Chapter R
Review of Basic Algebra
46. True, since −7 = −7 is true.
Exercise Set R.1 1
48. True, since −11 > −13 is true.
2
RC2. The correct answer is (b). See page R-2 in the text.
50. x < −1
RC4. The correct answer is (f). See page R-3 in the text.
⫺1 0
RC6. The correct answer is (d). See page R-4 in the text.
√ 52. x ≥ −1
2. 0, 1, 12, 25
⫺1 0
√ 12
4. −6, 0, 1, −4, 12, 25, −
3 54. x < 0
√
6. 3, 0.131331333133331 . . .
0
8. 12
56. x ≤ 0
11 7
10. −3.43, −11, 12, 0, ,−
34 13
0
12. All of them
58. 3
14. {s, o, l, v, e}
60. 16
16. {1, 3, 5, 7, 9, 11}
62. 127
18. {−3, −2, −1}
13
64.
20. {x|x is an integer greater than 3 and less than 11} 8
66. 16.4
22. {x|x is a rational number or x is an irrational number}, or
{x|x is a real number} 68. 465
24. {x|x is a real number and x ≤ 21}, or {x|x ≤ 21} 0
70. = |0| = 0
−15
26. 18 > 0
72. | − 5| ≥ | − 2|
28. 7 > −7
74. | − 8| ≤ |8|, or | − 8| ≥ |8| since | − 8| = |8|
30. 0 > −11
32. −6 < −3 Exercise Set R.2
34. −7 > −10
RC2. The opposite of a negative number is positive.
1 11
36. −13 <
5 250 RC4. The absolute value of a negative number is positive.
38. −13.99 < −8.45 RC6. The sum of 0 and a negative number is negative.
14 27
40. − = −0.933 . . . and − = −0.509433962 . . . , so RC8. The quotient of a negative number and a positive num-
15 53
14 27 ber is negative.
− <− .
15 53 2. −25
42. 7 > x
4. 2
2
44. t ≤ 10 6. −12
3
Copyright
c 2017 Pearson Education, Inc.
,2 Chapter R: Review of Basic Algebra
8. −3 55
68. −
12
10. −10
70. 3
12. −32
72. 432
14. 1.9
74. 71.04
16. −11.19 65
76.
2 14
18. −
7 64
78. −
5 125
20. −
4 80. −8
3 1 6 1 5 82. −9
22. − + = − + = −
4 8 8 8 8
5 7 20 21 41 84. 8
24. − + − =− + − =−
6 8 24 24 24 86. 0.7
26. 9
88. Not defined
28. 0
90. 0
2
30.
3 92. Not defined
32. 2x 10
94.
9
34. −5
6
96. −
36. −20 5
1
38. 46 98. −
65
40. 0 1
100.
42. 5 0.8
This can also be expressed as follows:
44. 44 1 1 10 10 5
= · = = , or 1.25
0.8 0.8 10 8 4
46. −17.7
102. 8x
48. −34.8 3 6 3 7 7
104. ÷ − = · − =−
13 5 7 5 6 10
50. −
5 12 3 12 10
106. − ÷ − =− · − =8
7 5 7 20 13 5 10 5 3
52. − − − =− + =
8 2 8 8 8 108. −3
2 4 10 12 22
54. − − = − + − =− 110. 110
3 5 15 15 15
4 5 36 35 1 112. −4000
56. − − − =− + =−
7 9 63 63 63 5 1 5 2 5
114. ÷ − = · − =−
58. −40 8 2 8 1 4
60. −45 5 5 5 6 2
116. − ÷ − =− · − =
9 6 9 5 3
62. 35 3 5 3 8 24
118. − ÷ − =− · − =
64. −152 5 8 5 5 25
5 5 5 6 3
66. 42.7 120. − ÷ − =− · − =
8 6 8 5 4
Copyright
c 2017 Pearson Education, Inc.
, Exercise Set R.3 3
122. 7 5
24.
2
−1.9 −1.9 10 19
124. = · =− , or − 0.095 26. 1
20 20 10 200
−17.8 −17.8 10 178 89 28. 1
126. = · =− = − , or − 5.5625
3.2 3.2 10 32 16
−3
128. Not defined 1 1 1 125
30. = 3 = =1· = 125
5 1 1 1
3 8 7 10 1
130. ,− ; − , ; 1, −1; 0, does not exist; 6.4, , or 5 125
8 3 10 7 −6.4
1 1
− ; −a, −4
6.4 a 5 1 1 16 16
32. = 4 = =1· =
132. 26 2 5 625 625 625
√ 2 16
134. 3, π, 4.57557555755557 . . .
1
136. All of them 34.
x6
3 36. y 7
138. 5 >
8
1 1
140. 123 > −10 38. =−
(−4)3 64
142. Use the definition of subtraction. The number that
40. 9−2
3 3
can be added to 11.7 to obtain −7 is −7 − 11.7, or
3 7 9
4 4 42. n−5
−7 − 11 = −19 , or −19.45.
4 10 20 44. (−8)−6
46. [6 − 4(8 − 5)] = [6 − 4 · 3]
Exercise Set R.3
= [6 − 12]
= −6
RC2. True; see page R-23 in the text.
48. 10[7 − 4(8 − 5)] = 10[7 − 4 · 3] = 10[7 − 12] =
RC4. False; see page R-22 in the text.
10[−5] = −50
RC6. False; see page R-20 in the text.
50. [9(7 − 4) + 19] − [25 − (7 + 3)] =
RC8. True; see page R-22 in the text. [9 · 3 + 19] − [25 − 10] = [27 + 19] − 15 =
2. 6 3 46 − 15 = 31
1 1
4. x4 52. [48 ÷ (−3)] ÷ − = −16 ÷ − = 64
4 4
6. t5
54. 30 · 10 − 18 · 25 = 300 − 450
8. (3.8) 5
= −150
3
4 56. (9 − 12)2 = (−3)2 = 9
10. −
5 92 − 122 = 81 − 144 = −63
12. 9 · 9 · 9 = 729 58. 7 · 8 − 32 − 23 = 7 · 8 − 9 − 8 = 56 − 9 − 8 = 39
14. (−7) · (−7) = 49 60. 43 + 20 · 10 + 72 − 23 = 64 + 20 · 10 + 49 − 23 =
16. (0.1)(0.1)(0.1)(0.1)(0.1)(0.1) = 0.000001 64 + 200 + 49 − 23 = 290
18. (−3)(−3)(−3)(−3) = 81 62. (9 · 8 + 3 · 3)2 = (72 + 9)2 = (81)2 = 6561
2 2 2 2 16 64. 5000 · (4 + 1.16)2 = 5000 · (5.16)2 =
20. · · · =
3 3 3 3 81 5000 · (26.6256) = 133, 128
√
22. 6
Copyright
c 2017 Pearson Education, Inc.
, 4 Chapter R: Review of Basic Algebra
66. (43 · 6 − 14 · 7)3 + (33 · 34)2 = 110. 2.3
(258 − 98) + (1122) = (160) + (1122) =
3 2 3 2
112. 900
4, 096, 000 + 1, 258, 884 = 5, 354, 884
114. −79
68. 18 − (2 · 3 − 9) = 18 − (6 − 9) = 18 − (−3) = 21
116. −79
70. (18 − 2)(3 − 9) = 16(−6) = −96
118. 23
72. [(−32) ÷ (−2)] ÷ (−2) = 16 ÷ (−2) = −8
2 15 2 · 15
74. 30 · 20 − 15 · 24 = 600 − 360 = 240 120. − − =
3 16 3 · 16
76. 16 ÷ (19 − 15)2 − 7 = 16 ÷ (4)2 − 7 = 2·3·5
=
3·2·8
16 ÷ 16 − 7 = 1 − 7 = −6
/·3
2 /·5
=
78. 53 + 20 · 40 + 82 − 29 = 125 + 20 · 40 + 64 − 29 = /·2
3 /·8
5
125 + 800 + 64 − 29 = 960 =
8
80. 4000 · (3 + 1.14)2 = 4000 · (4.14)2 = 122. 2(61 · 6−1 − 6−1 · 60 )
4000 · (17.1396) = 66, 558.4
1 1
= 2 6· − ·1
6 6
8(7 − 3) 8·4 32
82. = = =8 1
4 4 4 = 2 1−
6
43 64 5 10
84. = =8 = 2· =
8 8 6 6
86. 53 − 72 = 125 − 49 = 76 5
=
3
88. 10(−5) + 1(−1) = −50 − 1 = −51
124. 12345679 · 9 = 111, 111, 111
90. 14 − 2(−6) + 7 = 14 + 12 + 7 = 33 12345679 · 18 = 222, 222, 222
92. −32 − 8 ÷ 4 − (−2) = −32 − 2 − (−2) = 12345679 · 27 = 333, 333, 333
−32 − 2 + 2 = −32 According to this pattern, we have 12345679 · 36 =
444, 444, 444.
94. (3 − 8)2 = (−5)2 = 25 √
126. (π) 2 ≈ 5.047497267
96. 28 − 103 = 28 − 1000 √
( 2)π ≈ 2.970686424
= −972 √ √
Thus, (π) 2
is larger than ( 2)π .
98. 2 × 103 − 5000 = 2 × 1000 − 5000 =
2000 − 5000 = −3000
Exercise Set R.4
100. 6[9 − (3 − 4)] = 6[9 − (−1)] = 6[9 + 1] =
6[10] = 60 RC2. (c)
102. 1000 ÷ (−100) ÷ 10 = −10 ÷ 10 = −1 RC4. (e)
20 − 62 20 − 36 −16 8 RC6. (g)
104. = = =−
92 + 32 81 + 9 90 45
RC8. (f)
4|6 − 7| − 5 · 4 4| − 1| − 5 · 4 4·1−5·4
106. = = =
6 · 7 − 8|4 − 1| 6 · 7 − 8|3| 6·7−8·3 2. t + 11, or 11 + t
4 − 20 −16 8
= =− 4. d − 0.203
42 − 24 18 9
53 − 32 + 12 · 5 125 − 9 + 12 · 5 6. 18 + z, or z + 18
108. = =
−32 ÷ (−16) ÷ (−4) 2 ÷ (−4) 8. d + c, or c + d
125 − 9 + 60 176
= = −352 c
−
1
−
1 10. c ÷ h, or
2 2 h
Copyright
c 2017 Pearson Education, Inc.
Review of Basic Algebra
46. True, since −7 = −7 is true.
Exercise Set R.1 1
48. True, since −11 > −13 is true.
2
RC2. The correct answer is (b). See page R-2 in the text.
50. x < −1
RC4. The correct answer is (f). See page R-3 in the text.
⫺1 0
RC6. The correct answer is (d). See page R-4 in the text.
√ 52. x ≥ −1
2. 0, 1, 12, 25
⫺1 0
√ 12
4. −6, 0, 1, −4, 12, 25, −
3 54. x < 0
√
6. 3, 0.131331333133331 . . .
0
8. 12
56. x ≤ 0
11 7
10. −3.43, −11, 12, 0, ,−
34 13
0
12. All of them
58. 3
14. {s, o, l, v, e}
60. 16
16. {1, 3, 5, 7, 9, 11}
62. 127
18. {−3, −2, −1}
13
64.
20. {x|x is an integer greater than 3 and less than 11} 8
66. 16.4
22. {x|x is a rational number or x is an irrational number}, or
{x|x is a real number} 68. 465
24. {x|x is a real number and x ≤ 21}, or {x|x ≤ 21} 0
70. = |0| = 0
−15
26. 18 > 0
72. | − 5| ≥ | − 2|
28. 7 > −7
74. | − 8| ≤ |8|, or | − 8| ≥ |8| since | − 8| = |8|
30. 0 > −11
32. −6 < −3 Exercise Set R.2
34. −7 > −10
RC2. The opposite of a negative number is positive.
1 11
36. −13 <
5 250 RC4. The absolute value of a negative number is positive.
38. −13.99 < −8.45 RC6. The sum of 0 and a negative number is negative.
14 27
40. − = −0.933 . . . and − = −0.509433962 . . . , so RC8. The quotient of a negative number and a positive num-
15 53
14 27 ber is negative.
− <− .
15 53 2. −25
42. 7 > x
4. 2
2
44. t ≤ 10 6. −12
3
Copyright
c 2017 Pearson Education, Inc.
,2 Chapter R: Review of Basic Algebra
8. −3 55
68. −
12
10. −10
70. 3
12. −32
72. 432
14. 1.9
74. 71.04
16. −11.19 65
76.
2 14
18. −
7 64
78. −
5 125
20. −
4 80. −8
3 1 6 1 5 82. −9
22. − + = − + = −
4 8 8 8 8
5 7 20 21 41 84. 8
24. − + − =− + − =−
6 8 24 24 24 86. 0.7
26. 9
88. Not defined
28. 0
90. 0
2
30.
3 92. Not defined
32. 2x 10
94.
9
34. −5
6
96. −
36. −20 5
1
38. 46 98. −
65
40. 0 1
100.
42. 5 0.8
This can also be expressed as follows:
44. 44 1 1 10 10 5
= · = = , or 1.25
0.8 0.8 10 8 4
46. −17.7
102. 8x
48. −34.8 3 6 3 7 7
104. ÷ − = · − =−
13 5 7 5 6 10
50. −
5 12 3 12 10
106. − ÷ − =− · − =8
7 5 7 20 13 5 10 5 3
52. − − − =− + =
8 2 8 8 8 108. −3
2 4 10 12 22
54. − − = − + − =− 110. 110
3 5 15 15 15
4 5 36 35 1 112. −4000
56. − − − =− + =−
7 9 63 63 63 5 1 5 2 5
114. ÷ − = · − =−
58. −40 8 2 8 1 4
60. −45 5 5 5 6 2
116. − ÷ − =− · − =
9 6 9 5 3
62. 35 3 5 3 8 24
118. − ÷ − =− · − =
64. −152 5 8 5 5 25
5 5 5 6 3
66. 42.7 120. − ÷ − =− · − =
8 6 8 5 4
Copyright
c 2017 Pearson Education, Inc.
, Exercise Set R.3 3
122. 7 5
24.
2
−1.9 −1.9 10 19
124. = · =− , or − 0.095 26. 1
20 20 10 200
−17.8 −17.8 10 178 89 28. 1
126. = · =− = − , or − 5.5625
3.2 3.2 10 32 16
−3
128. Not defined 1 1 1 125
30. = 3 = =1· = 125
5 1 1 1
3 8 7 10 1
130. ,− ; − , ; 1, −1; 0, does not exist; 6.4, , or 5 125
8 3 10 7 −6.4
1 1
− ; −a, −4
6.4 a 5 1 1 16 16
32. = 4 = =1· =
132. 26 2 5 625 625 625
√ 2 16
134. 3, π, 4.57557555755557 . . .
1
136. All of them 34.
x6
3 36. y 7
138. 5 >
8
1 1
140. 123 > −10 38. =−
(−4)3 64
142. Use the definition of subtraction. The number that
40. 9−2
3 3
can be added to 11.7 to obtain −7 is −7 − 11.7, or
3 7 9
4 4 42. n−5
−7 − 11 = −19 , or −19.45.
4 10 20 44. (−8)−6
46. [6 − 4(8 − 5)] = [6 − 4 · 3]
Exercise Set R.3
= [6 − 12]
= −6
RC2. True; see page R-23 in the text.
48. 10[7 − 4(8 − 5)] = 10[7 − 4 · 3] = 10[7 − 12] =
RC4. False; see page R-22 in the text.
10[−5] = −50
RC6. False; see page R-20 in the text.
50. [9(7 − 4) + 19] − [25 − (7 + 3)] =
RC8. True; see page R-22 in the text. [9 · 3 + 19] − [25 − 10] = [27 + 19] − 15 =
2. 6 3 46 − 15 = 31
1 1
4. x4 52. [48 ÷ (−3)] ÷ − = −16 ÷ − = 64
4 4
6. t5
54. 30 · 10 − 18 · 25 = 300 − 450
8. (3.8) 5
= −150
3
4 56. (9 − 12)2 = (−3)2 = 9
10. −
5 92 − 122 = 81 − 144 = −63
12. 9 · 9 · 9 = 729 58. 7 · 8 − 32 − 23 = 7 · 8 − 9 − 8 = 56 − 9 − 8 = 39
14. (−7) · (−7) = 49 60. 43 + 20 · 10 + 72 − 23 = 64 + 20 · 10 + 49 − 23 =
16. (0.1)(0.1)(0.1)(0.1)(0.1)(0.1) = 0.000001 64 + 200 + 49 − 23 = 290
18. (−3)(−3)(−3)(−3) = 81 62. (9 · 8 + 3 · 3)2 = (72 + 9)2 = (81)2 = 6561
2 2 2 2 16 64. 5000 · (4 + 1.16)2 = 5000 · (5.16)2 =
20. · · · =
3 3 3 3 81 5000 · (26.6256) = 133, 128
√
22. 6
Copyright
c 2017 Pearson Education, Inc.
, 4 Chapter R: Review of Basic Algebra
66. (43 · 6 − 14 · 7)3 + (33 · 34)2 = 110. 2.3
(258 − 98) + (1122) = (160) + (1122) =
3 2 3 2
112. 900
4, 096, 000 + 1, 258, 884 = 5, 354, 884
114. −79
68. 18 − (2 · 3 − 9) = 18 − (6 − 9) = 18 − (−3) = 21
116. −79
70. (18 − 2)(3 − 9) = 16(−6) = −96
118. 23
72. [(−32) ÷ (−2)] ÷ (−2) = 16 ÷ (−2) = −8
2 15 2 · 15
74. 30 · 20 − 15 · 24 = 600 − 360 = 240 120. − − =
3 16 3 · 16
76. 16 ÷ (19 − 15)2 − 7 = 16 ÷ (4)2 − 7 = 2·3·5
=
3·2·8
16 ÷ 16 − 7 = 1 − 7 = −6
/·3
2 /·5
=
78. 53 + 20 · 40 + 82 − 29 = 125 + 20 · 40 + 64 − 29 = /·2
3 /·8
5
125 + 800 + 64 − 29 = 960 =
8
80. 4000 · (3 + 1.14)2 = 4000 · (4.14)2 = 122. 2(61 · 6−1 − 6−1 · 60 )
4000 · (17.1396) = 66, 558.4
1 1
= 2 6· − ·1
6 6
8(7 − 3) 8·4 32
82. = = =8 1
4 4 4 = 2 1−
6
43 64 5 10
84. = =8 = 2· =
8 8 6 6
86. 53 − 72 = 125 − 49 = 76 5
=
3
88. 10(−5) + 1(−1) = −50 − 1 = −51
124. 12345679 · 9 = 111, 111, 111
90. 14 − 2(−6) + 7 = 14 + 12 + 7 = 33 12345679 · 18 = 222, 222, 222
92. −32 − 8 ÷ 4 − (−2) = −32 − 2 − (−2) = 12345679 · 27 = 333, 333, 333
−32 − 2 + 2 = −32 According to this pattern, we have 12345679 · 36 =
444, 444, 444.
94. (3 − 8)2 = (−5)2 = 25 √
126. (π) 2 ≈ 5.047497267
96. 28 − 103 = 28 − 1000 √
( 2)π ≈ 2.970686424
= −972 √ √
Thus, (π) 2
is larger than ( 2)π .
98. 2 × 103 − 5000 = 2 × 1000 − 5000 =
2000 − 5000 = −3000
Exercise Set R.4
100. 6[9 − (3 − 4)] = 6[9 − (−1)] = 6[9 + 1] =
6[10] = 60 RC2. (c)
102. 1000 ÷ (−100) ÷ 10 = −10 ÷ 10 = −1 RC4. (e)
20 − 62 20 − 36 −16 8 RC6. (g)
104. = = =−
92 + 32 81 + 9 90 45
RC8. (f)
4|6 − 7| − 5 · 4 4| − 1| − 5 · 4 4·1−5·4
106. = = =
6 · 7 − 8|4 − 1| 6 · 7 − 8|3| 6·7−8·3 2. t + 11, or 11 + t
4 − 20 −16 8
= =− 4. d − 0.203
42 − 24 18 9
53 − 32 + 12 · 5 125 − 9 + 12 · 5 6. 18 + z, or z + 18
108. = =
−32 ÷ (−16) ÷ (−4) 2 ÷ (−4) 8. d + c, or c + d
125 − 9 + 60 176
= = −352 c
−
1
−
1 10. c ÷ h, or
2 2 h
Copyright
c 2017 Pearson Education, Inc.
