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Properties of Real Numbers

1. –(–7.2). 2. 1 – (–3). 3. –9 + (–4.5). 4. (–3.4)(–2). 2 5. 3 5. 6. – + ( – ). 5. –15 ÷ 3 . Properties of Real Numbers. ALGEBRA 2 LESSON 1-1. (For help, go to Skills Handbook page 845.). Simplify. 1-1. Solutions. 1. –(–7.2) = 7.2 3. –9 + (–4.5) = –13.5

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Properties of Real Numbers

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  1. 1. –(–7.2) 2. 1 – (–3) 3. –9 + (–4.5) 4. (–3.4)(–2) 2 5 3 5 6. – + (– ) 5. –15 ÷ 3 Properties of Real Numbers ALGEBRA 2 LESSON 1-1 (For help, go to Skills Handbook page 845.) Simplify. 1-1

  2. Solutions 1. –(–7.2) = 7.2 3. –9 + (–4.5) = –13.5 5. –15 ÷ 3 = –5 6. – + (– ) = = – = –1 2. 1 – (–3) = 1 + 3 = 4 4. (–3.4)(–2) = 6.8 2 5 3 5 –2 + (–3) 5 5 5 Properties of Real Numbers ALGEBRA 2 LESSON 1-1 1-1

  3. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 A pilot uses the formula d = a2 + 7290a to find the number of miles d that can be seen when flying at an altitude of a miles. Which set of numbers best describes the values for each variable? The pilot’s instrument panel will display the altitude a as a rational number in the form of a terminating decimal. The value of d obtained by replacing a with a rational number can be either irrational or rational (consider a = 1.47). So, the values of d are best described as real numbers. 1-1

  4. 3 4 – is between –1 and 0. Use a calculator to find that 7 2.65. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 3 4 Graph the numbers – , 7 , and 3.6 on a number line. 1-1

  5. 9 = 3, so – 9 = –3. –9 < – 9. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 Compare –9 and – 9. Use the symbols < and >. Since –9 < –3, it follows that 1-1

  6. Opposite: –(–3 ) = 3 1 7 1 7 1 1 7 22 Reciprocal: = = – 1 7 22 7 –3 – 1 4 Reciprocal: Properties of Real Numbers ALGEBRA 2 LESSON 1-1 Find the opposite and the reciprocal of each number. 1 7 a. –3 b. 4 Opposite: –4 1-1

  7. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 Which property is illustrated? a. (–7)(2 • 5) = (–7)(5 • 2) b. 3 • (8 + 0) = 3 • 8 The given equation is true because 2 • 5 = 5 • 2. The given equation is true because 8 + 0 = 8. So, the equation uses the Commutative Property of Multiplication. This is an instance of the Identity Property of Addition. 1-1

  8. 1 3 1 3 1 3 1 3 4 is 4 units from 0, so | 4 | = 4 . Properties of Real Numbers ALGEBRA 2 LESSON 1-1 1 3 Simplify | 4 |, |–9.2|, and |3 – 8|. –9.2 is 9.2 units from 0, so |–9.2| = 9.2. |3 – 8| = |–5| and –5 is 5 units from 0. So, |–5| = 5, and hence |3 – 8| = 5. 1-1

  9. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 Pages 8–10 Exercises 1. natural numbers, whole numbers, integers, rational numbers, real numbers 2. irrational numbers, real numbers 3. irrational numbers, real numbers 4. integers, rational numbers, real numbers 5. whole numbers, integers, rational numbers, real numbers 6. rational numbers, real numbers 7. rational numbers, real numbers 8. irrational numbers, real numbers 9. whole numbers, rational numbers 10. natural numbers, rational numbers 11. real numbers 12. 13. 14. 15. 1-1

  10. 5 18 1 200 Properties of Real Numbers ALGEBRA 2 LESSON 1-1 16. 17. > 18. = 19. > 20. < 21. < 22. = 23. > 24. < 25. – > – , – < – 26. 0.075 < 0.39, 0.39 > 0.075 27. –2.3 < 2.1, 2.1 > –2.3 28. –5.2 < –4.8, –4.8 > –5.2 29. 3.04 < 3.4, 3.4 > 3.04 30. 0.4 < 0.4, 0.4 > 0.4 31. –4 < – 4,– 4 > – 4 32. 5 < 7, 7 > 5 33. – 3 > – 5 – 5 < – 3 34. –200, 35. –3 , 36. 0.01, –100 1 4 1 3 1 3 1 4 3 5 1-1

  11. 2 7 – 1 3 3 3 or 1 2 1 – 3 50 117 Properties of Real Numbers ALGEBRA 2 LESSON 1-1 7 2 37. 38. – 3, 39. –2 , 40. 2.34, – 41. 3 – , 42. Identity Prop. Of Mult. , 43. Dist. Prop. 44. Comm. Prop. of Add. 45. Assoc. Prop. of Mult. 46. Comm. Prop. of Mult. 47. Identity Prop. of Add. 48. Inverse Prop. of Add. 49. Inverse Prop. of Mult. 50. Dist. Prop. 51. Comm. Prop. of Mult. 52. Assoc. Prop. of Add. 53. 10.3 54. 0.06 55. –25 56. 1.6 57. 58. 3 1 3 1-1

  12. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 1 3 59. 3 60. –2 61–68. Answers may vary. Samples are given. 61. –5 62. –3 63. –1 64. 65. 1 66. 3 67. 4 68. 4.8 69. natural numbers, whole numbers, integers, rational numbers, real numbers 70. irrational numbers, real numbers 71. irrational numbers, real numbers 72. rational numbers, real numbers 73. irrational numbers, real numbers 74. irrational numbers, real numbers 75. > 76. > 77. < 78. > 1 2 1 4 1 2 2 3 1-1

  13. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 79. < 80. < 81. < 82. < 83. Answers may vary. Sample: 4 is a whole number, but is not a whole number. 84. Answers may vary. Sample: 7 is a natural number, but –7 is not a natural number. 85. 0 is a whole number, and since –0 = 0, the opposite of 0 is a whole number. 86. Answers may vary. Sample: The integer –1 has –1 as its reciprocal, so –1 is an integer whose reciprocal is an integer. 87. Answers may vary. Sample: 2 and 2 are irrational numbers, but their product (2) is a rational number. 88. Check students’ work. 89. All except the Identity Prop. of Add. (since 0 is not in the set of natural numbers) and the inverse properties 1 4 1-1

  14. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 90. all except the Inverse Prop. of Mult. and the Inverse Prop. of Add. 91. all except the Inverse Prop. of Mult. 92. all the properties 93. Comm., Assoc., and Distr. Prop. 94. No; explanations may vary. Sample: The only pairs of integers that have a product of –12 are –1 and 12, –2 and 6, –3 and 4, –4 and 3, –6 and 2, and –12 and 1. None of these pairs has a sum of –3. 95. D 96. H 97. C 98. I 99. A 1-1

  15. 100. [2] The opposite of the reciprocal of 5 is the opposite of , or – , and the reciprocal of the opposite of 5 is the reciprocal of –5, or – . [1] includes a statement or example that a reciprocal is a multiplicative inverse or that a number times its reciprocal is 1, OR a statement or example that an opposite is an additive inverse or that a number plus its opposite is 0 101. [2] 1 and –1; for a number n to be equal to its reciprocal, n = , which means n2 = 1. So n = 1 or –1. [1] only includes answer 1 and –1 with no explanation 102. 1.9 103. –3.8 104. 27 105. 0 106. –0.4 107. 7 108. , or 2 109. , or 11 110. ,or 1 1 5 1 5 1 5 9 4 1 4 1 n 35 3 2 3 3 2 1 2 Properties of Real Numbers ALGEBRA 2 LESSON 1-1 1-1

  16. Properties of Real Numbers ALGEBRA 2 LESSON 1-1 111. 5 112. 38 113. 45 1-1

  17. 5 8 8 13 –1 , Properties of Real Numbers ALGEBRA 2 LESSON 1-1 7 5 1. Graph –2.3, 12, and on a number line. 2. Replace each with the symbol <, >, or = to make the sentence true. a. 2.03 2.8 b. 1 1.2 3. Fit the opposite and the reciprocal of 1 . 4. Name the property of real numbers illustrated by the equation. (–2)(3 + ) = (–2)( + 3) 5. Simplify |9 – (–5)|. < 2 9 = 5 8 Comm. Prop. of Add. 14 1-1

  18. 1. 8 • 3 – 2 • 4 2. 8 – 4 + 6 ÷ 3 3. 24 ÷ 12 • 4 ÷ 3 4. 3 • 82 + 12 ÷ 4 5. 27 + 18 ÷ 9 – 32 + 1 6. (40 + 24) ÷ 8 – (23 + 1) Algebraic Expressions ALGEBRA 2 LESSON 1-2 (For help, go to Skills Handbook page 845.) Use the order of operations to simplify each expression. 1-2

  19. Algebraic Expressions ALGEBRA 2 LESSON 1-2 1. 8 • 3 – 2 • 4 = (8 • 3) – (2 • 4) = 24 – 8 = 16 2. 8 – 4 + 6 ÷ 3 = 8 – 4 + (6 ÷ 3) = 8 – 4 + 2 = (8 – 4) + 2 = 4 + 2 = 6 3. 24 ÷ 12 • 4 ÷ 3 = (24 ÷ 12) • 4 ÷ 3 = 2 • 4 ÷ 3 = (2 • 4) ÷ 3 = 8 ÷ 3 = or 2 4. 3 • 82 + 12 ÷ 4 = 3 • 64 + 12 ÷ 4 + (3 • 64) + (12 ÷ 4) = 192 + 3 = 195 5. 27 + 18 ÷ 9 – 32 + 1 = 27 + 18 ÷ 9 – 9 + 1 = 27 + (18 ÷ 9) – 9 + 1 = 27 + 2 – 9 + 1 = (27 + 2) – 9 + 1 = 29 – 9 + 1 = (29 – 9) + 1 = 20 + 1 = 21 6. (40 + 24) ÷ 8 – (23 + 1) = (40 + 24) ÷ 8 – (8 + 1) = 64 ÷ 8 – 9 = (64 ÷ 8) – 9 = 8 – 9 = –1 Solutions 8 3 2 3 1-2

  20. Algebraic Expressions ALGEBRA 2 LESSON 1-2 Evaluate 7x – 3xy for x = –2 and y = 5. 7x – 3xy = 7(–2) – 3(–2) (5) Substitute –2 for x and 5 for y. = –14 – (–30) Multiply first. = –14 + 30 To subtract, add the opposite. = 16 Add. 1-2

  21. Algebraic Expressions ALGEBRA 2 LESSON 1-2 Evaluate (k – 18)2 – 4k for k = 6. (k – 18)2 – 4k = (6 – 18)2 – 4(6) Substitute 6 for k. = (–12)2 – 4(6) Subtract within parentheses. = 144 – 4(6) Simplify the power. = 144 – 24 Multiply. = 120 Subtract. 1-2

  22. 19 Algebraic Expressions ALGEBRA 2 LESSON 1-2 The expression –0.08y2 + 3y models the percent increase of Hispanic voters in a town from 1990 to 2000. In the expression, y represents the number of years since 1990. Find the approximate percent of increase of Hispanic voters by 1998. Since 1998 – 1990 = 8, y = 8 represents the year 1998. –0.08y2 + 3y = –0.08(8)2 + 3(8)  Substitute 8 for y The number of Hispanic voters had increased by about 19%. 1-2

  23. Algebraic Expressions ALGEBRA 2 LESSON 1-2 Simplify by combining like terms. 2h – 3k + 7(2h – 3k) 2h – 3k + 7(2h – 3k) = 2h – 3k + 14h – 21kDistributive Property = 2h + 14h – 3k – 21kCommutative Property = (2 + 14)h – (3 + 21)kDistributive Property = 16h – 24k 1-2

  24. c 2 c 2 P = c + + d + (d – c) + d + + c + d c 2 c 2 = c + + d + d – c + d + + c + d c 2 c 2 = + + c + 4d 2c 2 = + c + 4d Algebraic Expressions ALGEBRA 2 LESSON 1-2 Find the perimeter of this figure. Simplify the answer. = c + c + 4d = 2c + 4d 1-2

  25. Algebraic Expressions ALGEBRA 2 LESSON 1-2 Pages 15–17 Exercises 1. –30 2. 26 3. 368 4. –16 5. –16 6. 28 7. –70 8. –12 9. 1 ft 10. 4 ft 11. 64 ft 12. 1600 ft 13. 0.013 mm 14. 0.032 mm 15. 0.4 mm 16. 1.4 mm 17. $1210 18. $1331 19. $1464.10 20. $1610.51 21. 4a 22. 2s + 5 23. –9a + b 24. 6a + 3b 1-2

  26. Algebraic Expressions ALGEBRA 2 LESSON 1-2 41 4 25. 10r + 5s 26. 6w + 5z 27. 2x2 + x 28. xy +3x 29. –0.5x 30. 6x – 5 31. 10y + x – 4 32. –y – 6x 33. –3a + 15b 34. 4g – 2 35. –3x + 4y – z 36. 4a 37. 4a 38. 5 39. 17 40. 41 41. 66 42. –1 43. 44. 10 45. –765 46. a. about 180 million voters b. about 242 million voters; about 263 million voters c. –0.0078y2 + 1.265y + 65.27 d. about 87 million 1-2

  27. 5x2 2 7y2 12 Algebraic Expressions ALGEBRA 2 LESSON 1-2 3 4 47. – a2 + 2b2 48. 49. + 50. y 51. 3x + 6y 52. 4x + 2y 53. –2x2 + 2y2 54. F 55. C 56. A 57. G 58. B 59. H 60. E 61. I 62. D 63. Assoc. Prop. of Add., Comm. Prop. of Add., Assoc. Prop. of Add., Identity Prop. of Mult., Distr. Prop. of Add. 64. Distr. Prop.; addition; Distr. Prop. 65. Answers may vary. Sample: x3 + x2 + x – x3 – 2x, –x2 + 2x2 + 3x – 3x – x, x2 – 5x + 4x – x5 + x5, 3x2 – x2 – x2 + 7x – 8x 2y 15 1-2

  28. Algebraic Expressions ALGEBRA 2 LESSON 1-2 66. Answers may vary. Sample: 2(b – a) + 5(b – a) = (2 + 5) (b – a) Dist. Prop. = 7 (b – a) Addition = 7b – 7a Dist. Prop. 67.a. 18 b. 2x2, 18 c. Properties of operations were used to simplify the original expression. Those properties convert one expression into an equivalent expression, and equivalent expressions have equal values for all replacements of their variables. d. No; explanations may vary. Sample: Students should check each step in the simplification; they also should substitute more than one value for x in the original and simplified expressions. 68. B 69. G 1-2

  29. 1 10 1 16 Algebraic Expressions ALGEBRA 2 LESSON 1-2 70.[2] x = 0 or x = 3 since for 6x(x – 3) = 0 either 6x or x – 3 must be equal to 0. [1]x = 0 and x = 3 with no explanation 71. A 72. D 73. C 74. –1.5, – 2,–1.4, –0.5 75. –4.3, –|3.4|, |–3.4|, |–4.3| 76. – , – , – , 77. – , – , , 78. > 79. > 80. > 81. < 82. < 83. = 5 6 3 4 3 8 1 2 1 8 1 4 1-2

  30. Algebraic Expressions ALGEBRA 2 LESSON 1-2 1. Evaluate each expression for the given values of the variables. a. 4p – 3q + 8; p = –6, q = –10 b. –3m2 – (m + n)2; m = 2, n = –4 2. Simplify by combining like terms. a. 7 – 4(x + y) + 2(x – 3y) b. 3y2 + 8 – (2y2 + 12) c. –12(x2 + y) + 3(y + x2) 3. Find the perimeter. Simplify the answer. 14 –16 7 – 2x – 10y y2 – 4 –9x2 – 9y 8c + 4d 1-2

  31. 1. 5x – 9x + 3 2. 2y + 7x + y – 1 x 3 y 3 2y 3 3. 10h + 12g – 8h – 4g 4. + + – y 5. (x + y) – (x – y) 6. –(3 – c) – 4(c – 1) Solving Equations ALGEBRA 2 LESSON 1-3 (For help, go to Lesson 1-2.) Simplify each expression. 1-3

  32. x 3 y 3 2y 3 x 3 y 3 2y 3 3y 3 x + y + 2y – 3y 3 x + 0y 3 x 3 x + (1 + 2 – 3)y 3 Solving Equations ALGEBRA 2 LESSON 1-3 1. 5x – 9x + 3 = (5 – 9)x + 3 = –4x + 3 2. 2y + 7x + y – 1 = 7x + (2 + 1)y – 1 = 7x + 3y – 1 3. 10h + 12g – 8h – 4g = (12 – 4)g + (10 – 8)h = 8g + 2h 4. + + – y = + + – = = = = 5. (x + y) – (x – y) = (x + y) + (y – x) = (x – x) + (y + y) = 0 + 2y = 2y 6. –(3 – c) – 4(c – 1) = (c – 3) – 4c – 4(–1) = c – 3 – 4c + 4 = (1 – 4)c + (–3 + 4) = –3c + 1 Solutions 1-3

  33. Check: 7x + 3 = 2x – 12 7(–3) + 3 2(–3) – 12 –18 = –18 Solving Equations ALGEBRA 2 LESSON 1-3 Solve 7x + 3 = 2x – 12. 7x + 3 = 2x – 12 5x + 3 = –12 Subtract 2x from each side. 5x = –15 Subtract 3 from each side. x = –3 Divide each side by 5. 1-3

  34. 24 7 m = – Divide each side by 7. Solving Equations ALGEBRA 2 LESSON 1-3 Solve 4(m + 9) = –3(m – 4). 4(m + 9) = –3(m – 4) 4m + 36 = –3m + 12 Distributive Property 7m + 36 = 12 Add 3m to each side. 7m = –24 Subtract 36 from each side. 1-3

  35. A = 2( w + h + wh) A = 2 w + 2 h + 2whDistributive Property A – 2 h = 2 w + 2whSubtract 2 h from each side. A – 2 h = (2 + 2h)wDistributive Property = wDivide both sides by 2 + 2h. A – 2 h 2 + 2h Solving Equations ALGEBRA 2 LESSON 1-3 The formula for the surface area of a rectangular prism units long, w units wide, and h units high is A = 2( w + h + wh). Solve the formula for w. 1-3

  36. = / x a + 8 = b a() + a(8) = abMultiply each side by the least common denominator (LCD), a. x a The denominator cannot be zero, so a 0. Solving Equations ALGEBRA 2 LESSON 1-3 Solve + 8 = b for x. Find any restrictions on a and b. x a x + 8a = ab Simplify. x = ab – 8a Subtract 8a from each side. 1-3

  37. Relate: 2 • width + length = perimeter Define: Let w = the width. Then 3w = the length. Write: 2 w + 3w = 68 Solving Equations ALGEBRA 2 LESSON 1-3 Adrian will use part of a garage wall as one of the long sides of a rectangular rabbit pen. He wants the pen to be 3 times as long as it is wide. He plans to use 68 ft of fencing. Find the dimensions of the pen. 1-3

  38. 3 5 w = 13 Divide each side by 5. 4 5 3w = 40 Find the length. 3 5 4 5 The width is 13 ft and the length is 40 ft. Solving Equations ALGEBRA 2 LESSON 1-3 (continued) 5w = 68 Add. Check: Is the answer reasonable? Since the dimensions are about 14 ft by 41 ft and 14 + 14 + 41 = 69, the perimeter is about 69 ft. The answer is reasonable. 1-3

  39. Relate: Perimeter equals the sum of the lengths of the four sides. Define: Let x = the length of the shortest side. Then 2x = the length of the second side. Then 3x = the length of the third side. Then 6x = the length of the fourth side. Solving Equations ALGEBRA 2 LESSON 1-3 The sides of a quadrilateral are in the ratio 1 : 2 : 3 : 6. The perimeter is 138 cm. Find the lengths of the sides. 1-3

  40. Solving Equations ALGEBRA 2 LESSON 1-3 (continued) Write: 138 = x + 2x + 3x + 6x 138 = 12xCombine like terms. 11.5 = x 2x = 2(11.5) 3x = 3(11.5)6x = 6(11.5)Find the length of= 23 = 34.5 = 69 each side. Check:  Is the answer reasonable? Since 12 + 23 + 35 + 69 = 139, the answer is reasonable. The lengths of the sides are 11.5 cm, 23 cm, 34.5 cm, and 69 cm. 1-3

  41. Relate: distance first plane travels = distance second plane travels. Define: Let t = the time in hours for the second plane. Then t + = the time in hours for the first plane. 35 60 Write: 400t = 350 (t + ) 7 12 1225 6 400t = 350t + Distributive Property Solving Equations ALGEBRA 2 LESSON 1-3 A plane takes off from an airport and flies east at a speed of 350 mi/h. Thirty-five minutes later, a second plane takes off from the same airport and flies east at a higher altitude at a speed of 400 mi/h. How long does it take the second plane to overtake the first plane? 1-3

  42. 1225 6 50t = Solve for t. 1 12 t = 4 h or about 4 h 5 min Check:  Is the answer reasonable? In 4 h, the second plane travels 1600 mi. In 4 h, the first plane travels about 1600 mi. The answer is reasonable. 2 3 Solving Equations ALGEBRA 2 LESSON 1-3 (continued) 1-3

  43. Pages 21–24 Exercises = = / / 2A b 1. 23 2. 8 3. 4. –5 5. 6. – 7. 8. 9. 8 10. –6 11. 2 12. 0 13. – 14. –6 15. 2 16. 17. h = 18. g = 19. w = 20. r = 21. r = 22.h = 23. x = , a –b 24. x = , bc 2s t2 V lh 17 2 I pt 7 2 2 3 V r2 1 9 S 2 h 17 7 1 2 3 2 c a + b c c – b Solving Equations ALGEBRA 2 LESSON 1-3 1-3

  44. = = / / 5g 2 7 39 r1r2 r1 + r2 Rr1 r1 – R h + 5t2 t Solving Equations ALGEBRA 2 LESSON 1-3 27 5 2 3 2 3 2 3 2 5 25. x = a(c – b) or ac – ab, a 0 26. x = a(b + 5) or ab + 5a, a 0 27.x = 2(m + n) + 2 or2m + 2n + 2 28.x = – 1 29. 4 h 30. 300 mi/h, 600 mi/h 31. width = 4.5 cm, length = 7.5 cm 32. 4 in., 5 in., 6 in. 33. 11 cm, 11 cm, 16.5 cm, 16.5 cm 34. 7.5 cm, 10 cm, 12.5 cm 35.a.x + (x + 1) + (x + 2) = 90; 29, 30, 31 b. (x – 1) + x + (x + 1) = 90; 29, 30, 31 36. , or 1 37. 3 38. , or 5 39. 40. 30 41. , or 0.9875 42.R = 43.r2 = 44. h = 45.v = 3 2 79 80 S – 2 r2 2 r 46 39 1-3

  45. = = = = = = = = = = = / / / / / / / / / / / 2(v – s2) s cb 2da Solving Equations ALGEBRA 2 LESSON 1-3 10c a 46. h = 47.b2 = – b1 48. 40°, 140° 49. 34°, 56° 50. about 169.4 mi 51. 2.98 m 52. $360; $746.40 53. 43, 45, 47, 49 54. 34, 36, 38, 40 55. x = ab – b2 – a, b 0 56.x = , bd 57.x = , ac 58. x = , ab 59. x = , bc 60.x = , 3atcd 61.x = , 5bpaq 62. x = + 6, a, b, d 0 63. x = , a 0 64.x = + a, m 0, xa 65. a. t = b. about 40.9°F c.C = (F – 32) d. about 4.9°C c – a b – d a – c m 2A h b + d c – a s – 1055 1.1 3a – b – 8 a – b 3b + 2c – 5 b – c 5 9 2ab – 2c 3at – cd 4a – 3bc aq – 5bp 1-3

  46. Solving Equations ALGEBRA 2 LESSON 1-3 66. a. 10 cows, 30 chickens; Sample equation: 4c + 2(40 – c) = 100, where c is the number of cows. b. 80 legs; 20 legs; 2 legs; 10 cows c. Answers may vary. Sample: In all, a repair shop has 11 bicycles and tricycles to repair. These have a total of 26 wheels. How many bicycles and how many tricycles are there? 7 bicycles, 4 tricycles 67. a. If you solve ax – b = c for x, you get x = . Since b and c are integers, b + c is an integer. But a is a nonzero integer. So is the quotient of two integers and hence, by the definition of a rational number, is a rational number. b + c a b + c a b + c a 1-3

  47. 67. (continued) 73. –7 74. – , or –5 75. –20 76. – , or –5 77. 7x2 – 2x 78. ab – 6a 79. 2y – 7x 80. 11x – 6 81. –6x – 5 82. –2r – 5s = / c – b a b. Solutions are rational when 0, a 0, and is a perfect square (a whole number perfect square or, in simplest form, a fraction whose numerator and denominator are whole number perfect squares). 68. about 269.4 ft 69. 18.5 70. 20 71. 19.38° 72. 560 cm2 > 16 3 1 3 – c – b a 11 2 1 2 Solving Equations ALGEBRA 2 LESSON 1-3 1-3

  48. 11 7 – 7 8 61 mi Solving Equations ALGEBRA 2 LESSON 1-3 1. Solve 16x – 15 = –5x + 48. 2. Solve 5(1 – 3m) = 30 – 2(4m + 7). 3. Solve s = for b. 4. Mrs. Chern drove at a rate of 45 mi/h from her home to her sister’s house. She spent 1.5 hours having lunch with her sister. She then drove back home at a rate of 55 mi/h. The entire trip, including lunch, took 4 hours. How far does Mrs. Chern live from her sister? 5. Find three consecutive odd integers whose sum is 111. 3 a + b + c 2 b = 2s – a – c 35, 37, 39 1-3

  49. < < – – > > – – 1. 5 < 12 7. 3x + 3 = 2x – 3 8. 5x = 9(x – 8) + 12 2. 5 < –12 3. 5 12 4. 5 –12 5. 5 5 6. 5 5 Solving Inequalities ALGEBRA 2 LESSON 1-4 (For help, go to Lessons 1-1 and 1-3.) State whether each inequality is true or false. Solve each equation. 1-4

  50. < < – – > > – – Solving Inequalities ALGEBRA 2 LESSON 1-4 Solutions 1. 5 < 12, true 3. 5 12, false 5. 5 5, true 7. 3x + 3 = 2x – 3 3x – 2x = –3 – 3x = –6 2. 5 < –12, false 4. 5 –12, false 6. 5 5, true 8. 5x = 9(x – 8) + 12 5x = 9x – 72 + 12–4x = –60x = 15 1-4

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