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STUDENT’S SOLUTIONS MANUAL JUDITH A. PENNA Indiana University Purdue University Indianapolis COLLEGE ALGEBRA: GRAPHS AND MODELS FIFTH EDITION Marvin L. Bittinger Indiana University Purdue University Indianapolis Judith A. Beecher Indiana University Purdue University Indianapolis David J. Ellenbogen Community College of Vermont Judith A. Penna Indiana University Purdue University Indianapolis Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei TokyoChapter R Basic Concepts of Algebra 23. The endpoint −9 is included in the interval, so we use a bracket before the−9. The endpoint −4 is not included, Exercise Set R.1 so we use a parenthesis after the −4. Interval notation is √ 2 1 8 −9,−4). 3 1. Rational numbers: ,6, −2.45, 18.4, −11, 27, 5 , − , 3 6 7 √ 25. Both endpoints are included in the interval, so we use 0, 16 brackets. Interval notation is x,x +h. √ √ √ √ 6 5 3. Irrational numbers: 3, 26, 7.151551555...,− 35, 3 27. The endpoint p is not included in the interval, so we use a (Although there is a pattern in 7.151551555..., there is parenthesis before the p. The interval is of unlimited ex- no repeating block of digits.) tent in the positive direction, so we use the infinity symbol √ √ 3 ∞. Interval notation is (p,∞). 5. Whole numbers: 6, 27, 0, 16 29. Since 6 is an element of the set of natural numbers, the 7. Integers but not natural numbers: −11, 0 statement is true. 2 1 9. Rational numbers but not integers: , −2.45, 18.4, 5 , 31. Since 3.2 is not an element of the set of integers, the state- 3 6 8 ment is false. − 7 11 33. Since − is an element of the set of rational numbers, 5 11. This is a closed interval, so we use brackets. Interval no- the statement is true. tation is −5,5. √ 35. Since 11 is an element of the set of real numbers, the statement is false. 5 05 37. Since 24 is an element of the set of whole numbers, the 13. This is a half-open interval. We use a parenthesis on statement is false. the left and a bracket on the right. Interval notation is (−3,−1. 39. Since 1.089 is not an element of the set of irrational num- bers, the statement is true. 3 1 0 41. Since every whole number is an integer, the statement is 15. This interval is of unlimited extent in the negative direc- true. tion, and the endpoint−2 is included. Interval notation is 43. Since every rational number is a real number, the state- (−∞,−2. ment is true. 2 0 45. Since there are real numbers that are not integers, the statement is false. 17. This interval is of unlimited extent in the positive direc- 47. The sentence 3 + y = y + 3 illustrates the commutative tion, and the endpoint 3.8 is not included. Interval nota- property of addition. tion is (3.8,∞). 49. The sentence −3·1= −3 illustrates the multiplicative 0 3.8 identity property. 19. x7x,orxx 7. 51. The sentence 5·x = x·5 illustrates the commutative prop- erty of multiplication. This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is 53. The sentence 2(a+b)=(a+b)2 illustrates the commutative (7,∞). property of multiplication. 55. The sentence−6(m+n)=−6(n+m) illustrates the com- 07 mutative property of addition. 21. The endpoints 0 and 5 are not included in the interval, so 1 57. The sentence 8· = 1 illustrates the multiplicative inverse we use parentheses. Interval notation is (0, 5). 8 property. Copyright © 2013 Pearson Education, Inc. 2 Chapter R: Basic Concepts of Algebra 0 7 0+7 7 59. The distance of−8.15 from 0 is 8.15, so− 8.15=8.15. 9. z ·z = z = z ,or 0 7 7 7 z ·z =1·z = z 61. The distance 295 from 0 is 295, so295 = 295. √ √ √ √ 8 −6 8+(−6) 2 11. 5 · 5 =5 =5 ,or25 63. The distance of− 97 from 0 is 97, so− 97 = 97. −5 5 −5+5 0 13. m ·m = m = m =1 65. The distance of 0 from 0 is 0, so0=0. 1   3 −7 3+(−7) −4 15. y ·y = y = y ,or 5 5 5 5   4 y 67. The distance of from 0 is ,so   = . 4 4 4 4 4 −2 4+(−2) 2 17. (x+3) (x+3) =(x+3) =(x+3) 69. 14− (−8) =14+8 =22 = 22, or −3 8 −3+8+1 6 − 8− 14 =− 22=22 19. 3 · 3 ·3=3 =3 ,or729 3 2 3+2 5 71. − 3− (−9) =−3+9 =6=6, or 21. 2x · 3x =2· 3·x =6x − 9− (−3) =−9+3 =− 6=6 −5 −7 −5+(−7) −12 23. (−3a )(5a )=−3· 5·a =−15a ,or 73. 12.1− 6.7 =5.4=5.4, or 15 − 12 a 6.7− 12.1 =− 5.4=5.4       −3 5 2 −9 −3+2 5+(−9) 25. (6x y )(−7x y )=6(−7)x y =       3 15 6 15 21 21 75.       − − = − − = − = , or       42 4 8 8 8 8 8 −1 −4 −42x y ,or−         4   xy         15 3 15 3 15 6 21 21         − − = + = + = =         4 3 4 4 3 3 4+3 7 8 4 8 4 8 8 8 8 27. (2x) (3x) =2 ·x · 3 ·x =16· 27·x = 432x 77. − 7− 0 =− 7=7, or 3 2 3 3 2 2 3+2 29. (−2n) (5n) =(−2) n · 5 n =−8· 25·n = 0− (−7) =0+7 =7=7 5 −200n 79. Answers may vary. One such number is 35 y 35−31 4 31. = y = y 0.124124412444.... 31 y 1 −7 81. Answers may vary. Since− =0.0099 and b 1 −7−12 −19 101 33. = b = b ,or 12 19 1 b b − =−0.01, one such number is−0.00999. 100 2 −2 3 x y x 2−(−1) −2−1 3 −3 35. = x y = x y ,or 2 2 83. Since 1 +3 = 10, the hypotenuse of a right triangle with −1 3 x y y √ legs of lengths 1 unit and 3 units has a length of 10 units. −4 3 32x y 32 8x −4−(−5) 3−8 −5 37. = x y =8xy ,or 2 2 2 −5 8 5 ✏ 4x y 4 y ✏ c =1 +3 ✏ c ✏ ✏ 1 2 ✏ c =10 ✏ 2 4 4 2 4 4 2·4 4 8 4 ✏ 39. (2x y) =2 (x ) y =16x y =16x y √ 3 c = 10 3 5 5 3 5 5 3·5 15 41. (−2x ) =(−2) (x ) =(−2) x =−32x −1 −2 −2 −2 −1(−2) −2(−2) 43. (−5c d ) =(−5) c d = Exercise Set R.2 2 4 2 4 c d c d   = 2 1 1 (−5) 25 −7 −m 1. 3 = a = ,a=0 7 m 3 a 4 3 −5 4 3 12 4 −20 45. (3m ) (2m ) =3 m · 2 m = 3. Observe that each exponent is negative. We move each 432 12+(−20) −8 27· 16m = 432m ,or factor to the other side of the fraction bar and change the 8 m sign of each exponent.   3 −3 7 −3 7 3 3 −9 21 −5 4 2x y (2x y ) 2 x y x y 47. = = = = −1 −1 3 −3 −4 5 z (z ) z y x −9 21 21 3 8x y 8y z 5. Observe that each exponent is negative. We move each , or −3 9 z x factor to the other side of the fraction bar and change the   −5 sign of each exponent. 10 −8 7 24a b c 4 −5 2 −5 −5 −20 25 −10 −1 −12 6 6 49. =(2a b c ) =2 a b c , m n t t 6 −3 5 12a b c = ,or −6 1 12 12 t m n mn 25 b or 0 0 20 10 7. 23 = 1 (For any nonzero real number, a = 1.) 32a c Copyright © 2013 Pearson Education, Inc. Exercise Set R.2 3 7 −2 51. Convert 16,500,000 to scientific notation. 71. (4.2× 10 )(3.2× 10 ) 7 −2 We want the decimal point to be positioned between the =(4.2× 3.2)× (10 × 10 ) 1 and the 6, so we move it 7 places to the left. Since 5 =13.44× 10 This is not scientific notation. 16,500,000 is greater than 10, the exponent must be posi- 5 =(1.344× 10)× 10 tive. 6 =1.344× 10 Writing scientific notation 7 16,500,000=1.65× 10 −18 7 73. (2.6× 10 )(8.5× 10 ) 53. Convert 0.000000437 to scientific notation. −18 7 =(2.6× 8.5)× (10 × 10 ) We want the decimal point to be positioned between the −11 =22.1× 10 This is not scientific notation. 4 and the 3, so we move it 7 places to the right. Since −11 0.000000437 is a number between 0 and 1, the exponent =(2.21× 10)× 10 must be negative. −10 =2.21× 10 −7 0.000000437 = 4.37× 10 −7 −7 6.4× 10 6.4 10 75. = × 55. Convert 234,600,000,000 to scientific notation. We want 6 6 8.0× 10 8.0 10 the decimal point to be positioned between the 2 and the −13 =0.8× 10 This is not scientific 3, so we move it 11 places to the left. Since 234,600,000,000 notation. is greater than 10, the exponent must be positive. −1 −13 =(8× 10 )× 10 11 234,600,000,000 = 2.346× 10 −14 =8× 10 Writing scientific notation 57. Convert 0.00104 to scientific notation. We want the deci- mal point to be positioned between the 1 and the last 0, so −3 1.8× 10 77. we move it 3 places to the right. Since 0.00104 is a number −9 7.2× 10 between 0 and 1, the exponent must be negative. −3 1.8 10 −3 = × 0.00104 = 1.04× 10 −9 7.2 10 6 =0.25× 10 This is not scientific notation. 59. Convert 0.00000000000000000000000000167 to scientific −1 6 notation. =(2.5× 10 )× 10 5 We want the decimal point to be positioned between the =2.5× 10 1 and the 6, so we move it 27 places to the right. Since 0.00000000000000000000000000167 is a number between 0 79. The average number of pieces of trash per mile is the total and 1, the exponent must be negative. number of pieces of trash divided by the number of miles. −27 0.00000000000000000000000000167 = 1.67× 10 10 51.2 billion 5.12× 10 = 7 76 million 7.6× 10 5 61. Convert 7.6× 10 to decimal notation. 3 ≈ 0.6737× 10 The exponent is positive, so the number is greater than −1 3 ≈ (6.737× 10 )× 10 10. We move the decimal point 5 places to the right. 2 5 ≈ 6.737× 10 7.6× 10 = 760,000 2 On average, there are about 6.737× 10 pieces of trash on −7 63. Convert 1.09× 10 to decimal notation. each mile of roadway. The exponent is negative, so the number is between 0 and 81. The number of people per square mile is the total number 1. We move the decimal point 7 places to the left. of people divided by the number of square miles. −7 1.09× 10 =0.000000109 4 38,000 3.8× 10 = 10 65. Convert 3.496× 10 to decimal notation. −1 0.75 7.5× 10 5 The exponent is positive, so the number is greater than ≈ 0.50667× 10 10. We move the decimal point 10 places to the right. −1 5 ≈ (5.0667× 10 )× 10 10 3.496× 10 =34,960,000,000 4 ≈ 5.0667× 10 −8 4 67. Convert 5.41× 10 to decimal notation. There are about 5.0667× 10 people per square mile. The exponent is negative, so the number is between 0 and 83. We multiply the number of light years by the number of 1. We move the decimal point 8 places to the left. miles in a light year. −8 5.41× 10 =0.0000000541 12 12 4.22× 5.88× 10 =24.8136× 10 8 12 69. Convert 2.319× 10 to decimal notation. =(2.48136× 10)× 10 13 The exponent is positive, so the number is greater than =2.48136× 10 10. We move the decimal point 8 places to the right. The distance from Earth to Alpha Centauri C is 8 13 2.319× 10 = 231,900,000 2.48136× 10 mi. Copyright © 2013 Pearson Education, Inc. 4 Chapter R: Basic Concepts of Algebra 85. First find the number of seconds in 1 hour: 93. Since interest is compounded semiannually, n = 2. Substi- 60 min ✦ 60 sec tute 3225 for P,3.1% or 0.031 for i, 2 for n, and 4 for t 1 hour = 1✧ hr× × = 3600 sec in the compound interest formula. 1h✧r 1 min ✦   nt The number of disintegrations produced in 1 hour is the i A = P 1+ number of disintegrations per second times the number of n   seconds in 1 hour. 2·4 0.031 = 3225 1+ Substituting 37 billion× 3600 2 2·4 = 3225(1 + 0.0155) Dividing =37,000,000,000× 3600 2·4 10 3 = 3225(1.0155) Adding =3.7× 10 × 3.6× 10 Writing scientific 8 notation = 3225(1.0155) Multiplying 2 and 4 10 3 =(3.7× 3.6)× (10 × 10 ) ≈ 3225(1.130939628) Evaluating the 13 =13.32× 10 Multiplying exponential expression 13 ≈ 3647.2803 Multiplying =(1.332× 10)× 10 14 ≈ 3647.28 Rounding to the nearest cent =1.332× 10 14 One gram of radium produces 1.332× 10 disintegrations 95. Since interest is compounded quarterly, n = 4. Substitute in 1 hour. 4100 for P,2.3% or 0.023 for i, 4 for n, and 6 for t in the compound interest formula. 2 87. =5·3+8· 3 + 4(6− 2)   nt i 2 A = P 1+ =5·3+8· 3 +4· 4 Working inside parentheses n 2   =5·3+8·9+4· 4 Evaluating 3 4·6 0.023 = 4100 1+ Substituting = 15 + 72 + 16 Multiplying 4 4·6 = 4100(1 + 0.00575) Dividing = 87 + 16 Adding in order 4·6 = 4100(1.00575) Adding = 103 from left to right 24 = 4100(1.00575) Multiplying 4 and 6 89. 16÷ 4· 4÷ 2· 256 ≈ 4100(1.147521919) Evaluating the =4· 4÷ 2· 256 Multiplying and dividing exponential expression in order from left to right ≈ 4704.839868 Multiplying =16÷ 2· 256 ≈ 4704.84 Rounding to the nearest cent =8· 256 97. Substitute 250 for P, 0.05 for r and 27 for t and perform = 2048 the resulting computation. 2 4(8− 6) − 4·3+2· 8    91. 12·t r 1 0 3 +19 1+ − 1  12  2 S = P 4· 2 − 4·3+2· 8 r = Calculating in the 12 3+1 numerator and in     12·27 0.05 the denominator 1+ − 1 4· 4− 4·3+2· 8 12   = 250 = 0.05 4 12 16−12+16 = ≈ 170,797.30 4 4+16 = 99. Substitute 120,000 for S, 0.03 for r, and 18 for t and solve 4 for P. 20    12·t = r 4 1+ − 1  12  S = P =5 r 12    12·18 0.03 1+ − 1  12  120, 000 = P 0.03 12 216 (1.0025) − 1 120, 000 = P 0.0025 120, 000≈ P(285.94035) 419.67≈ P Copyright © 2013 Pearson Education, Inc. Exercise Set R.3 5 t 3t 2 4t 2 4t·2 8t 101. (x ·x ) =(x ) = x = x 19. (y− 3)(y+5) 2 = y +5y− 3y− 15 Using FOIL a+x x−a 4 2x 4 2x·4 8x 103. (t ·t ) =(t ) = t = t 2 = y +2y− 15 Collecting like terms a b 3 3a 3b 2 2 (3x y ) 27x y 105. = a b 2 2a 2b (−3x y ) 9x y 21. (x + 6)(x+3)   2 a b 2 = 3x y = x +3x+6x + 18 Using FOIL 2a 2b 2 =9x y = x +9x + 18 Collecting like terms 23. (2a + 3)(a+5) Exercise Set R.3 2 =2a +10a+3a + 15 Using FOIL 2 3 2 3 2 1. 7x − 4x +8x+5 =7x +(−4x )+8x+5 =2a +13a + 15 Collecting like terms 3 2 Terms: 7x ,−4x ,8x,5 25. (2x+3y)(2x +y) 3 The degree of the term of highest degree, 7x , is 3. Thus, 2 2 =4x +2xy+6xy+3y Using FOIL the degree of the polynomial is 3. 2 2 =4x +8xy+3y 4 3 3 4 3 3 3. 3a b− 7a b +5ab−2=3a b+(−7a b )+5ab+(−2) 2 27. (x+3) 4 3 3 Terms: 3a b,−7a b ,5ab,−2 2 2 = x +2·x·3+3 The degrees of the terms are 5, 6, 2, and, 0, respectively, 2 2 2 (A +B) = A +2AB +B so the degree of the polynomial is 6. 2 = x +6x+9 2 2 5. (3ab − 4a b− 2ab + 6)+ 2 2 2 29. (y− 5) (−ab − 5a b+8ab+4) 2 2 2 2 = y − 2·y·5+5 =(3− 1)ab +(−4− 5)a b+(−2+8)ab+(6+4) 2 2 2 2 2 (A−B) = A − 2AB +B =2ab − 9a b+6ab+10 2 = y − 10y+25 (2x+3y +z−7)+(4x− 2y−z + 8)+ 7. 2 (−3x +y− 2z− 4) 31. (5x− 3) 2 2 = (2+4− 3)x+(3−2+1)y+(1− 1− 2)z+ =(5x) − 2· 5x·3+3 2 2 2 (−7+8− 4) (A−B) = A − 2AB +B 2 =3x+2y− 2z− 3 =25x − 30x+9 2 3 2 3 2 9. (3x − 2x−x +2)− (5x − 8x−x +4) 33. (2x+3y) 2 3 2 3 2 2 =(3x − 2x−x +2)+(−5x +8x +x − 4) =(2x) + 2(2x)(3y)+(3y) 2 3 2 2 2 =(3− 5)x +(−2+8)x+(−1+1)x +(2− 4) (A+B) = A +2AB+B 2 =−2x +6x− 2 2 2 =4x +12xy+9y 4 2 3 2 11. (x − 3x +4x)− (3x +x − 5x+3) 2 2 35. (2x − 3y) 4 2 3 2 =(x − 3x +4x)+(−3x −x +5x− 3) 2 2 2 2 =(2x ) − 2(2x )(3y)+(3y) 4 3 2 = x − 3x +(−3− 1)x +(4+5)x− 3 2 2 2 (A−B) = A − 2AB +B 4 3 2 4 2 2 = x − 3x − 4x +9x− 3 =4x − 12x y+9y 2 4 2 4 13. (3a )(−7a ) = 3(−7)(a ·a ) 37. (n + 6)(n− 6) 6 =−21a 2 2 2 2 = n − 6 (A +B)(A−B)= A −B 3 4 2 4 3 2 2 15. (6xy )(9x y )=(6· 9)(x·x )(y ·y ) = n − 36 5 5 =54x y 39. (3y + 4)(3y− 4) 3 2 2 2 2 2 17. (a−b)(2a −ab+3b ) =(3y) − 4 (A +B)(A−B)= A −B 3 2 2 =(a−b)(2a )+(a−b)(−ab)+(a−b)(3b ) =9y − 16 Using the distributive property 41. (3x− 2y)(3x+2y) 4 3 2 2 2 3 =2a − 2a b−a b +ab +3ab − 3b 2 2 2 2 =(3x) − (2y) (A−B)(A+B)= A −B Using the distributive property 2 2 three more times =9x − 4y 4 3 2 2 3 =2a − 2a b−a b+4ab − 3b Collecting like terms Copyright © 2013 Pearson Education, Inc. 6 Chapter R: Basic Concepts of Algebra 3 2 43. (2x+3y + 4)(2x+3y− 4) 15. x −x − 5x+5 2 = (2x+3y) + 4(2x+3y)− 4 = x (x− 1)− 5(x− 1) 2 2 2 =(2x+3y) − 4 =(x− 1)(x − 5) 2 2 =4x +12xy+9y − 16 3 2 17. a − 3a − 2a+6 2 2 45. (x + 1)(x− 1)(x +1) = a (a− 3)− 2(a− 3) 2 2 2 =(x − 1)(x +1) =(a− 3)(a − 2) 4 = x − 1 2 19. w − 7w+10 n n n n n 2 n 2 47. (a +b )(a −b )=(a ) − (b ) We look for two numbers with a product of 10 and a sum 2n 2n of−7. By trial, we determine that they are−5 and−2. = a −b 2 w − 7w+10 =(w− 5)(w− 2) n n 2 n 2 n n n 2 49. (a +b ) =(a ) +2·a ·b +(b ) 2 2n n n 2n 21. x +6x+5 = a +2a b +b We look for two numbers with a product of 5 and a sum 2 3 51. (x− 1)(x +x + 1)(x +1) of 6. By trial, we determine that they are 1 and 5. 2 3 =(x− 1)x +(x− 1)x+(x− 1)· 1(x +1) 2 x +6x+5 =(x + 1)(x+5) 3 2 2 3 =(x −x +x −x +x− 1)(x +1) 2 3 3 23. t +8t+15 =(x − 1)(x +1) 3 2 2 We look for two numbers with a product of 15 and a sum =(x ) − 1 of 8. By trial, we determine that they are 3 and 5. 6 = x − 1 2 t +8t+15 =(t + 3)(t+5) a−b a+b 53. (x ) 2 2 25. x − 6xy− 27y (a−b)(a+b) = x We look for two numbers with a product of−27 and a sum 2 2 a −b = x of−6. By trial, we determine that they are 3 and−9. 2 2 2 x − 6xy− 27y =(x+3y)(x− 9y) 55. (a +b +c) =(a +b +c)(a +b +c) 2 2 27. 2n − 20n−48=2(n − 10n− 24) =(a +b +c)(a)+(a +b +c)(b)+(a +b +c)(c) 2 Now factor n −10n−24. We look for two numbers with a 2 2 2 = a +ab +ac +ab +b +bc +ac +bc +c product of−24 and a sum of−10. By trial, we determine 2 2 2 2 that they are 2 and−12. Then n − 10n− 24 = = a +b +c +2ab+2ac+2bc (n + 2)(n− 12). We must include the common factor, 2, to have a factorization of the original trinomial. Exercise Set R.4 2 2n − 20n−48=2(n + 2)(n− 12) 2 29. y − 4y− 21 1. 3x+18 =3·x+3·6=3(x+6) We look for two numbers with a product of−21 and a sum 3 2 2 2 2 3. 2z − 8z =2z ·z− 2z ·4=2z (z− 4) of−4. By trial, we determine that they are 3 and−7. 2 2 2 2 y − 4y− 21 = (y + 3)(y− 7) 5. 4a − 12a+16 =4·a − 4· 3a+4·4=4(a − 3a+4) 4 3 2 2 2 31. y − 9y +14y = y (y − 9y + 14) 7. a(b− 2) +c(b−2)=(b− 2)(a +c) 2 Now factor y − 9y + 14. Look for two numbers with a 3 2 9. 3x −x +18x− 6 product of 14 and a sum of−9. The numbers are−2 and 2 2 −7. Then y − 9y+14 = (y− 2)(y− 7). We must include = x (3x− 1) + 6(3x− 1) 2 2 the common factor, y , in order to have a factorization of =(3x− 1)(x +6) the original trinomial. 3 2 11. 4 3 2 2 y −y +2y− 2 y − 9y +14y = y (y− 2)(y− 7) 2 = y (y−1)+2(y− 1) 3 2 2 2 2 33. 2x − 2x y− 24xy =2x(x −xy− 12y ) 2 =(y− 1)(y +2) 2 2 Now factor x −xy− 12y . Look for two numbers with a 3 2 13. 24x − 36x +72x− 108 product of−12 and a sum of−1. The numbers are−4 and 2 2 3 2 3. Then x −xy−12y =(x−4y)(x+3y). We must include = 12(2x − 3x +6x− 9) the common factor, 2x, in order to have a factorization of 2 = 12x (2x− 3) + 3(2x− 3) the original trinomial. 2 = 12(2x− 3)(x +3) 3 2 2 2x − 2x y− 24xy =2x(x− 4y)(x+3y) Copyright © 2013 Pearson Education, Inc. Exercise Set R.4 7 2 35. 2n +9n− 56 4. Split the middle term using the numbers found in step (3): We use the FOIL method. y =4y− 3y 1. There is no common factor other than 1 or−1. 5. Factor by grouping. 2. The factorization must be of the form 2 2 (2n+)(n+). 2y +y−6=2y +4y− 3y− 6 3. =2y(y+2)− 3(y+2) Factor the constant term, −56. The possibilities are−1·56, 1(−56),−2·28, 2(−28),−4·16, 4(−16), =(y + 2)(2y− 3) −7· 8, and 7(−8). The factors can be written in 2 2 the opposite order as well: 56(−1),−56· 1, 28(−2), 43. 6a − 29ab+28b −28· 2, 16(−4),−16· 4, 8(−7), and−8· 7. We use the FOIL method. 4. Find a pair of factors for which the sum of the out- 1. There is no common factor other than 1 or−1. side and the inside products is the middle term, 2. The factorization must be of the form 9n. By trial, we determine that the factorization (6x+)(x+ ) or (3x+ )(2x+). is (2n− 7)(n + 8). 3. Factor the coefficient of the last term, 28. The pos- 2 37. 12x +11x+2 sibilities are 1· 28, −1(−28), 2· 14, −2(−14), 4· 7, We use the grouping method. and−4(−7). The factors can be written in the op- posite order as well: 28·1,−28(−1), 14·2,−14(−2), 1. There is no common factor other than 1 or−1. 7· 4, and−7(−4). 2. Multiply the leading coefficient and the constant: 4. Find a pair of factors for which the sum of the out- 12· 2 = 24. side and the inside products is the middle term, 3. Try to factor 24 so that the sum of the factors is the −29. Observe that the second term of each bino- coefficient of the middle term, 11. The factors we mial factor will contain a factor of b. By trial, we want are 3 and 8. determine that the factorization is (3a−4b)(2a−7b). 4. Split the middle term using the numbers found in 2 45. 12a − 4a− 16 step (3): We will use the grouping method. 11x=3x+8x 1. Factor out the common factor, 4. 5. Factor by grouping. 2 2 2 2 12a − 4a− 16 = 4(3a −a− 4) 12x +11x+2 = 12x +3x+8x+2 2 Now consider 3a − a− 4. Multiply the leading 2. =3x(4x+1)+2(4x+1) coefficient and the constant: 3(−4) =−12. =(4x + 1)(3x+2) 3. Try to factor −12 so that the sum of the factors is 2 the coefficient of the middle term, −1. The factors 39. 4x +15x+9 we want are−4 and 3. We use the FOIL method. 4. Split the middle term using the numbers found in 1. There is no common factor other than 1 or−1. step (3): 2. The factorization must be of the form −a =−4a+3a (4x+)(x+ ) or (2x+ )(2x+). 5. Factor by grouping. 3. Factor the constant term, 9. The possibilities are 2 2 3a −a−4=3a − 4a+3a− 4 1· 9,−1(−9), 3· 3, and−3(−3). The first two pairs of factors can be written in the opposite order as = a(3a−4)+(3a− 4) well: 9· 1,−9(−1). =(3a− 4)(a+1) 4. Find a pair of factors for which the sum of the out- We must include the common factor to get a factor- side and the inside products is the middle term, ization of the original trinomial. 15x. By trial, we determine that the factorization 2 12a − 4a− 16 = 4(3a− 4)(a+1) is (4x + 3)(x + 3). 2 2 2 2 47. z − 81 = z − 9 =(z + 9)(z− 9) 41. 2y +y− 6 2 2 2 We use the grouping method. 49. 16x −9=(4x) − 3 =(4x + 3)(4x− 3) 1. There is no common factor other than 1 or−1. 2 2 2 2 51. 6x − 6y =6(x −y )=6(x +y)(x−y) 2. Multiply the leading coefficient and the constant: 4 2 4 2 2(−6) =−12. 53. 4xy − 4xz =4x(y −z ) 2 2 2 3. Try to factor −12 so that the sum of the factors is =4x(y ) −z the coefficient of the middle term, 1. The factors we 2 2 =4x(y +z)(y −z) want are 4 and−3. Copyright © 2013 Pearson Education, Inc. 8 Chapter R: Basic Concepts of Algebra 4 4 4 4 2 55. 7pq − 7py =7p(q −y ) 87. x +9x+20 2 2 2 2 =7p(q ) − (y ) We look for two numbers with a product of 20 and a sum 2 2 2 2 of 9. They are 4 and 5. =7p(q +y )(q −y ) 2 2 2 x +9x+20= (x + 4)(x+5) =7p(q +y )(q +y)(q−y) 2 2 2 2 89. y − 6y+5 57. x +12x+36 = x +2·x·6+6 2 We look for two numbers with a product of 5 and a sum =(x+6) of−6. They are−5 and−1. 2 2 2 2 2 59. 9z − 12z+4= (3z) − 2· 3z·2+2 =(3z− 2) y − 6y+5=(y− 5)(y− 1) 2 2 2 2 61. 1− 8x+16x =1 − 2· 1· 4x+(4x) 91. 2a +9a+4 2 =(1− 4x) We use the FOIL method. 1. There is no common factor other than 1 or−1. 3 2 63. a +24a + 144a 2. 2 The factorization must be of the form = a(a +24a + 144) (2a+)(a+). 2 2 = a(a +2·a·12+12 ) 3. Factor the constant term, 4. The possibilities are 2 = a(a + 12) 1·4,−1(−4), and 2·2. The first two pairs of factors can be written in the opposite order as well: 4· 1, 2 2 65. 4p − 8pq+4q −4(−1). 2 2 =4(p − 2pq +q ) 4. Find a pair of factors for which the sum of the out- 2 =4(p−q) side and the inside products is the middle term, 9a. By trial, we determine that the factorization 3 3 3 67. x +64 = x +4 is (2a + 1)(a + 4). 2 =(x + 4)(x − 4x + 16) 2 93. 6x +7x− 3 3 3 3 69. m − 216 = m − 6 We use the grouping method. 2 =(m− 6)(m +6m + 36) 1. There is no common factor other than 1 or−1. 3 3 71. 8t +8=8(t +1) 2. Multiply the leading coefficient and the constant: 3 3 6(−3) =−18. =8(t +1 ) 2 3. Try to factor −18 so that the sum of the factors is =8(t + 1)(t −t+1) the coefficient of the middle term, 7. The factors we 5 2 2 3 73. 3a − 24a =3a (a − 8) want are 9 and−2. 2 3 3 =3a (a − 2 ) 4. Split the middle term using the numbers found in 2 2 step (3): =3a (a− 2)(a +2a+4) 7x=9x− 2x 6 2 3 3 75. t +1=(t ) +1 5. Factor by grouping. 2 4 2 =(t + 1)(t −t +1) 2 2 6x +7x−3=6x +9x− 2x− 3 2 2 77. 18a b− 15ab =3ab· 6a− 3ab· 5b =3x(2x+3)− (2x+3) =3ab(6a− 5b) =(2x + 3)(3x− 1) 3 2 2 2 2 2 79. x − 4x +5x− 20 = x (x−4)+5(x− 4) 95. y − 18y+81 = y − 2·y·9+9 2 2 =(x− 4)(x +5) =(y− 9) 2 2 2 2 2 81. 8x − 32 = 8(x − 4) 97. 9z − 24z+16 = (3z) − 2· 3z·4+4 2 =8(x + 2)(x− 2) =(3z− 4) 2 2 2 2 2 83. 4y − 5 99. x y − 14xy+49=(xy) − 2·xy·7+7 2 =(xy− 7) There are no common factors. We might try to factor this polynomial as a difference of squares, but there is no 2 2 101. 4ax +20ax− 56a=4a(x +5x− 14) integer which yields 5 when squared. Thus, the polynomial is prime. =4a(x + 7)(x− 2) 3 3 2 2 2 2 103. 3z −24=3(z − 8) 85. m − 9n = m − (3n) 3 3 =3(z − 2 ) =(m+3n)(m− 3n) 2 =3(z− 2)(z +2z+4) Copyright © 2013 Pearson Education, Inc. Exercise Set R.5 9 7 7 3 3 105. 16a b+54ab 129. (x +h) −x 6 6 2 2 =2ab(8a +27b ) =(x +h)−x(x +h) +x(x +h)+x 2 3 2 3 2 2 2 2 =2ab(2a ) +(3b ) =(x +h−x)(x +2xh +h +x +xh +x ) 2 2 4 2 2 4 2 2 =2ab(2a +3b )(4a − 6a b +9b ) = h(3x +3xh +h ) 2 3 2 131. (y− 4) +5(y− 4)− 24 107. y − 3y − 4y+12 2 2 = u +5u− 24 Substituting u for y− 4 = y (y− 3)− 4(y− 3) 2 =(u + 8)(u− 3) =(y− 3)(y − 4) =(y− 4 + 8)(y− 4− 3) Substituting y− 4 =(y− 3)(y + 2)(y− 2) for u 3 2 109. x −x +x− 1 =(y + 4)(y− 7) 2 = x (x−1)+(x− 1) 2n n n 2 n 133. x +5x −24=(x ) +5x − 24 2 =(x− 1)(x +1) n n =(x + 8)(x − 3) 4 4 111. 5m −20=5(m − 4) 2 135. x +ax +bx +ab = x(x +a)+b(x +a) 2 2 =5(m + 2)(m − 2) =(x +a)(x +b) 3 2 113. 2x +6x − 8x− 24 2m 2n n 137. 25y − (x − 2x +1) 3 2 =2(x +3x − 4x− 12) m 2 n 2 =(5y ) − (x − 1) 2 =2x (x+3)− 4(x + 3) m n m n =5y +(x − 1)5y − (x − 1) 2 =2(x + 3)(x − 4) m n m n =(5y +x − 1)(5y −x +1) =2(x + 3)(x + 2)(x− 2) 4 2 139. (y− 1) − (y− 1) 2 2 2 2 115. 4c − 4cd−d =(2c) − 2· 2c·d−d 2 2 =(y− 1) (y− 1) − 1 2 =(2c−d) 2 2 =(y− 1) y − 2y+1− 1 2 2 6 3 3 2 3 =(y− 1) (y − 2y) 117. m +8m −20=(m ) +8m − 20 2 = y(y− 1) (y− 2) We look for two numbers with a product of−20 and a sum of 8. They are 10 and−2. 6 3 3 3 m +8m −20=(m + 10)(m − 2) Exercise Set R.5 4 3 119. p− 64p = p(1− 64p ) 1. x−5=7 3 3 = p1 − (4p) x = 12 Adding 5 2 = p(1− 4p)(1+4p+16p ) The solution is 12. 4 2 121. y −84+5y 3. 3x+4 =−8 4 2 = y +5y − 84 3x =−12 Subtracting 4 2 2 = u +5u− 84 Substituting u for y x =−4 Dividing by 3 =(u + 12)(u− 7) The solution is−4. 2 2 2 =(y + 12)(y − 7) Substituting y for u 5. 5y−12=3 8 2 2 8 5y = 15 Adding 12 2 2 123. y − + y = y + y− 49 7 7 49 y = 3 Dividing by 5    4 2 The solution is 3. = y + y− 7 7   2 7. 6x−15=45 9 3 3 2 2 125. x +3x + = x +2·x· + 6x = 60 Adding 15 4 2 2   2 x = 10 Dividing by 6 3 = x + The solution is 10. 2   2 1 1 1 9. 5x−10=45 2 2 127. x −x + = x − 2·x· + 4 2 2 5x = 55 Adding 10   2 1 x = 11 Dividing by 5 = x− 2 The solution is 11. Copyright © 2013 Pearson Education, Inc. 10 Chapter R: Basic Concepts of Algebra 11. 9t+4 =−5 27. 24 = 5(2t+5) 9t =−9 Subtracting 4 24=10t+25 t =−1 Dividing by 9 −1=10t Subtracting 25 1 The solution is−1. − = t Dividing by 10 10 13. 8x+48=3x− 12 1 The solution is− . 10 5x+48 =−12 Subtracting 3x 5x =−60 Subtracting 48 29. 5y− 4(2y−10)=25 x =−12 Dividing by 5 5y− 8y+40=25 The solution is−12. −3y + 40 = 25 Collecting like terms −3y =−15 Subtracting 40 15. 7y−1=23− 5y y = 5 Dividing by−3 12y− 1 = 23 Adding 5y The solution is 5. 12y = 24 Adding 1 y = 2 Dividing by 12 31. 7(3x+6)=11− (x+2) The solution is 2. 21x+42 = 11−x− 2 21x+42 = 9−x Collecting like terms 17. 3x−4 = 5+12x 22x + 42 = 9 Adding x −9x− 4 = 5 Subtracting 12x 22x =−33 Subtracting 42 −9x = 9 Adding 4 3 x =− Dividing by 22 x =−1 Dividing by−9 2 The solution is−1. 3 The solution is− . 2 19. 5− 4a = a− 13 33. 4(3y− 1)−6=5(y+2) 5− 5a =−13 Subtracting a 12y− 4−6=5y+10 −5a =−18 Subtracting 5 12y−10=5y + 10 Collecting like terms 18 a = Dividing by−5 7y− 10 = 10 Subtracting 5y 5 18 7y = 20 Adding 10 The solution is . 5 20 y = Dividing by 7 21. 3m−7=−13 +m 7 20 2m−7=−13 Subtracting m The solution is . 7 2m =−6 Adding 7 2 35. x +3x−28=0 m =−3 Dividing by 2 (x + 7)(x− 4) = 0 Factoring The solution is−3. x+7=0 or x− 4 = 0 Principle of zero products 23. 11− 3x=5x+3 x =−7 or x=4 11− 8x = 3 Subtracting 5x The solutions are−7 and 4. −8x =−8 Subtracting 11 2 37. x +5x=0 x=1 x(x + 5) = 0 Factoring The solution is 1. x=0 or x + 5 = 0 Principle of zero products 25. 2(x+7)=5x+14 x=0 or x =−5 2x+14=5x+14 The solutions are 0 and−5. −3x + 14 = 14 Subtracting 5x 2 39. y +6y+9 = 0 −3x = 0 Subtracting 14 (y + 3)(y+3) = 0 x=0 The solution is 0. y+3=0 or y+3 = 0 y =−3 or y =−3 The solution is−3. Copyright © 2013 Pearson Education, Inc. Exercise Set R.5 11 2 2 41. x +100=20x 53. x −36=0 2 x − 20x + 100 = 0 Subtracting 20x (x + 6)(x−6)=0 (x− 10)(x−10)=0 x+6=0 or x−6=0 x−10=0 or x− 10 = 0 x =−6 or x=6 x=10 or x=10 The solutions are−6 and 6. The solution is 10. 2 55. z = 144 2 2 43. x − 4x−32=0 z −144=0 (x− 8)(x+4)=0 (z + 12)(z−12)=0 x−8=0 or x+4 = 0 z+12=0 or z−12=0 x=8 or x =−4 z =−12 or z =12 The solutions are 8 and−4. The solutions are−12 and 12. 2 2 45. 57. 3y +8y+4=0 2x −20=0 2 (3y + 2)(y+2)=0 2x =20 2 x =10 3y+2 = 0 or y+2=0 √ √ 3y =−2 or y =−2 x = 10 or x =− 10 Principle of square roots √ √ √ 2 The solutions are 10 and− 10, or± 10. y =− or y =−2 3 2 2 59. 6z −18=0 The solutions are− and−2. 2 3 6z =18 2 2 47. 12z +z =6 z =3 √ √ 2 12z +z−6=0 z = 3 or z =− 3 √ √ √ (4z + 3)(3z− 2) = 0 The solutions are 3 and− 3, or± 3. 4z+3=0 or 3z−2=0 1 61. A = bh 4z =−3 or 3z =2 2 3 2 2A = bh Multiplying by 2 on both sides z =− or z = 4 3 2A = b Dividing by h on both sides 3 2 h The solutions are− and . 4 3 63. P =2l+2w 2 49. 12a − 28 = 5a P − 2l =2w Subtracting 2l on both sides 2 12a − 5a− 28 = 0 P − 2l (3a + 4)(4a− 7) = 0 = w Dividing by 2 on both sides 2 3a+4=0 or 4a−7=0 3a =−4 or 4a=7 1 65. A = h(b +b ) 4 7 1 2 2 a =− or a = 3 4 2A 2 4 7 = b +b Multiplying by on both sides 1 2 The solutions are− and . h h 3 4 2A −b = b , or 1 2 51. 14 = x(x− 5) h 2 2A−b h 14 = x − 5x 1 = b 2 2 h 0= x − 5x− 14 4 0=(x− 7)(x+2) 3 67. V = πr 3 x−7=0 or x+2 = 0 3 3V =4πr Multiplying by 3 on both sides x=7 or x =−2 3V 3 = π Dividing by 4r on both sides The solutions are 7 and−2. 3 4r Copyright © 2013 Pearson Education, Inc. 12 Chapter R: Basic Concepts of Algebra 9 69. 83. x−3x− 2x− (5x− (7x− 1)) = x+7 F = C +32 5 x−3x− 2x− (5x− 7x + 1) = x+7 9 F − 32 = C Subtracting 32 on both sides x−3x− 2x− (−2x + 1) = x+7 5 5 5 x−3x− 2x+2x− 1 = x+7 (F − 32) = C Multiplying by on both sides 9 9 x−3x− 4x− 1 = x+7 x−3x− 4x+1 = x+7 71. Ax +By = C x−−x+1 = x+7 Ax = C−By Subtracting By on both sides x +x−1= x+7 C−By 2x−1= x+7 A = Dividing by x on both sides x x−1=7 73. 2w+2h +l = p x=8 2h = p− 2w−l Subtracting 2w and l The solution is 8. 2 2 p− 2w−l 85. (5x +6x)(12x − 5x−2)=0 h = Dividing by 2 2 x(5x + 6)(4x + 1)(3x−2)=0 75. 2x− 3y =6 x=0 or 5x+6 = 0 or 4x+1 = 0 or 3x−2=0 −3y =6− 2x Subtracting 2x x=0 or 5x =−6 or 4x =−1 or 3x=2 6− 2x 6 1 2 y = , or Dividing by−3 x=0 or x =− or x =− or x = −3 5 4 3 6 1 2 2x− 6 The solutions are 0,− ,− , and . 5 4 3 3 3 2 87. 3x +6x − 27x−54=0 77. a = b +bcd 3 2 3(x +2x − 9x−18)=0 a = b(1 +cd) Factoring 2 3x (x+2)− 9(x + 2) = 0 Factoring by grouping a = b Dividing by 1 +cd 2 3(x + 2)(x −9)=0 1+cd 3(x + 2)(x + 3)(x−3)=0 2 79. z = xy−xy 2 x+2=0 or x+3 = 0 or x−3=0 z = x(y−y ) Factoring z x =−2 or x =−3 or x=3 2 = x Dividing by y−y 2 y−y The solutions are−2,−3, and 3. 81. 35− 3(4−t)− 2 = 53(5t−4)+8− 26 Exercise Set R.6 35−12+3t− 2 = 515t−12+8− 26 3−7+3t− 2 = 515t− 4− 26 5 1. Since − is defined for all real numbers, the domain is −21+9t−2=75t− 20− 26 3 xx is a real number. 9t− 23 = 75t− 46 3x− 3 −66t− 23 =−46 3. x(x− 1) −66t =−23 The denominator is 0 when the factor x = 0 and 23 t = also when x −1=0, or x = 1. The domain is 66 xx is a real number and x =0 and x =1. 23 The solution is . x+5 x+5 66 5. = 2 x +4x− 5 (x + 5)(x− 1) We see that x + 5 = 0 when x =−5 and x−1=0 when x = 1. Thus, the domain is xx is a real number and x =−5 and x=1 . 7. We first factor the denominator completely. 2 2 7x − 28x+28 7x − 28x+28 = 2 2 (x −4)(x +3x−10) (x+2)(x−2)(x+5)(x−2) We see that x + 2 = 0 when x = −2, x− 2 = 0 when x = 2, and x + 5 = 0 when x =−5. Thus, the domain is xx is a real number and x= −2 and x =2 and x= −5. Copyright © 2013 Pearson Education, Inc. Exercise Set R.6 13 2 2 2 2 2 x −y x +xy +y x − 4 (x + 2)(x✏ −2) x+2 27. · 9. = = 3 3 2 2 2 x −y x +2xy +y x − 4x+4 (x− 2)(x✏ −2) x− 2 2 2 (x +y)(x−y)(x +xy +y ) 3 2 2 x − 6x +9x x(x − 6x+9) = 11. 2 2 = (x−y)(x +xy +y )(x +y)(x +y) 3 2 2 x − 3x x (x− 3) 2 2 1 (x +y)(x−y)(x +xy +y ) x /(x✏ −3)(x− 3) = · 2 2 = x +y (x +y)(x−y)(x +xy +y ) x /·x(x✏ −3) 1 x− 3 = · 1 Removing a factor of 1 = x +y x 1 2 2 6y +12y− 48 6(y +2y− 8) = 13. = x +y 2 2 3y − 9y+6 3(y − 3y+2) 2 2 2· / 3· (y + 4)(y✏ −2) (x−y) −z x−y +z 29. = ÷ 2 2 3 /(y− 1)(y✏ −2) (x +y) −z x +y−z 2 2 2(y+4) (x−y) −z x +y−z = = · 2 2 y− 1 (x +y) −z x−y +z (x−y +z)(x−y−z)(x +y−z) 4−x −1(x✏ −4) 15. = = 2 (x +y +z)(x +y−z)(x−y +z) x +4x− 32 (x✏ −4)(x+8) −1 1 (x−y +z)(x +y−z) x−y−z = , or − = · (x−y +z)(x +y−z) x +y +z x+8 x+8 2 2 2 2 x−y−z r−s r −s (r−s)(r −s ) =1· Removing a factor of 1 17. · = 2 2 x +y +z r +s (r−s) (r +s)(r−s) x−y−z (r✏ −s)(r✏ −s)(r✏ +s)· 1 = = x +y +z (r✏ +s)(r✏ −s)(r✏ −s) =1 7 3 7+3 31. + = 2 3 5x 5x 5x x +2x− 35 9x − 4x 19. · 10 3 2 3x − 2x 7x+49 = 5x (x✏ +7)(x− 5)(✧ x)(3x + 2)(3✘ x−2) / 5 · 2 = = x /·x(3✘ x−2)(7)(x✏ +7) / 5 ·x (x− 5)(3x+2) 2 = = 7x x 2 2 4x +9x+2 x − 1 21. 4 3a 4+3a · 2 2 33. + = x +x− 2 3x +x− 2 3a+4 3a+4 3a+4 (4x + 1)(x✏ +2)(x✏ +1)(x✏ −1) = 1 (4+3a=3a+4) = (x✏ +2)(x✏ −1)(3x− 2)(x✏ +1) 5 3 4x+1 35. − , LCD is 8z = 4z 8z 3x− 2 5 2 3 2 2 = · − m −n m−n 4z 2 8z 23. ÷ r +s r +s 10 3 = − 2 2 m −n r +s 8z 8z = · 7 r +s m−n = 8z (m +n)(m✏ −n)(r✏ +s) = (r✏ +s)(m✏ −n) 3 2 37. + 2 = m +n x+2 x − 4 2 3 2 3x+12 (x+4) = + , LCD is (x+2)(x−2) 25. ÷ 2 x+2 (x + 2)(x− 2) 2x− 8 (x− 4) 3 x− 2 2 2 3x+12 (x− 4) = · + = · x+2 x− 2 (x + 2)(x− 2) 2 2x− 8 (x+4) 3x− 6 2 3(x✏ +4)(x✏ −4)(x− 4) = + = (x + 2)(x− 2) (x + 2)(x− 2) 2(x✏ −4)(x✏ +4)(x+4) 3x− 4 3(x− 4) = = (x + 2)(x− 2) 2(x+4) Copyright © 2013 Pearson Education, Inc. 14 Chapter R: Basic Concepts of Algebra y 2 9x+2 7 39. 47. − + 2 2 2 y −y− 20 y+4 3x − 2x− 8 3x +x− 4 y 2 9x+2 7 = − , LCD is (y + 4)(y− 5) = + , (y + 4)(y− 5) y+4 (3x + 4)(x− 2) (3x + 4)(x− 1) y 2 y− 5 LCD is (3x + 4)(x− 2)(x− 1) = − · (y + 4)(y− 5) y+4 y− 5 9x+2 x− 1 7 x− 2 = · + · (3x+4)(x−2) x− 1 (3x+4)(x−1) x− 2 y 2y− 10 = − 2 (y + 4)(y− 5) (y + 4)(y− 5) 9x − 7x− 2 7x− 14 = + y− (2y− 10) (3x+4)(x−2)(x−1) (3x+4)(x−1)(x−2) = 2 (y + 4)(y− 5) 9x − 16 = y− 2y+10 (3x + 4)(x− 2)(x− 1) = (y + 4)(y− 5) (3✘ x+4)(3x− 4) = −y+10 (3✘ x+4)(x− 2)(x− 1) = (y + 4)(y− 5) 3x− 4 = (x− 2)(x− 1) 3 x− 5y 41. + 2 2 x +y x −y 5a ab 4b 49. + + 3 x− 5y 2 2 a−b a −b a +b = + , LCD is (x +y)(x−y) x +y (x +y)(x−y) 5a ab 4b = + + , 3 x−y x− 5y a−b (a +b)(a−b) a +b = · + x +y x−y (x +y)(x−y) LCD is (a +b)(a−b) 3x− 3y x− 5y 5a a +b ab 4b a−b = + = · + + · (x +y)(x−y) (x +y)(x−y) a−b a +b (a +b)(a−b) a +b a−b 4x− 8y 2 2 5a +5ab ab 4ab− 4b = = + + (x +y)(x−y) (a+b)(a−b) (a+b)(a−b) (a+b)(a−b) 2 2 y 2 5a +10ab− 4b 43. + = y− 1 1−y (a +b)(a−b) y −1 2 = + · 7 x+8 3x− 2 y− 1 −1 1−y 51. − + 2 2 x+2 4−x 4− 4x +x y −2 = + 7 x+8 3x− 2 y− 1 y− 1 = − + , 2 x+2 (2 +x)(2−x) (2−x) y− 2 2 = LCD is (2 +x)(2−x) y− 1 2 7 (2−x) x+8 2−x = · − · + x y 2 2+x (2−x) (2 +x)(2−x) 2−x − 45. 2x− 3y 3y− 2x 3x− 2 2+x · x −1 y 2 (2−x) 2+x = − · 2x− 3y −1 3y− 2x 2 2 2 28−28x+7x −(16−6x−x )+3x +4x−4 x −y = 2 = − (2 +x)(2−x) 2x− 3y 2x− 3y 2 2 2 28− 28x+7x −16+6x +x +3x +4x− 4 x +y = = x− (−y)= x +y 2 (2 +x)(2−x) 2x− 3y 2 2 11x − 18x+8 11x − 18x+8 = , or 2 2 (2 +x)(2−x) (x + 2)(x− 2) Copyright © 2013 Pearson Education, Inc. Exercise Set R.6 15 2 8 c 8 2 c + c· + 1 x x +2 2 2 2 c c c 53. 59. + + = 2 2 c 2 x+1 2−x x −x− 2 1+ 1· + 2 c c c 1 x x +2 = + + 3 c +8 x+1 2−x (x + 1)(x− 2) 2 2 c = 1 −1 x x +2 c+2 = + · + x+1 −1 2−x (x + 1)(x− 2) c 2 1 −x x +2 3 c +8 c = + + , = · x+1 x− 2 (x + 1)(x− 2) 2 c c+2 LCD is (x + 1)(x− 2) 2 (c✏ +2)(c − 2c+4)/c 2 = 1 x− 2 −x x+1 x +2 /c·c(c✏ +2) = · + · + x+1 x− 2 x− 2 x+1 (x + 1)(x− 2) 2 c − 2c+4 2 2 = x− 2 −x −x x +2 c = + + (x+1)(x−2) (x+1)(x−2) (x+1)(x−2) 2 2 x− 2−x −x +x +2 2 2 2 2 x +xy +y x +xy +y = 61. = (x + 1)(x− 2) 2 2 2 2 x y x x y y − · − · 0 y x y x x y = (x + 1)(x− 2) 2 2 x +xy +y = =0 3 3 x −y xy a−b xy 2 2 a−b ab b =(x +xy +y )· 55. = · 3 3 2 2 x −y 2 2 a −b b a −b 2 2 (x +xy +y )(xy) ab = 2 2 a−b ab (x−y)(x +xy +y ) = · 2 2 b (a +b)(a−b) x +xy +y xy = · 2 2 x +xy +y x−y ab✧(a✏ −b) xy = /( b a +b)(a✏ −b) =1· x−y a xy = = a +b x−y x y x y − − 1 a 1 xy y x y x 57. −1 a− a· − = · , LCM is xy a−a 1 1 1 1 63. a a a xy = = + + −1 1 a 1 a +a y x y x a + a· +   a a a x y − (xy) 2 a − 1 y x =   1 1 a = + (xy) 2 a +1 y x 2 2 a x −y = 2 a − 1 a x +y = · 2 a a +1 (x✏ +y)(x−y) = 2 a − 1 (x✏ +y)· 1 = 2 a +1 = x−y Copyright © 2013 Pearson Education, Inc. 16 Chapter R: Basic Concepts of Algebra 1 2 1 x+3 2 x− 3 + · + · 3 3 3 2 2 3 3 (x +h) −x x +3x h+3xh +h −x x− 3 x+3 x− 3 x+3 x+3 x− 3 65. 73. = = 3 4 3 x+2 4 x− 1 h h − · − · x− 1 x+2 x− 1 x+2 x+2 x− 1 2 2 3 3x h+3xh +h = x+3+2(x− 3) h 2 2 (x− 3)(x+3) h /(3x +3xh +h ) = = 3(x+2)− 4(x− 1) h /· 1 (x− 1)(x+2) 2 2 =3x +3xh +h x+3+2x− 6     5 5 x+1 (x+1)+(x− 1) (x− 3)(x+3) +1 =     75. x− 1 x− 1 3x+6− 4x+4 =     x+1 (x+1)− (x− 1) (x− 1)(x+2) − 1 x− 1 x− 1 3x− 3 5 2x x− 1 (x− 3)(x+3) = · = x− 1 2 −x+10 5 (x− 1)(x+2) / 2x(x✦ −1) = 1· 2 /(x✦ −1) 3x− 3 (x− 1)(x+2) = · 5 = x (x− 3)(x+3) −x+10 (3x− 3)(x− 1)(x+2) n(n + 1)(n+2) (n + 1)(n+2) = , or 77. + (x− 3)(x + 3)(−x + 10) 2· 3 2 2 3(x− 1) (x+2) n(n + 1)(n+2) (n + 1)(n+2) 3 = + · , (x− 3)(x + 3)(−x + 10) 2· 3 2 3 LCD is 2· 3 a 1+a a a 1+a 1−a + · + · 1−a a 1−a a a 1−a 67. = n(n + 1)(n+2)+3(n + 1)(n+2) 1−a a 1−a 1+a a a = + · + · 2· 3 a 1+a a 1+a 1+a a (n + 1)(n + 2)(n+3) 2 2 a +(1−a ) = Factoring the num- 2· 3 a(1−a) erator by grouping = 2 2 (1−a )+a 2 2 2 a(1 +a) x − 9 5x − 15x+45 x +x 79. · + 3 2 x +27 x − 2x− 3 4+2x 1 a /(1 +a) = · 2 2 a /(1−a) 1 (x + 3)(x− 3)(5)(x − 3x+9) x +x = + 2 1+a (x + 3)(x − 3x + 9)(x− 3)(x+1) 4+2x = 2 2 1−a (x + 3)(x− 3)(x − 3x+9) 5 x +x · + = 2 (x + 3)(x− 3)(x − 3x+9) x+1 4+2x 1 2 1 1 2 1 + + + + 2 2 2 2 2 2 a b 2 a ab b a ab b 5 x +x 69. = · , 2 2 =1· + 1 1 1 1 a b x+1 4+2x − − 2 2 2 2 a b a b 2 5 x +x 2 2 = + LCM is a b x+1 2(2 +x) 2 2 b +2ab +a = 2 5· 2(2 +x)+(x +x)(x+1) 2 2 b −a = 2(x + 1)(2 +x) (b✏ +a)(b +a) = 3 2 20+10x +x +2x +x (b✏ +a)(b−a) = 2(x + 1)(2 +x) b +a = 3 2 x +2x +11x+20 b−a = 2(x + 1)(2 +x) 2 2 2 2 2 (x +h) −x x +2xh +h −x 71. = h h Exercise Set R.7 2 2xh +h =  2 h 1. (−21) =− 21=21 h /(2x +h)   = 2 2 3. 9y = (3y) =3y=3y h /· 1  =2x +h 2 5. (a− 2) =a− 2 Copyright © 2013 Pearson Education, Inc. Exercise Set R.7 17 √  √ √ √ √ 3 3 3 3 5 2+3 32 = 5 2+3 16· 2 7. −27x = (−3x) =−3x 45. √ √ √  =5 2+3· 4 2 4 8 4 2 4 2 2 9. 81x = (3x ) =3x =3x √ √ =5 2+12 2 √ √ √ 5 5 5 11. 32 = 2 =2 = (5 + 12) 2 √ √ √ √ √ √ =17 2 13. 180 = 36·5= 36· 5=6 5 √ √ √ √ √ √ √ √ √ √ √ 6 20− 4 45 + 80=6 4· 5− 4 9·5+ 16· 5 47. 15. 72 = 36·2= 36· 2=6 2 √ √ √ =6· 2 5− 4· 3 5+4 5 √ √ √ √ √ 3 3 3 3 3 √ √ √ 17. 54 = 27·2= 27· 2=3 2 =12 5− 12 5+4 5 √ √ √ √ √ 2 4 2 4 2 2 = (12−12+4) 5 19. 128c d = 64c d ·2=8cd 2=8 2cd √   √ =4 5 4 4 4 6 4 4 4 2 2 21. 48x y = 16x y · 3x =2xy 3x = √ √ √ √ 2 2 4 49. 8 2x − 6 20x− 5 8x 2 2xy 3x √ √ √ 2 =8x 2− 6 4· 5x− 5 4x · 2 √  2 2 √ √ √ 23. x − 4x+4 = (x− 2) =x− 2 =8x 2− 6· 2 5x− 5· 2x 2 √ √ √ √ √ √ √ √ 2 =8x 2− 12 5x− 10x 2 25. 15 35 = 15· 35 = 3· 5· 5·7= 5 · 3·7= √ √ √ √ √ 2 =−2x 2− 12 5x 5 · 3·7=5 21 √ √ √ √  √ √ √ √ √ 2 51. 8+2 5 8− 2 5 27. 8 10 = 8· 10 = 2· 4· 2·5= 2 · 4·5= √   √  √ √ 2 2 = 8 − 2 5 2· 2 5=4 5    √ √ =8− 4· 5 3 4 2 4 2 2 29. 2x y 12xy = 24x y = 4x y ·6=2x y 6 =8− 20    √ √ 3 3 3 3 2 3 3 3 31. 3x y 36x = 108x y = 27x · 4y =3x 4y =−12 √ √ √ √    3 3 4 3 5 53. 33. (2 3+ 5)( 3− 3 5) 2(x+4) 4(x+4) = 8(x+4) √ √ √ √ √ √ √ √  3 3 2 =2 3· 3− 2 3· 3 5+ 5· 3− 5· 3 5 = 8(x+4) · (x+4) √ √  3 2 =2· 3− 6 15 + 15− 3· 5 =2(x+4) (x+4) √ √ =6− 6 15 + 15− 15   √   8 2 3 16 24 2 3 m n m n m n =−9− 5 15 8 8 35. = = 8 √ √ √ 2 2 2 2 2 2 55. ( 2− 5) =( 2) − 2· 2·5+5  √ √  =2− 10 2+25 40xy 40xy 37. √ = = 5y √ 8x 8x =27− 10 2  √ √ √ √ √ √ √  3 2 2 2 2 2 57. ( 5− 6) =( 5) − 2 5· 6+( 6) 3x 3x 1 1 3 3 √ 39. = = = √ 3 5 3 5 24x 8x 2x 24x =5− 2 30+6 √   4 3 =11− 2 30 64a 64·a ·a 3 3 41. = 3 3 27b 27·b 59. We use the Pythagorean theorem. We have a = 47 and √ √ 3 3 3 64a a b = 25. = √ 3 2 2 2 3 c = a +b 27b √ 3 2 2 2 4a a c =47 +25 = 2 3b c = 2209 + 625   2 3 2 7x 7·x ·x c = 2834 43. = 6 6 36y 36·y c≈ 53.2 √ √ 2 x 7x The distance across the pond is about 53.2 yd. =  6 36y √ x 7x = 3 6y Copyright © 2013 Pearson Education, Inc. 18 Chapter R: Basic Concepts of Algebra √ √ √ √   2 1− 2 1− 2 2 3+ 6 a 2 2 √ √ = √ √ · √ √ 61. a) 73. h + = a Pythagorean theorem 2 3− 6 2 3− 6 2 3+ 6 2 √ √ √ √ 2 2 3+ 6− 2 6− 12 a 2 2 = h + = a 4· 3− 6 4 √ √ √ √ 2 3a 2 3+ 6− 2 6− 2 3 2 h = = 4 12− 6  √ √ 2 − 6 6 3a h = = , or − 6 6 4 √ √ √ a 6 6 m + n h = 3 √ √ = √ √ ·√ √ 2 75. m− n m− n m + n b) Using the result of part (a) we have √ √ 6( m + n) = √ √ 1 2 2 A = · base · height ( m) − ( n) 2 √ √ 6 m+6 n   √ 1 a a a = A = a· 3 + = a m−n 2 2 2 2 √ √ √ √ 2 √ 50 50 2 100 10 a √ √ √ A = 3 77. = · = = 4 3 3 2 3 2 3 2    √ 63. 3 x 2 2 4 8 8 2 3 3 3 √ √ 79. = · = = = 3 3 ❅ 5 5 4 20 20 20 ❅ √ √ √ √ √ ❅ 8 2 xx 11 11 11 121 11 ❅ 81. √ = √ ·√ = √ = √ ❅ 3 3 11 33 33 ❅ √ √ √ ❅ 9− 5 9− 5 9+ 5 83. √ = √ · √ x 3− 3 3− 3 9+ 5 √ √ 2 2 2 2 2 x +x =(8 2) Pythagorean theorem 9 − ( 5) = √ √ √ 2 2x = 128 27+3 5− 9 3− 15 2 81− 5 x =64 = √ √ √ 27+3 5− 9 3− 15 x=8 76    √ √ √ √ √ = 3 3 7 21 21 21 27+3 5− 9 3− 15 65. = · = = √ = √ √ √ 7 7 7 49 7 49 √ √ √ a + b a + b a− b 85. √ √ √ √ √ = · √ √ 3 3 3 3 3 3a 3a 7 7 5 35 35 a− b 67. √ = √ · √ = √ = √ √ 3 3 3 3 2 2 5 25 25 5 125 ( a) − ( b) = √ √    √ 3 3a( a− b) 16 16 3 48 48 3 3 3 √ 69. = · = = = a−b 3 9 9 3 27 27 √ = √ √ √ 3a a− 3a b 3 3 8· 6 2 6 = 3 3  5/6 6 5 √ 87. y = y 2 2 3+1 √ = √ ·√ 71. √  3 4 3/4 1/4 3 3 3− 1 3− 1 3+1 89. 16 = (16 ) = 16 =2 =8 √ 2( 3+1) = 1 1 1 −1/3 3− 1 91. 125 = = √ = 3 1/3 √ 125 5 125 2( 3+1) √ = √  4 5/4 5 4 2 a a a a a 5/4 −3/4 4 √ √ √ 93. a b = = = , or a 3/4 4 4 3 3 3 b b b b 2 /( 3+1) = √ √ √ √ / 2· 1 3 3 3 3 5/3 7/3 2 5 7 5 7 2 95. m n = m n = m n = mn m n √ = 3+1 √ 5 3/5 3 97. 17 =17 √  4 5 4/5 99. 12 =12 Copyright © 2013 Pearson Education, Inc. Chapter R Review Exercises 19 √  √ √    3 1/3 a √ 1/2 1/3 1/6    101. 11 = 11 = (11 ) =11 √ a √ a/2 a/2 a 127. a = a = a √ √ 3 1/2 1/3 1/2+1/3 5/6 103. 5 5=5 · 5 =5 =5 √ 5 2 2/5 1/5 2 2 105. 32 =32 = (32 ) =2 =4 Chapter R Review Exercises 3/2 1/2 3/2+1/2 2 107. (2a )(4a )=8a =8a     6 6 3 3 2 1/2 1/2 x x x x b 1. True 109. = = , or −4 2 −4 −2 9b 3 b 3b 3 3. True 2/3 5/6 x y √ 2/3−(−1/3) 5/6−1/2 1/3 3 √ 111. = x y = xy = x y 4 3 12 3 −1/3 1/2 x y 5. Rational numbers: −7, 43,− ,0, 64, 4 , , 102 9 4 7 1 2 5 2 1/2 5/2 2/3 · · 1/3 5/3 √ 2 3 2 3 3 113. (m n ) = m n = m n = 7. Integers: −7, 43, 0, 64, 102 √ √ √ √ 3 3 3 3 5 5 2 m n = mn = n mn √ 3 9. Natural numbers: 43, 64, 102 3/4 2/3 4/3 3/4+2/3 3/4+4/3 a (a +a )= a +a = 115. √ √ 11. (−4,7 12 12 17/12 25/12 17 25 a +a = a + a =   √ √ 12   2 12 5 7 7 7 7 a a +a a   13. The distance of− from 0 is ,so − = .   8 8 8 8 √ √ 3 1/3 1/2 2/6 3/6 117. 6 2=6 2 =6 2 2 15. 3· 2− 4· 2 + 6(3− 1) 2 3 1/6 =(6 2 ) 2 √ =3· 2− 4· 2 +6· 2 Working inside 6 = 36· 8 parentheses √ 6 2 = 288 =3· 2− 4·4+6· 2 Evaluating 2  √ =6− 16 + 12 Multiplying 3 2 1/4 2 1/3 3/12 2 4/12 119. 4 xy x y =(xy) (x y) =(xy) (x y) =−10 + 12 Adding in order   1/12 3 2 4 = (xy) (x y) = 2 from left to right   1/12 3 3 8 4 = x y x y −5  17. Convert 8.3× 10 to decimal notation. 12 11 7 = x y The exponent is negative, so the number is between 0 and  √ √   3 1/3 121. 4 3 4 3 4 3/2 1/3 1. We move the decimal point 5 places to the left. a a = a a =(a a ) −5 11/2 1/3 8.3× 10 =0.000083 =(a ) 11/6 = a 19. Convert 405,000 to scientific notation. √ 6 11 = a We want the decimal point to be positioned between the √ 6 5 4 and the first 0, so we move it 5 places to the left. Since = a a 405,000 is greater than 10, the exponent must be positive.   3 3/2 2/3 3 2 5 (a +x) (a +x) (a +x) (a +x) 405,000 = 4.05× 10 123. √ = 4 1/4 a +x (a +x) 5 −3 21. (3.1× 10 )(4.5× 10 ) 26/12 (a +x) 5 −3 = =(3.1× 4.5)× (10 × 10 ) 3/12 (a +x) 2 =13.95× 10 This is not scientific notation. 23/12 =(a +x) 2  =(1.395× 10)× 10 12 23 = (a +x) 3  =1.395× 10 Writing scientific notation 12 11 =(a +x) (a +x)  4 −5 −2 4+(−2) −5+1 23. 1 (−3x y )(4x y)=−3(4)x y = 2 125. 1+x +√ 2 2 2 1+x −12x 12x 2 −4 √ −12x y , or , or − 4 4 2  2 y y 1+x 1 1+x 2 = 1+x · +√ ·√ 2 2 2 1+x 1+x 1+x √ √ √ √ 4 4 4 2 2 2 25. 81 = 3 =3 (1 +x ) 1+x 1+x = + 2 2 1+x 1+x √ 2 2 (2 +x ) 1+x = 2 1+x Copyright © 2013 Pearson Education, Inc.

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