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Solution Manual for Elementary and Intermediate Algebra Concepts and Applications, 7th Edition

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    S OLUTIONS M ANUAL E LEMENTARY AND I NTERMEDIATE A LGEBRA C ONCEPTS AND A PPLICATIONS S EVENTH E DITION Marvin L. Bittinger Indiana University Purdue University Indianapolis David J. Ellenbogen Community College of Vermont Barbara L. Johnson Ivy Tech Community College of Indiana

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    CONTENTS CHAPTER 1 INTRODUCTION TO ALGEBRAIC EXPRESSIONS..................... 1 CHAPTER 2 EQUATIONS, INEQUALITIES, AND PROBLEM SOLVING ..... 33 CHAPTER 3 INTRODUCTION TO GRAPHING ................................................ 90 CHAPTER 4 POLYNOMIALS ........................................................................... 151 CHAPTER 5 POLYNOMIALS AND FACTORING .......................................... 198 CHAPTER 6 RATIONAL EXPRESSIONS AND EQUATIONS ....................... 247 CHAPTER 7 FUNCTIONS AND GRAPHS........................................................ 311 CHAPTER 8 SYSTEMS OF LINEAR EQUATIONS AND PROBLEM SOLVING ................................................................... 347 CHAPTER 9 INEQUALITIES AND PROBLEM SOLVING ............................. 422 CHAPTER 10 EXPONENTS AND RADICALS................................................... 467 CHAPTER 11 QUADRATIC FUNCTIONS AND EQUATIONS ........................ 518 CHAPTER 12 EXPONENTIAL FUNCTIONS AND LOGARITHMIC FUNCTIONS ..................................................... 592 CHAPTER 13 CONIC SECTIONS ........................................................................ 645 CHAPTER 14 SEQUENCES, SERIES, AND THE BINOMIAL THEOREM ................................................................ 684

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    Chapter 1 Introduction to Algebraic Expressions Exercise Set 1.1 1. In the expression 4 + x , the number 4 is a constant . 2. In the expression 4 + x , the symbol + indicates the operation of addition. 3. To evaluate an algebraic expression, we substitute a number for each variable and carry out the operations. 4. An equation contains an equal sign. 5. 10 1 n  does not contain an equals sign, so it is an expression. 6. 3 21 x  contains an equals sign, so it is an equation. 7. 2 5 9 x   contains an equals sign, so it is an equation. 8. 5( 2) x  does not contain an equals sign, so it is an expression. 9. 45 1 a   contains an equals sign, so it is an equation. 10. 4 5 a b  does not contain an equals sign, so it is an expression. 11. 2 3 8 x y   contains an equals sign, so it is an equation. 12. ( 7) 5 r t   does not contain an equals sign, so it is an expression. 13. Substitute 9 for a and multiply. 5 5 9 45 a    14. 11 7 77   15. Substitute 4 for r and subtract. 12 4 8   16. 8 2 8 10 t     17. 45 5 9 a b   18. 14 13 27 9 3 3    19. 2 14 16 4 4 4 4 x y      20. 54 6 9  21. 55 20 35 5 7 7 7 p q      22. 9 9 6 54 3 18 18 m q     23. 5 5 9 45 3 15 15 z y     24. 20 8 12 6 2 2    25. 2 (6 ft)(4 ft) (6)(4)(ft)(ft) 24 ft , or 24 square feet bh    26. 27,000 24 hr 1125  27. 2 2 1 2 1 (5 cm)(6 cm) 2 1 = (5)(6)(cm)(cm) 2 5 = 6 cm 2 15 cm , or 15 square centimeters A bh     28. (a) 3(30 sec) 90 sec;  (b) 3(90 sec) 270 sec;  (c) 3(2 min) 6 min  29. 2 (67 ft)(12 ft) =(67)(12)(ft)(ft) =804 ft , or 804 square feet A bh   30. 8 0.571 14 h a   31. Let r represent Ron’s age. Then we have 5, r  or 5 . r  32. 4 , or 4 a a  33. 6 , or 6 b b  34. Let p represent Patti’s weight. Then we have 7, p  or 7 . p  35. 9 c  36. 4 d  37. 6 , or 6 q q   38. 11 , or 11 z z   39. p t 

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    2 Chapter 1: Introduction to Algebraic Expressions 40. n m  41. y x  42. Let a represent Kurt’s age. Then we have 2. a  43. , or x x w w  44. Let s and t represent the numbers. Then we have , s t  or . s t 45. Let l and h represent the box’s length and height, respectively. Then we have , or . l h h l   46. , or d f f d   47. 9 2 , or 2 9 m m   48. Let a represent Abby’s speed and w represent the wind speed. Then we have 2 . a w  49. Let y represent “some number.” Then we have 1 13, or 13. 4 4 y y   50. Let n represent the number; 10 4 n  51. Let a and b represent the two numbers. Then we have 5( ). a b  52. Let x and y represent the numbers. Then we have 1 ( ), or . 3 3 x y x y   53. Let w represent the number of women attending. Then we have 64% of w , or 0.64 w . 54. Let y represent “a number.” Then we have 38% of y , or 0.38 y . 55. Let x represent the number. Translating: What number added to 73 is 201? 73 201 x                73 201 x   56. Let x represent the number. 7 1596 x  57. Let x represent the number. Rewording: 42 times what number is 2352? Translating: 42 2352 x            42 2352 x  58. Let x represent the number. 345 987 x   59. Let s represent the number of unoccupied squares. The number of Rewording: added to 19 is 64. unoccupied squares Translating: 19 64 s             19 64 s   60. Let h represent the number of hours the carpenter worked. $35 $3640 h  61. Let x represent the total amount of waste generated, in millions of tons. the total 87 million Rewording: 34.5% of amount is tons. of waste Translating: 34.5% 87 x                   34.5% 87, or 0.345 87 x x    62. Let t represent the length of the average commute in Fort Bliss, in minutes. 51.2 59.8 t   63. We look for a pattern in the data. We try subtracting. 8 3 5 11 6 5 9 4 5 12 7 5 10 5 5 13 8 5             The amount is the same, 5, for each pair of numbers. Let a represent the age of the child and f represent the number of grams of dietary fiber. We reword and translate as follows:  dietary child's Rewording: is added to 5 fiber age Translating: 5 f a            5 f a   64. Let c represent the cost of tuition and h represent the hours of classes. 100 c h  65. We look for a pattern in the data. We try subtracting. 6.59 4.17 2.42 7.18 4.76 2.42 8.76 6.34 2.42       The amount is the same, 2.42, for each pair of numbers. Let n represent the nonmachinable cost and n represent the machinable cost. We reword and translate as follows:  nonmachinable machinable is added to 2.42 cost cost 2.42 n m             2.42 n m  

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    Exercise Set 1.1 3 66. Let r represent the amount received and s represent the amount spent. 3 r s   67. We look for a pattern in the data. We try dividing. 10,000 30,000 10,000 10,000 1 3 20,000 40,000 10,000 10,000 2 4     The amount is the same, 10,000, for each pair of numbers. Let v represent the number of vehicle miles traveled and d represent the number drivers. We reword and translate as follows: number of number Rewording: is 10,000 times miles traveled of drivers Translating: 10,000 v d                   10,000 v d  68. Let w represent the depth of water and s represent the depth of snow. 10 w s   69. The sum of two numbers m and n is , m n  and twice the sum is 2( ). m n  Choice (f) is the correct answer. 70. Five less than a number x is 5. x  If this expression is equal to 12, we have the equation 5 12. x   Choice (h) is the correct answer. 71. Twelve more than a number t is 12. t  If this expression is equal to 5, we have the equation 12 5. t   Choice (d) is the correct answer. 72. The product of two numbers a and b is . a b  Half of this product is 1 2 . a b   Choice (c) is the correct answer. 73. The sum of a number t and 5 is 5, t  and 3 times the sum is 3( 5). t  Choice (g) is the correct answer. 74. The sum of two numbers x and y is , x y  and twice this sum is 2( ). x y  If this expression is equal to 48, we have the equation 2( ) 48. x y   Choice (b) is the correct answer. 75. The product of two numbers a and b is ab , and 1 less than this product is 1. ab  If this expression is equal to 48, we have the equation 1 48. ab   Choice (e) is the correct answer. 76. The quotient of two numbers x and y is , x y and 6 more than this quotient is 6. x y  Choice (a) is the correct answer. 77. Writing Exercise . A variable is a letter that is used to stand for any number chosen from a set of numbers. An algebraic expression is an expression that consists of variables, constants, operation signs, and/or grouping symbols. A variable expression is an algebraic expression that contains a variable. An equation is a number sentence with the verb =. The symbol = is used to indicate that the algebraic expressions on either side of the symbol represent the same number. 78. Writing Exercise . To evaluate an algebraic expression means to find the value of the expression when its variables are given values. 79. Writing Exercise . No; for a square with side s , the area is given by . A s s   The area of a square with side 2 s is given by (2 )(2 ) 4 s s s s     4 2 . A A  80. Writing Exercise . Answers may vary. Juliet was born in 2006. Find her age in 2014. 81. Area of sign: 2 1 2 (3 ft)(2.5 ft) 3.75 ft A   Cost of sign: $120(3.75) $450  82. The shaded area is the area of a rectangle with dimensions 20 cm by 10 cm less the area of a triangle with base 20 cm – 4 cm – 5 cm, or 11 cm, and height 7.5 cm. We perform the computation 2 2 2 1 (20 cm)(10 cm) (11 cm)(7.5) 2 200 cm 41.25 cm 158.75 cm , or 158.75 square centimeters     83. When x is twice y , then y is one-half x , so 12 6. 2 y   12 6 6 2 3 3 3 x y      84. 6, 2 2 6 12 6 12 18 9 2 2 2 x y x x y           85. When a is twice b , then b is one-half a , so 16 2 8. b   16 8 24 6 4 4 4 a b      86. When a is three times b , then b is one-third a , so 18 3 6. b   18 6 12 4 3 3 3 a b      87. The next whole number is one more than 3 : w  3 1 4 w w     88. The preceding odd number is 2 less than 2 : d  2 2 d d    89. , or 2 2 l w l w l w     90. , or 4 s s s s s    91. If t is Molly’s race time, then Dion’s race time is 3 t  and Ellie’s race time is 3 5 8. t t    

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    4 Chapter 1: Introduction to Algebraic Expressions 92. 2 7, or 9 a a    93. Writing Exercise . Yes; the area of a triangle with base b and height h is given by 1 2 . A bh  The area of a triangle with base b and height 2 h is given by   1 1 2 2 (2 ) 2 2 . b h bh A   Exercise Set 1.2 1. Equivalent expressions represent the same number. 2. Changing the order of multiplication does not affect the answer. This is an example of a commutative law. 3. The result of addition is called a sum . 4. The numbers in a product are called factors . 5. Commutative 6. Associative 7. Distributive 8. Commutative 9. Commutative 10. Distributive 11. 11 Changing the order t  12. 2 a  13. 8 4 x  14. c ab  15. 3 9 y x  16. 7 3 b a  17. 5(1 ) a  18. 9(5 ) x  19. 7 Changing the order x  20. yx 21. ts 22. 13 m  23. 5 ba  24. 3 x y   25. ( 1)5 a  26. ( 5)9 x  27. (8 ) x y   28. 5 ( ) m r   29. ( ) 7 u v   30. ( 2) x y   31. ( ) ab c d   32. ( ) m np r   33. 10( ) x y 34. 4( ) uv 35. (2 ) a b 36. (9 7) r  37. (3 2)( ) a b   38. (5 )(2 ) x y  39. ( 6) ( ) 6 ( 6) s t s t t s         40. 7 ( ) ( ) 7 ( 7) 7 ( ) (7 ) ( 7) v w v w v w v w v w v w                 Answers may vary. 41. (17 ) 17( ) Using the associative law a b ab  (17 ) (17 ) Using the commutative law ( 17) Using the commutative law again a b b a b a   Answers may vary. 42. (3 ) (3 ) 3( ) (3 ) ( 3) (3 ) x y y x yx x y x y x y      43. (1 ) 2 ( 1) 2 Commutative law (1 2) Associative law 3 Simplifying x x x x           44. (2 )4 4(2 ) Commutative law (4 2) Associative law 8 Simplifying a a a a     45. ( 3)7 (3 7) Associative law 21 Simplifying 21 Commutative law m m m m       46. 4 (9 ) (4 9) Associative law 13 Simplifying 13 Commutative law x x x x          47. 1 x xyz   The term are separated by plus signs. They are x , xyz , and 1. 48. 9, 17 , a abc 49. 2 5 3 a a b b   The terms are separated by plus signs. They are 2 a , , 3 a b and 5 b .

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    Exercise Set 1.2 5 50. 4 3 , 20, a xy b 51. 4 , 4 x y 52. 14, 2 y 53. 5 5 n n   The factors are 5 and n . 54. uv u v   The factors are u and v . 55. 3( ) 3 ( ) x y x y     The factors are 3 and ( ). x y  56. ( )12 ( ) 12 a b a b     The factors are ( ) a b  and 12. 57. The factors are 7, a and b . 58. The factors are m , n and 2. 59. ( )( ) ( ) ( ) a b x y a b x y       The factors are ( ) and ( ). a b x y   60. (3 )( ) (3 ) ( ) a b c a b c       The factors are (3 ) and ( ). a b c   61. 2( 15) 2 2 15 2 30 x x x        62. 3 15 x  63. 4(1 ) 4 1 4 4 4 a a a        64. 7 7 y  65. 10(9 6) 10 9 10 6 90 60 x x x        66. 54 63 m  67. 5( 2 3 ) 5 5 2 5 3 5 10 15 r t r t r t            68. 20 32 12 x p   69. ( )2 (2) (2) 2 2 a b a b a b      70. 7 14 x  71. ( 2)5 (5) (5) 2(5) 5 5 10 x y x y x y         72. 12 6 6 a b   73. 2 2 Using the distributive law 2( ) The common factor is 2. a b a b      Check: 2( ) 2 2 2 2 a b a b a b        74. 5 5 5( ) y z y z    Check: 5( ) 5 5 y z y z    75. 7 7 7 1 7 The common factor is 7. 7(1 ) Using the distributive law y y y        Check: 7(1 ) 7 1 7 7 7 y y y        76. 13 13 13(1 ) x x    Check: 13(1 ) 13 13 x x    77. 32 2 2 16 2 1 2(16 1) x x x        Check: 2(16 1) 2 16 2 1 32 2 x x x        78. 20 5 5(4 1) a a    Check: 5(4 1) 20 5 a a    79. 5 10 15 5 5 2 5 3 5( 2 3 ) x y x y x y            Check: 5( 2 3 ) 5 5 2 5 3 5 10 15 x y x y x y            80. 3 27 6 3(1 9 2 ) b c b c      Check: 3(1 9 2 ) 3 27 6 b c b c      81. 7 35 7 7 5 7( 5 ) a b a b a b        Check: 7( 5 ) 7 7 5 7 35 a b a b a b        82. 3 24 3( 8 ) x y x y    Check: 3( 8 ) 3 24 x y x y    83. 44 11 22 11 4 11 11 2 11(4 2 ) x y z x y z x y z            Check: 11(4 2 ) 11 4 11 11 2 44 11 22 x y z x y z x y z            84. 14 56 7 7(2 8 1) a b a b      Check: 7(2 8 1) 14 56 7 a b a b      85. 3(2 ) 3( 2) Commutative law of addition 3 3 2 Distributive law 3 6 Multiplying x x x x          86. ( 4)5 5( 2) Commutative law of multiplication 5 5 4 Distributive law 5 20 Multiplying y y y y          87. 7(2 3 ) 7(2 ) 7(3 ) Distributive law (7 2) (7 3) Associative law of multiplication 14 21 Multiplying x y x y x y x y          88. (4 2)8 8(4 2) Commutative law of multiplication 8(4 ) 8(2) Distributive law (8 4) 8(2) Associative law of multiplication 32 16 Multiplying a a a a a           89. Writing Exercise . No; in general, when subtracting, the result depends on the order in which the operation is performed. 90. Writing Exercise . No; in general, when dividing, the result depends on the grouping. 91. Writing Exercise . Terms are added; factors are multiplied. Examples will vary.

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    6 Chapter 1: Introduction to Algebraic Expressions 92. Writing Exercise . 2(3 4 ) 2(3 ) 2(4 ) Distributive law (2 3) (2 4) Associative law 6 8 x y x y x y x y          93.               2( 1) 3 2 2 1) 3 Distributive law = 2 2 3 Multiplying 2 2 3 Associative law of addition 2 3 2 Commutative law of addition 2 3 2 Associative law of addition (2 3) 2 Distributive l x x x x x x x x x x x x x                        aw 5 2 Adding 5 2 x x     94. 12 4( 5) (4 3) 4( 5) Writing 12 as a product 4(3 ) 4( 5) Associative law of multiplication 4(3 ) 4( ) 4(5) Distributive law 4(3 5) Distributive law a b a b a b a b a b                95. The expressions are equivalent by the distributive law. 8 4( ) 8 4 4 4(2 ) a b a b a b         96. The expressions are not equivalent. Let 2 and 3. a b   Then we have: 5(2 3) 5 6 30, but 5 2 5 3 10 5 3 50 3 150.              97. The expressions are not equivalent. Let 1. m  Then we have: 7 7 7 3 1 1 , but 3 3 3 1 3 7 3 7 . 7           98. The expressions are equivalent by the commutative law of multiplication and the distributive law. ( )5 5( ) 5 ( ) 5 ( ) rt st rt st t r s t r s         99. The expressions are not equivalent. Let 1 and 0. x y   Then we have: 30 0 1 15 0 15 15, but 5[2(1 3 0)] 5[2(1)] 5 2 10.             100. The expressions are equivalent by the commutative and associative laws of multiplication and the distributive law. [ (2 3 )]5 5[ (2 3 )] 5 (2 3 ) 10 15 c b c b c b c bc        101. Writing Exercise . 3(2 ) 3(2 0) 3 2 6 x       ; 6 6 0 6 x     The result indicates that 3(2 ) x  and 6 x  are equivalent when 0. x  (By the distributive law, we know they are not equivalent for all values of . x ) 102. Writing Exercise . 15 40 5(3 8) 15 4 40 60 40 100 5(3 4 8) 5(12 8) 5 20 100 x x                Although the expressions 15 40 x  and 5(3 8) x  are equivalent when 4 x  , this result does not guarantee that the factorization is correct. (See Exercise 101.) 103 . Answers may vary. a . Let x represent the number of overtime hours worked in one week. Aidan: 10(1.5 40) x  Beth: 10 40 10(1.5) x   Cody: 15 400 x  b . We simplify each expression. Aiden: 10(1.5 40) 10(1.5 ) 10(40) 15 400 x x x      Beth: 10 40 10(1.5) 400 15 x x     Cody: 15 400 x  All expressions are equivalent to 15 400. x  Exercise Set 1.3 1. The top number in a fraction is called the numerator . 2. A prime number has exactly two different factors. 3. To divide two fractions, multiply by the reciprocal of the divisor. 4. We need a common denominator in order to add fractions. 5. Since 35 5 7   , choice (b) is correct. 6. Since 60 3 20   , choice (c) is correct. 7. Since 65 is an odd number and has more than two different factors, choice (d) is correct. 8. The only even prime number is 2, so choice (a) is correct. 9. 9 is composite because it has more than two different factors. They are 1, 3, and 9. 10. 15 is composite because it has more than two different factors. They are 1, 3, 5, and 15. 11. 41 is prime because it has only two different factors, 41 and 1. 12. 49 is composite because it has more than two different factors. They are 1, 7, and 49. 13. 77 is composite because it has more than two different factors. They are 1, 7, 11, and 77. 14. 37 is prime because it has only two different factors, 37 and 1. 15. 2 is prime because it has only two different factors, 2 and 1.

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    Exercise Set 1.3 7 16. 1 is not prime because it does not have two different factors. It is not composite because it does not have more than two different factors. Thus 1 is neither prime nor composite. 17. The terms “prime” and “composite” apply only to natural numbers. Since 0 is not a natural number, it is neither prime nor composite. 18. 16 is composite because it has more than two factors. They are 1, 2, 4, 8, and 16. 19. Factorizations: 1 50, 2 25, 5 10    List all of the factors of 50: 1, 2, 5, 10, 25, 50 20. Factorizations: 1 70  , 2 35  , 5 14  , 7 10  Factors: 1, 2, 5, 7, 10, 14, 35, 70 21. Factorizations: 1 42  , 2 21  , 3 14  , 6 7  List all of the factors of 42: 1, 2, 3, 6, 7, 14, 21, 42 22. Factorizations: 1 60  , 2 30  , 3 20  , 4 15  , 5 12  , 6 10  Factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 23. 39 3 13   24. 2 17  25. We begin factoring 30 in any way that we can and continue factoring until each factor is prime. 30 2 15 2 3 5      26. 2 7 7   27. We begin by factoring 27 in any way that we can and continue factoring until each factor is prime. 27 3 9 3 3 3      28. 2 3 3 3    29. We begin by factoring 150 in any way that we can and continue factoring until each factor is prime. 150 2 75 2 3 25 2 3 5 5          30. 2 2 2 7    31. 31 has exactly two different factors, 31 and 1. Thus, 31 is prime. 32. 180 2 2 3 3 5      33. 210 2 105 2 3 35 2 3 5 7          34. 79 is prime. 35. 115 5 23   36. 11 13  37. 21 7 3 Factoring numerator and denominator 35 7 5 7 3 Rewriting as a product of two fractions 7 5 3 7 1 1 5 7 3 Using the identity property of 1 5          38. 2 10 20 10 26 2 13 13     39. 16 2 8 2 8 2 2 1 56 7 8 7 8 7 7         40. 8 9 72 8 27 3 9 3     41. 12 1 12 Factoring and using the identity 48 4 12 property of 1 to to write 12 as 1 12 1 12 4 12 1 1 1 4 4          42. 6 3 18 3 84 6 14 14     43. 52 13 4 4 1 4 13 13 1 1       44. 12 11 132 11   1 11  12  45. 19 1 19 Factoring and using the identity 76 4 19 property of 1 to write 19 as 1 19 19 1 4       19 19 Removing a factor equal to 1: 1 19 1 4   46. 1 17 17 51   3 17  1 3  47. 150 6 25 Factoring and using the identity 25 1 25 property of 1 to write 25 as 1 25 25 6 1       25 25 Removing a factor equal to 1: 1 25 6 1 6 Simplifying    48. 5 36 180 36   1 36  5 

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    8 Chapter 1: Introduction to Algebraic Expressions 49. 42 2 21 Factoring the numerator 50 2 25 and the denominator 2 21 2 Removing a factor equal to 1: 21 25 1 2 25 2         50. 5 15 75 15 80 5 16 16     51. 120 2 60 Factoring 82 2 41 2 60 2 Removing a factor equal to 1: 1 2 41 2 60 41         52. 5 15 75 45   3 15  5 3  53. 210 2 7 15 Factoring 98 2 7 7 2 7 15 2 7 Removing a factor equal to1: 1 7 7 5 1 2 7 2 7               54. 2 2 5 7 140 2 350 2 5 5 7 5         55. 1 3 1 3 Multiplying numerators and 2 5 2 5 denominators 3 10      56. 11 2 4 11 8 11 8 44 10 5 10 5 2 5 5 25           57. 3 3 2 2 9 4 9 4 6 2 3 2 3 2 3           58. By the commutative law of multiplication, 11 12  and 12 11  are equivalent. Thus, the product of the numerators will be equivalent to the product of the denominators, so the result is a number divided by itself, or 1. 11 12 11 12 1 12 11 12 11      59. 1 3 1 3 Adding numerators; keeping 8 8 8 the common denominator 4 8 1 4 1 Simplifying 2 4 2         60. 1 7 8 2 4 4 10 10 10 2 5 5       61. Using 18 as the common 4 13 4 2 13 denominator 9 18 9 2 18 8 13 18 18 21 18 7 3 7 Simplifying 6 3 6            62. 4 8 12 8 20 4 5 15 15 15 15 3      63. Multiplying numerators 3 3 and denominators 7 7 b b a a   64. 5 5 y xy x z z   65. Adding numerators; keeping 4 6 10 the common denominator n n n   66. 9 5 4 x x x   67. Using 30 as the 3 8 3 3 8 2 common denom 9 16 30 30 2 inator 10 15 10 3 15 2 5 30 5 5 5 Simplifying 6 5 6             68. 7 5 21 10 31 8 12 24 24 24     69. 11 4 7 1 7 7 7    70. 12 2 10 2 5 5 5    71. Using 18 as the 13 4 13 4 2 common denominator 1 18 9 18 3 8 18 18 8 9 2 5 1        72. 13 11 39 11 28 15 45 45 45 45     73. Using 90 as the 11 2 11 3 2 10 common denominato 33 20 90 90 13 9 r 30 9 30 9 10 0 3         74. 5 5 15 10 5 14 21 42 42 42    

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    Exercise Set 1.3 9 75. Multiplying by the reciprocal of 7 3 7 5 the divisor 6 3 8 5 5 6 3 1     76. 7 10 7 3 21 5 3 5 10 50     77. 4 3 9 4 12 9 12 27 9 1 4 1 4         78. 9 1 9 9 1 1 9 4 4 9 4 9 4        79. Note that we have a number divided by itself. Thus, the result is 1. We can also do this exercise as follows: 7 7 7 13 7 13 1 13 13 13 7 7 13        80. 1 1 1 5 5 1 10 5 10 1 10 2      81. 2 7 2 5 2 3 2 3 6 5 7 3 7 5 7 5 35 3         82. 3 8 3 5 15 1 8 1 8 5    83. 9 1 9 2 9 2 9 18 1 2 1 1 1 1 2         84. 3 7 3 1 1 6 7 6 14    85. Writing Exercise . If the fractions have the same denominator and the numerators and/or denominators are very large numbers, it would probably be easier to compute the sum of the fractions than their product. 86. Writing Exercise . If the fractions have different denominators, it would probably be easier to compute the product of the fractions than their sum. 87. Writing Exercise . Bryce is canceling incorrectly. The number 2 is not a common factor of both terms in the numerator, so it cannot be canceled. For example, let 1 x  . Then (2 1)/8 3/8   but (1 1)/4 2/4 1/2    . The expressions are not equivalent. 89. Product 56 63 36 72 140 96 168 Factor 7 7 2 36 14 8 8 Factor 8 9 18 2 10 12 21 Sum 15 16 20 38 24 20 29 90. We need to find the smallest number that has both 6 and 8 as factors. Starting with 6 we list some numbers with a factor of 6, and starting with 8 we also list some numbers with a factor of 8. Then we find the first number that is on both lists. 6, 12, 18, 24, 30, 36, … 8, 16, 24, 32, 40, 48, … Since 24 is the smallest number that is on both lists, the carton should be 24 in. long. 91. 4 4 3 3 2 2 16 9 4 2 15 8 12 3 5 2 4 3 4 5                 92. 9 2 4 1 9 8 1 2 36 2 4 9 x y xy xy x y              93. 5 9 45 5 9 9 p q r s pqrs q prst p r s t t            94. 13 19 247 323   17 19  13 17  95. 15 15 4 9 6 25 15 xy x y      2 2 3 3 2 3 25 15 x y x           6 25 y   96. 10 10 12 25 2 30 20 x y z x y      2 2 3 5 5 2 3 10 x y z           5 2 2 2 5 z x y       97. 27 25 15 27 18 27 18 15 25 15 18 25 27 25 3 9 5 5 15 18 3 5 2 9 5 2 ab np mn ab bc ab bc mn np mn bc np ab np a b n p mn bc m n b c ap cm                        98. 45 45 24 32 30 24 30 32 3 15 xyz xyz ab ac xz ab xz ac     2 2 8 3 8 2 15 x y z a c a b              2 cy b x z    99. 3 23 23 5 23 9 4 4 4 1 9 9 4 2 4 2 2 2 23 2 23 2 4 9 4 9 23 2 23 2 2 9 18 rs rs rs rs st st st st rs rs st st r s r s t t                   100. 5 26 3 7 7 26 5 4 14 7 14 2 5 5 2 13 5 65 7 2 7 49 mn mn mn np np np m n m n p p             

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    10 Chapter 1: Introduction to Algebraic Expressions 101.        2 4 7 m m 5 9 4 7 (m)(m) 5 9 28 28 m , or square meters 45 45 A lw     102.       2 2 2 1 1 10 5 m m 2 2 7 4 1 10 5 (m)(m) 2 7 4 1 10 5 m 2 7 4 1 2 5 5 m 2 7 4 25 25 m , or square meters 28 28 A bh                103.   5 32 128 4 4 3 m 4 m m, 9 9 9 2 or 14 m 9 P s      104.           4 7 2 2 2 m 2 m 5 9 8 14 m m 5 9 8 14 m 5 9 8 9 14 5 m 5 9 9 5 72 70 m 45 45 142 7 m, or 3 m 45 45 P l w                105. There are 12 edges, each with length 3 10 2 cm. We multiply to find the total length of the edges. 3 23 12 2 cm 12 cm 10 10 12 23 cm 10 2 6 23 cm 2 5 138 3 cm, or 27 cm 5 5           Exercise Set 1.4 1. Since 3 0.15, 20  we can write 3 20 as a terminating decimal. 2. If a number is an integer , it is either a whole number or the opposite of a whole number. 3. 0 is the only whole number that is not a natural number. 4. A number like 5, which cannot be written precisely in fraction notation or decimal notation, is an example of an irrational number . 5. The opposite of 1 is –1. 6. When two numbers are opposites, they have the same absolute value . 7. – n is the opposite of n . 8. x 9. 10 x   10. 6 y  11. The real number –9500 corresponds to borrowing $9500. The real number 5000 corresponds to the award of $5000. 12. –150; 65 13. The real number 100 corresponds to 100°F. The real number –80 corresponds to 80°F below zero. 14. –1312; 29,035 15. The real number –777.68 corresponds to a 777.68-point fall. The real number 936.42 corresponds to a 936.42- point gain. 16. The real number 10,000 corresponds to a $10,000 grant, and the real number –4500 corresponds to $4500 spent. 17. The real number 8 corresponds to an 8-yd gain, and the real number –5 corresponds to a 5-yd loss. 18. 4, –5 19. 20. 21. The graph of –4.3 is 3 10 of a unit to the left of –4. 22. 23. Since 10 1 3 , 3 3  its graph is 1 3 of a unit to the right of 3. 24. 17 2 3 3.4 5 5     

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    Exercise Set 1.4 11 25. From lowest to highest: points-per-game 26. From lowest to highest: field-goal-percentage 27. 7 8 means 7 8,  so we divide. 0.875 8 7.000 6 4 60 56 40 40 0 We have 7 0.875. 8  28. 0.125 8 1.000 8 20 16 40 40 0 1 1 0.125, so 0.125. 8 8     29. We first find decimal notation for 3 . 4 Since 3 4 means 3 4,  we divide. 0.75 4 3.00 2 8 20 20 0 Thus, 3 3 4 4 0.75, so 0.75.     30. ... 1.833 6 11.000 6 50 48 20 18 20 11 1.83. 6  31. 7 6 means 7 6,  so we divide. ... 1.166 6 7.000 6 10 6 40 36 40 Thus 7 7 1.16, so 1.16. 6 6     32. ... 0.4166 12 5.0000 4 8 20 12 80 72 80 72 8 5 5 0.416, so 0.416. 12 12     33. 2 means 2 3, 3  so we divide. ... 0.666 3 2.000 18 20 18 20 18 2 We have 2 0.6. 3  34. 0.25 4 1.00 8 20 20 0 1 0.25. 4  35. We first find decimal notation for 1 . 2 Since 1 2 means 1 2,  we divide. 0.5 2 1.0 1 0 0 Thus, 1 1 0.5, so 0.5. 2 2    

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    12 Chapter 1: Introduction to Algebraic Expressions 36. ... 0.111 9 1.000 9 10 9 10 9 1 1 1 0. 1, so 0. 1. 9 9     37. Since the denominator is 100, we know that 13 0.13. 100  We could also divide 13 by 100 to find this result. 38. 0.45 20 9.00 80 100 100 0 9 9 0.45, so 0.45. 20 20     39. 40. 41. 42. 43. Since 5 is to the right of 0, we have 5 0.  44. 8 8   45. Since –9 is to the left of 9, we have 9 9.   46. 0 7   47. Since –8 is to the left of –5, we have 8 5.    48. 4 3    49. Since –5 is to the right of –11, we have 5 11.    50. 3 4    51. Since 12.5  is to the left of –10.2, we have 12.5 10.2.    52. 10.3 14.5    53. We convert to decimal notation. 5 11 0.416 and 0.44. 12 25   Thus, 5 11 . 12 25  54. 27 14 0.82353 and 0.77143. 17 35       Thus, 27 14 . 17 35    55. 2 x   has the same meaning as 2. x   56. 9 a  57. 10 y  has the same meaning as 10. y  58. 12 t   59. 5 83, 4.7, 0, , 2.16, 62 9   60. 62 61. –83, 0, 62 62. , 17  63. All are real numbers. 64. 0, 62 65. | 58| 58   since –58 is 58 units from 0. 66. 47 67. | 12.2| 12.2   since –12.2 is 12.2 units from 0. 68. 4.3 69. | 2 | 2  since 2 is 2 units from 0. 70. 456 71. 9 9 7 7   since 9 9 is 7 7  units from 0. 72. 3 73. |0| 0  since 0 is 0 units from itself. 74. 3 4 75. | | | 8 | 8 x    76. | | | 5 | 5 a    77. Writing Exercise . Yes; every integer can be written as n /1, a quotient of the form a / b where 0. b  78 . Writing Exercise . No; for instance, –2 is an integer and –2 is not a natural number. 79. Writing Exercise . No; |0| 0  which is neither positive nor negative.

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    Mid-Chapter Review 13 80. Writing Exercise . There are infinitely many rational numbers between 0 and 1. Consider only rational numbers of the form 1 , n where n is an integer greater than 1. There are infinitely many integers greater than 1, so there are infinitely many numbers 1 , n all between 0 and 1. (These numbers are a subset of the rational numbers between 0 and 1.) 81. Writing Exercise . No; every positive number is nonnegative, but zero is nonnegative and zero is not positive. 82. List the numbers as they occur on the number line, from left to right: –17, –12, 5, 13 83. List the numbers as they occur on the number line, from left to right: –23, –17, 0, 4 84. 3 5 3 2 1 1 , , , , , 3 2 4 6 8 6    can be written in decimal notation as 0.666, 0.5, 0.75, 0.833, 0.375,    0.166, respectively. Listing from least to greatest (in fractional form), we have 5 3 2 1 3 1 , , , , , . 6 4 3 6 8 2    85. Converting to decimal notation, we can write 4 4 4 4 4 4 4 , , , , , , 5 3 8 6 9 2 3  as 0.8, 1.33, 0.5, 0.66, 0.44, 2, 1.33,  respectively. List the numbers (in fractional form) as they occur on the number line, from left to right: 4 4 4 4 4 4 4 , , , , , , 3 9 8 6 5 3 2  86. | 5| 5 and | 2| 2, so | 5| | 2|.        87. |4| 4 and | 7| 7, so |4| | 7|.      88. | 8| 8 and |8| 8, so | 8| |8|.      89. |23| 23 and | 23| 23, so |23| | 23|.      90. | | 19 x  x represents a number whose distance from 0 is 19. Thus, 19 or 19. x x    91. x represents an integer whose distance from 0 is less than 3 units. Thus, 2, 1, 0, 1, 2. x    92. 2 | | 5 x   x represents an integer whose distance from 0 is greater than 2 and also less than 5. Thus, 4, 3, 3, 4. x    93. 1 0.33 3 1 1 1 0.1 1 3 3 3 3 9      94. 1 3 0.99 3(0.33) 3 3 3     95. 1 50 5.55 50(0.1 1) 50 9 9     (See Exercise 93.) 96. 1 70 7.77 70(0.1 1) 70 9 9     (See Exercise 94.) 97. a < 0 98. Nonpositive numbers include zero. Thus, x < 0. 99. | | 10 x  100. Distance from zero can go in the positive or negative direction. Thus, | | 20. t  101. Writing Exercise . The number entered by hand is an approximation of 2 while the value that is squared immediately after being calculated is actually regarded by the calculator as 2. 102. Writing Exercise . The statement 2 | | a a  for any real number a is true. If a is nonnegative, then 2 . a a  If a is negative, 2 a a   since 2 a must be nonnegative. Thus, for a nonnegative number, the result is the number and, for a negative number, the result is the opposite of the number. This describes the absolute value of a number. Mid-Chapter Review 1. 22 10 Substituting 3 3 12 Subtracting 3 4 Dividing x y      2. 14 7 7 2 7 1 7(2 1) x x x        3. 3 12 Substituting 15 x y     4. 2 2 10 4 5 5 a    5. d – 10 6. Let h represent the number of hours worked; 8 h 7. Let s represent the number of students originally enrolled; 5 27 s   8. No; 13 13 8 104 94 t     9. 10 x + 7 10. (3 a ) b 11. 4(2 x + 8) = 8 x + 32 12. 3(2 m + 5 n + 10) = 6 m + 15 n + 30 13. 18 x + 15 = 3(6 x ) + 3(5) = 3(6 x + 5) 14. 9 c + 12 d + 3 = 3(3 c + 4 d + 1)

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