ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS

CHAPTER ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS Although people tody re mking greter use of deciml frctions s they work with clcultors, computers, nd the metric system, common frctions still surround us. We use common frctions in everydy mesures: 4 -inch nil, -yrd gin in footbll, pint of crem, cups of flour. We buy dozen eggs, not 0. dozen eggs. We describe minutes s 4 hour, not 0. hour. Items re sold t third A B off, or t frction of the originl price. Frctions re lso used when shring. For emple, Andre designed some beutiful Ukrinin eggs this yer. She gve onefifth of the eggs to her grndprents.then she gve one-fourth of the eggs she hd left to her prents. Net, she presented her unt with one-third of the eggs tht remined. Finlly, she gve one-hlf of the eggs she hd left to her brother, nd she kept si eggs. Cn you use some problem-solving skills to discover how mny Ukrinin eggs Andre designed? In this chpter, you will lern opertions with lgebric frctions nd methods to solve equtions nd inequlities tht involve frctions. 4 CHAPTER TABLE OF CONTENTS 4- The Mening of n Algebric Frction 4- Reducing Frctions to Lowest Terms 4- Multiplying Frctions 4-4 Dividing Frctions 4- Adding or Subtrcting Algebric Frctions 4-6 Solving Equtions with Frctionl Coefficients 4-7 Solving Inequlities with Frctionl Coefficients 4-8 Solving Frctionl Equtions Chpter Summry Vocbulry Review Eercises Cumultive Review 9

40 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions 4- THE MEANING OF AN ALGEBRAIC FRACTION EXAMPLE A frction is quotient of ny number divided by ny nonzero number. For emple, the rithmetic frction 4 indictes the quotient of divided by 4. An lgebric frction is quotient of two lgebric epressions. An lgebric frction tht is the quotient of two polynomils is clled frctionl epression or rtionl epression. Here re some emples of lgebric frctions tht re rtionl epressions: 4c 4 d The frction b mens tht the number represented by, the numertor, is to be divided by the number represented by b, the denomintor. Since division by 0 is not possible, the vlue of the denomintor, b, cnnot be 0.An lgebric frction is defined or hs mening only for vlues of the vribles for which the denomintor is not 0. Find the vlue of for which 9 is not defined. Solution The frction 9 is not defined when the denomintor, 9, is equl to 0. 9 0 9 Answer EXERCISES Writing About Mthemtics. Since ny number divided by itself equls, the solution set for is the set of ll rel numbers. Do you gree with this sttement? Eplin why or why not. b b b. Aron multiplied by (equl to ) to obtin the frction b b. Is the frction b b b b b equl to the frction b b for ll vlues of b? Eplin your nswer. Developing Skills In, find, in ech cse, the vlue of the vrible for which the frction is not defined.. 4. 6. 6. y 7. y 0 y 8. y 9. 0. 4y. 4. 7 4

Reducing Frctions to Lowest Terms 4 Applying Skills In 7, represent the nswer to ech problem s frction.. Wht is the cost of one piece of cndy if five pieces cost c cents? 4. Wht is the cost of meter of lumber if p meters cost 980 cents?. If piece of lumber 0 0 centimeters in length is cut into y pieces of equl length, wht is the length of ech of the pieces? 6. Wht frctionl prt of n hour is m minutes? 7. If the perimeter of squre is y inches, wht is the length of ech side of the squre? 4- REDUCING FRACTIONS TO LOWEST TERMS A frction is sid to be reduced to lowest terms or is lowest terms frction when its numertor nd denomintor hve no common fctor other thn or. Ech of the frctions 0 nd cn be epressed in lowest terms s. The rithmetic frction 0 is reduced to lowest terms when both its numertor nd denomintor re divided by : 0 0 4 4 The lgebric frction is reduced to lowest terms when both its numertor nd denomintor re divided by, where 0: 4 4 Frctions tht re equl in vlue re clled equivlent frctions. Thus, 0 nd re equivlent frctions, nd both re equivlent to, when 0. The emples shown bove illustrte the division property of frction: if the numertor nd the denomintor of frction re divided by the sme nonzero number, the resulting frction is equl to the originl frction. In generl, for ny numbers, b, nd, where b 0 nd 0: b b 4 4 b When frction is reduced to lowest terms, we list the vlues of the vribles tht must be ecluded so tht the originl frction is equivlent to the reduced form nd lso hs mening. For emple: 4 4 4 4 4 (where 0) cy dy cy 4 y dy 4 y d c (where y 0, d 0)

4 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions When reducing frction, the division of the numertor nd the denomintor by common fctor my be indicted by cncelltion. Here, we use cncelltion to Here, we use cncelltion to divide the numertor nd the divide the numertor nd the denomintor by : denomintor by ( ) : ( ) 8 ( ) 8 6 9 9 6 (where ) By re-emining one of the emples just seen, we cn show tht the multipliction property of one is used whenever frction is reduced: ( )? ( ) 8? 6? ( ) ( ) 6? 6 6 However, when the multipliction property of one is pplied to frctions, it is referred to s the multipliction property of frction. In generl, for ny numbers, b, nd, where b 0 nd 0: b b? b? b ( )( ) ( ) Procedure To reduce frction to lowest terms: METHOD. Fctor completely both the numertor nd the denomintor.. Determine the gretest common fctor of the numertor nd the denomintor.. Epress the given frction s the product of two frctions, one of which hs s its numertor nd its denomintor the gretest common fctor determined in step. 4. Use the multipliction property of frction. METHOD. Fctor both the numertor nd the denomintor.. Divide both the numertor nd the denomintor by their gretest common fctor.

Reducing Frctions to Lowest Terms 4 EXAMPLE Reduce to lowest terms. 4 Solution METHOD METHOD? 4 7? 4 7? 7?? 7? 7 7 Answer ( 0) 7 EXAMPLE Epress s lowest terms frction. Solution METHOD METHOD Answer ( 0) EXAMPLE 6 0 6 ( ) 0?? ( ) ( )? Reduce ech frction to lowest terms.. 6 b. 4 4 8 Solution. Use Method : b. Use Method : 6 ( 4)( 4) 4 ( )( 4) 4? 4 4 4? 4 6 ( ) 0 0 4 Answers. (, 4) b. 4 ( ) ( ) 0 ( ) 4 8 4( ) ( ) 4( ) 4

44 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions EXERCISES Writing About Mthemtics 4. Kevin used cncelltion to reduce 8 to lowest terms s shown below. Wht is the error in Kevin s work? 4 8 4 8 4. Kevin let 4 to prove tht when reduced to lowest terms, 8. Eplin to Kevin why his resoning is incorrect. Developing Skills In 4, reduce ech frction to lowest terms. In ech cse, list the vlues of the vribles for which the frctions re not defined. 4 7y. 4. 4c 9r 6y. 6d 6. 0r b y 7. 8. y bc cb 9. 0. 6by 9y 4bc.. 7 8y. 4. 4 6 4 y. b 0 6. y 7. b y 8c 8. 90y 48 b 4 y 6 8y 8m 9. 0... 40m 4 6 8m b. 4. 0 b b. 6 6. y 9y 6 b y 8b 7. 0b 4 7d y 8. 9. 0. 9b 4 8 7d 4 y. 4 b.. 4. 9 6 b 7r s b 9. b s r 6. 7. 8. b 9 s r 6 b( b) r 9. 40. y 4. 4. r 6 8 y b 9 r 9 y 4. 7 44. 4. 46. 6 4 4 y y 4 47. 48. 6 49. 6 0. 0 9 7 6 8 r. 4r. 48 8. 7 4. 7y y r r ( ) y 0y

Multiplying Frctions 4.. Use substitution to find the numericl vlue of, then reduce ech numericl frction to lowest terms when: () 7 () 0 () 0 (4) () 4 (6) 0 b. Wht pttern, if ny, do you observe for the nswers to prt? c. Cn substitution be used to evlute when? Eplin your nswer. d. Reduce the lgebric frction to lowest terms. e. Using the nswer to prt d, find the vlue of, reduced to lowest terms, when 8,76. ( f. If the frction is multiplied by to become ) ( ), will it be equivlent to? Eplin your nswer. 4- MULTIPLYING FRACTIONS The product of two frctions is frction with the following properties:. The numertor is the product of the numertors of the given frctions.. The denomintor is the product of the denomintors of the given frctions. In generl, for ny numbers, b,, nd y, when b 0 nd y 0: We cn find the product of methods. METHOD. 7 7 b? y by 9 nd 4 in lowest terms by using either of two 7 7 9 4 7 7 9 4 08 6 7 9 9 7 9 9 7 7 7 METHOD. 7 9 4 7 7 9 4 7 Notice tht Method requires less computtion thn Method since the reduced form of the product ws obtined by dividing the numertor nd the denomintor by common fctor before the product ws found. This method my be clled the cncelltion method. The properties tht pply to the multipliction of rithmetic frctions lso pply to the multipliction of lgebric frctions.

46 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions To multiply 4y by 7y nd epress the product in lowest terms, we my use either of the two methods. In this emple, 0 nd y 0. METHOD. 7y? 4y? 4y 7y? 70 y 0 y y? y y y? y METHOD. 7? 4y 7y? y 4y y (the cncelltion method) While Method is longer, it hs the dvntge of displying ech step s property of frctions. This cn be helpful for checking work. Procedure To multiply frctions: METHOD. Multiply the numertors of the given frctions.. Multiply the denomintors of the given frctions.. Reduce the resulting frction, if possible, to lowest terms. METHOD. Fctor ny polynomil tht is not monomil.. Use cncelltion to divide numertor nd denomintor by ech common fctor.. Multiply the resulting numertors nd the resulting denomintors to write the product in lowest terms. EXAMPLE Multiply nd epress the product in reduced form: 9b? 6b Solution METHOD () Multiply the numertors nd denomintors of the given frctions: () Reduce the resulting frction to lowest terms: 9b? 6b? 6b 0 b 9b? 9 b 0? b b 0? 0

Multiplying Frctions 47 METHOD () Divide the numertors nd denomintors by the common fctors b nd : () Multiply the resulting numertors nd the resulting denomintors: 9b? 6b 9b? 0 6b 0 Answer ( 0, b 0, 0) EXAMPLE Multiply nd epress the product in simplest form:? 8 Solution Think of s.? 8? 8 6 8 4 4? 9? 9 9 9 Answer ( 0) EXAMPLE Multiply nd simplify the product: 6? 4 Solution 6? Answer 6 ( 0, ) 4 () ( )? 4 ( ) 6 EXERCISES Writing About Mthemtics. When reduced to lowest terms, frction whose numertor is equls. Wht is the denomintor of the frction? Eplin your nswer.. Does for ll vlues of nd z? Eplin your nswer. z z? z z z

48 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions Developing Skills In 4, find ech product in lowest terms. In ech cse, list ny vlues of the vrible for which the frctions re not defined. 8. 4. 6? y. 6. 7.? 0 6 9y? 0 d? d 6? 0 8 4 m 8. 9. 0.. 6r. 8? s? 0rs y? y y? 4y mn? m n 8 6 m 6r 0m. 4 4. b 7 9 y y. 6. 7.? 8n? m 6n? c 8? 4? 7c b 8 y 4 b 8. 9. 0.?? b b b? b b r 7s 8. r? r 0. s? s 4. 6? 4.. 9 6.??? 6 6 4 ( b) 7. 4b ( ) 8. 6b 9. 7 8??? 4b ( b ) 4b 4 8 y 4 6 6 0. 6 y. y 4c??. 4 8? 9y c y y. 4. 4 8. 8 0y 90?? 4 9? 6 8 4 (y 9) y 4 8 8 6 6. 7. 8.?? 6? 8 b d 9. 8 8 b 40. 4. 6 6? 4 d? d 0? b 8 8 b d 6 6 4. Wht is the vlue of 4 when 6,908? 6? 4 6 4-4 DIVIDING FRACTIONS We know tht the opertion of division my be defined in terms of the multiplictive inverse, the reciprocl. A quotient cn be epressed s the dividend times the reciprocl of the divisor. Thus: 8 4 8 8 8 nd We use the sme rule to divide lgebric frctions. In generl, for ny numbers, b, c, nd d, when b 0, c 0, nd d 0: b 4 d c b? d c d bc 8 7 4 8 7 7 8 4 Procedure To divide by frction, multiply the dividend by the reciprocl of the divisor.

Dividing Frctions 49 EXAMPLE Solution 6c Divide: 4 4c4 d 4d How to Proceed () Multiply the dividend by the reciprocl of the divisor: () Divide the numertors nd denomintors by the common fctors: 6c 4 4c4 6c? 4d d 4d d 4c 4 6c d d? 4d 4c 4 c () Multiply the resulting numertors nd the resulting denomintors: 4d 9c 4d Answer 9c (c 0, d 0) EXAMPLE Solution 8 4 Divide: 4 4 8 How to Proceed () Multiply the dividend by the reciprocl of the divisor: () Fctor the numertors nd denomintors, nd divide by the common fctors: () Multiply the resulting numertors nd the resulting denomintors: 8 4 4 4 8 4 8? 8 4 8 ( ) ( ) ( )? ( ) ( ) 4 ( ) ( ) ( ) Answer ( ) ( 0,,, ) Note: If,, or, the dividend nd the divisor will not be defined. If 0, the reciprocl of the divisor will not be defined.

0 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions EXERCISES Writing About Mthemtics. Eplin why the quotient 4 is undefined for nd for.. To find the quotient ( 4) 4 4, Ruth cnceled ( 4) in the numertor nd denomintor nd wrote 4?. Is Ruth s nswer correct? Eplin why or why not. Developing Skills In 7, find ech quotient in lowest terms. In ech cse, list ny vlues of the vribles for which the quotient is not defined. 7. 4.. 8 4 6. 9 4 4 4b 0 4 7 y 7b 7. 8. y 9. 6 0. b 8c 4 b y 4 0cd 4 4b y 4 y c d y 4 4 y.. 9y. 4. 8 4 y 9 4 8 7 4 4 b 4b 0. 4 4 6. 9 0 b 7. b 6 4 b 4 4 0 4 8 b b 8. b y 6 ( ) 9. 0. 4 4 ( b ) 8 4 (y y ) 4 6 4 6. y 8y 0y. 4 4. (9 y ) 4 y 8y 4 6 4 ( ) 6y y y 0 y 4.. 6. 6 9 4? y? ( y)? 4 4 4 4 y 4 y 4 4 ( b) 7. 4 b b b? b ( b) 8. For wht vlue(s) of is 4 undefined? y 9. Find the vlue of 6y 9 0y 0 4 when y 70. y 9 y y 0. If y nd y 4 z, wht is the vlue of z? 4- ADDING OR SUBTRACTING ALGEBRAIC FRACTIONS We know tht the sum (or difference) of two rithmetic frctions tht hve the sme denomintor is nother frction whose numertor is the sum (or difference) of the numertors nd whose denomintor is the common denomintor of the given frctions. We use the sme rule to dd lgebric frctions tht hve the sme nonzero denomintor. Thus:

Adding or Subtrcting Algebric Frctions Arithmetic frctions 7 7 7 6 7 7 7 7 4 7 Algebric frctions b b b b Procedure To dd (or subtrct) frctions tht hve the sme denomintor:. Write frction whose numertor is the sum (or difference) of the numertors nd whose denomintor is the common denomintor of the given frctions.. Reduce the resulting frction to lowest terms. EXAMPLE Add nd reduce the nswer to lowest terms: 4 4 9 Solution 4 4 9 4 9 4 4 7 7 Answer ( 0) EXAMPLE Subtrct: 4 7 6 6 4 Solution 4 7 6 4 (4 7) ( 4) 4 7 4 6 6 6 6 Answer 6 ( 0) Note: In Emple, since the frction br is symbol of grouping, we enclose numertors in prentheses when the difference is written s single frction. In this wy, we cn see ll the signs tht need to be chnged for the subtrction. In rithmetic, in order to dd (or subtrct) frctions tht hve different denomintors, we chnge these frctions to equivlent frctions tht hve the sme denomintor, clled the common denomintor. Then we dd (or subtrct) the equivlent frctions. For emple, to dd 4 nd 6, we use ny common denomintor tht hs 4 nd 6 s fctors. METHOD. Use the product of the denomintors s the common denomintor. Here, common denomintor is 4 6, or 4. 4 6 4 6 6 6 4 4 8 4 4 4 4 Answer

Algebric Frctions, nd Equtions nd Inequlities Involving Frctions METHOD. To simplify our work, we use the lest common denomintor (LCD), tht is, the lest common multiple of the given denomintors. The LCD of 4 nd 6 is. 4 6 4 6 9 Answer To find the lest common denomintor of two frctions, we fctor the denomintors of the frctions completely. The LCD is the product of ll of the fctors of the first denomintor times the fctors of the second denomintor tht re not fctors of the first. 4 6 LCD Then, to chnge ech frction to n equivlent form tht hs the LCD s the denomintor, we multiply by, where is the number by which the originl denomintor must be multiplied to obtin the LCD. 4 A B 4 A B 9 6 A B 6 A B Note tht the LCD is the smllest possible common denomintor. Procedure To dd (or subtrct) frctions tht hve different denomintors:. Choose common denomintor for the frctions.. Chnge ech frction to n equivlent frction with the chosen common denomintor.. Write frction whose numertor is the sum (or difference) of the numertors of the new frctions nd whose denomintor is the common denomintor. 4. Reduce the resulting frction to lowest terms. Algebric frctions re dded in the sme mnner s rithmetic frctions, s shown in the emples tht follow.

Adding or Subtrcting Algebric Frctions EXAMPLE Solution Add: b b How to Proceed () Find the LCD of the frctions: () Chnge ech frction to n equivlent frction with the lest common denomintor, b : () Write frction whose numertor is the sum of the numertors of the new frctions nd whose denomintor is the common denomintor: b b b b b LCD b b b b b b? b b b? b b b b b b Answer ( 0, b 0) b EXAMPLE 4 Subtrct: Solution LCD 4 EXAMPLE Epress s frction in simplest form: y y Solution LCD y y y Answer y (y ) 4 y y? A y y B 4 4 4?? 4 8 0 6 (8 0) ( 6) 8 0 6 6 Answer y y y y y y y y

4 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions EXAMPLE 6 Subtrct: 6 4 Solution 4 ( )( ) ( ) LCD ( )( ) Answer (, ) 6 4 6 ( )( ) ( ) ( )( ) 6 ( ) ( )( ) 6 6 ( )( ) 6 ( )( ) ( ) ( )( ) EXERCISES Writing About Mthemtics. In Emple, the nswer is 6. Cn we divide nd 6 by to write the nswer in lowest terms s? Eplin why or why not.. Joey sid tht if. Do you gree with Joey? Eplin why or why not. Developing Skills In 4, dd or subtrct the frctions s indicted. Reduce ech nswer to lowest terms. In ech cse, list the vlues of the vribles for which the frctions re not defined.. 4c 4c 4c 6 4. r t s t. 6y 4 6. 7. 4y 7 y 4y 8. 6 r 9. 0. 4r r r 6 8 r 6 r r 6. y.. 6 y y 6 b 4. b 4 8. 4 7 0 6. 6 4 7. 7 4 b 9 8. 4 9. 8 0. 8b 9 4b 6 0c 0c 9 0c 9d 6 d 7d d

Adding or Subtrcting Algebric Frctions y 4. d.. y d 7 6 4 b y 4 c 7 4.. y b b 0b 6. c 4y y c 6c 7. 8. 9. 7 y 6 9 0 0... 6 7 y 4y 4 4. 4.. 6 y 6 y 9 8 8 4 4 y y 4 6. 7. y y 4 8. 6 4 8 y 9. 40. 4. y 6 y y 4y 7 4. 4. 7 4 ( )( ) ( )( ) 7 0 Applying Skills In 44 46, represent the perimeter of ech polygon in simplest form. 7 44. The lengths of the sides of tringle re represented by,, nd 0. 4 4. The length of rectngle is represented by 4, nd its width is represented by. 46. Ech leg of n isosceles tringle is represented by 7, nd its bse is represented by 6 8. In 47 nd 48, find, in ech cse, the simplest form of the indicted length. 7 47. The perimeter of tringle is 4, nd the lengths of two of the sides re 8 nd. Find the length of the third side. 4 48. The perimeter of rectngle is, nd the mesure of ech length is. Find the mesure of ech width. 49. The time t needed to trvel distnce d t rte of speed r cn be found by using the formul t d r.. For the first prt of trip, cr trvels miles t 4 miles per hour. Represent the time tht the cr trveled t tht speed in terms of. b. For the reminder of the trip, the cr trvels 0 miles t 60 miles per hour. Represent the time tht the cr trveled t tht speed in terms of. c. Epress, in terms of, the totl time for the two prts of the trip. 0. Ernesto wlked miles t miles per hour nd then wlked miles t ( ) miles per hour. Represent, in terms of, the totl time tht he wlked.. Frn rode her bicycle for miles t 0 miles per hour nd then rode ( ) miles frther t 8 miles per hour. Represent, in terms of, the totl time tht she rode.

6 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions 4-6 SOLVING EQUATIONS WITH FRACTIONAL COEFFICIENTS The following equtions contin frctionl coefficients: 0 60 60 6 0 6 Ech of these equtions cn be solved by finding n equivlent eqution tht does not contin frctionl coefficients. This cn be done by multiplying both sides of the eqution by common denomintor for ll the frctions present in the eqution. We usully multiply by the lest common denomintor, the LCD. Note tht the eqution 0. 0 cn lso be written s 0, since deciml frction cn be replced by common frction. Procedure To solve n eqution tht contins frctionl coefficients:. Find the LCD of ll coefficients.. Multiply both sides of the eqution by the LCD.. Solve the resulting eqution using the usul methods. 4. Check in the originl eqution. EXAMPLE Solve nd check: 8 Solution How to Proceed Check () Write the eqution: () Find the LCD: 8 LCD 8? 8 Answer () Multiply both sides of the eqution by the LCD: (4) Use the distributive property: () Simplify: 0 (6) Solve for : 8 0 A B (8) A B A B (8)? 8 8 8

Solving Equtions with Frctionl Coefficients 7 EXAMPLE Solution Solve:. 4 0 4 b. 7 6 9 0. 4 0 4 b. 7 6 0 9 LCD 4 LCD 0 4 A 4 B 4 A 0 4 B 4 A 4 B 4(0) 4 A 4 B 80 80 In Emple, the check is left to you. 0 A 7 6 9 0 B 0() 0 A 7 6 B 0 A 9 0 B 0() ( 7) ( 9) 90 0 6 + 7 90 40 Answer 4 6 90 4 8 7 Answer EXAMPLE A womn purchsed stock in the PAX Compny over months. In the first month, she purchsed one-hlf of her present number of shres. In the second month, she bought two-fifths of her present number of shres. In the third month, she purchsed 4 shres. How mny shres of PAX stock did the womn purchse? Solution Let totl number of shres of stock purchsed. Then number of shres purchsed in month, number of shres purchsed in month, 4 number of shres purchsed in month. The sum of the shres purchsed over months is the totl number of shres. month month month totl 4 0 A 4 B 0() 4 40 0 9 40 0 40

8 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions Check month (40) 70, month (40) 6, month 4 70 6 4 40 Answer 40 shres EXAMPLE 4 In child s coin bnk, there is collection of nickels, dimes, nd qurters tht mounts to $.0. There re times s mny qurters s nickels, nd more dimes thn nickels. How mny coins of ech kind re there? Solution Let the number of nickels. Then the number of qurters, nd the number of dimes. Also, 0.0 the vlue of the nickels, 0.() the vlue of the qurters, nd 0.0( ) the vlue of the dimes. Write the eqution for the vlue of the coins. To simplify the eqution, which contins coefficients tht re deciml frctions with denomintors of 00, multiply ech side of the eqution by 00. The totl vlue of the coins is $.0. 0.0 0.() 0.0( ).0 00[0.0 0.() 0.0( )] 00(.0) () 0( ) 0 7 0 0 0 90 0 0 90 70 Check There re nickels, () 9 qurters, nd 8 dimes. The vlue of nickels is $0.0() $0. The vlue of 9 qurters is $0.(9) $. The vlue of 8 dimes is $0.0(8) $0.80 $.0 Answer There re nickels, 9 qurters, nd 8 dimes.

Solving Equtions with Frctionl Coefficients 9 Note: In problem such s this, chrt such s the one shown below cn be used to orgnize the informtion: Coins Number of Coins Vlue of One Coin Totl Vlue Nickels 0.0 0.0 Qurters 0. 0.() Dimes 0.0 0.0( ) EXERCISES Writing About Mthemtics. Abby solved the eqution 0. 0.84 s follows: 0. 0.84 0. 0. 0.84 0. 8.4 Is Abby s solution correct? Eplin why or why not.. In order to write the eqution 0. 0.84 s n equivlent eqution with integrl coefficients, Heidi multiplied both sides of the eqution by 0. Will Heidi s method led to correct solution? Eplin why or why not. Compre Heidi s method with multiplying by 00 or multiplying by,000. Developing Skills In 7, solve ech eqution nd check.. 7 4. 6 t 8. 8 6. 4 6 m 7. 9 8. y 0 9. 7 0 0. 4. y.. 4 44 y 6 9 4. r. 7 6. 6 r 7. 7y 8. 9. 4 y 4 6 0. t 6 t 4 m 4 6 m r 6 4 m 7 8 t 4 6 y 4 y t... 0.0y. 8.7 4. 0.4 0.08 4.4. c 0.c 0 6. 0.08y 0.9 0.0y 7..7 0 0. 8. 0.0( ) 8 9. 0.0( 8) 0.07

60 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions 0. 0.4( 9) 0.( 4). 0.06( ) 0.04( 8). 0.04 0.0(,000 ) 7. 0.0 0.04(,00 ) 48 4. 0.0 0 0.06( 0). 0.08 0.0( 00) 4 0.4 0. 6. 0. 4 7. 6 0. 4 8. The sum of one-hlf of number nd one-third of tht number is. Find the number. 9. The difference between one-fifth of positive number nd one-tenth of tht number is 0. Find the number. 40. If one-hlf of number is incresed by 0, the result is. Find the number. 4. If two-thirds of number is decresed by 0, the result is 0. Find the number. 4. If the sum of two consecutive integers is divided by, the quotient is 9. Find the integers. 4. If the sum of two consecutive odd integers is divided by 4, the quotient is 0. Find the integers. 44. In n isosceles tringle, ech of the congruent sides is two-thirds of the bse. The perimeter of the tringle is 4. Find the length of ech side of the tringle. 4. The lrger of two numbers is less thn times the smller. If the smller number is equl to one-third of the lrger number, find the numbers. 46. The lrger of two numbers eceeds the smller by 4. If the smller number is equl to three-fifths of the lrger, find the numbers. 47. Seprte 90 into two prts such tht one prt is one-hlf of the other prt. 48. Seprte 0 into two prts such tht one prt is two-thirds of the other prt. Applying Skills 49. Four vegetble plots of unequl lengths nd of equl widths re rrnged s shown. The length of the third plot is one-fourth the length of the second plot. 4 The length of the fourth plot is one-hlf the length of the second plot. The length of the first plot is 0 feet more thn the length of the fourth plot. If the totl length of the four plots is 00 feet, find the length of ech plot. 0. Sm is now one-sith s old s his fther. In 4 yers, Sm will be one-fourth s old s his fther will be then. Find the ges of Sm nd his fther now.. Robert is one-hlf s old s his fther. Twelve yers go, he ws one-third s old s his fther ws then. Find their present ges.

Solving Equtions with Frctionl Coefficients 6. A coch finds tht, of the students who try out for trck, 6% qulify for the tem nd 90% of those who qulify remin on the tem throughout the seson. Wht is the smllest number of students who must try out for trck in order to hve 0 on the tem t the end of the seson?. A bus tht runs once dily between the villges of Alpc nd Down mkes only two stops in between, t Billow nd t Comfort. Tody, the bus left Alpc with some pssengers. At Billow, one-hlf of the pssengers got off, nd si new ones got on. At Comfort, gin onehlf of the pssengers got off, nd, this time, five new ones got on. At Down, the lst pssengers on the bus got off. How mny pssengers were bord when the bus left Alpc? 4. Slly spent hlf of her money on present for her mother. Then she spent one-qurter of the cost of the present for her mother on tret for herself. If Slly hd $6.00 left fter she bought her tret, how much money did she hve originlly?. Bob plnted some lettuce seedlings in his grden. After few dys, one-tenth of these seedlings hd been eten by rbbits. A week lter, one-fifth of the remining seedlings hd been eten, leving 6 seedlings unhrmed. How mny lettuce seedlings hd Bob plnted originlly? 6. My hs times s mny dimes s nickels. In ll, she hs $.40. How mny coins of ech type does she hve? 7. Mr. Jntzen bought some cns of soup t $0.9 per cn, nd some pckges of frozen vegetbles t $0.9 per pckge. He bought twice s mny pckges of vegetbles s cns of soup. If the totl bill ws $9.4, how mny cns of soup did he buy? 8. Roger hs $.0 in dimes nd nickels. There re more dimes thn nickels. Find the number of ech kind of coin tht he hs. 9. Bess hs $.80 in qurters nd dimes. The number of dimes is 7 less thn the number of qurters. Find the number of ech kind of coin tht she hs. 60. A movie theter sold student tickets for $.00 nd full-price tickets for $7.00. On Sturdy, the theter sold 6 more full-price tickets thn student tickets. If the totl sles on Sturdy were $,07, how mny of ech kind of ticket were sold? 6. Is it possible to hve $4.0 in dimes nd qurters, nd hve twice s mny qurters s dimes? Eplin. 6. Is it possible to hve $6.00 in nickels, dimes, nd qurters, nd hve the sme number of ech kind of coin? Eplin. 6. Mr. Symms invested sum of money in 7% bonds. He invested $400 more thn this sum in 8% bonds. If the totl nnul interest from these two investments is $7, how much did he invest t ech rte? 64. Mr. Chrles borrowed sum of money t 0% interest. He borrowed second sum, which ws $,00 less thn the first sum, t % interest. If the nnul interest on these two lons is $0.0, how much did he borrow t ech rte?

6 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions 4-7 SOLVING INEQUALITIES WITH FRACTIONAL COEFFICIENTS In our modern world, mny problems involve inequlities. A potentil buyer my offer t most one mount for house, while the seller will ccept no less thn nother mount. Inequlities tht contin frctionl coefficients re hndled in much the sme wy s equtions tht contin frctionl coefficients. The chrt on the right helps us to trnslte words into lgebric symbols. Procedure Words is greter thn b is less thn b is t lest b is no less thn b is t most b is no greter thn b Symbols b b b b To solve n inequlity tht contins frctionl coefficients:. Find the LCD, positive number.. Multiply both sides of the inequlity by the LCD.. Solve the resulting inequlity using the usul methods. EXAMPLE Solution. Solve the inequlity, nd grph the solution set on number line:. 6. b. 6 A 6 B. 6() 6 A B 6 A 6 B.. Since no domin ws given, use the domin of rel numbers. Answer: 6.. 0 4 6 8 0 4 6 b. y 8 4y 7 # Since no domin ws given, use the domin of rel numbers. Answer: y 0 y 8 4y 7 # 4 A y 8 4y 7 B # 4() 4 A y B 4 A 8 4y 7 B # 4 y 6 8y # 4 y # 6 y #

Solving Inequlities with Frctionl Coefficients 6 EXAMPLE Two boys wnt to pool their money to buy comic book. The younger of the boys hs one-third s much money s the older. Together they hve more thn $.00. Find the smllest possible mount of money ech cn hve. Solution Let the number of cents tht the older boy hs. Then the number of cents tht the younger boy hs. The sum of their money in cents is greter thn 00. 00 A B. (00) + 600 4 600 0 0 The number of cents tht the younger boy hs must be n integer greter thn 0. The number of cents tht the older boy hs must be multiple of tht is greter thn 0. The younger boy hs t lest cents. The older boy hs t lest cents. The sum of nd is greter thn 00. Answer The younger boy hs t lest $0. nd the older boy hs t lest $.. EXERCISES Writing About Mthemtics. Eplin the error in the following solution of n inequlity.. A B.(). 6. In Emple, wht is the domin for the vrible? Developing Skills In, solve ech inequlity, nd grph the solution set on number line.. 4. 0 9 4. y y,. 6 c. c y y 6. 7. 8. 9 y 6 $ y 4. 6 4 8 # 8 t 9. 0 # 4 t 0. $...7 4

64 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions y 0. y 0.y. 7. 4. 7 # 0 4c m. d 4, 7d 6. 7. $ 7 m 7 9 $ c 7 6 4 0 6 y 8. 9. 0. y 6,. 7 0, 0 r t 4. r #. $ t 6 4 t 9. # 6 4. If one-third of n integer is incresed by 7, the result is t most. Find the lrgest possible integer.. If two-fifths of n integer is decresed by, the result is t lest 4. Find the smllest possible integer. 6. The sum of one-fifth of n integer nd one-tenth of tht integer is less thn 40. Find the gretest possible integer. 7. The difference between three-fourths of positive integer nd one-hlf of tht integer is greter thn 8. Find the smllest possible integer. 8. The smller of two integers is two-fifths of the lrger, nd their sum is less thn 40. Find the lrgest possible integers. 9. The smller of two positive integers is five-siths of the lrger, nd their difference is greter thn. Find the smllest possible integers. Applying Skills 0. Tlk nd Tell Answering Service offers customers two monthly options. OPTION Mesured Service OPTION Unmesured Service bse rte is $ bse rte is $0 ech cll costs $0.0 no dditionl chrge per cll Find the lest number of clls for which unmesured service is cheper thn mesured service.. Pul erned some money mowing lwns. He spent one-hlf of this money for book, nd then one-third for CD. If he hd less thn $ left, how much money did he ern?. Mry bought some cns of vegetbles t $0.89 per cn, nd some cns of soup t $0.99 per cn. If she bought twice s mny cns of vegetbles s cns of soup, nd pid t lest $0, wht is the lest number of cns of vegetbles she could hve bought?. A coin bnk contins nickels, dimes, nd qurters. The number of dimes is 7 more thn the number of nickels, nd the number of qurters is twice the number of dimes. If the totl vlue of the coins is no greter thn $7.0, wht is the gretest possible number of nickels in the bnk? 4. Rhod is two-thirds s old s her sister Alice. Five yers from now, the sum of their ges will be less thn 60. Wht is the lrgest possible integrl vlue for ech sister s present ge?

Solving Frctionl Equtions 6. Four yers go, Bill ws 4 times s old s his cousin Mry. The difference between their present ges is t lest. Wht is the smllest possible integrl vlue for ech cousin s present ge? 6. Mr. Drew invested sum of money t 7 % interest. He invested second sum, which ws $00 less thn the first, t 7% interest. If the totl nnul interest from these two investments is t lest $60, wht is the smllest mount he could hve invested t 7 %? 7. Mr. Lehtimki wnted to sell his house. He dvertised n sking price, but knew tht he would ccept, s minimum, nine-tenths of the sking price. Mrs. Ptel offered to buy the house, but her mimum offer ws seven-eighths of the sking price. If the difference between the seller s lowest cceptnce price nd the buyer s mimum offer ws t lest $,000, find:. the minimum sking price for the house; b. the minimum mount Mr. Lehtimki, the seller, would ccept; c. the mimum mount offered by Mrs. Ptel, the buyer. 8. When pcking his books to move, Philip put the sme number of books in ech of boes. Once pcked, the boes were too hevy to lift so Philip removed one-fifth of the books from ech bo. If t lest 00 books in totl remin in the boes, wht is the minimum number of books tht Philip originlly pcked in ech bo? 4-8 SOLVING FRACTIONAL EQUATIONS An eqution is clled n lgebric eqution when vrible ppers in t lest one of its sides. An lgebric eqution is frctionl eqution when vrible ppers in the denomintor of one, or more thn one, of its terms. For emple, d 6d 4 y y y re ll frctionl equtions. To simplify such n eqution, cler it of frctions by multiplying both sides by the lest common denomintor of ll frctions in the eqution. Then, solve the simpler eqution. As is true of ll lgebric frctions, frctionl eqution hs mening only when vlues of the vrible do not led to denomintor of 0. KEEP IN MIND When both sides of n eqution re multiplied by vrible epression tht my represent 0, the resulting eqution my not be equivlent to the given eqution. Such equtions will yield etrneous solutions, which re solutions tht stisfy the derived eqution but not the given eqution. Ech solution, therefore, must be checked in the originl eqution.

66 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions EXAMPLE Solve nd check: Solution Multiply both sides of the eqution Check by the lest common denomintor, 6. 6? 6 A B 6 A B 6 6? 6 A B 6 A B 6 A B 6? 6 6 Answer 6 EXAMPLE Solve nd check: 0 7 Solution Multiply both sides of the eqution by Check the lest common denomintor,. 0 7 0 7 () 0? 7 0 ( ) A B ( )(7) 0 0 0? 7 0 7 4 0 0 7 4 The only possible vlue of is vlue for which the eqution hs no mening becuse it leds to denomintor of 0. Therefore, there is no solution for this eqution. Answer The solution set is the empty set, or { }. EXAMPLE Solve nd check: 6 4

Solving Frctionl Equtions 67 Solution METHOD Multiply both sides of the eqution by the LCD, 4: 4 A B 4 A 6 6 8 0 ( )( 4) 0 4 B 8 (6 ) 8 6 0 4 0 4 METHOD Use the rule for proportion: the product of the mens equls the product of the etremes. 6 4 (6 ) 8 6 8 6 8 0 6 8 0 ( )( 4) 0 0 4 0 4 Check 4 6 4 6 4 6 4 4 6 4 4 EXAMPLE 4 Answer or 4 Solve nd check: ( ) Solution Multiply both sides of the eqution by the LCD, ( )( ) : Answer 0 ( )( ) A B A ( ) B ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( ) Check ( ) ( ) 0 0? (0 ) 4 0? () 0

68 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions EXERCISES Writing About Mthemtics. Nthn sid tht the solution set of r r 0 is the set of ll rel numbers. Do you gree with Nthn? Eplin why or why not. y. Pm multiplied ech side of the eqution by (y + )(y ) to obtin the y y eqution y y, which hs s its solution y 0. Pru sid tht the eqution y is proportion nd cn be solved by writing the product of the mens equl y y to the product of the etremes. She obtined the eqution (y ) (y ), which hs s its solution 0 nd. Both girls used correct method of solution. Eplin the difference in their nswers. Developing Skills In 6, eplin why ech frctionl eqution hs no solution. 6 4 4. 4.. 4 6. In 7 4, solve ech eqution, nd check. 0 7. 8. y 9. 0. 4 8 0. 8 9. y y 4 9 0 y. 7 4.. y 7 8 8 y 9 6. 7. y y 8. b 6 9. b 6 b 4 7 0. 8 4.. 6 0. 4. 6. 4 6. 4 4 7. 4 4z 8. 9. 7 z r 0. r. y y. 7 4. m m y 4. 4. 7 6. 8y y 7. 8. y y 9. 9 40. 6 4. b b 4. b b 4. b 6 b 44. 9 8 4. 4

In 46 49, solve ech eqution for in terms of the other vribles. t t b 46. k 0 47. k k 48. c 49. 0. If by c, y c, b c, 0, nd c 0, is it possible to know the numericl vlue of without knowing numericl vlues of, b, c, nd y? Eplin your nswer Applying Skills Chpter Summry 69 d d e. If 4 is divided by number, the result is 6. Find the number.. If 0 is divided by number, the result is 0. Find the number.. The sum of 0 divided by number, nd 7 divided by the sme number, is 9. Find the number. 4. When the reciprocl of number is decresed by, the result is. Find the number.. The numertor of frction is 8 less thn the denomintor of the frction. The vlue of the frction is. Find the frction. 6. The numertor nd denomintor of frction re in the rtio : 4. When the numertor is decresed by 4 nd the denomintor is incresed by, the vlue of the new frction, in simplest form, is. Find the originl frction. 7. The rtio of boys to girls in the chess club is 4 to. After boys leve the club nd girls join, the rtio is to. How mny members re in the club? 8. The length of Emily s rectngulr grden is 4 feet greter thn its width. The width of Srh s rectngulr grden is equl to the length of Emily s nd its length is 8 feet. The two grdens re similr rectngles, tht is, the rtio of the length to the width of Emily s grden equls the rtio of the length to the width of Srh s grden. Find the possible dimensions of ech grden. (Two nswers re possible.) CHAPTER SUMMARY An lgebric frction is the quotient of two lgebric epressions. If the lgebric epressions re polynomils, the frction is clled rtionl epression or frctionl epression. An lgebric frction is defined only if vlues of the vribles do not result in denomintor of 0. Frctions tht re equl in vlue re clled equivlent frctions. A frction is reduced to lowest terms when n equivlent frction is found such tht its numertor nd denomintor hve no common fctor other thn or. This frction is considered lowest terms frction.

70 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions Opertions with lgebric frctions follow the sme rules s opertions with rithmetic frctions: Multipliction Division ( 0, y 0) 4 b y? y b y? y b y b b ( 0, y 0, b 0) Addition/subtrction with the sme denomintor c b c b c (c 0), c b c c b (c 0) Addition/subtrction with different denomintors (first, obtin the common denomintor): b d c b? d d d c? b b bd d bc d bc bd bd (b 0, d 0) A frctionl eqution is n eqution in which vrible ppers in the denomintor of one or more thn one of its terms. To simplify frctionl eqution, or ny eqution or inequlity contining frctionl coefficients, multiply both sides by the lest common denomintor (LCD) to eliminte the frctions. Then solve the simpler eqution or inequlity nd check for etrneous solutions. VOCABULARY 4- Algebric frction Frctionl epression Rtionl epression 4- Reduced to lowest terms Lowest terms frction Equivlent frctions Division property of frction Cncelltion Multipliction property of frction 4- Cncelltion method 4- Common denomintor Lest common denomintor 4-8 Algebric eqution Frctionl eqution Etrneous solution REVIEW EXERCISES. Eplin the difference between n lgebric frction nd frctionl epression.. Wht frctionl prt of centimeter is millimeters? y. For wht vlue of y is the frction y 4 undefined? 4. Fctor completely: 7

In 8, reduce ech frction to lowest terms. 8bg 4d. 6. 7. 60 bg 8. 7d In 9, in ech cse, perform the indicted opertion nd epress the nswer in lowest terms. 9. y 4? 9 8 0. y?. 6c 4 c. 7b 4 8 m. 6 m 6 9 4. k k k 4. 4 6. y yz 4 7. 8. b y 7? 9. 0. y b 0 b 4. 4 0 4 c. c 8 9. 4 ( 9 ) b b b 4. If the sides of tringle re represented by, 6, nd, epress the perimeter of the tringle in simplest form. b. If, b, nd c 4, wht is the sum of c? In 6, solve ech eqution nd check. k 6. 0 4 7. 0 4 8. 6 9. m 0 m t 0. t 0. Review Eercises 7 y y 6 4 8y y 8y 0 In 4, solve ech eqution for r in terms of the other vribles. S c.. r p h pr 4. r n 0. Mr. Vromn deposited sum of money in the bnk. After few yers, he found tht the interest equled one-fourth of his originl deposit nd he hd totl sum, deposit plus interest, of $,400 in the bnk. Wht ws the originl deposit? 6. One-third of the result obtined by dding to certin number is equl to one-hlf of the result obtined when is subtrcted from the number. Find the number. 7. Of the totl number of points scored by the winning tem in bsketbll gme, one-fifth ws scored in the first qurter, one-sith ws scored in the second qurter, one-third ws scored in the third qurter, nd 7 ws scored in the fourth qurter. How mny points did the winning tem score? 8. Ross drove 00 miles t r miles per hour nd 60 miles t r 0 miles per hour. If the time needed to drive 00 miles ws equl to the time needed to drive 60 miles, find the rtes t which Ross drove. (Epress the time needed for ech prt of the trip s t d r.)

7 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions 9. The totl cost, T,of n items tht cost dollrs ech is given by the eqution T n.. Solve the eqution T n for n in terms of T nd. b. Use your nswer to to epress n, the number of cns of sod tht cost $.00 if ech cn of sod costs dollrs. c. Use your nswer to to epress n, the number of cns of sod tht cost $.00 if ech cn of sod costs dollrs. d. If the number of cns of sod purchsed for $.00 is 4 less thn the number purchsed for $.00, find the cost of cn of sod nd the number of cns of sod purchsed. 40. The cost of two cups of coffee nd bgel is $.7. The cost of four cups of coffee nd three bgels is $4.. Wht is the cost of cup of coffee nd the cost of bgel? 4. A piggybnk contins nickels, dimes, nd qurters. The number of nickels is 4 more thn the number of dimes, nd the number of qurters is times the number of nickels. If the totl vlue of the coins is no greter thn $8.60, wht is the gretest possible number of dimes in the bnk? Eplortion Some rtionl numbers cn be written s terminting decimls nd others s infinitely repeting decimls. () Write ech of the following frctions s deciml:, 4,, 8, 0, 6, 0,, 0, 00 () Wht do you observe bout the decimls written in ()? () Write ech denomintor in fctored form. (4) Wht do you observe bout the fctors of the denomintors? () Write ech of the following frctions s deciml:, 6, 9,,,, 8,, 4, 0 Wht do you observe bout the decimls written in ()? (7) Write ech denomintor in fctored form. (8) Wht do you observe bout the fctors of the denomintors? (9) Write sttement bout terminting nd infinitely repeting decimls bsed on your observtions.

Cumultive Review 7 CUMULATIVE REVIEW CHAPTERS 4 Prt I Answer ll questions in this prt. Ech correct nswer will receive credits. No prtil credit will be llowed.. The product of nd is () 0 () 7 () 8 0 (4) 8 7. In the coordinte plne, the point whose coordintes re (, ) is in qudrnt () I () II () III (4) IV. In deciml nottion,.7 0 is () 0.07 () 0.007 () 7. (4) 7 4. The slope of the line whose eqution is y is () () () (4). Which of the following is n irrtionl number? (). () () "9 (4) " 6. The fctors of 7 8 re () 9 nd () ( 9) nd ( ) () 9 nd (4) ( 9) nd ( ) 7. The dimensions of rectngulr bo re 8 by by 9. The surfce re is () 60 cubic units () 7 squre units () 60 squre units (4) 4 squre units 8. The length of one leg of right tringle is 8 nd the length of the hypotenuse is. The length of the other leg is () 4 () 4" () 4" (4) 80 9. The solution set of is () {, } () {} () {0, } (4) {, } 0. In the lst n times bsebll plyer ws up to bt, he got hits nd struck out the rest of the times. The rtio of hits to strike-outs is n n () n () n () n (4) Prt II Answer ll questions in this prt. Ech correct nswer will receive credits. Clerly indicte the necessry steps, including pproprite formul substitutions, digrms, grphs, chrts, etc. For ll questions in this prt, correct numericl nswer with no work shown will receive only credit.

74 Algebric Frctions, nd Equtions nd Inequlities Involving Frctions. Mrs. Kniger bought some stock on My for $,00. By June, the vlue of the stock ws $,640. Wht ws the percent of increse of the cost of the stock?. A furlong is one-eighth of mile. A horse rn 0 furlongs in. minutes. Wht ws the speed of the horse in feet per second? Prt III Answer ll questions in this prt. Ech correct nswer will receive credits. Clerly indicte the necessry steps, including pproprite formul substitutions, digrms, grphs, chrts, etc. For ll questions in this prt, correct numericl nswer with no work shown will receive only credit.. Two cns of sod nd n order of fries cost $.60. One cn of sod nd two orders of fries cost $.80. Wht is the cost of cn of sod nd of n order of fries? 4.. Drw the grph of y. b. From the grph, determine the solution set of the eqution. Prt IV Answer ll questions in this prt. Ech correct nswer will receive 4 credits. Clerly indicte the necessry steps, including pproprite formul substitutions, digrms, grphs, chrts, etc. For ll questions in this prt, correct numericl nswer with no work shown will receive only credit.. If the mesure of the smllest ngle of right tringle is nd the length of the shortest side is 6. centimeters, find the length of the hypotenuse of the tringle to the nerest tenth of centimeter. 6. The re of grden is 0 squre feet. The length of the grden is foot less thn twice the width. Wht re the dimensions of the grden?