y + 9y) 0) (p q m ) ) (0z + 0y + ) ) pq( q + p + 6r) ) {0} ) {, 9} ) { 6) { 6} ) {0} } ) {0} 9) {0} 0) {0} ) {0} ) {0} ) {0, 0} ) {0, } ) {} 6) {0} ) {6} ) {} 9) {0} 0) {} ) { } ) { 0, }

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