Chapter 1
Introduction to Real Numbers and Algebraic
Expressions
Exercise Set 1.1
2. Substitute 34 for n: 600(34) = 20,400, so $20,400 is col-
6. A=1
2·30 in. ·26 in. = 390 in2
10. 6y=6·7=42
22. d−14
40. 7c−6
42. 8x+5,or5+8x
44. 5x−8y
48. 55tmiles
56. 2·5−3·1
7·6=10 −3
7·6=7/ ·1
7/ ·6=1
6
Exercise Set 1.2
2. 750; −125
14. 0.1875
16 3.0000
16
14 0
16. 5
3means 5 ÷3, so we divide.
18. 0.4166…
20. 1.22
22. 0.25
4 1.00
24. 1.125
8 9.000
8
10
26. 0.35
20 7.00
34. −4<−3
16 =−0.8125 and −5
9=−0.5,so −13
16 <−5
9.
48. 8>x
60. 6
Exercise Set 1.3 3
72. w
4y=52
4·13 =52
52 =1
Exercise Set 1.3
2. −3
24. −2
44. −7
16 +7
8=−7
16 +14
16 =7
16
58. 27
60. −1
68. −10
72. Let V= the final value of the stock.
80.
0.625
8 5.000
84. When xis negative, the inverse of x, −x, is positive, so −x
4 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
Exercise Set 1.4
RC2. −18 −(−6) = −18 + 6; the correct answer is (b).
2. −5
24. −1
30. −25
48. 3
9−9
9=3
9+−9
9=−6
9=−2
3
64. 1500
66. 4.94
70. −3
8−−1
2=−3
8+1
2=−3
8+4
8=1
8
86. 5
92. −11
16 +5
32 −−1
4+7
8=−11
16 +5
32 +1
4+7
8=
100. Let D= the number of degrees the temperature dropped.
D=49
◦F−(−54◦F) = 103◦F
Chapter 1 Mid-Chapter Review 5
Chapter 1 Mid-Chapter Review
5. −x=−(−4) = 4;
15. 7
3means 7 ÷3, so we divide.
2.33
3 7.00
18. −8≥−5 is false because neither −8>−5 nor −8=−5is
true.
19. −4≤−4 is true because −4=−4 is true.
32. 7+(−9) The absolute values are 7 and 9. The difference
−8+(−9) = −17
36. 2
3+−9
8The absolute values are 2
3and 9
8. The dif-
6 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
41. 5−(−11) = 5 + 11 = 16
=9
49. −4+(−10) −(−3) −12 = −4+(−10)+3+(−12)
51. −9+(−3)+16−(−10) = −9+(−3)+16+10
52. Let T= the difference in the temperatures.
53. Let P= the sum of the gains and losses.
55. Answers may vary. Three examples are π,−√7, and
Exercise Set 1.5
2. −15
10. 18
22. −6.6
Exercise Set 1.6 7
30. 1
4
48. −518.4
56. 151,200
62. For w=1
10:
64. For s=1:
For x=−3:
74. T=5(−$3) = −$15
78. First find the total amount aof the checks written.
82. d−e
3d=5−1
3·5=4
15
Exercise Set 1.6
8 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
6. 9
18. 0
26. −1
10
42. −47.3·1
21.4,or 47.3·−1
21.4
9÷−5
6=−5
9·−6
5=30
45 =2
3
62. 21
25
74. Amount of increase = $63.87 −$59.33 = $4.54
80. 4·(−6) ·(−2) ·(−1) = −24 ·2=−48
86. First we find 10
3.
88. There are none. For a=0,−aand ahave opposite signs
Exercise Set 1.7 9
Exercise Set 1.7
RC6. 3·5=5·3 illustrates the commutative law of multipli-
6. 4
9x=4
9x·y
y=4y
9xy
20. t+rs, sr +t,ort+sr
44. 20x+32+12p
54. −2
3a−2
3b+8
62. 8x, −1.4y
72. 5(5m−1)
74. 10(x−5)
90. c(x+y−z)
116. 99t
10 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
128. 11
4x+2
3y−4
5x−1
6y+12
130. −4.9
142. Since 0 is to the right of −4, 0 >−4 is true.
Exercise Set 1.8
4. −a+b
22. 4a−b−4(5a−7b+8c)
=4a−b−20a+28b−32c
=−16a+27b−32c
30. 10[7 −4(7 −5)] = 10[7 −4·2] = 10[7 −8] = 10[−1] = −10
32. [7(8 −4) + 16] −[15 −(7 + 8)]
34. [9(x+5)−7] + [4(x−12)+9]
=13x−1
36. [6(x+4)−12] −[5(x−8) + 14]
38. 4{[8(x−3) + 9] −[4(3x−2)+6]}
=4{8x−15 −12x+2}
42. 8−(2 ·3−9) = 8 −(6 −9) = 8 −(−3)= 8+3=11
Chapter 1 Summary and Review: Study Guide 11
78. 52−43−3
92−22−15=25 −64 −3
81 −4−1=−39 −3
77 −1=−42
76 =
82. |3−5|2−|7−13|
|12 −9|+|11 −14|=|−2|2−|−6|
|3|+|−3|=22−6
3+3 =
4−6
6=−2
6=−1
3
94. x−{x−1−[x−2−(x−3−{x−4−
96. 33+3
3+3
3=3·33=3·3·3·3=3
4
Chapter 1 Vocabulary Reinforcement
Chapter 1 Concept Reinforcement
verse is 1.
4. False; a<bhas the same meaning as b>a.
Chapter 1 Study Guide
4. −5.6+(−2.9) Two negative numbers. Add the absolute
9. 45y
27y=5·9y
3·9y=5
3·9y
9y=5
3·1=5
3
11. 27x+9y−36z=9·3x+9·y−9·4z= 9(3x+y−4z)
‚
12 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
Chapter 1 Review Exercises
1. Substitute 17 for xand 5 for yand carry out the compu-
5. The distance of 126 from 0 is 126, so |126|= 126.
7. The graph of 8
9is 8
9of the way from 0 to 1.
9. Since −1 is to the right of −6, we have −1>−6.
13. −9≤11 is true since −9<11 is true.
so the answer is negative. 4 + (−7) = −3
22. 6+(−9)+(−8)+7
a) Add the negative numbers: −9+(−8) = −17
c) Add the results: −20.1+15.1=−5
24. −3−(−7)+7−10 = −3+7+7+(−10)
25. −9
10 −1
2=−9
10 −5
10 =−9
10 +−5
10=−14
10 =
27. −9·(−6) = 54
34. 2(−3.4−12.2) −8(−7) = 2(−15.6) −8(−7)
Chapter 1 Summary and Review: Review Exercises 13
36. −16 ÷4−30 ÷(−5) = −4−(−6)
20 =−40
20 =−2
38. Let t= the total gain or loss. We represent the gains as
39. Let a= Chang’s total assets after he borrows $2500.
Total
40. First we multiply to find the total drop din the price:
42. 5(3x−7) = 5 ·3x−5·7=15x−35
46. 2x−14 = 2 ·x−2·7=2(x−7)
51. 7x−3y−9x+8y=7x−9x−3y+8y
52. 6x+3y−x−4y=6x−x+3y−4y
53. −3a+9b+2a−b=−3a+2a+9b−b
56. 3[11 −3(4 −1)] = 3[11 −3·3] = 3[11 −9] = 3 ·2=6
59. 5{[6(x−1)+7]−[3(3x−4)+8]}
mn +5=5+mn by the commutative law of addition.
and associative laws, so answer B is correct.
14 Chapter 1: Introduction to Real Numbers and Algebraic Expressions
63. (|2.7−3|+3
2−|−3|)÷(−3)
64. 2000 −1990
+ 1980 −1970
+… +20−10
65. Note that the sum of the lengths of the three horizontal
Chapter 1 Discussion and Writing Exercises
4. Consider a
Chapter 1 Test
1. Substitute 10 for xand 5 for yand carry out the compu-
tations.
4. Since −1
8. The distance of −7 from 0 is 7, so |−7|=7.
12. The opposite of −1.4 is 1.4 because −1.4+1.4=0.
15. If x=−8, then −x=−(−8) = 8.
16. 3.1−(−4.7) = 3.1+4.7=7.8
40
19. 2−(−8)=2+8=10
=7
8
22. 4·(−12) = −48
23. −1
27. −2(16) −|2(−8) −53|=−2(16) −|2(−8) −125|
changed over the five week period.
Total
30. Let D= the difference in the elevations.
We carry out the subtraction.
31. First we multiply to find the total decrease din the popu-
Now we subtract to find the new population p.
32. First we subtract to find the total drop in temperature t.
33. 3(6 −x)=3·6−3·x=18−3x
38. 5x−(3x−7) = 5x−3x+7=2x+7
44. |−27−3(4)|−|−36|+|−12|
45. a−{3a−[4a−(2a−4a)]}
=a−{−3a}