Solutions for Algebra for College Students 9th Us Edition by Lial by 6alsm

Chapter R Review of the Real Number System

R.1 Fractions, Decimals, and Percents

Classroom Examples, Now Try Exercises

1. (a) 12343433 1 20545455

(b) 8811 4868616

N1. (a) 30565655 1 42767677

(b) 101011 70710717

2. The fraction bar represents division. Divide the numerator of the improper fraction by the denominator. 3 1037 30 7

Thus, 377 3. 1010 

N2. The fraction bar represents division. Divide the numerator of the improper fraction by the denominator. 18 592 5 42 40 2

Thus, 922 18. 55 

3. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction. 5315  and 15419 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 419 3. 55 

N3. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction. 31133  and 33235 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 235 11. 33 

4. To multiply two fractions, multiply their numerators and then multiply their denominators. Then simplify and write the answer in lowest terms.

N4. To multiply two fractions, multiply their numerators and then multiply their denominators. Then simplify and write the answer in lowest terms.

5. (a) To divide fractions, multiply by the reciprocal of the divisor.

(b) Change both mixed numbers to improper fractions. Then multiply by the reciprocal of the second fraction.

N5. (a) To divide fractions, multiply by the reciprocal of the divisor.

(b) To divide fractions, multiply by the reciprocal of the divisor.

6. (a) To find the sum of two fractions having the same denominator, add the numerators and keep the same denominator.

Write 25 90 in lowest terms. 25555 9018518

(b) Since 30235  and 45335,  the least common denominator must have one factor of 2 (from 30), two factors of 3 (from 45), and one factor of 5 (from either 30 or 45), so it is 233590.

Write each fraction with a denominator of 90.

Now add. 7221421425 304590909090

(c) Since 1025  and 422,  the least common denominator is 22520.  Write each fraction with a denominator of 20. 3326 1010220

and 1155 45 420

Now subtract. 31651 102020420 

(d) Write each mixed number as an improper fraction. 82 31273 31 82

The least common denominator is 8. Write each fraction with a denominator of 8. 27 8 remains unchanged, and 33412 2248 

Now subtract. 32712271215 28 , 888 8 27

or 1 7 8

N6. (a) To find the sum of two fractions having the same denominator, add the numerators and keep the same denominator. 1313 888 4 8 14 24 1 2

(b) Since 12223  and 8222,  the least common denominator must have three factors of 2 (from 8) and one factor of 3 (from 12), so it is 222324.  Write each fraction with a denominator of 24. 55210 1212224  and 339 832 3 84  Now add. 5310910919 12824242424 

(c) Since 1111  and 933,  the least common denominator is 331199.  Write each fraction with a denominator of 99. 55945 1111999

and 221122 991199 

Now subtract. 52452223 119999999

(d) Write each mixed number as an improper fraction.

151317 42 3636

The least common denominator is 6. Write each fraction with a denominator of 6. 17 6 remains unchanged, and 1313226 3326  Now subtract. 1317261726179 366666

Now reduce. 9333 622 , 3

 or 1 1 2

7. (a) 15 0.15 100  (b) 9 0.009 1000  (c) 525 2.52 1010 

N7. (a) 8 0.8 10  (b) 431 0.431 1000  (c) 58258 2.582 100100

N8. (a) 68.900 42.720 8.973 120.593  (b) 351.800 2.706 349.094

9. (a) 30.21decimal place 0.0523decimal places 604 1510134 1.57044decimal places 

(b) 0.062decimal places 0.122decimal places 12 6224

0.00724decimal places 

(c) To change the divisor 0.37 into a whole number, move each decimal point two places to the right. Move the decimal point straight up and divide as with whole numbers. 14.8 37547.6

5.4760.3714.8. 

(d) To change the divisor 3.1 into a whole number, move each decimal point one place to the right. Move the decimal point straight up and divide as with whole numbers. 1.21 3137.60

We carried out the division to 2 decimal places so that we could round to 1 decimal place. Therefore, 3.763.11.2. 

N9. (a) 9.322decimal places 1.41decimal place 3728 932213

13.0483decimal places  

(b) 0.61decimal place 0.0043decimal places 24134

0.00244decimal places 

(c) To change the divisor 1.3 into a whole number, move each decimal point one place to the right. Move the decimal point straight up and divide as with whole numbers. 5.641 1373.340 65 83 78 54 52 20 13 7

We carried out the division to 3 decimal places so that we could round to 2 decimal places. Therefore, 7.3341.35.64. 

10. (a) Move the decimal point three places to the right.

19.5100019,500 

(b) Move the decimal point one place to the left.

960.11096.01 

N10. (a) Move the decimal point one place to the right.

294.72102947.2 

(b) Move the decimal point two places to the left. Insert a 0 in front of the 4 to do this.

4.7931000.04793 

11. (a) Divide 3 by 50. Add a decimal point and as many 0s as necessary. 0.06

503.00 300 0

Therefore, 3 0.06. 50 

(b) Divide 11 by 1. Add a decimal point and as many 0s as necessary.

0.090909...

111.000000...

Note that the pattern repeats. Therefore, 1 0.09, 11  or about 0.091.

N11. (a) Divide 20 by 17. Add a decimal point and as many 0s as necessary. 0.85 2017.00 160 100 100 0

Therefore, 17 0.85. 20 

(b) Divide 2 by 9. Add a decimal point and as many 0s as necessary. 0.222... 92.000... 18 20 18 20 18 2

Note that the pattern repeats. Therefore, 2 0.2, 9  or 0.222.

12. (a) 85%0.85 

(b) 110%1.10,  or 1.1

(d) 0.16516.5% 

N12. (a) 52%0.52 

(b) 2%02%0.02 

(d) 3.53.50350% 

13. (a) 65 65% 100  In lowest terms, 6513513 10020520 

(b) 1.51.510153 1.5% 100100101000200

N13. (a) 20 20% 100  In lowest terms, 201201 1005205  (b) 160 160% 100  In lowest terms, 16082083 , or 1 10052055

14. (a) 33 100% 5050 3100 % 501 3502 % 50 6%

N14. (a) 66 100% 2525 6100 % 251 6254 % 25 24%

(b) 77 100% 99 7100 % 91 700 % 9 7 77%, or 77.7% 9

Exercises

1. True; the number above the fraction bar is called the numerator and the number below the fraction bar is called the denominator.

2. True; 5 divides the 31 six times with a remainder of one, so 311 6. 55 

3. False; this is an improper fraction. Its value is 1.

4. False; the reciprocal of 6 3 2  is 21 . 63 

5. 16282 24383

 Therefore, C is correct.

6. Simplify each fraction to find which are equal to 5 9 . 355 3 15 2979

655 6 30 5949

22020 237 40 7437

1155 11 55 999 9

Therefore, C is correct.

7. 8181811 1 16282822

8. 141411 1 12343433 4

9. 15353555 1 18363666

10. 164444 2 44 1 545 05 45

11. 903303303 1 15053053055

12. 1005 1407 5205205 1 7207207

13. 1811811811 1 9051851855

14. 1611611611 1 64416444 16

15. 14462462466 1 12052452455

16. 132121112111212 1 7117117 777

17. 1 712 7 5 Therefore, 125 1. 77 

18. 1 916 9 7 Therefore, 167 1.

20. 5 1367 65 2 Therefore, 672 5. 1313 

21. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction.

5210  and 10313 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 313 2. 55 

22. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction.

7535  and 35641 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 641 5. 77 

23. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction.

31236  and 36238 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 238 12. 33 

24. Multiply the denominator of the fraction by the natural number and then add the numerator to obtain the numerator of the improper fraction.

51050  and 50151 

The denominator of the improper fraction is the same as the denominator in the mixed number.

Thus, 151 10. 55  25. 464624 575735

29. 1121121266 10510525525

30. 1101101255 8724728 87

31. 158158 425425 3542 455 32 5 61 , or1 55

32.

33. 3213 21 717 213 17 373 17 33 9 1

35. Change both mixed numbers to improper fractions. 12135 31 4343

36. Change both mixed numbers to improper fractions. 23 33 3 88 21 55 88 5 644 , or

37. To divide fractions, multiply by the reciprocal of the divisor.

38. To divide fractions, multiply by the reciprocal of the divisor.

39. To divide fractions, multiply by the reciprocal of the divisor.

40. To divide fractions, multiply by the reciprocal of the divisor.

41. To divide fractions, multiply by the reciprocal of the divisor.

45. To divide fractions, multiply by the reciprocal of the divisor.

42. To divide fractions, multiply by the reciprocal of the divisor.

46. To divide fractions, multiply by the reciprocal of the divisor.

43. To divide fractions, multiply by the reciprocal of the divisor.

47. Change the first number to an improper fraction, and then multiply by the reciprocal of the divisor.

44.

48. Change the first number to an improper fraction, and then multiply by the reciprocal of the divisor.

49. Change both mixed numbers to improper fractions, and then multiply by the reciprocal of the divisor.

50. Change both mixed numbers to improper fractions, and then multiply by the reciprocal of the divisor.

53. 747411 15151515

54. 25257 9999 

55. 7171 121212 8 12 24 34 2 3

56. 5351 16161 38 6162 

57. Since 933,  and 3 is prime, the LCD (least common denominator) is 339.  1133 3339 

51. Change both mixed numbers to improper fractions, and then multiply by the reciprocal of the divisor.

52. Change both mixed numbers to improper fractions, and then multiply by the reciprocal of the divisor.

Now add the two fractions with the same denominator. 51538 93999 

58. To add 4 15 and 1 , 5 first find the LCD. Since 1535  and 5 is prime, the LCD is 15. 41413 553 43 1515 1515 43 7 15 15

    

59. Since 8222  and 623,  the LCD is 22224. 3  3339 88324

 and 55420 66424 

Now add fractions with the same denominator. 3592029 8624244 , 2  or 5 1 24

60. Since 623  and 933,  the LCD is 23318.  553 3 15 6618

 and 2 22 992 4 18 

Now add fractions with the same denominator. 515 , 691818 2419 18

 or 1 1 18

61. Since 933  and 1644,  the LCD is 3344144.  1680 991 55 6144

 and 33927 16169144 

Now add fractions with the same denominator. 538027 916144144 107 144



62. Since 422  and 2555,  the LCD is 2255100.  332575 4425100

 and 66424 25254100 

Now add fractions with the same denominator. 36752499 425100010010 

63. 1124125 33 88888 11819 22 44444 11259 32 8484 



Since 8222  and 422,  the LCD is 222  or 8. 112592 32 84842 2518 88 433 , or 5 88

64. 212214 44 1112113 22 2 33333 6 6666  

Since 623,  the LCD is 6. 3 6 2114213 4 6 2 6326 2813 415 , 66 or 6 

65. 72725 9999 

66. 8383 11111 5 111

67. 133133 151515 10 15 252 353

 

68. 113113 121212 12 24 4 8 2 33

69. Since 1243  (12 is a multiple of 3), the LCD is 12. 144 3412

Now subtract fractions with the same denominator. 71743131 123121212434

70. Since 632  (6 is a multiple of 2), the LCD is 6. 1 2 3 3 36

Now subtract fractions with the same denominator. 5132 5 6666 121 23 23

71. Since 12223  and 933,  the LCD is 22336. 3

77321 1212336  and 144 9436

Now subtract fractions with the same denominator. 7121417 129363636 

72. The LCD of 12 11111314 and 16 is 48. 1612163124 334 4848 29 48



73. 3316319 44 44444 22527 11 55555



Since 422,  and 5 is prime, the LCD is 22520.  3219574 41 454554 9528 2020 677 , or 3 2020

74. Change both numbers to improper fractions then add, using 45 as the common denominator. 4419 31 199135 5995 17165 45 13 5959 45 106 4 16 , o 4 r 55 2  



75. 74 88 12 4 22722 99999 22214 33333 4



Since 933,  and 3 is prime, the LCD is 339.  143 84 33 2274 939 7442 99 35 ,or 3 9 2 9

76. 5584589 1212121212 5524529 66 7 6 44 66 7   Since 12223  and 623,  the LCD is 22312.  5589 12612 8 292 74 62 58 12 7 ,or 2 12 9 12 31 12

77. Observe that there are 24 dots in the entire figure, 6 dots in the triangle, 12 dots in the rectangle, and 2 dots in the overlapping region.

(a) 121 242  of all the dots are in the rectangle.

(b) 61 244  of all the dots are in the triangle.

(d) 21 126  of the dots in the rectangle are in the overlapping region.

78. (a) 12 is 1 3 of 36, so Maureen got a hit in exactly 1 3 of her at-bats.

(b) 5 is a little less than 1 2 of 11, so Chase got a hit in just less than 1 2 of his at-bats.

(d) 8 is 1 2 of 16, and 10 is 1 2 of 20, so Joe and Greg each got hits 1 2 of the time they were at bat.

79. 367.9412

(a) Tens: 6

(b) Tenths: 9

(d) Ones: 7

(e) Hundredths: 4

80. Answers will vary. One example is 5243.0164.

81. 46.249

(a) 46.25

(b) 46.2 (c) 46 (d) 50

82. (a) 0.889 (b) 0.444 (c) 0.976 (d) 0.865

83. 4 0.4 10 

0.6

8053805 3.8053 10001000

1665166

106. 34.042decimal places 0.562decimal places 20424 17020224 19.06244decimal places

107. 22.412decimal places 330decimal places 6723 6723202 739.532decimal places

108. 55.762decimal places 720decimal places 11152 39032202 4014.722decimal places

109. 0.21decimal place 0.032decimal places 6123 0.0063decimal places

110. 0.072decimal places 0.0043decimal places 28235 0.000285decimal places

113. To change the divisor 11.6 into a whole number, move each decimal point one place to the right. Move the decimal point straight up and divide as with whole numbers. 2.8 116324.8 232 092 0 000 8 928 0 Therefore, 32.4811.62.8  .

114. To change the divisor 17.4 into a whole number, move each decimal point one place to the right. Move the decimal point straight up and divide as with whole numbers.

4.9 174852.6 696 156 000 6 1566 0 Therefore, 85.2617.44.9 

115. To change the divisor 9.74 into a whole number, move each decimal point two places to the right. Move the decimal point straight up and divide as with whole numbers. 2.05 9741996.70 1948 4870 4870 0 Therefore, 19.9679.742.05. 

116. To change the divisor 5.27 into a whole number, move each decimal point two places to the right. Move the decimal point straight up and divide as with whole numbers. 8.44 5274447.88 4216 2318 2108 2108 2108 0 Therefore, 44.47885.278.44. 

117. Move the decimal point one place to the right. 123.26101232.6

118. Move the decimal point one place to the right.

785.91107859.1 

119. Move the decimal point two places to the right. 57.1161005711.6 

120. Move the decimal point two places to the right.

82.0531008205.3 

121. Move the decimal point three places to the right.

0.094100094 

122. Move the decimal point three places to the right.

0.025100025 

123. Move the decimal point one place to the left.

1.62100.162 

124. Move the decimal point one place to the left.

8.04100.804 

125. Move the decimal point two places to the left. 124.031001.2403 

126. Move the decimal point two places to the left. 490.351004.9035

127. Move the decimal point three places to the left. 23.291000023.2910000.02329 

128. Move the decimal point three places to the left. 59.81000059.810000.0598 

129. Convert from a decimal to a percent. 0.010.01100%1%

130. Convert from a percent to a decimal.

131. Convert from a percent to a fraction. 5 5%

In lowest terms, 5151 10020520

in Lowest Terms

132. Convert to a decimal first. Divide 1 by 10. Move the decimal point one place to the left. 1100.1

Convert the decimal to a percent. 0.10.1100%10%

in Lowest Terms (or Whole Number)

133. Convert the decimal to a percent. 0.1250.125100%12.5%

in Lowest Terms

134. Convert the percent to a decimal first. 20%0.20,  or 0.2 Convert from a percent to a fraction. 20 20% 100  In lowest terms, 201201 1005205

in Lowest Terms (or Whole Number) Decimal

135. Convert to a decimal first. Divide 1 by 4. Add a decimal point and as many 0s as necessary.

137. Convert the percent to a decimal first. 50%0.50,  or 0.5

Convert from a percent to a fraction.

Convert the decimal to a percent.

136. Convert to a decimal first. Divide 1 by 3. Add a decimal point and as many 0s as necessary.

138. Divide 2 by 3. Add a decimal point and as many 0s as necessary.

Note that the pattern repeats. Therefore, 2 0.6.

Note that the pattern repeats. Therefore, 1 0.3.

Convert the decimal to a percent.

139. Convert the decimal to a percent first. 0.750.75100%75%

Convert from a percent to a fraction.

140. Convert the decimal to a percent. 1.01.0100%100%

143. Divide 9 by 4. Add a decimal point and as many 0s as necessary.

141. Divide 21 by 5. Add a decimal point and as many 0s as necessary.

144. Divide 15 by 4. Add a decimal point and as many 0s as necessary.

145. Divide 3 by 8. Add a decimal point and as many 0s as necessary.

142. Divide 9 by 5. Add a decimal point and as many 0s as necessary.

146. Divide 7 by 8. Add a decimal point and as many 0s as necessary.

147. Divide 5 by 9. Add a decimal point and as many 0s as necessary. 0.555... 95.000... 45 50 45 50 45 5

Note that the pattern repeats. Therefore, 5 0.5, 9  or about 0.556.

148. Divide 8 by 9. Add a decimal point and as many 0s as necessary. 0.888... 98.000...

150. Divide 5 by 6. Add a decimal point and as many 0s as necessary. 0.833... 65.000... 48 20 18 20 18 2

Note that the pattern repeats. Therefore, 5 0.83, 6  or about 0.833.

151. 54%0.54 

152. 39%0.39  153. 7%07%0.07 

154. 4%04%0.04 

155. 117%1.17  156. 189%1.89 

157. 2.4%02.4%0.024  158. 3.1%03.1%0.031 

Note that the pattern repeats. Therefore, 8 0.8, 9  or about 0.889.

149. Divide 1 by 6. Add a decimal point and as many 0s as necessary. 0.166... 61.000... 6 40 36 40 36 4

Note that the pattern repeats. Therefore, 1 0.16, 6  or about 0.167.

159. 1 6%6.25%06.25%0.0625 4 

160. 1 5%5.5%05.5%0.055 2 

161. 0.7979%  162. 0.8383% 

163. 0.022%  164. 0.088% 

165. 0.0040.4% 

166. 0.0050.5%  167. 1.28128%  168. 2.35235% 

169. 0.4040%  170. 0.60.6060%  171. 51 51% 100 

172. 47 47% 100 

173. 15 15%

100  In lowest terms, 15353 10020520

174. 35 35%

100  In lowest terms, 35757 10020520

175. 2 2%

100  In lowest terms, 2121 10050250

176. 8 8%

100  In lowest terms, 8242 10025425

177. 140 140%

100  In lowest terms, 1407207 , 1005205  or 2 1 5

178. 180 180% 100  In lowest terms, 1809209 , 1005205  or 4 1 5

179. 7.57.51075 7.5% 100100101000

In lowest terms, 753253 1000402540

180. 2.52.51025 2.5% 100100101000

In lowest terms, 251251 1000402540

181. 4441004520 100%%%80% 55515

182. 33 100% 2525 3100 % 251 3425 % 25 12%

183. 77 100% 5050 7100 % 501 7250 % 50 14%

184. 99 100% 2020 9100 % 201 9520 % 20 45%

185. 22 100% 1111 2100 % 111 200 % 11 18.18%

444100400 100%%%44.4% 99919

9991009425 100%%%225% 44414

188. 8881008520 100%%%160% 55515

189. 1313 100% 66 13100 % 61 13250 % 23 216.6%

R.2 Basic Concepts from Algebra

Classroom Examples, Now Try Exercises

1.  | xx is a natural number greater than 12} {13, 14, 15, … } is the set of natural numbers greater than 12.

N1.  | pp is a natural number less than 6} {1, 2, 3, 4, 5} is the set of natural numbers less than 6.

2. One answer is  | xx is a whole number less than  6.

N2. One answer is  | xx is a natural number between 8 and  13.

3. (a) 7 is an integer, a rational number 7 since 7, 1

and a real number.

(b) 3.14, a terminating decimal, is a rational number and a real number.

N3. (a) The whole numbers in the given set are in the set {0, 5}.

(b) The rational numbers in the given set are in the set 1 2.4, 1, , 0, 0.3, 5. 2   

4. (a) The statement “Some integers are whole numbers” is true since 0 and the positive integers are whole numbers. The negative integers are not whole numbers.

(b) The statement “Every real number is irrational” is false. Some real numbers such as 4 , 9 0.3, and 16 are rational. Others such as 2 and  are irrational.

N4. (a) The statement “All integers are irrational numbers” is false. In fact, all integers are rational.

(b) The statement “Every whole number is an integer” is true since the integers include the whole numbers, the negatives of the whole numbers, and 0.

5. (a) |3|3 

(b) |3|(3)3 

|3|(3)3 

(d) |3|(3)3 

(e) Evaluate each absolute value, and then add.

|8||1|819 

(f) |81||7|7 

N5. (a) |7|(7)7 

(b) Evaluate the absolute value. Then find the additive inverse.

|7|(7)7 

|7|(7)7 

(d) Evaluate each absolute value, and then subtract.

|4||4|440 

(e) 4400 

6. Physicians assistants: 37.437.4 

Locomotive firers: 78.678.6

Since 78.637.4,  locomotive firers will show the greater change.

N6. Home health aides: 46.746.7 

Parking enforcement workers: 35.335.3

Locomotive firers: |78.6|78.6 

Since 35.3 is less than 46.7 and 78.6, parking enforcement workers will show the least change.

7. (a) 84 is false, since 8 is to the left of 4 on a number line, 8 is less than, not greater than, 4.

(b) 25 is true, since 2 is to the left of 5 on a number line.

N7. (a) 51 is false, since 5 is to the left of 1 on a number line, 5 is less than, not greater than, 1.

(b) 76 is true, since 7 is to the left of 6 on a number line.

8. (a) 23 is false, since 2 is to the right of 3 on a number line, it must be greater.

(b) 19 reads “1 is greater than or equal to 9”. This is true, since 1 is to the right of 9 on a number line.

(d) Since 3(4)12  and 2612,  the statement becomes 1212,  which is false.

N8. (a) Since 6 is to the left of 5 on a number line, 6 is less than, not greater than 5. The statement is false

(b) Since 10 is to the left of 2 on a number line, the statement is true.

(d) Since 10660  and 8(7)56,  the statement becomes 6056,  which is true.

Exercises

1. Yes, the choice of variables is not important. The sets have the same description so they represent the same set.

2. (a) The only integer between 6.75 and 7.75 is 7.

(b) There are an infinite number of rational numbers between 1 4 and 3 ; 4 one example is 1 2

(d) The opposite of any natural number is an integer that is not a whole number; one example is 1.

(e) There are an infinite number of irrational numbers between 4 and 9; one example is 5.

(f) One example is . Other answers are possible.

3. The set of natural numbers is {1, 2, 3, … }, so the set of natural numbers less than 6 is {1, 2, 3, 4, 5}.

4. The set is{1, 2, 3, 4, 5, 6, 7, 8}.

5. The set of integers is {,3,2,1,0,1,2,3,},  so the set of integers greater than 4 is {5, 6, 7, 8,… }.

6. The set is {9, 10, 11, 12, …}.

7. The set of integers is {,3,2,1,0,1,2,3,}, so the set of integers less than or equal to 4 is ,4 {,1,0,1,2,3}. 

8. The set is {,2,1,0,1,2}.

9. The set of even integers is {,2,0,2,4,},  so the set of even integers greater than 8 is {10, 12, 14, 16, …}.

10. The set is {,7,5,3,1}.

11. This is the set of numbers that lie a distance of 4 units from 0 on the number line. Thus, the set of numbers whose absolute value is 4 is {4,4}.

12. The set is {7,7}.

13.  | xx is an irrational number that is also rational} Irrational numbers cannot also be rational numbers. The set of irrational numbers that are also rational is .

14. The set is , the empty set.

15. {2, 4, 6, 8} can be described by  | xx is an even natural number less than or equal to  8.

16. {11, 12, 13, 14} can be described by  | xx is an integer between 10 and  15.

17. {4, 8, 12, 16, …} can be described by  | xx is a multiple of 4 greater than  0.

18. {,6,3,0,3,6,}  can be described by  | xx is an integer multiple of 3 .

19. Place dots for 4,2,0,3, and 5 on a number line.

20. Place dots for 3,1,0,4, and 6 on a number line.

21. Place dots for 61 1.2,0.25,0, 54  513 0.83,3.25,5.2, 64  and 11 5.5, 2  on a number line.

22. Place dots to indicate the points 2 0.6,0, 3  4129 0.8,2.4,4.5, 552  and 4.8 on a number line.

23. (a) The elements 8, 13, and 75 5 (or 15) are natural numbers.

(b) The elements 0, 8, 13, and 75 5 are whole numbers.

(d) The elements 6 9,0.7,0, ,4.6, 7 21 8,,13, 2 and 75 5 are rational numbers.

(e) The elements 6 and 7 are irrational numbers.

(f) All the elements are real numbers.

24. (a) The elements 5, 17, and 40 2 (or 20) are natural numbers.

(b) The elements 0, 5, 17, and 40 2 are whole numbers.

(d) The elements 313 42 8,0.6,0, ,5,, 17, and 40 2 are rational numbers.

(e) The elements 5,3, and  are irrational numbers.

(f) All the elements are real numbers.

25. The statement “Every integer is a whole number” is false. Some integers are whole numbers, but the negative integers are not whole numbers.

26. The statement “Every natural number is an integer” is true. The integers include the natural numbers, their negatives, and zero.

27. The statement “Every irrational number is an integer” is false. Irrational numbers have decimal representations that neither terminate nor repeat, so no irrational numbers are integers.

28. The statement “Every integer is a rational number” is true. Every integer q can be written as 1 q

29. The statement “Every natural number is a whole number” is true. Whole numbers consist of the natural numbers and zero.

30. The statement “Some rational numbers are irrational” is false. No rational numbers are irrational numbers.

31. The statement “Some rational numbers are whole numbers” is true. Every whole number is rational.

32. The statement “Some real numbers are integers” is true. Every integer is a real number.

33. The statement “The absolute value of any number is the same as the absolute value of its additive inverse” is true. The distance on a number line from 0 to a number is the same as the distance from 0 to its additive inverse.

34. The statement “The absolute value of any nonzero number is positive” is true. Every nonzero number is a positive distance from 0.

35. (a) (4)4 (Choice A)

(b) |4|4  (Choice A)

(Choice B)

(d) |(4)||4|4

(Choice B)

36. In general ||xa  (with 0 a  ), is true for two values of a; namely, a and .a In this case, ||4 x  is true for 4 x  and 4. x 

37. (a) The additive inverse of 6 is 6.

(b) 60,  so |6|6. 

38. (a) The additive inverse of 9 is 9.

(b) 90,  so |9|9. 

39. (a) The additive inverse of 12 is (12)12.

(b) 120, so 12(12)12. 

40. (a) The additive inverse of 14 is (14)14.

(b) 140, so 14(14)14. 

41. (a) The additive inverse of 6 5 is 6 5

(b) 6 0, 5  so 66 . 55 

42. (a) The additive inverse of 0.16 is 0.16.

(b) 0.160,  so 0.160.16. 

43. Use the definition of absolute value. 80, so 8(8)8.

44. 19(19)19 

45. 3 0, 2  so 33 . 22 

46. 33 44 

47. 5(5)5

48. 12(12)12

49. 2[(2)](2)2 

50. 6[(6)](6)6 

51. 4.5(4.5)4.5

52. 12.4(12.4)12.4

53. 23235 

54. 1614161430

55. 109(1)1

56. 126(6)6

57. 93936 

58. |10|71073 

59. 123123 33 0

60. 731073100 

61. (a) The greatest absolute value is 13.913.9.  Therefore, New Orleans had the greatest change in population. The population increased by 13.9%.

(b) The least absolute value is 3.03.0. Therefore, Toledo had the least change in population. The population decreased by 3.0%.

62. (a) The greatest absolute value is 35,23035,230. For China, imports exceeded exports by $35,230 million.

(b) The least absolute value is 230230.  For New Zealand, exports exceeded imports by $230 million.

63. Compare the depths of the bodies of water. The deepest, that is, the body of water whose depth has the greatest absolute value, is the Pacific Ocean  14,04014,040 followed by the Indian Ocean  12,80012,800, the Caribbean Sea  84488448, the South China Sea  48024802, and the Gulf of California  23752375.

64. Compare the heights of the mountains. The shortest, that is, the smallest value, is Point Success  14,16414,164  followed by Rainier  14,41014,410,  Matlalcueyetl

 14,63614,636,  Steele

 16,62416,624,  and Denali

 20,31020,310  .

65. True; the absolute value of the depth of the Pacific Ocean is 14,04014,040 which is greater than the absolute value of the depth of the Indian Ocean, 12,80012,800.

66. False; the absolute value of the depth of the Gulf of California is 23752375 which is

less than the absolute value of the depth of the Caribbean Sea, 84488448.

67. True; since 6 is to the left of 1 on a number line, 6 is less than 1.

68. True; since 4 is to the left of 2 on a number line, 4 is less than 2.

69. False; since 4 is to the left of 3 on a number line, 4 is less than 3, not greater.

70. False; since 3 is to the left of 1 on a number line, 3 is less than 1, not greater.

71. True; since 3 is to the right of 2 on a number line, 3 is greater than 2.

72. True; since 6 is to the right of 3 on a number line, 6 is greater than 3.

73. False; since 33,3 is not greater than 3.

74. False; since 55,5 is not less than 5.

75. The inequality 62  can also be written 26.  In each case, the inequality symbol points toward the smaller number so both inequalities are true.

76. 51  is equivalent to 15. 

77. 94 is equivalent to 49. 

78. 61 is equivalent to 16. 

79. 510 is equivalent to 105.

80. 712 is equivalent to 127.

81. “7 is greater than 1 ” can be written as 71. 

82. “4 is less than 10” can be written as 410.

83. “5 is greater than or equal to 5” can be written as 55. 

84. “6 is less than or equal to 6” can be written as 66.

85. “133 is less than or equal to 10” can be written as 13310. 

86. “514  is greater than or equal to 19” can be written as 51419. 

87. “50  is not equal to 0” can be written as 500. 

88. “67  is not equal to 13” can be written as 6713. 

89. 05False 

90. 110False

91. 77True 

92. 1010True 

93. ? 673 610True  

The last statement is true since 6 is to the left of 10 on a number line.

94. ? 741 75True  

95. ? 2546 1010True  

96. ? 8735 1515True  

97. ? 33 33True  

98. ? 44 44True  

99. ? 86 86False  

100. ? 104 104False  

101. 2016: IA13,608, OH9546, PA8212 In 2016, Iowa (IA), Ohio (OH), and Pennsylvania (PA) had production greater than 6000 million eggs.

102. OH: 95469708;FL: 23642463 For Ohio (OH) and Florida (FL), egg production was less in 2016 than in 2015.

103. 2016: TX5572,OH9546 xy  Since 55729546, xy is true.

104. 2015: IA12,471,PA7778 xy 

Two inequalities that compare the production in these two states are xy  and yx 

R.3 Operations on Real Numbers

Classroom Examples, Now Try Exercises

1. (a) 6(15)(615)21

(b) 1.1(1.2)(1.11.2)2.3

N1. (a) 4(9)(49)13

(b) 7.25(3.57)(7.253.57)

2. (a) Subtract the lesser absolute value from the greater absolute value (73), and take the sign of the larger.

3(7)(73)4 

(b) 37734 

(d) 3131232 8484288 321 888

N2. (a) Subtract the lesser absolute value from the greater absolute value (157), and take the sign of the larger.

157(157)8 

(b) 5121257 

(d) 5257293518 9797796363 351817 636363 

3. (a) 5125(12)7  (b) 12(5)12517  (c) 11.5(6.3)11.56.3 (11.56.3) 5.2 

(d) 32323324 43434334 9817 121212

N3. (a) 9159(15)6  (b) 4114(11)15  (c) 5.67(2.34)5.672.34 (5.672.34) 3.33

(d) 43434539 95959559 20277 454545

4. (a) 6(2)816281 4(8)1 121 13 

(b) 3[(7)15]6386 116 17 

N4. 4(27)124(9)12 4912 512 7

5. 1(8)1877  N5. 10(7)1071717 

6. (a) 7(2)14  The numbers have different signs, so the product is negative.

(b) 0.9(15)13.5

(d) 0(3)0(3)0

The multiplication property of 0 states that any product of any real number and 0 is 0.

N6. (a) 3(10)30

The numbers have the same sign, so the product is positive.

(b) 0.7(1.2)0.84

The numbers have different signs, so the product is negative.

(c)

(d) 14(0)1400

The multiplication property of 0 states that any product of any real number and 0 is 0.

Exercises

1. The sum of two positive numbers is a positive number. For example, 18624. 

2. The sum of two negative numbers is a negative number. For example, 7(21)28. 

3. The sum of a positive number and a negative number is negative if the negative number has the greater absolute value. For example, 1495.

4. The sum of a positive number and a negative number is positive if the positive number has the greater absolute value. For example, 15(2)13. 

5. The difference of two positive numbers is negative if the number with the greater absolute value is subtracted from the one with the lesser absolute value. For example, 5127. 

6. The difference of two negative numbers is negative if the number with lesser absolute value is subtracted from the one with greater absolute value. For example, 15(3)12. 

7. The product of two numbers with the same sign is positive. For example, (2)(8)16  and 2816. 

8. The product of two numbers with different signs is negative. For example, 5(15)75.

9. The quotient formed by any nonzero number divided by 0 is undefined, and the quotient formed by 0 divided by any nonzero number is 0. For example, 17 0 is undefined and 0 42 is equal to 0.

10. The sum of a positive number and a negative number is 0 if the numbers are additive inverses. For example, 4(4)0. 

11. 6(13)(613)19  12. 8(16)(816)24  13. 156(156)9  14. 179(179)8 

15. 13(4)1349 

16. 19(13)19136 

17. 172222175 

18. 121616124 

19. 7328919 34121212 

20. 541587 69181818 

21. The difference between 4.5 and 2.8 is 1.7. The number with the greater absolute value, 4.5 is positive, so the answer is positive. Thus, 2.84.51.7.

22. 3.86.26.23.8 2.4  

23. 494(9)5 

24. 373(7)4 

25. 656(5)(65)11 

26. 8178(17)(817)25 

27. 8(13)81321 

28. 12(22)122234 

29. 16(3)16313 

30. 21(6)21615 

31. 12.31(2.13)12.312.13 (12.312.13) 10.18 

32. 15.88(9.42)15.889.42 (15.889.42) 6.46

33. 9494274067 103103303030

34. 333362127 144144282828

35. 8614(14)14

36. 71522(22)22

37. 4955

38. 5611 

39. 24242(4)6 

40. 1613161316(13)3 

41. 759(75)92911 

42. 12141821816 

43. 6(2)8628880 

44. 7(4)11741111110 

45. 8(12)2681226 2026 186 12    

46. 3(14)6431464 1764 114 7    

47. 94(3)6(94)36 1336 106 4

   



48. 106(12)910(6)129 16129 49 5

  

49. 0.3840.623.620.62 3  

50. 2.9580.055.050.05 5  

51. 5211582 436121212 232 1212 217 ,or 124

 

 

52.

53.

57. 11572 11572 672 12 1

58. 6348 6348 948 58 3

59. 4 A  and 2. B  The distance between A and B is the absolute value of their difference.

4266

60. 4 A  and 5. C  The distance between A and C is the absolute value of their difference.

4599

61. 6 D  and 1 . 2 F  The distance between D and F is the absolute value of their difference.

1121 6 222 13 2 131 ,or6 22

62. 3 2 E  and 5. C  The distance between E and C is the absolute value of their difference.

3310 5 222 13 2 131 ,or6 22

63. It is true for multiplication (and division). It is false for addition and subtraction when the number to be subtracted has the lesser absolute value. A more precise statement is, "The product or quotient of two negative numbers is positive."

64. Because the reciprocal of a number is the quotient of 1 (a positive number) and the number, the reciprocal always has the same sign as the number.

65. The product of two numbers with the same sign is positive, so 8(5)40.

66. The product of two numbers with the same sign is positive, so 20(4)80.

67. The product of two numbers with different signs is negative, so 5(7)35.

68. The product of two numbers with different signs is negative, so 6(9)54.

69. 4(0)400

The multiplication property of 0 states that any product of any real number and 0 is 0.

70. 0(8)080 

The multiplication property of 0 states that any product of any real number and 0 is 0.

71. 11 22(0)00 

The multiplication property of 0 states that any product of any real number and 0 is 0.

72. 0(4.5)04.50 

The multiplication property of 0 states that any product of any real number and 0 is 0.

73. The product of two numbers with different signs is negative, so 33(16)4412. 44

74. The product of two numbers with different signs is negative, so 33(35)5721. 55

75. The product of two numbers with the same sign is positive, so 11 10252. 55

76. The product of two numbers with the same sign is positive, so 11 22(18)299. 

77. The product of two numbers with the same sign is positive, so 5125266 . 2252555

78. The product of two numbers with the same sign is positive, so 9219373 7367494

79. The product of two numbers with the same sign is positive, so 324338 1. 8989

80. The product of two numbers with the same sign is positive, so 2222211 1. 1141122

81. The product of two numbers with the same sign is positive, so

0.80.50.4.

82. The product of two numbers with the same sign is positive, so

0.50.60.3.

83. The product of two numbers with different signs is negative, so 0.06(0.4)0.024. 

84. The product of two numbers with different signs is negative, so 0.08(0.7)0.056. 

85. The quotient of two nonzero real numbers with different signs is negative, so 1411 14277. 222

86. The quotient of two nonzero real numbers with different signs is negative, so 3911 393133. 131313

87. The quotient of two nonzero real numbers with the same sign is positive, so 2411 24466. 444

88. The quotient of two nonzero real numbers with the same sign is positive, so 4511 45595. 999

89. The quotient of two nonzero real numbers with different signs is negative, so 10011 1004254. 252525

90. The quotient of two nonzero real numbers with different signs is negative, so 15011 1505305. 303030

91. 01 00 88

92. 01 00 1414

93. Division by 0 is undefined, so 5 0 is undefined.

94. Division by 0 is undefined, so 13 0 is undefined.

95. The quotient of two nonzero real numbers with the same sign is positive, so 101210525525 17517121726102

96. 2233225211510 23523332331169

97. 12 124123 13 4 133134 3 3439 13413

1742117 912363636

106. 11343935 1216484848

108. 114332053 159454545

109. 723214274422 3045109090909045

110. 8733270929 1562060606060

111. 8524522 251255345315

112. 973373721 20154535455100

113. 594533223 6105232555

114. 495223351 320123225434

163482

121.  8.623.7518.623.75132.351  122.  37.813.58237.813.58251.382  123. 2.5(0.8)(1.5)(2)(1.5)3  124. 1.6(0.5)(2.5)(0.8)(2.5) 2   125. 24.8464.14  126. 32.8484.105  127. 2496(0.52)4800  128. 1875(0.25)7500 

129. 1001001 100100100 1 0.01100 100 10,000  

130. 60606100 6060 6 0.061006 100 101001000 

131. 14.29.814.39 

132. 89.4121.32568.085 

133. To find the difference between these two temperatures, subtract the lowest temperature from the highest temperature.

100(80)10080 180  

The difference is 180°F.

134. 120(29)12029 149  

The difference is 149°F.

(b) 222.7375.0032.06179.79 

Her balance is $179.79.

139. (a) 4.9316.278.41(28.46)30.18 31.33 

The sum for 2014–2018 was 31.33%.

(b)  28.468.4136.87 

The difference between the percents from 2008 and 2007 was 36.87%.

The difference between the percents from 2009 and 2008 was 58.64%.

140. (a) 14222518577475 

The total is $475 thousand, which represents a loss of $475 thousand.

(b)  7718577185 262  

The difference was $262 thousand dollars.

135. 48.3535.9920.0028.5066.27



12.3620.0028.5066.27



7.6428.5066.27

 

36.1466.27

30.13 

His balance is $30.13.

The difference was $83 thousand dollars.

141. Year Difference (in billions)

2001 $1991$1863$128 

2006 $2407$2655$248 

2011 $2303$3603$1300 

136. 37.6025.9919.3425.0058.66





11.6119.3425.0058.66

7.7325.0058.66

32.7358.66

25.93

Her balance is $25.93.

137. 382.4525.1034.5045.0098.17 466.02

His balance is $466.02.

(a) To pay off the balance, his payment should be $466.02.

(b) 466.0230024.66190.68  His balance is $190.68.

138. 237.5947.2512.3920.00222.73  Her balance is $222.73.

(a) To pay off her balance, her payment should be $222.73.

2016 $3268$3853$585 

142. There was a surplus in the year 2001. There were defecits in the years 2006, 2011, and 2016. A positive difference between receipts and outlays indicates a surplus, while a negative difference indicates a deficit.

R.4 Exponents, Roots, and Order of

Operations

Classroom Examples, Now Try Exercises

1. (a) 4 22222 77777

(b) (10)(10)(10)(10)3 

N1. (a) (3)(3)(3)(3)3

(b) 2 (0.5)(0.5)(0.5) 

2. (a) 5 22222232 

(b) 2 (0.5)(0.5)(0.5)0.25 

(d) 4 (4)(4)(4)(4)(4)256

N2. (a) 3 5555125

(b) 4 2222216 55555625

(d) 3 (3)(3)(3)(3)27

3. (a) 4 33333 933 27381

(b) 4 (3)(3)(3)(3)(3)81

N3. (a) 2 77749 

(b) 2 (7)(7)(7)49

4. (a) 497  because 7 is positive and 2 749. 

(b) 12111 819  because 2 11121 , 981

so 12111 . 819 

(d) 49 is not a real number.

N4. (a) 12111  because 2 11121. 

(b) 10010 93  because 2 10100 . 39

(d) 121 is not a real number.

5. (a) 5924458Multiply. 53Add.  

(b) 41242432Divide. 46Multiply. 2Subtract.

N5. 1534215122Multiply. 32Subtract. 5Add.

6. (a) Add and subtract inside parentheses.

2 2 (42)4(83) 645 

Evaluate the power. 6165

105Subtract from left to right. 15

(b)  52 61218 43 512218 6 4131 6036 6Multiply. 43



 Subtract left from right. 61512 21

N6. (a) Work inside the absolute value bars.

2 510537  Subtract inside the absolute value bars.

2 2 51054 51054Take absolute value. 251054Evaluate the power. 2524Divide. 234Add. 27Subtract.

(b)  25 6916 38 29516 6 3181 1880 6Multiply 38 6610Subtract from left 10to right.

N8. (a)   323724 218 218 29 yx

(b) 22(4)36 33(4)(7) 166 3(4)(7) 166 12(7) 166 84 105 8442 xz xy

Exercises

1. 6 7(777777) 117,649, 6 (7)(7)(7)(7)(7)(7)(7) 117,649

N7. 2 2 3643 28328 649 Evaluate powers and roots. 48328 636 Multiply. 42428 30 Add and subtract. 2828 30 Undefined. 0

8. (a) 5 5(12)(3)64 1121 5(12)3(8) 121 6024 121 8484 1313 xzy x

(b) 2323 242(3) 162(27) 16(54) 38 wz

Thus, the original statement, 66 7(7), is false. Note that the statement 66 7(7) is true

2. 5(5)77 is true. (5)577  since (5)7 has an odd number of negative factors.

3. The statement “25 is a positive number” is true. The symbol always gives a positive square root provided the radicand is positive.

4. 3583(58)  is true. The order of operations says that multiplication should be done before addition.

5. The statement “(6)7 is a negative number” is true. 7 (6) gives an odd number of negative factors, so the product is negative.

6. The statement “(6)8 is a positive number” is true. (6)8 gives an even number of negative factors, so the product is positive.

7. The statement “The product of 10 positive factors and 10 negative factors is positive” is true. The product of an even number of negative factors is positive.

8. The statement “The product of 5 positive factors and 5 negative factors is positive” is false. The product of 5 positive factors is positive and the product of 5 negative factors is negative, so the product of these two products is negative.

9. The statement “In the exponential expression 5 2, the base is 2 ” is false. The base is 2, not 2. If the problem were written 5 (2), then 2 would be the base.

10. The statement “ a is positive for all positive numbers a” is true.

11. 4 1010101010

15. (9)(9)(9)(9)3

16. 4 (4)(4)(4)(4)(4)

18. (0.1)(0.1)(0.1)(0.1)(0.1)(0.1)(0.1)6 

19. 7 zzzzzzzz

20. 5 aaaaaa

21. (a) 2 864 

2 8(88)64

(d) 2 (8)(64)64

22. (a) 3 464

(b) 3 4(444)64

3 (4)(4)(4)(4)64

(d) 3 (4)(64)64

23. 2 44416

24. 2 66636

25. 3 (0.3)(0.3)(0.3)(0.3)0.027  26. 3 (0.1)(0.1)(0.1)(0.1)0.001 

3 11111 5555125

11111 66666 11 66661296

3 7777343 101010101000

31. 3 (5)(5)(5)(5)125 

5 (3)(3)(3)(3)(3)(3)243  33. 8 (2)(2)(2)(2)(2)(2)(2)(2)(2) 256 

34. 6 (3)(3)(3)(3)(3)(3)(3)729

35. 6 3(333333)729 

36. 6 4(444444)4096

37. 4 8(8888)4096

38. 3 10(101010)1000

39. (a) 2 7 is negative because the negative sign is not part of the base.

(b) 2 (7) is positive because the negative sign is part of the base, and the base is squared, so there are an even number of factors.

(d) (7)3 is negative because the negative sign is part of the base, and the base is cubed, so there are an odd number of factors.

(e) 4 7 is negative because the negative sign is not part of the base.

(f) 4 (7) is positive because the negative sign is part of the base, and the base is raised to the fourth power, so there are an even number of factors.

40. (a) 14412,  which is choice B.

(b) 144 is not a real number, which is choice C.

41. 819  since 9 is positive and 2 981. 

42. 648  since 8 is positive and 2 864. 

43. 16913  since 13 is positive and 2 13169. 

44. 22515  since 15 is positive and 2 15225. 

45.  400400(20)20 

46.  900900(30)30 

47. 10010 12111  since 10 11 is positive and 2 10100 11121 

48. 22515 16913  since 15 13 is positive and 2 15225 13169

49. 2 0.49(0.7)(0.7)0.7

50. 2 0.64(0.8)(0.8)0.8

51. There is no real number whose square is negative, so 36 is not a real number.

52. There is no real number whose square is negative, so 121 is not a real number.

53. If a is a positive number, then a is a negative number. Therefore, a is not a real number, and aisnotarealnumber.

54. If a is a positive number, then a is also a positive number, and a is a negative number.

55. (a) Since 91539514,  the grandson's answer was correct.

(b) The reasoning was incorrect. The operation of division must be done first, and then the addition follows. The grandson's “Order of Process rule” is not correct. It just happens coincidentally in this problem that he obtained the correct answer the wrong way.

56. 77777771497 8497 577 50 

57. 12341212Multiply. 24Add.

58. 15521510Multiply. 25Add.

59. 6312418124Multiply. 183Divide. 15Subtract. 

60. 94823682Multiply. 364Divide. 32Subtract.

61. 10302310153Divide. 1045Multiply. 55Add.

  



62. 1224321282Divide. 1216Multiply. 28Add.

 

63. 2 3(5)(2)(8) 3(25)(2)(8)Evaluatepower. 7516Multiply. 91Subtract.







64. 2 9(2)(3)(2) 9(4)(3)(2)Evaluate power. 366Multiply. 42Subtract.

  

65. 573(2)3 573(8)Evaluate power. 5218Multiply, change sign. 168Subtract. 8Add. 







66. 435(3)3



435(27)Evaluate power. 41527Multiply. 1927Subtract. 8Add.

67.  736(2)(3)

7(6)(2)(3)Evaluate root. 426Multiply. 48Subtract.

68.  864(3)(7)



8(8)(3)(7)Evaluate root. 6421Multiply. 85Subtract.

69. Simplify within absolute value bars. 645243 61243

6(1)243Take absolute value. 68Multiply and divide. 2Subtract. 

70. Simplify within absolute value bars. 42482 4282

4(2)82Take absolute value. 816Multiply. 8Add.

71. Simplify within absolute value bars.

2 2 2 65(8)3 11(8)3

11(8)3Take absolute value. 11(8)9Evaluate power. 889Multiply. 97Subtract.

72. Simplify within absolute value bars. 2 2 2 23(9)4 5(9)4



5(9)4Take absolute value. 5(9)16Evaluate power. 4516Multiply. 61Subtract.

73. Simplify within parentheses. 2 2 2 1845(37) 1845(4) 18454Simplify parentheses. 181654Evaluate power. 254Subtract. 74Add. 11Add.

  



74. Simplify within parentheses. 2 2 2 1029(18) 1029(7) 10297Simplify parentheses. 10497Evaluate power. 697Subtract. 157Add. 22Add.

75. 25 6(9)166(6)10Multiply. 38 010Add. 10Subtract.  

76. 35 7(8)127610Multiply. 46 1310Add. 23Add.



 

77. 2 14(2610) 7 4(1210)Multiply. 42Work inside parentheses. 2Divide.

78.   3 12653 4 9303Multiply. 910Work inside parentheses. 1Subtract.

79. Evaluate root and power in numerator, and subtract in the denominator.

(54)22 52 (52)4 7 (3)4 Work inside parentheses. 7 7 Subtract. 7 1Divide.

80. Evaluate root and power in numerator, and subtract in the denominator.

(916)32 61 (94)9 7 (5)9 Work inside parentheses. 7 14 Subtract. 7 2Divide.

81. 2 2(5)(3)(2) 831 106 Evaluate power, multiply. 891 4 Add. 11 4 Subtract. 0

Since division by 0 is undefined, the given expression is undefined.

82. 3 3(4)(5)(8) 226 1240 Evaluate power, multiply. 826 28 Add, Subtract. 66 28 Subtract. 0

Since division by 0 is undefined, the given expression is undefined.

83. 33(3)64 3(3)8 98 1 ab

84. 22(3)64 68 2 ab

85. 646(3) 863 14317

86. 646(3) 86(3) 2(3)1 bca

87. 33 424(3)2(6) 4(27)12 1081296

88. 44 333(3)3(6) 3(81)18 24318261

89. 22 2 2()2(36)(3)(6) 2(9)(18) 2(81)18 16218

  

180

acca 

90. 22 2 2 4()4(3)(6)(6(3)) 4(3)(6)(63) 4(3)(6)(9) 4(3)(6)81 7281

153

91. 22 4644(3) (6) 84(3) 36 812 36 20 36 5 9 ba c

92. 332(3)2(6) 64 272(6) 8 2712 8 15 8 15 8 ac b

93. 33 22(6)(3) 464(64)6(3) 12(27) 256(18) 1515 238238

94. 22 33(6)(3) 262(64)6(6) 189 12836 27 92 ca ba

95. 13 848 24 26 8 wyx

96. 1 124(1.25)12 2 56 1 wzy

4 4 311 422 31 816 615 161616 xyy

99. 23 31 4231.25 42 33 41.25 22 41.25 2.75

101. (0.54854850)100031.44 (150,0000.54854850)100031.44 (82,2754850)100031.44

77,425100031.44

77.42531.44

2434.2422434

The owner would pay $2434 in property taxes.

102. (0.54854850)100031.44 (200,0000.54854850)100031.44 (109,7004850)100031.44 104,850100031.44 104.8531.44 3296.4843296

The owner would pay $3296 in property taxes.

103. (a) numberofoz%alcohol0.075body weight inlbhrofdrinking0.015  

483.20.07519020.015  (b) 483.20.07519020.015





153.60.07519020.015



11.5219020.015



0.06120.015

0.0610.03

0.031

104. numberofoz%alcohol0.075body weightinlbhrofdrinking0.015  

0.035 

364.00.07513530.015

1440.07513530.015

10.813530.015

0.0830.015

0.080.015

105. (a) BAC for the man

483.20.075(19025)20.015

0.024

BAC for the woman

364.00.075(13525)30.015

0.023

Increased weight results in lower BACs.

(b) Decreased weight will result in higher BACs.

BAC for the man

0.040 

483.20.075(19025)20.015

BAC for the woman

364.00.075(13525)30.015 0.053

106. BAC for the man

483.20.07519010.015

0.046

BAC for the woman

0.05 

364.00.07513520.015

Decreased time results in higher BACs.

107. 2.4934962 x

(a) 2.493(2003)4962$31.5  billion

(b) 2.493(2010)4962$48.9  billion

(d) The amount spent on pets roughly doubled from 2003 to 2016.

108. (a) Year Average Price (in dollars)

(b) The average price of a movie ticket in the United States increased by $8.68$5.40 $3.28 from 2000 to 2016.

R.5 Properties of Real Numbers

Classroom Examples, Now Try Exercises

1. (a) 4(5)4[(5)] 4()(4)(5) 420 pp p p  

(b) Use the second form of the distributive property.

62(62) 4 mmm m 

(d) 5(42)5(4)5(2)5() 20105 pqrpqr pqr  

N1. (a) 2(3)2[3()] 2(3)(2)() 62 xyxy xy xy 

(b) Use the second form of the distributive property.

412[4(12)] 8 kkk k  

2. (a) 313Identity property (13)Distributive property 2 xxxx x x    (b) (34) 1(34)Identity property 1(3)(1)(4)Distributive property 3(4) 34 p p p p p 



 



N2. (a) 771Identity property (71)Distributive property 8 xxxx x x   

(b) (53)

  

1(53)Identity prop. 1(5)(1)(3)Distributive prop. 53 pq pq pq pq

3. (See Example 3 for more detailed steps.) 129471 124791 (1247)91 98 bbb bbb b b   

N3. (See Example 3 for more detailed steps.) 71034 731104 (731)104 96 xxx xxx x x 

4. (a) 12957 (12957) 1 bbbb b b b  

(b) 6(27)3

6273Distributive property 2673Commutative prop. 24Combine like terms. x x x x

[4(2)]Associative property

[4(2)]Associative property [4(2)]Commutative property [(42)]Associative property (8)Multiply. (8)Associative property 8 mn mn mn mn mn mn mn mn

(d) 2(3)5(21)

26105Distributive property 21065Commutative prop. 811Combine like terms. xx xx xx x  

N4. (a) 3(4)15

31215Distributive prop. 31215Commutative prop.

427Combine terms. tt tt tt t

742 7142Identity property 742Distributive property

32Subtract. xx xx xx x

[5(6)]Associative property

[5(6)]Associative property [5(6)]Commutative property [(56)]Associative property (30)Multiply. 30()Associative property 30 xy xy xy xy xy xy xy xy

(d) 3(57)8(4)

1521832Distributive property 1582132Commutative prop. 753Combine like terms. xx xx xx x

Exercises

1. The identity element for addition is 0 since, for any real number a, 00. aaa  Choice B is correct.

2. The identity element for multiplication is 1 since, for any real number a, 11. aaa  Choice C is correct.

3. The additive inverse of a is a since, for any real number a, ()0aa and 0. aa  Choice A is correct.

4. The multiplicative inverse of a is 1 a since, for any nonzero real number a, 1 1 a a  and 1 1. a a  Choice D is correct.

5. The distributive property provides a way to rewrite a product such as ()abc  as the sum .abac 

6. The commutative property is used to change the order of two terms or factors.

7. The associative property is used to change the grouping of three terms or factors.

8. Like terms are terms with the same variables raised to the same powers.

9. When simplifying an expression, only like terms can be combined.

10. The numerical coefficient in the term 2 7 yz is 7.

11. Using the distributive property, 2()22. mpmp 

12. 3()33 abab 

13. Using the distributive property, 12()12[()] 12()(12)() 1212. xyxy xy xy 

14. 10()10[()] 1010 pqpq pq  

15. Use the second form of the distributive property.

53(53) 8 kkk k  

16. 65(65) 11 aaa a  

17. 797(9) [7(9)] 2 rrrr r r   

18. 464(6) [4(6)] 2 nnnn n n   

19. Since there is no common variable factor here, we cannot use the distributive property to simplify the expression.

20. Since there is no common variable factor here, we cannot use the distributive property to simplify the expression.

21. Use the identity property, then the distributive property.

717 (17) 8 aaaa a a   

22. 919Identity property (19) 10 ssss s s   

23. 11 (11) 2 xxxx x x   

24. 11 (11) 2 aaaa a a   

25. (2)1(2) 1(2)(1)() 2 dfdf df df   

26. (3)1(3) 1(3)(1)() 3 mnmn mn mn   



27. ()1() 1()(1)() xyxy xy xy

 



28. (34)1(34) 1(3)(1)(4) 34 xyxy xy xy

 

29. 2(32)22(3)2(2) 264 xyzxyz xyz  

30. 8(35)8(3)88(5) 24840 xyzxyz xyz 







31. 12432 (12432) 3 yyyy y y







32. 5985 (5985) 11



rrrr r r





33. 654611 641156 (6411)11 111or11 ppp ppp p pp









34. 812359 835129 (835)3 103 xxx xxx x x





35. 3(2)563 36563 35663 (35)663 215 kk kk kk k k

36. 5(3)624 515624 562154 (562)11 911 rrr rrr rrr r r

37. 10(48) 101(48) 1048 42 y y y y

38. 6(95) 61(95) 695 91 y y y y

39.

10(3)()10(3) 30 30 xyxy xy xy

40.

8(6)()8(6) 48 48 xyxy xy xy

41.

42.

22 (12)(7)(12)(7) 33 8(7) 56 wzwz wz wz

55 (18)(5)(18)(5) 66 15(5) 75 wzwz wz wz

43. 3(4)2(1) 31222 32122 14 mm mm mm m

44. 6(5)4(6) 630424 643024 254 aa aa aa a

45. 0.25(84)0.5(62)

0.25(8)0.25(4)(0.5)(6)(0.5)(2) 23 23 (11)23 01 1 pp pp pp pp p p

46. 0.4(105)0.8(510)

 

0.4(10)0.4(5)(0.8)(5)(0.8)(10) 4248 10 xx xx xx x







47. (25)3(24)2 256122 (262)(512) 27 ppp ppp p p









48. (712)2(47)6 1(712)2(47)6 7128146 526 mmm mmm mmm m







 

49. 23(25)3(46)8 261512188 612215188 (612)13188 6318 639 zz zz zz z z z











50. 44(43)6(28)7 4161212487 1612412487 457 kk kk kk k





51. 58(58)13Distributive prop. xxxx 

52. 96(96)3Distributiveproperty yyyy 

53. 5(9)(59)45Associative prop. rrr 



54. 4(128)(412)8Assoc. prop. 8816



55. 5995Commutative prop. xyyx 

56. 577(5)35Commutative prop.

57. 177Identity property 

58. 12012Identity property xx

59. 11 0Inverse prop. 44tyty  A number plus its opposite equals 0.

60. 98 1Inverse prop. 89

61. 8(4)8(4)8Distributive prop. 328 xx x

62. 3()333Distributive prop. xyzxyz

63. Use the multiplication property of 0. 0(0.8759)0 xy Zero times any quantity equals 0.

64. Use the multiplication property of 0. 0(0.3512)0 tu

65. Answers will vary. One example of commutativity is washing your face and brushing your teeth. The activities can be carried out in either order. An example of noncommutativity is putting on your socks and putting on your shoes.

66. Answers will vary. One example is waking up and going to sleep. Another example is tying and untying your shoes. These activities are inverses.

67. 9619419(964)19 (100)19 1900

68. 2760274027(6040) 27(100) 2700

69. 333 588(588) 222 3503 (50) 212 150 75 2

70. 888 (17)(13)(1713) 555 830 (30)8 55 8648

71. 8.75(15)8.75(5)8.75(155) 8.75(10) 87.5

72. 4.31(69)4.31(31)4.31(6931) 4.31(100) 431

73. The terms have been grouped using the associative property of addition.

74. The terms have been regrouped using the associative property of addition.

75. The order of the terms inside the parentheses has been changed using the commutative property of addition.

76. The terms have been regrouped using the associative property of addition.

77. The common factor, x, has been factored out using the distributive property.

78. The numbers in parentheses have been added using arithmetic facts to simplify the expression.

Chapter R Test

3. 13.250 6.417 6.833

4. 0.71decimal place 0.042decimal places 28123

0.0283decimal places  

5. Convert the percent to a decimal first. 4%0.04  Convert from a decimal to a fraction. 41 0.04 10025 

Fraction in Lowest Terms (or Whole Number) Decimal Percent 1 25 0.04 4%

6. Convert the fraction to a decimal first. 5 0.83 6  Convert from a decimal to a percent. 0.830.83100%83.3% 

Fraction in Lowest Terms (or Whole Number) Decimal Percent 5 6 0.83 83.3%

7. Convert the decimal to a percent first. 1.51.5100%150%  Convert from a decimal to a fraction. 13 1.51 22 

Fraction in Lowest Terms (or Whole Number) Decimal Percent 3 2

8. 5 {3,0.75,,5,6.3} 3

Place dots at 5 3,0.75,1.6,5, 3  and 6.3.

9. The elements 0,3,25(or 5), and 24 2 (or 12) are whole numbers.

10. The elements 1,0,3,25(or 5), and 24 2 (or 12) are integers.

11. The elements 1,0.5,0,3,25(or 5),7.5, and 24 2 (or 12) are rational numbers.

12. All the elements in the set are real numbers except 4. 13. 614(11)(3) 8(11)3 330 

14. 51025102 793793 59107221 7997321 457042 636363 457042 63 17 63

 

15. 10436(4) 1012(24) 2(24)26



16. 22 742(6)(4) 7161216 19

  

17. 2[3(12)2] 9(3)(2) 2[3(3)2] 3(3)(2) 2[8] 9(2) 1616 77

18. 2 3 84352(1) 3224

84952(1) 3824 32452 2424 11 Undefined 0

19. 19614,  because 14 is positive and 2 14196. 

20. 225(225)(15)15 

21. Since there is no real number whose square is 16,16 is not a real number.

22. Let3, 3,and25kmr  2 2 82 2

24. When simplifying (38)(46), rr the subtraction sign in front of (46) r  changes the sign of the terms 4 r and 6. (38)(46)

25. (a) The answer is B, Inverse property The sum of 6 and its inverse, 6, equals zero.

(b) The answer is D, Associative property The order of the terms is the same, but the grouping has changed.

(d) The answer is F, Multiplication property of 0. Multiplication by 0 always equals 0.

(e) The answer is C, Identity property The addition of 0 to any number does not change the number.

(f) The answer is C, Identity property Multiplication of any number by 1 does not change the number.

(g) The answer is E, Commutative property The order of the terms a and b is reversed.

8(3)2(3) 252

8(3)2(9) 23 2418 23 66 or 2323 km r 

23. 3(24)4(35)24 3(2)(3)(4)4(3) 4(5)24 612122024 612412202 1010 kkk kk k kkk kkk k

Chapter 1

Linear Equations, Inequalities, and Applications

1.1 Linear Equations in One Variable

Classroom Examples, Now Try Exercises

1. (a) 9100 x  is an equation because it contains an equality symbol.

(b) 910 x  is an expression because it does not contain an equality symbol.

N1. (a) 2173xx  is an expression because it does not contain an equality symbol.

(b) 2173xx  is an equation because it contains an equality symbol.

2. 489171

xxx x  



? ? ? 5112133 5(4)112(4)133(4)Let 4. 201181312 92112 99True



 

121710Combine terms. 1217171017Subtract 17. 510Combine terms.

510 Divide by 5. 55 2 xxx xx xxxxx x x x 

Check by substituting 2 for x in the original equation.

The solution set is {4}. 3.  3453 12353Distributiveprop. 12253Combineterms. 1222532Subtract 2. 1233Combineterms. 123333Add3. 153Combineterms. 153 Divideby3. 33 5 xxx xxx xx xxxxx x x x x x

We will use the following notation to indicate the value of each side of the original equation after we have substituted the proposed solution and simplified.

Check 5:275253True x 

The solution set is   5.

? ? ? 489171

xxx x 

4(2)8(2)917(2)1Let2. 8169341

24251 2424True

The solution set is   2. N2. 5112133













51113Combine terms. 51113Add. 61113Combine terms. 611111311Subtract11. 624Combine terms.

624 Divide by6. 66 4 xxx xx xxxxx x x x x x

Check by substituting 4 for x in the original equation.

N3.  541232 xx  5201232Distributiveprop. 53232Combineterms. 5322322Add2. 7323Combineterms. 73232332Add32. 735Combineterms. 735 Divideby7. 77 5 xx xx xxxxx x x x x x

  

We will use the following notation to indicate the value of each side of the original equation after we have substituted the proposed solution and simplified.

Check 5:512310True x 

The solution set is   5.

4. 6(4)82(35) 648610Distributiveprop. 2210Combineterms. 2210Add. 2310Combineterms. 21031010Add10. 123Combineterms. 123 Divideby3. 33 4 xxx xxx xx xxxxx x x x x x

Check 4:23234True x 

The solution set is   4.

N4.  23264136 26184436Dist.prop. 418440Combineterms. 4181844018Add 18. 42240Combineterms. 440224040Subtract 40. 4422Combineterms. 4422 Divideby22. 2222 2 xx xx xx xxxxx x x x x x





Check 2:230436True x 

The solution set is   2.

5. Multiply each side by the LCD, 4, and use the distributive property 131 242 131 444 242 2(1)1(3)2 2232 352 33Subtract5. 1Divideby3. xx xx xx xx x x x  

Check 21 1:0True 42 x 

The solution set is {1}.

N5. Multiply each side by the LCD, 8, and use the distributive property. 424 5 48 424 888(5) 48 2(4)1(24)40 282440 4440 444Add4. 11Divideby4.

Check 713 11:5True 44 x 

The solution set is   11.

6. Multiply each term by 100. 0.02(60)0.040.03(50) 2(60)43(50) 12041503 4303 30 xx xx xx xx x     

Check 30:1.21.22.4True x 

The solution set is   30.

N6. Multiply each term by 100. 0.080.12(4)0.03(5) 812(4)3(5) 81248315 448315 4633 637 9 xxx xxx xxx xx xx x x 





   

Check 9:0.720.600.12True x 

The solution set is   9.



 

7. (a) 5(2)2(1)31 5102231 3831 383313 81False xxx xxx xx xxxx

 

Since the result, 81,  is false, the equation has no solution and is called a contradiction. The solution set is .

(b) Multiply each side by the LCD, 3, and use the distributive property. 121 333 121 333 333 1231 3131 xx x xx x xxx xx

This is an identity. Any real number will make the equation true. The solution set is {all real numbers}.

This is a conditionalequation.

Check 0:5(1)05True x 

The solution set is   0.

N7. (a) 93(4)6(2) 9312612 612612 xxx xxx xx

This is an identity. Any real number will make the equation true. The solution set is   all real numbers.



(b) 3(21)23 6323 833 933Subtract. 90Subtract3. 0Divideby9. xxx xxx xx xx x x

This is a conditionalequation.

Check 0:3(1)3True x 

The solution set is   0.

Since the result, 2110, is false, the equation has no solution and is called a contradiction. The solution set is 

Exercises

1. A collection of numbers, variables, operation symbols, and grouping symbols, such as  2815, x is an algebraic expression. While an equation does include an equality symbol, there isnot an equality symbol in an algebraic expression.

2. A linear equation in one variable (here x) can be written in the form 0, axb with 0. a  Another name for a linear equation is a first-degree equation, because the greatest power on the variable is one.

3. If we let 2 x  in the linear equation 259, x  a true statement results. The number 2 is a solution of the equation, and  2 is the solution set.

4. A linear equation with one solution in its solution set, such as 259, x  is a conditional equation.

5. A linear equation with an infinite number of solutions is an identity. Its solution set is   all real numbers.

6. A linear equation with no solution is a contradiction. Its solution set is the empty set .

7. 320 xx can be written as 42, x  so it is linear.

949 x  is in linear form. Choices A and C are linear.

8. 2 12 x  in choice B is nonlinear because the variable is squared.

326 xy in choice D is nonlinear because there are two variables. Choices B and D are nonlinear.

9. 324xx is an equation because it contains an equality symbol.

10. 3244 xx  is an equation because it contains an equality symbol.

11. 4(3)2(1)10 xx is an expression because it does not contain an equality symbol.

12. 4(3)2(1)10 xx is an expression because it does not contain an equality symbol.

13. 101243 xx  is an equation because it contains an equality symbol.

14. 1012430 xx  is an equation because it contains an equality symbol.

15. A sign error was made when the distributive property was applied. The left side of the second line should be 846. xx

822337

84637Distributive property 4637Combine like terms.

xx xx

1Subtract 3 and 6.

The correct solution is 1.

16. The first step should have been to distribute 2 to (31). x  122(31)11

126211Distributive property

61011Combine terms. 610101110Subtract 10. 61 1 Divide by 6. 6

The correct solution is 1 6 .

17. (a) The true statement 7 = 7 indicates that the solution set is {all real numbers}, choice B.

(b) The final line is x = 0, which is the form of solution for a conditional equation. The solution set is {0}, choice A.

(c) The false statement 7 = 0 indicates that the equation has no solution. The solution set is the empty set , choice C.

18. An equation will have the solution set {all real numbers} if it is an identity. The only equation listed here that is not an identity is choice C: 43 430 0 xx xx x

19. 781

78818Subtract8. 77 77 Divideby7. 77 1 x x x x x

We will use the following notation to indicate the value of each side of the original equation after we have substituted the proposed solution and simplified.

Check 1:781True x 

The solution set is {1}.

20. 5421 544214Add4. 525 525 Divideby5. 55 5 x x x x x     

Check 5:25421True x 

The solution set is   5.

21. 5236 523363Subtract3. 226 22262Subtract2. 28 28 Divideby2. 22 4 xx xxxxx x x x x x       

Check 4:202126True x 

The solution set is {4}.

22. 9179 917797Subtract7. 219 21191Subtract1. 210 210 Divideby2. 22 5 xx xxxxx x x x x x  

   



Check 5:451359True x  The solution set is {5}.

23. 75158 2158Combineterms. 2158Subtract. 158Combineterms. 1515815Subtract15. 7Combineterms. xxx xx xxxxx x x x     



Check 7:49351578True x  The solution set is {7}.

24. 2445

345Subtract4. 39Subtract4. 3Divideby3. xxx xx xx x x

445Combineterms.

Check 3:643125True x 

The solution set is   3.

25. 121595359

27434Combine terms. 3044.Add3. 300Add4. 0Divideby30. www ww ww w w 

Check 0:9559True w 

The solution set is   0.

26. 458464

464Combine terms. 544Subtract6. 50Add4. 0Divideby5. xxx xx xx x x 

Check 0:844True x 

The solution set is {0}.

27. 3(24)202 612202Distributiveproperty 81220Add2. 832Add12. 4Divideby8. tt tt tt t t 

Check 4:3(4)208True t

The solution set is   4.

28. 2(32)4 644Distributiveproperty 654Subtract. 510Subtract6. 2Divideby5. xx xx xx x x

Check 2:2(1)24True x

The solution set is   2.

29. 5(1)3264



553264Distributive prop. 2364Combineterms. 384Add2. 78Subtract4. 7 Divideby8. 8 xxx xxx xx xx x x

Check: Substitute 7 8 for x and show that both sides equal 1.25. The screen shows a typical Check on a calculator.

The solution set is 7 8 

30. 5(3)4542

5154542Distributive prop. 91042Combineterms. 11104Add2. 116Subtract10. 6 Divideby11. 11 xxx xxx xx xx x x

Check 6 : 11 13524554412 True 1111111111 x  

The solution set is 6 . 11     

31. 2593(4)5 393125 39317 917False xxx xx xx

The equation is a contradiction. The solution set is 

32. 62112(23)4 411464 411410 1110False xxx xx xx    

The equation is a contradiction. The solution set is .

33. 2(3)6(7) 26642 4642 436 9 xx xx x x x

Check

9:2662True x 

The solution set is   9.

34. 4(9)8(3) 436824 43624 460 15 xx xx x x x

Check 15:4(24)8(12)True x

The solution set is {15}.

35. 3(21)2(2)5 63245 475 42 21 42 xx xx x x x

Check 15:3(0)25True 22 x

The solution set is 1 . 2

36. 4(2)2(3)6 48266 626 68 84 63 xx xx x x x

Check 4213 :426True 333 x

The solution set is 4 3

37. 23(4)2(3) 231226 51226 36 6 2 3 xxx xxx xx x x

Check 2:43(2)2(1)True x

The solution set is {2}.

38. 63(52)4(1) 615644 9644 510 10 2 5 xxx xxx xx x x

Check 2:123(8)4(3)True x 

The solution set is {2}.

39. 64(32)5(4)10 612852010 1412530 918 2 xxx xxx xx x x

Check 2:124(7)5(6)10True x 

The solution set is {2}.

40. 23(42)2(3)2 2126262 41224 28 4 xxx xxx xx x x  

Check 4:83(4)2(1)2True x 

The solution set is {4}.

41. 2(3)43(4)2 2643122 310310 xxx xxx xx 





The equation is an identity. The solution set is {all real numbers}.

42. 4(27)2253(21) 82822563 828828 xxx xxx xx

  

The equation is an identity.

The solution set is {all real numbers}.

43. 2[(24)3]2(1) 2[243]2(1) 2[1]2(1) 11Divideby2. 121Add. 22Subtract1. 1Divideby2. xxx xxx xx xx xx x x  







Check 1:2[123]0True x 

The solution set is {1}.