6. y = 4_. 8. D: x > 0; R: y > 0; 9. x 1 (3)(12) = (9) y 2 36 = 9 y 2. 9 = _ 9 y = y x 1 (12)(60) = x 2.

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1 INVERSE VARIATION, PAGES 676 CHECK IT OUT! PAGES 686 a. No; the product y is not constant. b. Yes; the product y is constant. c. No; the equation cannot be written in the form y k. y 5. D: > ; R: y > ; y y y ()(6) (4) y 4 y 4 4 y 4 y 5. y y (4.)(6) (.) y 58. y 58.. y y The child s weight is 8.65 lb. THINK AND DISCUSS, PAGE 6. Possible answer: The function rule will have the form y k, and y will be the same nonzero constant for each ordered pair.. EXERCISES, PAGES 66 GUIDED PRACTICE, PAGE 6. The graph of an inverse variation consists of disconnected branches.. Yes; the product y is constant.. No; the product y is not constant. 4. No; the equation cannot be written in the form y k. 5. Yes; y equals the constant y 4 8. D: > ; R: y > ; 9. y y ()() (9) y 6 9 y y 9 4 y. y y ()(6) (45) The gear has 6 teeth. 7. y 6. y y ()(4) (6) PRACTICE AND PROBLEM SOLVING, PAGES 66. No; the product y is not constant.. Yes; the product y is constant. 4. Yes; the function can be written in the form y k. 5. No; the function cannot be written in the form y k. 6. y 7. y Copyright by Holt, Rinehart and Winston. 9 Holt Algebra

2 8. D: > ; R: y > 9. y y ()(4) (6) y y (.5)(5) (4.5)y y y 4.5 y June can buy about yd. 6 a. y k Helen puts psi pressure. b. y k The area is 57.5 i n.. y y (7)(9) (6) y 6 6 y y 6.5 y. direct; 8 4. inverse; 4 5. neither 6. direct; 5 7. inverse; 8. neither 9. inverse; 5. direct; 5. d ; inverse. direct n. neither 4. inverse 5. y ; D: natural numbers; R: y > 6. y k 6π An inverse variation function, y k and k. So neither nor y can be. 8. Substitute the known values of and y into y k and solve for k. Use this value of k in the equation y k. 9. Inverse operations, additive inverse, etc.; all of the terms involve moving in opposite directions; inverse variation describes a relationship in which the variables move in opposite directions (one increases while the other decreases). 4a. y k 75 k (75)() k 5 k b. y k 5 4. C; when, y 4 and when, y C; the product of y is not constant. c. 4. D; y k k 7 ()(7) k 4 k CHALLENGE AND EXTEND, PAGE the linear function y 46. ft in. 6 in. y y (6)() (6) y 7 6 y y 6 y The force created at P is lb. 47. y y ( 4 )() ( 6 ) y 6 y 44. y y ()(.5) () Brad can buy tickets. 6 y y The strength of the signal is about watts. SPIRAL STANDARDS REVIEW, PAGE D: {}; R: {4,,, }; no 49. D: {4,,,, 4}; R: {,, 5}; yes ( + 6 ) ± ±9 or 5 Copyright by Holt, Rinehart and Winston. 9 Holt Algebra

3 5. d 6d 7 d 6d 7 d 6d (d ) 6 d ± 6 d ±4 d 7 or 5. y + 6y 5 y + y 5 4 y + y ( y + ) y + ± y + ± y 5 or a 4, b, c 6 Discriminant b 4ac ( ) 4(4)(6) 9 (96) 5 > has solutions. RATIONAL FUNCTIONS, PAGES 6464 CHECK IT OUT! PAGES 6467 a. y b. y 4 The ecluded value is. The ecluded value is ; y c. y a. y The ecluded value is 4. b. y c. y ; y 5 a. b ; y 5 4a. D: > ; R: natural numbers > b. THINK AND DISCUSS, PAGE 68. Yes; the function is undefined at 5.. To find the vertical asymptote, set + 9 equal to and solve for. The vertical asymptote is 9. To find the horizontal asymptote, look at the constant term. The horizontal asymptote is y 5.. EXERCISES, PAGES 6964 GUIDED PRACTICE, PAGE 69. ecluded value. y 4 The ecluded value is. + + No solutions, therefore no ecluded values. y 4. y The ecluded value is. 6. y ; y ; y 8. y 5. y The ecluded value is y ; y 9. y ; y Copyright by Holt, Rinehart and Winston. 94 Holt Algebra

4 .. 4a. D: > ; R: < y < b... PRACTICE AND PROBLEM SOLVING, PAGES y 7 6. y 4 4 The ecluded value is. 4 The ecluded value is y 8. y 5 5 The ecluded value is. 5 The ecluded value is ; y 9. y. y ; y ; y. y ; y 9. y a. D: > ; R: natural numbers > 5 b y 4 9. y The ecluded value is.. y The ecluded value is y The ecluded value is The ecluded value is.. 5. Copyright by Holt, Rinehart and Winston. 95 Holt Algebra

5 7. y ; y 9. y + 5 ; y 5 4. B 4. A 4. C 8. y 5 ; y 5 4. y ; y Student B is incorrect. The student identified the value at which there is a vertical asymptote ( ). 45. D: > 5; R: nonnegative values 46a. b. y 6 c. y It will take 5 hours. 5. y > 5 + > 5 > > 5 D: > I and III; II and IV 56a. y 5 c. 54. y > 7 > > 7 > 7 D: > 7 b. D: natural numbers 57. The graph of y k is the reflection of the graph of y k across the ais. 47. translated 6 units right 48. translated 7 units left 49. translated 4 units up 5. y > > > D: > 5. translated units right and 9 units down 5. y + + > + > > D: > 58. A 59. D 6. f() + Copyright by Holt, Rinehart and Winston. 96 Holt Algebra

6 CHALLENGE AND EXTEND, PAGE 64 6a. Yes; the function is a quotient of polynomials. b. D: all real numbers c. R: < y d. no 6. No; the graph of f() has an ecluded value at. The graph of g() has no ecluded values. 6. y + + SPIRAL STANDARDS REVIEW, PAGE t + 5 < (t + ) 4t + 5 < t + 9 4t < t + 4 t < j + < 4j 9 j + 9 < 4j 9 < j < j j > 68. c 5 > c + 7 c > c > c c < 7. 4 ( + )( ) + or or The zeros are ± ( )( ) or The only zero is. 65. (r + ) r 6 r + r 6 r r 8 r (g + ) g 5 5g + g 5 5g g 5 g 5 g (m ) < (6 m) 6m < m 6m < 4 m 8m < 4 m < ( + )( ) + or The zeros are and. 7. Let s represent side length of square piece, in inches. Then dimensions of new piece are (s ) in. by (s + ) in. A lw A (s )(s + ) 78 s 78 s s s 784 6(49) (7) 8 Dimensions of new piece are 8 6 in. by 8 + in. SIMPLIFYING RATIONAL EXPRESSIONS, PAGES CHECK IT OUT! PAGES a. b. b t + 5 c. t + 5 t 5 There are no ecluded values. k k + 7k + k + 7k + (k + )(k + 4) k + or k + 4 k or k 4 The ecluded values are and 4. a. 5 m 5m 5 m 5 m m m m ; m c. n n n n ; n b. b. b 5 b + b + 5 (b + 5)(b 5) (b + 5 ) b 5 b ( ) ( )( + ) ( ) ( +)( ) + b + 5b b + 5b b(b + 5) b or b + 5 b 5 The ecluded values are and 5. b. 6 p + p p + 6p ( p + ) p + 6p r + a. r + 7r + r + (r + )(r + 5) r ( 4) (4 )(4 + ) ( 4) ( 4)(4 + ) 4a. (4 + ) 4 + c. ( ) ( + )( ) + 5. The barrel cactus with a radius of inches has less of a chance to survive. Its surfaceareatovolume ratio is greater than for a cactus with a radius of 6 inches. Copyright by Holt, Rinehart and Winston. 97 Holt Algebra

7 THINK AND DISCUSS, PAGE Possible answer: The epression has the ecluded values and, but cannot be identified as an ecluded value in the simplified form, +.. EXERCISES, PAGES GUIDED PRACTICE, PAGE Both the numerator and denominator are polynomials.. 5 m m The ecluded value is.. p p p 5 p p 5 (p 5)(p + ) p 5 or p + p 5 or p The ecluded values are 5 and a 8a 4 a 4 a 7. a a a ; a y + y + ; y ( 8) or 8 8 The ecluded values are and d + d d + 6 d(d + 6) d + 6 d; d y 5 y ; y 5 9. h h + 4 h (h + ) h h + ; h b + 4. b + 5b + 4 b + 4 (b + 4)(b + ) b c + 5c + 6 (c + )(c 4) (c + )(c + ) (c + )(c 4) c + c 4 j 5 j + j 5 (j + 5)(j 5) (j )(j + 5) j 5 j n 6 64 n (n 8) (8 n)(8 + n) (n 8) (n 8)(8 + n) (8 + n) 8 + n 9. 5r r + 4r 5(r ) (r + 6)(r ) 5 (r + 6). a. 5q 5 ( q ) 5(q ) ( q)( + q) 5(q ) (q )( + q) 5 ( + q) 5 + q b h b + b h ( + 4). 6 ( + 4) + 4 ;. s 4 s + 4s + 4 (s + )(s ) (s + ) s s + ( )( + ) ( )( + ) ( + )( + ) + p ( h ) b ( h ) ( b + b ) p 4p 5 p + (p 5)(p + ) p ( ) ( )( 4) 4( ) ( )( 4) ( 7) (7 )(7 + ) ( 7) ( 7)(7 + ) (7 + ) a a + a 5 ( a) (a + 5)(a ) (a ) (a + 5)(a ) a + 5 b b + b Copyright by Holt, Rinehart and Winston. 98 Holt Algebra

8 b. ( b )h b h ( b ) + ( b ) ( b + b )h b b + b h They will be the same. PRACTICE AND PROBLEM SOLVING, PAGES c 5. c + c c + c c(c + ) The ecluded value is. c or c + c The ecluded values are and ( 5)( + ) 5 or + 5 or The ecluded values are 5 and. n n 7n 4 n 7n 4 (n + )(n 4) n + or n 4 n or n 4 The ecluded values are and d + 4 d d + 4 d (d + ) d + 4 d ; d. y 4 y 5 y 4 y 5 y 4 y. 5 y ; y q 6 q 9q + 8 q 6 (q )(q 6) q 4. t t 5t + 6 t (t )(t ) t 9. m m 4 m m 4 ; m 4. t 6t t 8t t 8t t 8 ; t. z z + z (z ) (z + )(z ) z z + 5. p 6p 7 p 4p 5 (p 7)(p + ) (p 5)(p + ) p 7 p ( + )( ) ( + )( + ) + 5 4(5 ) ( 5)( + 5) 4( 5) 8. 4 ( 5)( + 5) v 6 44 v (v ) ( v)( + v) (v ) (v )( + v) ( + v) + v 4a. S V lw + lh + wh lwh ( ) ( )( 4) 4 b b + 8b ( b) (b + 7)(b ) (b ) (b + 7)(b ) b + 7 (lw + lh + wh) lwh b. S A ( ) 84 V A ; S B ( ) V B 8 6 Bo A; the surface area of bo A is 84 i n and the surface area of bo B is i n. The volumes are the same, so the company should use the bo with less surface area, bo A ; ( + 4) 5 + 5( + 4) n + n + 5n n + 5n 46. n ( n + n + 5) n ( n + 5) n + n + 5 n + 5 j 5 j 5 j 5 (j + 5)(j 5) j p + p + 6 p + 7 (p + 6 ) (p + 6) p + 6 a a + a a a w + w 7 6w (w )(w + 7) (w ) w + 7 Copyright by Holt, Rinehart and Winston. 99 Holt Algebra

9 48. n n 56 n 6n + 64 (n + 7)(n 8) (n 8 ) n + 7 n ( + 5 ) already simplified 5a. Crew Size () Workdays Workers b. y 5 S 5a. V 6 s 6 s s ( + ) + + ( + ) ( + ) 5 (5 )(5 + ) ( 5)( + ) ( 5)(5 + ) ( 5)( + ) Construction Days (y) c. The ecluded value is. b. s, 6 s 6 c. s 6, 6 s Set the denominator equal to and solve for the variable. 55. Possible answer: D; 4 or 4 or + ( 4)( + ) ( 4 ) 4 4 The ecluded value is A; bh bh CHALLENGE AND EXTEND, PAGE Sometimes; the rational epression + has as an ecluded value, but the rational epression has no ecluded values Never; the numerator must be a polynomial. 6. Sometimes; the rational function y + has as an asymptote, but the rational function y + has no asymptotes v 6 v 4 v 4v v( v) (v )(v + ) v(v ) (v )(v + ) v v +.5y..5 y.4 (.5y.) (.5 y.4) 5y 5 y 4 5(5y ) (5y + )(5y ) 5 5y ( + ) ( ) The ecluded value is a 7a + a + 9a 5 (a )(a ) (a )(a + 5) a a ( + ) or + The ecluded values are and. SPIRAL STANDARDS REVIEW, PAGE D: {,,, 9}; R: {, 4, 5, 6} ± 96 ±4 The ecluded values are ± D: {4, 5, 5, 9}; R: {7,, } 7. Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

10 h 5 h 75. b 4 already simplified 74. s t 6 t 6 s 76. v w 4 w 4 v TECHNOLOGY LAB: GRAPH RATIONAL FUNCTIONS, PAGE 649 TRY THIS, PAGE 649. When the denominator of y is set ( )( 4) equal to and solved for, then equals or 4. However, when the denominator of y is set 4 equal to, the ecluded value seems to be only 4.. No; for the value of, y is undefined, while y 4 is defined as. a. asymptote b. hole CONCEPT CONNECTION, PAGE 65. It would take the crew 5 construction days.. y ; the number of people in the crew; y the number of construction days; inverse variation. As the crew size increases, the number of construction days decreases. 4. y 6.5 It would take the crew 6.5 construction days. 5. y There are 6 people in the crew. 6. D: natural numbers; R: y > 7. y ; and y READY TO GO ON? PAGE 65. No; the product y is not constant.. Yes; the product y is constant.. yes; y 4. no; cannot be written in the form y k 5. yes; constant product 6. no; cannot be written in the form y k 7. y 6 9. y y (6)(4) (8) y 44 8 y 8 y She can buy 8 calculators.. ; and y. ; and y 8. y 4. ; and y. ; and y Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

11 4. D: > ; R: natural numbers > n n The ecluded value is ( + )( + 4) + or + 4 or 4 The ecluded values are and t t + t t + t t(t + ) t or t + t The ecluded values are and ; s +. s 4s 5 s + (s 5)(s + ) ; s and s 5 s 5 p p 8 p 8 p 8 The ecluded value is 8.. n n n n n(n ) n ; n and n (4 ) ( 4 ) ( 4) ( 4 ) 4 ; 4 S. cone S πrl + π r πr(l + r) cylinder πrh + π r πr(h + r) 4 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS, PAGES CHECK IT OUT! PAGES a. (c 4) 5 45(c 4) 45 (4c + 6) 5(4c + 6) 45(c 4) 5 (4)(c 4) 9 5(c 4) 4 5(c 4) 9 4. m 5 m + 6 m 4m m 5 m + 6 m 4m m 5 (m + ) (m 6)(m + ) (m 5) m 6 m 5 m 6 a. n 5 n + 8n + 6 n + 4n n n n 5 n(n + 4) (n + 4 ) (n 5)(n + ) n + 4 n(n + ) n + 4 n + n p + 4 b. p p p + p p + 6 p + 4 (p 5)(p + ) p(p + ) p + 6 (p + 4)(p 5) p ( p + 6) p p p + 6p 4a. 5 ( 5) ( 5) b. 5 y 5 z 4 y y z 4y 4 y 7 z y z 5 y 4 6 b. 8v w v 4 6v w 4 8v w w 4 6v v 4 6v w 6 8 v 4 w 6 v Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

12 c. + ( + ) ( ) + ( + )( ) ( + )( + ) (5 ) 5. P(blue, blue) ( 5 + )(5 +). The probability is about.. THINK AND DISCUSS, PAGE 655. Possible answer: Dividing by a polynomial is the same as multiplying by the reciprocal of the polynomial.. EXERCISES, PAGES GUIDED PRACTICE, PAGE h j h k j h k h 4 j k h j k 6h 5j k ( ) 4( + ) 6( + ) ( ) ( + ) ( + ) ( ) c 4 d c 5a c d 5a c 4 d c 4 d a 6. 4y y z y z ab c a c a b c 6. p q 5 p 4 q 5p q 8 p 6 q 4 6pq p 5 q 7. ( 4y + 8 y 4) 4y + 8 y 4 (y + )(y ) 4(y + ) (y ) y (5 + ) ( + )(5 + ) m ( m 7m ) 6m + 8 m 6m + 8 m 7m m (m )(m + ) 6(m + ) m(m ) m m 4p. ( 8p + 6 p 5p 4) 4p 8p + 6 p 5p 4 4p (p 7)(p + ) 8(p + ) p(p 7) p 7p. c ( c c ) 4c + 4 c 4c + 4 c c c (c )(c + ) 4(c + ) c(c ) 4 c + c ( + )( + 4) ( 4)( + ) 4 ( + )( + 4) + j 5. j 5j + 6 j 4j + j 4 j (j )(j ) (j )(j ) (j ). a ( a + a + 5) a a ( a + a + 5) a + a + 5a. a + 6ab b 5 + a a b + 5ab a(a + 6b) 5 + a b ab(a + 5) a + 6b b Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

13 6. p + 4pq p 6 q 8 q p ( p + 4q) p ( p + 4q) ( q 4) ( q 4) q q p q + q 4 4 p 6q q 7. r + 5r + 4 r + 8 r 6 r + (r + )(r + 4) (r + 4) (r + 4)(r 4) r + (r + 4) r 4 r + 8 r 4 9. a 4 b a a c 8 c 4 a 4 b 4 8 c a c a c 4 a 4 b c 4 4 a 4 c 4 b c ( 5) ( )( + 5) ( ) 8. y 8 y y + y 49 (y 8)(y + ) ( y ) ( y 49) y 6y 6 y 4 5 y m + m 4 m + 4 m m m m + m m m m 4 m + 4 m ( m + ) m m(m ) (m ) 4(m ) m m 4 ( + 5)( 5) 4 ( m + ) ( 5) 5 a. Let be the number of black tiles, then there are + white tiles, and + ( + ) + tiles in total. P(black, white) ( + ) ( + )( + 9) 5(5 + ) b. P(black, white) ( 5 + )( 5 + 9). The probability is about.. PRACTICE AND PROBLEM SOLVING, PAGES p 6 q p 7 r r p 8 q 7 r 4 4. r t r s t 6s t 8 r 4 s 6 r 4 s t 48 r 4 s t 8 5. y + 5 y + (y + ) (y + 5) y + y m 8m m + 6m 6 4 m 8m 6. ( m + 7m 8) m + 7m 8 m + 6m 6 4m(m ) (m )(m + 8) (m )(m + 8) 4m(m ) 4 m 4m ( 4 6 ) ( a + 4a + ) a + 6 a + 6 a + 4a + (a + )(a + ) (a + ) (a + ) a ( )( + ) ( ) ( + ) + 6 n + 8n 9. n n + 9n + 8 n + 6 6n(n + ) (n )(n + ) (n + )(n + 8) (n + ) n(n ) n + 8 n n n + 8. a b a + 4b 5 a + a b 6 a b + 6 a b a b (a + b) 5 a (a + b) 6 a b(a + b) 5 a (a + b) 5 a + 5 a b. t 5t t (t + )(t ) 5(t + ) 5 (t ). 6 j k 5 4 j k 5j j 6 j k 5 j 5j 4 j k 8 j k 5 j 4 k 9 k j Copyright by Holt, Rinehart and Winston. 44 Holt Algebra

14 . a 4 ( 8a a ) a a 4 8a a a a 4 a 8a a a 4 a a(4 a) a ( + )( ) 6 ( + ) 4( ) 4 + 5a. P(red, blue) ( + 4) ( + ) ( + 4) ( + ) 4( + ) 4( + ) 4(4 + 8 ) b. P(red, blue) 9 4(4 () + 8() ) 6 The probability is Student A is incorrect because individual terms (rather than factors) were divided out y 4 ( ) ( ) 8 y y 6 y y 6 y y 4 y y 4 4 y 4 y y 5 y 4 y y 5 y 4 y 5 y y 4 y y y 4 4 y + y ( + )( ) y ( + ) y 4( ) b; the other epressions are all equivalent to y, but b simplifies to 4. y 8. 5 p q 9. 6 m 8m m 9 p q p m + m m + 4m + p q 6 m 8m + 4m + p 4 q q p m m + m m 9 6m(m ) (m + )(m + ) m (m + ) (m + )(m ) m ( )( ) 4( ) 5 ( 4 6 ) ( + )( ) 4 4( + )( ) 6 4. m m 6m w 4 7 w m 4 m 4m w m m m 4 m 4m w 6m 66 w m( m) m(m 4) (m + ) 6(m ) m (m 4) 4(m + ) m 4 m 4m Multiply by 4m. Then divide out the common m factor m to get a final answer of 4. 45a. I O y b. I I 6 4 I (6)(4) I 64 The height of the image is 64 cm. I O y 64 6 y 4 y (4)() y 8 y The distance between is 8 cm. c. M I 64 cm O 6 cm 4 The magnification of the lens is A; t + 4 t (t + 4) (t + 4) 9 (t + 4 ) C; b a 6a b 5 a b 4 b a ab 5b 7 4 Copyright by Holt, Rinehart and Winston. 45 Holt Algebra

15 48. C; ( + 5)( ) ( 8) ( 4)( 6) ( 4) ( 6)( + ) ( + ) + 9 CHALLENGE AND EXTEND, PAGE ( + 4) ( ) + ( ) ( + 4)( ) ( ) ( + )( ) + + ( + ) 5. c + 5 c 4 c + 6c + 5 c + c + 5 c + 6c + 5 c 4 c + c + 5 c + c 4 c + 6c + 5 c + 5 (c + )(c ) c + (c + )(c + 5) (c )(c + ) c c 5. y z y z y y z z y z z y 4 yz y z z a + a + 6a a + a + 5 a + a + a + 6a + 5 a + 5 a + a + 5 a + 6a + 5 a + a + (a + )(a + 5) a + 5 (a + ) a + SPIRAL STANDARDS REVIEW, PAGE Let represent number of etra laps (a + ) m + m m 9 Since money cannot be negative, m y 7 and y (both have slope ) 59. y 5 7 and y (both have slope 5); y + 5 and y 7 (both have slope ) 6. y 7 and y (both have slope 7) ( + )( ) ( ) Ecluded values: 4 4 ± ( 5) ( 4) ( + )( ) Ecluded values: ± ( ) 6. ( + 4) already simplified Ecluded values: Copyright by Holt, Rinehart and Winston. 46 Holt Algebra

16 5 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS, PAGES CHECK IT OUT! PAGES a. n n + n n n + n 4n n n a. 5a + a 4 a 4 a 4 5a + (a 4) y b. y + + y y + y + y y(y + ) y + y + y a + 6 a 4 a 4 (a + ) (a )(a + ) a b + 4 b. b + b + b 4 b + b 4 b + 4 (b + ) 4b + b + b 4 b + b 4 a. 5 f h 5 f f h 5f h 5 f h h LCM 5 f f h h 5 f h b. 4 ( 6)( + ) ( 6)( + 5) ( 6)( + 5) LCM ( 6)( + )( + 5) 4a. 4 d d b. a + 4a + 8 d 4 d( d d d ) ( a + a 8 a a(a + 4) d ) (a )(a + 4) + 8 a 8 d 6d a a + 8 a 6 d 6 d a d 6d a 6 d d(4d ) 6 d 4d d 5. Katy s rate against the current is 5, or 4. Katy s rate with the current is 5 +, or 6. total time ( ) + 6( ) + 5 When, 5 5 () 5 4 It will take Katy 5 h or.5 min to kayak the round 4 trip. THINK AND DISCUSS, PAGE 66. Possible answer: Factor the denominators. Then write each factor the greatest number of times it appears in the denominator.. EXERCISES, PAGES GUIDED PRACTICE, PAGE 66 y. + 5y y y y + 5y 6y y y y 4m +. m m + 8m + 5 m + 5 4m + + m + 8m + 5 m + m + 5 m + 5 m + 5 (m + 5)(m + 7) m + 5 m ( + 4)( 4) a 5a 6 a + a + a + a + 7a (5a 6) a + 4 a + a + a + a + (a + ) (a + )(a + ) a ( + ) ( + ) + 7. y y y 6 yz y z LCM y y z 6 y z Copyright by Holt, Rinehart and Winston. 47 Holt Algebra

17 ( + 4)( + 5) ( + 5)( 4) ( + 5)( 4) LCM ( + 5)( + 4)( 4) 9. y 6 (y + 4)(y 4) (y + 9)(y 4) (y + 9)(y 4) LCM (y + 9)(y + 4)(y 4). c 4 c c ( ) 4 c 9 c 4 c 5 c ( ) a. total time r + 4.5r r ( ) r r r ( + ) ( + )( + ) ).5r( 6 r b. total time 6 6 r The total travel time is 6 h. 4 6 PRACTICE AND PROBLEM SOLVING, PAGES y + 4y y y 4y + 4y 8y 8 y y 5. a a + + a a + a + a a + a y a + a + (a + )(a ) a + a ( ) ( )( ) 4 7. m m 6 6m m 6 m 6m m(m 6) m 6 m 6 m c + 8. c c 5 4 c 5 c + (c + 8) c 5 4 c 5 4 c 5 c 5 (c + 5)(c 5) 9. a 9a a c a 4a + a a 9a (5 a 4a + ) a a 5a a (a + )(a ) a a +. 4j k 4 m j k k k k m 5jm 5 5 j m LCM 5 5 j k k k k m j k 4 m. a + 4a 4a(a + ) 7a + 9 9(a + ) LCM 6a(a + ). p p p(p ) pq r p q r r LCM p q r r (p ) pq r (p ). 5 y z 5 y y z y 5 y y y LCM 5 y y y z y z ( ) LCM 5 7 ( ) 5 ( ) 5. y + 7y + (y + )(y + 5) y + 9y + (y + 4)(y + 5) LCM (y + )(y + 4)(y + 5) ( ) (6 + 5) ) ( 5 y y 7. y y 4y + y 9 y(y ) (y )(y ) y y 9 y y y y 9 y y ( ) y (y ) y (y ) y (y ) y (y ) (y ) y + (y ) Copyright by Holt, Rinehart and Winston. 48 Holt Algebra

18 8. t t 4 t t t t 4 t t ( ) t (t + 4) t 4 t 4 t + (t + 4) t 4 t + 4 t ( ) + + ( ) 5 + ( + ) ( ) ( ) 5 ( + ) ( ) 4 ( ). m 4m 8 m m 4m + 4 m 4(m )( m 9. m ) m z + 4 z 7z z + 4 7z z( 7 7z( 7) z + z 9 z ( 4 (m ) m(m ) 4 m 4(m ) 4(m ) m(m ) 4 m 4(m ) m 6m 4 m 4(m ) m 6m 4(m ) a. total time w +.85w w( 7) +.85w( ) 7 7w + 7w 7 7w b. When w, 7 7w 7 7() 7 5 It took him 7 h or 44min to complete his 5 walking. a. total time r r 5 5 5) + 45 r 5 r( r r 7 r b. When r 5, 7 7 r 5 4 The total travel time is 4 h. c. Divide the total distance (5 mi) by the total time. 4) ) y + y y c 49 c 49 c + y (5 + y) 7 c 5 + y 5 + y 7 c 49 c (7 c)(7 + c) 7 + c 6. 6a a b a + 6a a + 4 a( b ) b + b 6a a + b 4 a + ( 6a 4 b b ) a 8. r + r r + r + 9 r + r + r (r + 9) r r 9 r + b + b b (r + )(r ) r + r (8 5) ( 5 ) ( 5) y 4 y y ( y y 4 y y ( y y) y 4 y y y 4 y y y + 4) ( + ( + 4) + ) ( + )( + 4) + 6( + ) (+ )( + 4) ( + 4) + 6( + ) ( + )( + 4) 8 + ( + )( + 4) y 4. y 9 y + y 9 y (y )( y + y + ) y + (y + )(y )( ) y(y + ) (y + )(y ) (y + ) (y + )(y ) y(y + ) (y + ) (y + )(y ) y + y (y + )(y ) 4. Student A is incorrect. The student may have subtracted and then simplified before setting the denominator equal to. 4 is also an ecluded value. Copyright by Holt, Rinehart and Winston. 49 Holt Algebra

19 44. P + π () () + π () ( + π) 9 + π 9.46 The probability of winning a prize is about Possible answer: a. He subtracted from both sides of the equation. b. He subtracted from by finding a common denominator () ; 8 ; In this case, either binomial can be used as the least common denominator. After selecting the binomial for the denominator, change the other binomial to match by multiplying it by. 49. A; 6 p + 6 (p + ) p + 5. A; 4 5. D; ( ) ( + )( ) ( ) ( + )( ) ( + )( ) ( ) ( + )( ) + 4 ( + )( ) ( + ) ( + )( ) 5a. r + 5 ; time to post office: r r ) + 5 r 9 r + 5 r b. r + 5 r r ( 4 r c. When r, 4 ; time to library: 5 r r 4 () 4 9 It took her 4 h or h min to bike. 9 CHALLENGE AND EXTEND, PAGE y + y y + y( y y) + y ( + y)( y) ( y) ( + y)( y) + y ( + y)( y) ( y) ( + y) ( + y)( y) 4y ( + y)( y) ; ±y 54. m m 5m m( 5m 5m) + 4 ( m ) + 5m( m m) 5m m m m m 5m m m 9m + 4 ; m m 55. a y + b z + c yz a y( z + z) z( b y y) + yz( c az yz + by yz + c yz az + by + c ; yz, y and z ) 56. y y y y y ( SPIRAL STANDARDS REVIEW, PAGE 665 y ) a + a a 6 5 a a + 6a ( a 5 a + a ) ( 5) + ( 5) (polynomial does not ( + )( 5) factor further) 59. 5h + 5 h h 6 5 h 5h 6 5( h h ) 5(h 4)(h + ) (since (4)(), 4 + ) 6. s + 8 s 4( s + s ) 4 s (s + ) 6. + m + m 5m ( m 5) + m( m 5) (m + )( m 5) (polynomial does not factor further) t 4 t 4t 4(4 t t ) 4t(4 t t ) 4t(4 t 4t + t ) 4t[4t(t ) + (t )] 4t(4t + )(t ) Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

20 6. d 4d (d 6)(d + ) d 6 or d + d 6 or d 64. g 9g 4 g 9g + 4 (g )(g 4) g or g 4 g or g ( + ) t 8 t 4 ( t 4) t 4 ; t ± n + 5n n + n 6 n(n + 5) ( 4) (n )(n + 5) ( + 4)( 4) n ; n 5 and n n + 4 ; ±4 ALGEBRA LAB: MODEL POLYNOMIAL DIVISION, PAGE 666 TRY THIS, PAGE There is no arrangement of tiles that will form a rectangle with a side length of + because + is not a factor of DIVIDING POLYNOMIALS, PAGES CHECK IT OUT! PAGES a. (8 p 4 p + p) (4 p ) 8 p 4 p + p 4 p 8 p 4 p + p 4 p 4 p 4 p p + p b. (6 + 5) a. + 7k + k k + (k + )(k + 5) k + k + 5 c. s + s + 6 s + 6 (s + 6 ) s + 6 s + 6 b. ( a 8a + ) (a 6) a 6 a 8a + a a 6 a 8a + ( a 6a) a + (a + ) a 4a. ( m + 4m ) (m + ) m + m + 4m m 5 m + m + 4m ( m + 9m) 5m (5m 5) m 5 + m + b. ( y + y + ) (y ) y y + y + y + 6 y y + y + ( y y) 6y + (6y 8) y y b. b 49 b + 7 (b + 7)(b 7) b + 7 b 7 a. ( y 5y ) (y ) y y 5y y + y y 5y ( y 6y) y (y ) y + Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

21 5a. ( 4 + ) ( ) ( 4 + ) ( ) ( ) + ( + 4) 4 + (4 + 8) b. (4p + p ) (p + ) ( p + 4p ) (p + ) p + p + p + 4p p p + 6 p + p + p + 4p ( p + p ) p + 4p ( p p) 6p (6p + 6) 7 p p p + THINK AND DISCUSS, PAGE 67. Possible answer: It means the binomial is a factor of the polynomial.. ; the denominator of the remainder cannot equal zero.. EXERCISES, PAGES 6767 GUIDED PRACTICE, PAGE 67. ( 4 ) 4 4. ( 6 a 4 4 a ) 4a 6 a 4 4 a 4a 6 a 4 4a 4 a 4a 4 a a. ( b 4b + 4) b 4. (8 r r + 6) 6r b 4b r r + 6 b b b 4b b + 6r 4 8 r b 7b r r 6r + 6 6r r + b r 5. ( ) (5 m m ) 5 m 5 m m 5 m 5 m m 5 m 5 m 5 m m + m m 8. a a a 4 (a 4)(a + ) a 4 a ( )( + ) y + y y (y )(y + 5) y y + 5. t 6t t (t )(t 4) ( + )( + 5) t t +. p p. ( c + 7c + ) (c + 4) p + 4 (p + 4)(p 5) c + 4 c + 7c + p + 4 c + p 5 c + 4 c + 7c + ( c + 4c) c + (c + ) c + 4. ( s s 5 ) (s 5) s s s s + s 5 s s ( s 5s ) s 5 ( + 7) 4 (s 5) ( 4) s + Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

22 ( ) 6 (6 ) ( r + r + 5) (r ) r r + r + 5 r + 7 r r + r + 5 ( r 6r ) 7r + 5 (7r 5) 56 r r 9. ( n + 8n + 5) (n + 4) n + 4 n + 8n + 5 n + 4 n + 4 n + 8n + 5 ) ( n + 4n) 4n + 5 (4n + 6) n n + 4. ( t t + 4) (t ) t t t + 4 t + t t t + 4) ( t t ) t + 4 (t ) 5 t t. ( 8 n 6n 7 ) (n + ) n + 8 n 6n 7 4n 5 n + 8 n 6n 7 (8 n + 4n) n 7 (n 5) 4n 5 + n + 7. ( a + 4a + ) (a + ) a + a + 4a + a + a + a + 4a + ( a + a) a + (a + 4) a + + a +. ( b b + ) (b + ) b + b b + b b + b b + ( b + b) b + (b 6) 7 b + 7 b +. ( ) ( + ) ( + ) ( + ) ( 6 ) 6 + (6 + 8) 5 (5 45) ( p p 4) (p ) p p p + p 4 p + 4p + 8 p p p + p 4 ( p 6 p ) 4 p + p (4 p 8p) 8p 4 (8p 6) p + 4p p 5. ( m + ) (m ) m m + m + m + m m + m + ( m + m) m + (m ) m + + m Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

23 6. ( + 4 5) (5 + ) (4 + 5) ( + 5) (4 + ) 7 + ( 7 85 ) 85 5 ( ) ( 4 k k 8 ) (k + ) k + 4 k k 8 4 k 4k + k + 4 k + k k 8 (4 k + 4 k ) 4 k k ( 4 k 4k ) k 8 (k + ) 4 k 4k + + k + 8. ( j + 6j + ) (j + 4) j + 4 j + j + 6j + j 4j + j + 4 j + j + 6j + ( j + 4 j ) 4 j + 6j (4 j 6j) j + (j + 88) 86 j 4j j + 4 PRACTICE AND PROBLEM SOLVING, PAGES (9 t + t 6t) t. (5 n n + 5) 5n 9 t + t 6t 5 n n + 5 t 5n 9 t + t 6t 5 n t t t 5n n 5n + 5 5n t + 4 n + n t. (6 p p + 8) 4 p 6 p p p 6 p p 4 p 4p + + p p 4 p. 8 t + t t (4t + )(t ) t 4t + 5. ( 5 + 6) ( ) ( ) + 6 ( + 6) 6. ( m + 8m + 8) (m + ) m + m + 8m + 8 m + 4 m + m + 8m + 8 ( m + 4m) 4m + 8 (4m + 8) m (6 a + 7a ) (a + ) a + 6 a + 7a a a + 6 a + 7a (6 a + 9a) a (a ) a 8. ( 8 ) ( 4) ( ) 8 ( 8) +. 4 r 9r + r (4r )(r ) r 4r 4. g + 7g 6 g + (g )(g + ) g + g Copyright by Holt, Rinehart and Winston. 44 Holt Algebra

24 9. ( + 6) ( ) ( 6 ) (4 8) ( m + 5m + 8) (m + ) m + m + 5m + 8 m + m + m + 5m + 8 ( m + m) m + 8 (m + ) 5 m m + 4. ( 6 ) ( ) ( 6 ) ( ) ( m 4m ) (m ) m m + m 4m m + 5m + m m + m 4m ( m m ) m 4m ( m 5m ) 46m (46m ) m + 5m + + m 4. (6 t + t + 9) (t + 9) t t + t + t + 9 t 6t + 5 t t + t + t + 9 (6 t + 8 t ) 8 t + t ( 8 t 54t ) 75t + 9 (75t + 5) 6 t 6t t ( p 4 7 p + p + ) (p ) p p 4 + p 7 p + p + p + p + p + 7 p p 4 + p 7 p + p + ( p 4 p ) p 7 p ( p 9 p ) p + p ( p 6p) 7p + (7p ) p + p + p p 45. Apply long division n n ( 4) 5 + n (5 ) n 47. ( ) ( + ) ( + ) 5 (5 5) ( ) ( ) + ( 4 ) ( ) + Copyright by Holt, Rinehart and Winston. 45 Holt Algebra

25 48. ( ) ( + ) + + ( + ) 4 (4 ) 49a. The values of y are negative and decreasing. b. The values of y are postitive and decreasing. c. The function is not defined at ( + 5 ) ( + 5)( 5) ( 5)( 9) ( 9)( + 5) (.88) 9 5. A B m + m + m +.5m + (m + )(m + ) (m + ) m + 5. Student B is incorrect. The second term should be positive. 5a. y y y b. The function is undefined at. 54. The binomial is not a factor of the polynomial ( + ) (4 + 6) Possible answer: + ; 76 ; when, +, so the results are equal ( + ) + 8 ( + 8) Yes; there is no remainder when you divide by +, so + must be a factor of C 58. B; ( ) ( + ) ( + ) ( 4) B; 6. A; ( 5) ( 5) + ( 5) ( + )( 5) CHALLENGE AND EXTEND, PAGE (6 y + 4 y ) ( y) 6 y + 4 y y 6 y y y y + y + 4 y y ( ) ( ) ( ) + ( ) 6 + (6 6) ( ) ( ) ( ) ( + + ) + + Copyright by Holt, Rinehart and Winston. 46 Holt Algebra

26 6. ( + ) ( ) + ( + ) ( + ) ( + ) ( + ) ( ) ( + ) ( + 8) ( + ) ( + ) ( + 4) ( ) ( + 4) So the height is ( + )( + ) ( + ) + base height ( + 4) ( + ) The base is m longer than the height. 66a. (π ( )) (π ( + + )) ( ) ( + + ) ( + + ) ( ) H V B + The height of the cylinder is ( + ). b. B π r π ( + + ) r + + r ( + ) r + The radius of the base is ( + ). SPIRAL STANDARDS REVIEW, PAGE Let w represent number of weeks. 8 + w + 4w 6 + w 4w 6 w w It takes weeks. 68. Let represent price. 5 5 or Maimum price is $55; minimum price is $ ( ) y ( ) 4 ( ) + 4( + )( + ) + 4( + ) 7. k + 4 k k + k ( + k) k + ( + k)(k + ) k + + k + 4k k + 5k + ( + ) 4 + k + k + k (k + )(k + ) k 8y 4 7 y 6 7 y 6 7 SOLVING RATIONAL EQUATIONS, PAGES CHECK IT OUT! PAGES b. 4 h + h 4h (h + ) 4h h + h a. n n + 4 (n + 4) n n + 4 n n c. 7 ( 7) 7 6 a. a + + a + 4 a a(a + ) ( a b. j + 4 j j 6 j + j j j(j + ) ( 6 j + a) a + ) a(a + ) ( 4 a + a 4(a + ) a 4a + 4 a 4 j ) j ) j(j + ) ( 6j (j + ) (j + ) 6j j j + 4 6j 4 j 4 Copyright by Holt, Rinehart and Winston. 47 Holt Algebra

27 a. 7 7 ( 7) ( )( 7) ( 5)( 7) 5 or 7 5 or The only solution is 5, so 7 is an etraneous solution. b. + 4 ( + )( ) 4( ) ( )( 5) or 5 or c The solutions are and 5, and there are no etraneous solutions ( + ) ( 4) or 4 or () (4) The only solution is 4, so is an etraneous solution. THINK AND DISCUSS, PAGE 676. Possible answer: Some solutions may be etraneous and make the epressions undefined.. and. Possible answer: Because the denominators are the same, the epressions would only be equal when 4. Since is an ecluded value, there is no solution. 4. EXERCISES, PAGES GUIDED PRACTICE, PAGE 676. rational equation. etraneous solution s 6 4 s ( + 4) 5s 4(s 6) + 8 5s 4s 4 8 s 4 5. p + p (p) (p + ) 4p p 5p p ( ) 9( 4) ( + ) (. 8 d d(d + ) ( j 8. + ) ( + ) ( d + d j + 4(j + ) j 4j + 8 j j 8 j ( ) a a 9 5 a 9 ( 5) a 5 + ) ( + ) + ( + ) d) d(d + ) ( 8(d + ) d (d + ) 8d + 6 d d 6 d d 5 d) d + Copyright by Holt, Rinehart and Winston. 48 Holt Algebra

28 . s 6 4 s + s s(s 6) ( 4. 7 r + r(r ) ( 7 5. a 4 s + s) 6s 8(s 6) + (s 6) 6s 8s 48 + s 6 s 54 s 8 s 6) s(s 6) ( 4 r r r + 4(r ) + 4r (r ) 4r 4 + 4r r + 9r 5 a a r ) r(r ) ( r 5 9 (a ) a(a 4) a 6 a 4a a 7a + 6 (a )(a 6) a or a 6 a or a 6 r 5 6. r 6 6r ( r r ) 6r ( 5 6) r 5r r 5r (r )(r + 4) r or r + 4 r or r n 7 n 8. n ( 6 n n) ( 7 n ) 6n 7 n n + 6n 7 (n )(n + 7) n or n + 7 n or n ( )( + ) 4 6 ( )( + ) or + or r ) a a + a ( 5 ) a ( 4 a a + ) 5 4a + a a 4a 5 (a 5)(a + ) a 5 or a + a 5 or a. p + p p ( p) ( p p ) + p + p p p (p + )(p ) p + or p p or p. c 4 c c 4 (c 4) (c )(c 4) c c 5c + 4 c 8c + 6 (c 4 ) c 4 c 4 4 c 4 c c c 4 There is no solution, and 4 is etraneous.. w + w w w ( w ) ( w + w w w ) ( w ) () (w + ) w(w + ) w w + w w w w + w 4 (w )(w + 4) w or w + 4 w or w 4 w + w w w w + w w w + ( 4 ) ( 4 ) The only solution is 4, and is etraneous. Copyright by Holt, Rinehart and Winston. 49 Holt Algebra

29 ( 5) ( ) ( 5) 8 ( 5) ( 7) + ( 5) ( 5)( + 6) 5 or or (5) (6) The only solution is 6, and 5 is etraneous. PRACTICE AND PROBLEM SOLVING, PAGES n n 8( + ) ( ) n (n ) n 9n 6 n n ( ) ( + 4) s s 5 4 s 5 4 5s s c c 4 c c(c ) ( 7 c 7(c ) (c ) 4c 7c 7 c + 4c c 5. 9 m m 5 m m ( 9 m m) m ( 5 m ) 8 5 There is no solution; is etraneous ( + 5) ( + 4) 8 ( 8) There is no solution; is etraneous. c ) c(c ) ( 4 c ). ( ) ( ) ; is etraneous.. r 6 r 6 r ( r ) r ( r ) r 9r 8 r 9r + 8 (r )(r 6) r or r 6 r or r ( 6 ) ( + ) + + ( )( + 4) or + 4 or 4 8 ( 8 ) ( ) ( )( 4) or 4 or ( + 4)( ) ( + ) ( 5)( + ) 5 or ; and are etraneous 7. + ( + ) ( + ) + + ; 4 is etraneous ( ) ( 5 ) ) ( ) ( ( ) () There is no solution, and is etraneous. Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

30 9. t t t + 4 t t(t ) (t + 4)(t ) t 9t t + t t t + (t )(t ) t or t t or t t t t + 4 t t t t + 4 t () + 4 () The only solution is, and is etraneous ( ) ( + ) + ( )( + ) or + or () The only solution is, and is etraneous ( 4) There is no solution, and 4 is etraneous ( + )( + ) ( + 7) ( + 5)( ) 5 or ; is etraneous ( ) ( + ) ; is etraneous b. 9 + f % 9% 45 + f 9 + f 45 + f f.9(45 + f) 9 + f f.f.5 f 5 Clancy would have to make 5 free throws. 45. Karla Andrew Books 8 Stacks Books per Stack ( ) + 6()( ) 8() ( 4) 6( 4)( + ) 4 or ( and 6 are etraneous) Since >, 4. Karla has 4 stacks of books. 46a y b y y ( 5) y ( 4 + y ) 8y 5y + y y 4 The image will apear 4 cm aways from the lens. c y 7y ( 8) 7y ( 4 + y ) 4y y + 7 y 7 The distance between will increase to 7 cm. 47. No; you can only use cross multiplication if the equation is in the form of a proportion ; possible answer: first add the fractions on the left to get 4. Then cross multiply and solve to find that 4. This method may be easier than finding the LCM. 44a. 9 % % Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

31 49. B; ( 4) ( 6) ( + 4) ( + 6) ( + 4) 8 or + 4 or (4) (4) The only solution is, and 4 is etraneous. 5. A; + ( ) ( + ) ( ) ( ) ( + ) ( ) ( + ) or () + + () () The only solution is, and is etraneous. 5. D; 5 + ( 5 ) ( + ) ( )( + 5) or + 5 or 5 CHALLENGE AND EXTEND, PAGE Statements Reasons a. ( + 4) 6 Cross Products Property b. + 6 Distributive Property c. Subtraction Property of Equality d. 4 Division Property of Equality a 7 a ( + 4)( a) 7( a) + (4 a) 4a 7 7a ( + a) + a ( )( a) or a or a Since the equation has no solution, a (j)(j + ) j + j j + 4 (j)(j + ) + 4(j)(j + ) j + j + + j 4 j + 4j 4 j 8j j 9j 5 (j + )(j 5) j, 5 6 a + 8 a + 4 6(a)(a + ) + 8(a)(a + ) 4(a)(a + ) a (a + ) 6a a 4 a + 4a 4 a a 6 a 5a 8 (a + )(a 8) a, 8 Since age must be positive, j 5 and a There is a year difference between Jill s and Angela s ages. Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

32 SPIRAL STANDARDS REVIEW, PAGE y + and y are parallel. 56. y and y + 4 are perpendicular. 57. y and y + are parallel; y is perpendicular to both y and y + and y + are perpendicular. 59. y () + y () () + (): y 6 + ( + y ) Substitute in (): (4) y 7 y 7 y (4, 7) 6. y 7 () 4 y () () + (): 9 + 6y 8 + (8 6y 4) Substitute in (): (5) y 7 5 y 7 y y 6 (5, 6) a 7, b 5, c discriminant (5 ) 4(7)() > solutions 6. y 8 () + y 5 () () (): 4y 6 ( + y 5) 7y y Substitute in (): () (, ) 8 APPLYING RATIONAL EQUATIONS, PAGES CHECK IT OUT, PAGES Let m represent number of minutes. Cindy s part + Sara s part whole lawn 5 m + 4 m ( 5 m + 4 m ) () 4m + 5m 9m m 9 min, or about min s 9. Alcohol (ml) Total (ml) Original 5 5 New 5 + a 5 + a 5 + a 5 + a a.8(5 + a) 5 + a 4 +.8a.a 5 a 75 The chemist should add 75 ml of alcohol.. Distance (mi) Time (h) Rate (mi/h) Ryan t + t + Maya t t t + t (t) (t + ) (t)(t + ) (t) (t + ) (t)(t + ) t t + t t t + t (t 5)(t + 6) t 5, 6 Maya makes the trip in 5 h. THINK AND DISCUSS, PAGE 68. Danielle can weed the garden by herself in h, so with Omar s help it makes sense that they finish in less time than h.. EXERCISES, PAGES GUIDED PRACTICE, PAGE 68. Let h represent number of hours. Summer s part + Louise s part whole room h + 5 h 5 ( h + 5 h ) 5() 5h + h 5 8h 5 h h, or about h 5 min 8 Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

33 . Chicken (qt) Vegetable (qt) Original New + a + a ( + a) () + a 6 a 4 The chef should add 4 qt of chicken stock.. Distance (mi) Time (h) Rate (mi/h) Connor t t Matt t t + t t + t t t ( + t)(t ) (t) t + t t t t t (t 4)(t + ) t 4, It takes Connor 4 h. PRACTICE AND PROBLEM SOLVING, PAGES Let m represent number of minutes. old vacuum s part + new one s part whole room 9 h + 45 m 9 ( 9 m + 45 m ) 9() m + m 9 m 9 m It takes min. 5. Let h represent number of hours. ft pipe + ft pipe whole reservoir h + 6 h ( h + 6 h ) () h + h h h 4 It takes 4 h to fill the reservoir. 7. Distance (mi) Time (h) Rate (mi/h) Local 6 t + 6 t + Epress 6 t 6 t t + t 6(t) + 5(t)(t + ) 6(t + ) 4t + (t)(t + ) 4(t + ) 4t + t + t 4t + 48 t + t 48 (t + 8)(t 6) t 8, 6 The epress takes 6 h, so it travels at 6 6 mi/h 6 8a. Distance (mi) Rate (mi/h) Time (h) Biking r r Running r 4 r 4 b. + r r 4 4 c. (r 4) + (r) 4(r)(r 4) 5(r 4) + (r) (r)(r 4) 5r + r r 4r r r + (r )(r ) r, r is etraneous since r 4 >. Jen s rate while biking is mi/h. 9a. Area ( m ) Length (m) Width (m) Rect. A 96 l 96 l Rect. B 96 l 96 l 48 l b. 96 l 48 l + 4 c l 48 4l l Length l m Width 96 l 96 8 m 6. Orange Juice (c) Total (c) Original 8 New + a 8 + a + a 8 + a.4 + a.4(8 + a) + a. +.4a.6a. a Maria should add c. Copyright by Holt, Rinehart and Winston. 44 Holt Algebra

34 . machine A: min; machine B: 5 min; machine C: 8 min; machine D: min Let represent number of minutes with machines A and D. machine A + machine D whole job + 6 ( + ) 6() Machines A and D take 7.5 min. Let y represent number of minutes with machines B and C. machine A + machine D whole job 5 y + 8 y 9 ( 5 y + 8 y ) 9() 6y + 5y 9 y 9 y 8. Machines B and C take about 8. min. Machines A and D are faster.. Let n represent the number. n + n 5 4 n 4n (n + 4n n) ( 5 4 n ) 4 n n 4 n ± n The number is or.. B is incorrect. The correct rational equation is h 4 + h 6. a. f + with f, y y y + y (y )(y) (y )(y) + (y ) y y y y y 4y y 4y + (y )(y 4) y, 4 Since 4 6, y. b. (y)(y + ) (y) + (y + ) y + y y + y + y 4y (y )(y + 6) y, 6 Image is cm from lens. (y )(y) y 4. It will take half the time it takes them to mow the lawn individually, or 4 min. This makes sense since they split the work evenly. 5. Student responses will vary. 6. A; New solution has (4 + w) ml of water and (8 + w) ml total volume; so 4 + w 8 + w.7 7. B; m 9 ( m + m + m 45 45) 9() m + m 9 5m 9 m 8 CHALLENGE AND EXTEND, PAGE Let h represent number of hours. drain A + drain B + drain C whole tank h + 6 h + 4 h ( h + 6 h + 4 h ) () 4h + h + h It takes 9h h 9 4 h, or h min. 9. Rate ( job Time (h) h ) Luke t t Eddie t t Ryan t + t + Luke s part + Ryan s part + Eddie s part whole job ( t ) 4 + ( t) 4 + ( t + ) 4 4 t + t + 4 (t + ) t(t + ) ( 4 t + t + 4 t(t + )() (t + )) 4(t + ) + (t + ) + 4t t(t + ) 4t t + + 4t t + t t 7t 6 (t + )(t ) t, Luke: t h; Eddie: t () 6 h; Ryan: t + () + 4 h Copyright by Holt, Rinehart and Winston. 45 Holt Algebra

35 . Cranberry (c) Total (c) Original V V New V + 6 V + 6 V + 6 (V + 6) 6 ( V + 6 ) 6 ( (V + 6) ) V + 6 4V + 4 V There were c of punch to start with. SPIRAL STANDARDS REVIEW, PAGE y 4 4() + (5) yes. 4 + y 4 4() + () yes y 4 4(9) + () yes () + ()() + () ( + ) yes () + ()() + () ( ) yes. 4 + y 4 4() + (8) no y 4 4(5) + () no y 4 4(4) + () no (4 ) + (4)() + ( ) ( ) 9 (4 ) 8 no no; 49 (7), and no; 5 (5), and 7 is not a factor of 4 5 is not a factor of no; 6 (6 ), and 6 is not a factor of ( + )( + ) + + ( ) ( 8)( + ) 8 + ( 8) ( + )( + 4) ( ) ( )( 6) 6 ( ) ( 7)( + ) 7 + ( 7) ( + 5)( 8) ( 5) CONCEPT CONNECTION, PAGE y y ( ) y ( + y ) y y + ( )y y. y Undefined The yvalues are positive for >.. 4. M 7.5 cm 5 cm.5 The magnification of the lens is Magnification remains the same. Copyright by Holt, Rinehart and Winston. 46 Holt Algebra

36 READY TO GO ON? PAGE 685. n + n 5 ( n 5n) n + n 5 n 5n n + n(n 5) n 5 n(n + ) n + n. h 6h g h ( h ) g h 4 g 6h 5 g 4g g h g 4g g ( h ). 6 y 6 y y y 6 m 8m + 6 m 4. m + m m m 8 (m + )(m ) (m 4)(m + ) (m 4 ) (m ) m 4 5. n 6 n n 5 5 n n n 6 n n b c b c 8 b c b c 8 b c b c 8 b c b c b + (4 b + 4b) 4 b + 4b 4 b + 4b 4b(b + ) 9. m + 5 m 4 m 5 4 m 5 m + 5 m 4 m 5 8 m 4 m 5 m 4 y ( + )( + ) ( ) ( + ) ( + )( ) p p 5 p p p ( + 4) ( )( + 7) + 7. t + 4 t t t + t t t 5 t ) t ( ( ) 4 4. (6 d + 4d) d 6 d + 4d d 6 d d + 4d d d + 6. ( 7 4 ) ( + ) ( + )( 4) + 4 m + 6m + 5 m + 5 (m )(m + ) (m + 5)(m + ) m + 5 m m + 5 m + 5 m m + 5 m 4 m + 5. m m 5. (5 4 + ) ( ) ( a + a ) (a ) 8. (4 y 9) (y ) a a + a y 4 y + y 9 a + 5 y + a a + a y 4 y + y 9 ( a a) (4 y 6y) 5a 6y 9 (5a ) (6y 9) a + 5 y + 9. ( + 5 8) ( + ) ( ) ( + 4) 8 ( + ) Copyright by Holt, Rinehart and Winston. 47 Holt Algebra

37 . ( ) or. t The only solution is, and is etraneous. t 4 t + t(t + ) ( t + t) 4 t(t + ) 4 ( t + ) 6(t + ) + 4(t + ) t 6t + + 4t + 8 t t t. 4 n 7 n + n ( 4 n ) n ( 7 n + ) 4 7n + n n + 7n 4 (n )(n + 4) n or n d + n or n 4 d d + 8 (d + )(d + 8) 6(d + 8) d + d + 6 6d 48 d + 6d + 64 (d + 8 ) d + 8 d 8 d + d d There is no solution, and 8 is etraneous ( 6)( 4) 4 ( 6 ) ( ) 5 or or 6. h + h 6 ( h + h ) 6() h + h 6 5h 6 h or. 5 It will take them. h to shovel the walk together. 7. current:.6() 8 ml water, ml total new: (8 + w) ml water, ( + w) ml total 8 + w + w w.7( + w) 8 + w +.7w.w w. Student must add ml of water. STUDY GUIDE: REVIEW, PAGES rational epression. rational function. rational equation 4. inverse variation 5. discontinuous function LESSON, PAGE Yes; the product y is constant. 7. No; the product y is not constant. 8. y 4 9. y. y y (5)(6) () y y 5 y 4. y ; ; y 4. y y ()() (5) 66 5 The price will be $,. LESSON, PAGE 687. y. y ; 4; y ; ; y 5. y ; 7 4 ; y 5 Copyright by Holt, Rinehart and Winston. 48 Holt Algebra

38 8.. D: > ; R: y > 9. LESSON, PAGE p r 7 5p r 7 p The ecluded value is. r 7 The ecluded value is 7.. t t t t t t(t ) t or t ( 5)( + ) 5 or + t 5 or The ecluded values are and. The ecluded values are 5 and ( 5)( + 5) 5 or or 5 The ecluded values are 5 and ( 4)( 7) 4 or 7 4 or 7 The ecluded values are 4 and r r 7 r 7 r r r r r r The ecluded value is. 8. k 6 k 9 k k k (k ) k 6 k 9 k k (k ) k or k k or k The ecluded values are and ( )( + 6) + 4 ( )( + 6) or + 6 or 6 The ecluded values are and 6. 9 ( ) ( )( + ) ( ). 6 ( )( + ) + 9 ( )( + ) or + or The ecluded values are and ( + 5) ( )( + 5) ( )( + 5) or + 5 or 5 The ecluded values are and ( + )( + 6) ( 5)( + 6) ( 5)( + 6) 5 or or 6 The ecluded values are 5 and 6. Copyright by Holt, Rinehart and Winston. 49 Holt Algebra

39 . A square A () circle π 4 4 π π LESSON 4, PAGE b 5. b 6 ( b b ) b b 6 b b b (b )(b + ) (b ) b(b + ) b + b 6. 5a b ab a b a b 5 a b 4 a b 5 b b + 8. b + b 4 b + b b 6 b + (b 4)(b + 6) b(b + 6) (b 4)(b + 4) b ( 9) ( + )( ) ( + ) 4( ) n 4 m n mn 6 n 4 m n mn 6 n mn 4 m n 48m n 4 4 m n n m b(b + 4) b + b + 8b ( )( + ) 4 ( + )( ) ( )( + ) 4( + )( ) LESSON 5, PAGE a b 5 a a b a b 5 a b b LCM 5 a a b b a b 4. 6 ( ) 5 5 5( ) LCM 5 ( ) ( ) 4. b b + 8 b b + 8 b 8p 44. p 4p + 8p p 4p n 5 n + 5 n n n 5 (n + 5) n n p 4p h + h h 5 h 5 h h + h h 5 h 5 h ( ) h + h h 5 h h 5 h + h ( h) h 5 h + 5h h 5 LESSON 6, PAGE (5 + 5) n n 5 n 5 (n + )(n 5) n 5 n b b 5 b 7 b b + 4 (5 b) 7 b 5b 7 b 47. 5m + m + m 5m( m m) + m + m 6m + m + m m 6m + m + m 7m + m 49. r + r r ( ) + r r + r + r 4 r 5. 8 ( + )( ) + 5. ( ) ( + ) ( + ) + 6 ( + 6) + Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

40 54. ( + ) ( 5) ( 5) 6 (6 ) ( b 4b + ) (b ) b b + b 4b + b + 6b + 8 b b + b 4b + ( b 6 b ) 6 b 4b ( 6 b b ) 8b + (8b 6) 8 b + 6b b LESSON 7, PAGE r 4r r 4 r, so there are no etraneous solutions b b y y 6( + b) 5b 8 + 6b 5b b 8 8 b b or, so there are no etraneous solutions. 6. ( ) ( ) ( )( + ) or + or 7y 6 y 6 y + 7y y(6y + 7) y or 6y + 7 y 7 6 y, so the only solution is 7, and is 6 an etraneous solution. ; so the only solution is, and is an etraneous solution ( )( + ) or + or, so there are no etraneous solutions ( + 4 ) + 6 6, so there are no etraneous solutions ) ( 6 ( + 4) ( ) 4 + ±, so there are no etraneous solutions. 64. b + 4 b b b 4 b 4 b b b, so there are no etraneous solutions ( 6) 8( 4) ( 4)( + ) 4 or + 4 or ±4, so the only solution is, and 4 is an etraneous solution. 5 + (5 ) ( + ) ( 4)( 5) 4 or 5 4 or 5, so there are no etraneous solutions. Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

41 ( + ) 4( ) ( )( + ) or + or, so there are no etraneous solutions m m 5 7 m 5 9m m 5 + m 5 7 9m + m 5 7 9m + 7(m 5) 9m + 7m 5 m 8 m 9 m 5, so there are no etraneous solutions ( 4)( ) ( 4) ( ) or or ±, so the only solution is, and is an etraneous solution. LESSON 8, PAGE Let h represent number of hours. pipe A + pipe B whole tank h + 8 h 6 ( h + 8 h ) 6() h + h 6 Tank fills in 7 5 5h 6 h h or 7 h min. 7. original:.4(4) 6 ml water, 4 ml total new: (6 + w) ml water, (4 + w) ml total 6 + w 4 + w w.5(4 + w) 6 + w +.5w.5w 4 w 8 Chemist should add 8 ml of water. CHAPTER TEST, PAGE 69. y 8. y + + ; and y 5. y + + ; and y ( + 4)( 4) ( + 4)( ) 4. y y (.6)(5) (.5) y 9.5 y y The club can buy posters. + 5 ; and y 5 4. y + 4 ( )( + 4) or + 4 or 4 The ecluded values are and ( + 5)( ) ( + 5)( 5) 5 5 ( 5)( + 5) 5 or or 5 The ecluded values are ±5. 6. b 4 b b b b b 4 b b b The ecluded value is. 8. b b 5 5 b (b + )(b 5) (b 5) (b + ) b 5 b b 5 The ecluded value is 5. Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

42 . 4 ( ) ( + )( ) 4 +. a b 5 a 5 b 8 a 4 a 4 b 4 a 4 b 4 b. 4 y 4 y y 5 y 4 y 4 5 y y y 6 5 y 6 6 y 5 y ( + )( 4) ( + 4)( ) ( + 4)( 4) ( + )( + ) ( + )( ) ( + )( + ) b 6b b 4 b + b 8b + b 6b 8b + b + b b 4 b(b ) 4(b + ) b (b + ) 6 b ( + 5)( ) ( + )( + ) ( + )( ) ( + 5)( 5) ( + )( + ) ( + )( 5) m m + m m + m 4 m m (b ) 6. b b 5b b + + 5b b + 7 5b 8. 5 (5 ) ) + (. y + 4 y + y y y + 4 y + y y( ) y + 4 y + y y y + 4 y y 4 y ( 4)( + ) ( w + 5w ) (w + 4) w + 4 w + 5w w w + 4 w + 5w ( w + 8w) w (w ) w 6. ( 4 + 9) ( + ) ( + ) (6 ) a. ( ) ( ) ( ) + ( ) ( ) + +. ( 8 t t ) t 8 t t t 8 t t t t 4t 4. k k 5 k + 5 (k 7)(k + 5) k + 5 k 7 Copyright by Holt, Rinehart and Winston. 4 Holt Algebra

43 b The width is 4 cm n ( ) 9( ) or, so there are no etraneous solutions.. n + n 4 n n + 4 (n + 4) n(n ) n + n n n 4n (n 6)(n + ) n 6 or n + n 6 or n n or 4, so there are no etraneous solutions. n 4 ( n 4) (n 4)(n + ) n 8 n n 8 n + n n(n + ) n or n + n n ±, so the only solution is, and is an etraneous solution.. Let h be the number of hours needed. h + h h 6 + h 6 5h 6 h 5 Therefore, they will need 5 h. Copyright by Holt, Rinehart and Winston. 44 Holt Algebra

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