SOLUTIONS MANUAL
PREALGEBRA AND INTRODUCTORY ALGEBRA 6TH EDITION BY ELAYNE MARTIN -GAY 2025 INSTRUCTORS SOLUTIONS MANUL C TRIMBLE AND ASSOCIATES Contents
Chapter 1
1
Chapter 2
42
Chapter 3
71
Chapter 4
101
Chapter 5
164
Chapter 6
209
Chapter 7
253
Chapter 8
284
Chapter 9
324
Chapter 10
368
Chapter 11
414
Chapter 12
454
Chapter 13
507
Chapter 14
569
Chapter 15
613
Chapter 16
647
Appendix
698
Practice Final Exam
704
Chapter 1 5. In a whole number, each group of 3 digits is called a period.
Section 1.2 Practice Exercises 1. The place value of the 8 in 38,760,005 is millions.
6. The place value of the digit 4 in the whole number 264 is ones.
2. The place value of the 8 in 67,890 is hundreds. 7. hundreds 3. The place value of the 8 in 481,922 is tenthousands.
8. To read (or write) a number, read from left to right.
4. 54 is written as fifty-four. 9. 80,000 5. 678 is written as six hundred seventy-eight. 10. Dachsund 6. 93,205 is written as ninety-three thousand, two hundred five.
Exercise Set 1.2
7. 679,430,105 is written as six hundred seventynine million, four hundred thirty thousand, one hundred five.
2. The place value of the 5 in 905 is ones.
8. Thirty-seven in standard form is 37.
6. The place value of the 5 in 79,050,000 is tenthousands.
4. The place value of the 5 in 6527 is hundreds.
9. Two hundred twelve in standard form is 212. 10. Eight thousand, two hundred seventy-four in standard form is 8,274 or 8274.
8. The place value of the 5 in 51,682,700 is tenmillions. 10. 316 is written as three hundred sixteen.
11. Five million, fifty-seven thousand, twenty-six in standard form is 5,057,026. 12. 4,026,301 = 4,000,000 + 20,000 + 6000 + 300 + 1 13. a. Find Australia in the “Country” column. Read from left to right until the “bronze” column is reached. Australia won 22 bronze medals. b. Find the countries for which the entry in the “Total” column is greater than 60. The United States, China, ROC, and Great Britain won more than 60 medals. Vocabulary, Readiness & Video Check 1.2 1. The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... are called whole numbers. 2. The number 1286 is written in standard form. 3. The number “twenty-one” is written in words. 4. The number 900 + 60 + 5 is written in expanded form.
12. 5445 is written as five thousand, four hundred forty-five. 14. 42,009 is written as forty-two thousand, nine. 16. 3,204,000 is written as three million, two hundred four thousand. 18. 47,033,107 is written as forty-seven million, thirty-three thousand, one hundred seven. 20. 254 is written as two hundred fifty-four. 22. 105,447 is written as one hundred five thousand, four hundred forty-seven. 24. 1,110,000,000 is written as one billion, one hundred ten million. 26. 11,239 is written as eleven thousand, two hundred thirty-nine. 28. 202,700 is written as two hundred two thousand, seven hundred. 30. Four thousand, four hundred sixty-eight in standard form is 4468.
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
32. Seventy-three thousand, two in standard form is 73,002. 34. Sixteen million, four hundred five thousand, sixteen in standard form is 16,405,016.
74. answers may vary 76. A quadrillion in standard form is 1,000,000,000,000,000. Section 1.3 Practice Exercises
36. Two million, twelve in standard form is 2,000,012.
1.
4135 252 4387
2.
47, 364 135,898 183, 262
38. Six hundred forty thousand, eight hundred eighty-one in standard form is 640,881. 40. Two hundred thirty-four thousand in standard form is 234,000.
1 1 11
42. One thousand, eight hundred fifteen in standard form is 1815. 44. Two hundred sixty million, one hundred thirtyeight thousand, five hundred sixty-nine dollars in standard form is $260,138,569. 46. Seven hundred fifteen in standard form is 715. 48. 789 = 700 + 80 + 9 50. 6040 = 6000 + 40 52. 20,215 = 20,000 + 200 + 10 + 5
3. Notice 12 + 8 = 20 and 4 + 6 = 10. 12 + 4 + 8 + 6 + 5 = 20 + 10 + 5 = 35 12 2
4.
6432 789 54 28 7303
5. a.
54. 99,032 = 90,000 + 9000 + 30 + 2
b. 20 8 = 12 because 12 + 8 = 20
56. 47,703,029 = 40,000,000 + 7,000,000 + 700,000 + 3000 + 20 + 9
c.
93 93 = 0 because 0 + 93 = 93.
d.
42 0 = 42 because 42 + 0 = 42.
58. Mount Baker erupted in 1792, which is in standard form.
6. a.
9143 122 9021
Check: 9021 122 9143
b.
978 851 127
Check:
7. a.
69 7 49 64 8
60. Mount Shasta and Mount St. Helens have each had two eruptions listed. 62. Mount St. Helens has an eruption listed in 1980. All other eruptions listed in the table occurred before this one. 64. More Golden retrievers are registered than German shepherds.
68. The maximum height of an average-size standard dachsund is 9 inches. 70. The largest number is 77,753. 72. Yes
127 851 978
8 17
66. French bulldogs are second in popularity. 24 is written as twenty-four.
2
14 6 = 8 because 8 + 6 = 14.
Check: 648 49 697
2 12
b.
326 245 81
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Check:
81 245 326
ISM: Prealgebra and Introductory Algebra
c.
1234 822 412
Check:
412 822 1234
40 0 1 6 4 2 3 6
7. 865 95 = 770 8. 76 27 = 49 9. 147 38 = 109
9 3 1010
8. a.
Chapter 1: The Whole Numbers
Check:
236 164 400
10. 366 87 = 279 11. 9625 647 = 8978 12. 10,711 8925 = 1786
9
Vocabulary, Readiness & Video Check 1.3
910 10
b.
10 0 0 762 238
Check:
238 762 1000
9. 2 cm + 8 cm + 15 cm + 5 cm = 30 cm The perimeter is 30 centimeters. 10. 647 + 647 + 647 = 1941 The perimeter is 1941 feet. 11.
15, 759 458 15, 301 The radius of Neptune is 15,301 miles.
12. a.
The country with the fewest threatened amphibians corresponds to the shortest bar, which is Peru.
b. To find the total number of threatened amphibians for Brazil, Peru, and Mexico, we add. 110 78 191 379 The total number of threatened amphibians for Brazil, Peru, and Mexico is 379. Calculator Explorations
1. The sum of 0 and any number is the same number. 2. In 35 + 20 = 55, the number 55 is called the sum and 35 and 20 are each called an addend. 3. The difference of any number and that same number is 0. 4. The difference of any number and 0 is the same number. 5. In 37 19 = 18, the number 37 is the minuend, the 19 is the subtrahend, and the 18 is the difference. 6. The distance around a polygon is called its perimeter. 7. Since 7 + 10 = 10 + 7, we say that changing the order in addition does not change the sum. This property is called the commutative property of addition. 8. Since (3 + 1) + 20 = 3 + (1 + 20), we say that changing the grouping in addition does not change the sum. This property is called the associative property of addition. 9. To add whole numbers, we line up place values and add from right to left.
2. 76 + 97 = 173
10. We cannot take 7 from 2 in the ones place, so we borrow one ten from the tens place and move it over to the ones place to give us 10 + 2, or 12.
3. 285 + 55 = 340
11. triangle; 3
4. 8773 + 652 = 9425
12. To find the additional money needed to purchase, subtract the savings account amount from the purchase price.
1. 89 + 45 = 134
5. 985 + 1210 + 562 + 77 = 2834 6. 465 + 9888 + 620 + 1550 = 12,523
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
Exercise Set 1.3 2.
27 31 58
4.
37 542 579
6.
23 45 30 98
8.
236 6243 6479
111
20.
26 582 4 763 62, 511 67,882 1 11 2 1 2
22.
504, 218 321, 920 38, 507 594, 687 1, 459, 332
24.
957 257 700
Check:
700 257 957
26.
55 29 26
Check:
26 29 55
28.
674 299 375
Check:
375 299 674
30.
300 149 151
Check:
151 149 300
32.
5349 720 4629
Check: 4629 720 5349
34.
724 16 708
Check: 708 16 724
36.
1983 1914 69
Check:
1
1
10.
17, 427 821, 059 838, 486
12.
3 5 8 5 7 28
11
11
22
14.
64 28 56 25 32 205 11 2
16.
16 1056 748 7770 9590 1 111
18.
4
6789 4321 5555 16, 665
1
1
1
Copyright © 2025 Pearson Education, Inc.
69 1914 1983
ISM: Prealgebra and Introductory Algebra
11 1 1
38.
40, 000 23, 582 16, 418
Check:
6050 1878
Check:
16, 418 23, 582 40, 000 111
40.
4172
4172 1878 6050 11 11
42.
62, 222 39,898 22, 324
44.
986 48 938
Check:
22, 324 39,898 62, 222
80 93 17 9 2 201
1
802 6487 7289 The sum of 802 and 6487 is 7289. 62. “Find the total” indicates addition. 12
89 45 2 19 341 496 The total of 89, 45, 2, 19, and 341 is 496.
66. “Increased by” indicates addition. 712 38 750 712 increased by 38 is 750.
48. 10, 000 1786 8214
68. “Less” indicates subtraction. 25 12 13
1 11
50.
60. “Find the sum” indicates addition.
64. “Find the difference” indicates subtraction. 16 5 11 The difference of 16 and 5 is 11.
2
46.
Chapter 1: The Whole Numbers
12, 468 3 211 1 988 17, 667
25 less 12 is 13.
52. 3 + 4 + 5 = 12 The perimeter is 12 centimeters. 54. Opposite sides of a rectangle have the same length. 9 + 3 + 9 + 3 = 12 + 12 = 24 The perimeter is 24 miles.
70. “Subtracted from” indicates subtraction. 90 86 4 86 subtracted from 90 is 4.
56. 6 + 5 + 7 + 3 + 4 + 7 + 5 = 37 The perimeter is 37 inches.
72. Subtract 7953 million from 8687 million. 8687 7953 734 The world’s projected population increase is 734 million.
58. The unknown vertical side has length 3 + 5 = 8 feet. The unknown horizontal side has length 8 + 4 = 12 feet. 8 + 3 + 4 + 5 + 12 + 8 = 40 The perimeter is 40 feet.
74. Subtract the discount from the regular price. 276 69 207 The first-year price is $207.
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
76.
78.
164, 000 40, 000 204, 000 The total U.S. land area drained by the Ohio and Tennessee sub-basins is 204,000 square miles.
94. North Carolina has the fewest Aldi locations.
189, 000 75, 000 114, 000 The Upper Mississippi sub-basin drains 114,000 square miles more than the Lower Mississippi sub-basin.
98. The total number of Aldi locations in the states listed is 1335. 1335 923 2258 There are 2258 Aldi locations in the United States.
80. Opposite sides of a rectangle have the same length. 60 + 45 + 60 + 45 = 210 The perimeter is 210 feet. 82.
84.
59, 320 55, 492 3 828 They traveled 3828 miles on their trip. 299, 345 259, 516 558,861 The total number of F-Series trucks and Silverados sold during the first six months of 2022 was 558,861.
96. 211 + 92 + 203 + 98 + 92 + 123 + 91 + 152 + 146 + 127 = 1335 The total number of Aldi locations in the ten states listed is 1335.
11 1
100.
102. The minuend is 2863 and the subtrahend is 1904. 104. The minuend is 86 and the subtrahend is 25. 106. answers may vary 21
108.
86. The shortest bar corresponds to the quietest reading. Leaves rustling is the quietest. 88.
90.
100 70 30 The difference in sound intensity between live rock music and loud television is 30 dB.
19 214 49 651 933 The given sum is incorrect, the correct sum is 933. 11
112.
6
773 659 481 1913 The given sum is correct. 1 2
110.
119 99 20 The difference in volume between the mid-size and a sub-compact car is 20 cubic feet.
92. Opposite sides of a rectangle have the same length. 18 + 12 + 18 + 12 = 60 The perimeter of the puzzle is 60 inches.
605 9779 10, 384 The total highway mileage in Kansas is 10,384 miles.
389 89 478 The given difference is correct.
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114.
Chapter 1: The Whole Numbers
3. a.
To round 3474 to the nearest hundred, observe that the digit in the tens place is 7. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 3474 rounded to the nearest hundred is 3500.
b.
To round 76,243 to the nearest hundred, observe that the digit in the tens place is 4. Since this digit is less than 5, we do not add 1 to the digit in the hundreds place. The number 76,243 rounded to the nearest hundred is 76,200.
c.
To round 978,965 to the nearest hundred, observe that the digit in the tens place is 6. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 978,865 rounded to the nearest hundred is 979,000.
7168 547 7715 The given difference is incorrect. 7615 547 7068
116.
10, 244 8 534 1 710
118. answers may vary Section 1.4 Practice Exercises 1. a.
To round 57 to the nearest ten, observe that the digit in the ones place is 7. Since the digit is at least 5, we add 1 to the digit in the tens place. The number 57 rounded to the nearest ten is 60.
b. To round 641 to the nearest ten, observe that the digit in the ones place is 1. Since the digit is less than 5, we do not add 1 to the digit in the tens place. The number 641 rounded to the nearest ten is 640. c.
2. a.
To round 325 to the nearest ten observe that the digit in the ones place is 5. Since the digit is at least 5, we add 1 to the digit in the tens place. The number 325 rounded to the nearest ten is 330. To round 72,304 to the nearest thousand, observe that the digit in the hundreds place is 3. Since the digit is less than 5, we do not add 1 to the digit in the thousands place. The number 72,304 rounded to the nearest thousand is 72,000.
b.
To round 9222 to the nearest thousand, observe that the digit in the hundreds place is 2. Since the digit is less than 5, we do not add 1 to the digit in the thousands place. The number 9222 rounded to the nearest thousand is 9000.
c.
To round 671,800 to the nearest thousand, observe that the digit in the hundreds place is 8. Since this digit is at least 5, we add 1 to the digit in the thousands place. The number 671,800 rounded to the nearest thousand is 672,000.
4. 49 25 32 51 98
rounds to rounds to rounds to rounds to rounds to
5.
3785 2479
6.
11 16 19 31
50 30 30 50 + 100 260
rounds to rounds to
rounds to rounds to rounds to rounds to
4000 2000 2000 10 20 20 + 30 80
The total distance is approximately 80 miles. 7.
18, 617 1 607
rounds to rounds to
18,600 1 600 17, 000 The difference in the reported number of pertussis cases between 2019 and 2021 was 17,000.
Vocabulary, Readiness & Video Check 1.4 1. To graph a number on a number line, darken the point representing the location of the number. 2. Another word for approximating a whole number is rounding.
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
3. The number 65 rounded to the nearest ten is 70, but the number 61 rounded to the nearest ten is 60. 4. An exact number of products is 1265, but an estimate is 1000. 5. 3 is in the place value we’re rounding to (tens), and the digit to the right of this place value is 5 or greater, so we need to add 1 to the 3. 6. On a number line, 22 is closer to 20 than to 30. Thus, 22 rounded to the nearest ten is 20. 7. Each circled digit is to the right of the place value being rounded to and is used to determine whether or not we add 1 to the digit in the place value being rounded to. Exercise Set 1.4 2. To round 273 to the nearest ten, observe that the digit in the ones place is 3. Since this digit is less than 5, we do not add 1 to the digit in the tens place. The number 273 rounded to the nearest ten is 270. 4. To round 846 to the nearest ten, observe that the digit in the ones place is 6. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 846 rounded to the nearest ten is 850. 6. To round 8494 to the nearest hundred, observe that the digit in the tens place is 9. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 8494 rounded to the nearest hundred is 8500. 8. To round 898 to the nearest ten, observe that the digit in the ones place is 8. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 898 rounded to the nearest ten is 900. 10. To round 82,198 to the nearest thousand, observe that the digit in the hundreds place is 1. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. The number 82,198 rounded to the nearest thousand is 82,000. 12. To round 42,682 to the nearest ten-thousand, observe that the digit in the thousands place is 2. Since this digit is less than 5, we do not add 1 to the digit in the ten-thousands place. The number 42,682 rounded to the nearest ten-thousand is 40,000.
8
14. To round 179,406 to the nearest hundred, observe that the digit in the tens place is 0. Since this digit is less than 5, we do not add 1 to the digit in the hundreds place. The number 179,406 rounded to the nearest hundred is 179,400. 16. To round 96,501 to the nearest thousand, observe that the digit in the hundreds place is 5. Since this digit is at least 5, we add 1 to the digit in the thousands place. The number 96,501 rounded to the nearest thousand is 97,000. 18. To round 99,995 to the nearest ten, observe that the digit in the ones place is 5. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 99,995 rounded to the nearest ten is 100,000. 20. To round 39,523,698 to the nearest million, observe that the digit in the hundred-thousands place is 5. Since this digit is at least 5, we add 1 to the digit in the millions place. The number 39,523,698 rounded to the nearest million is 40,000,000. 22. Estimate 7619 to a given place value by rounding it to that place value. 7619 rounded to the tens place is 7620, to the hundreds place is 7600, and to the thousands place is 8000. 24. Estimate 7777 to a given place value by rounding it to that place value. 7777 rounded to the tens place is 7780, to the hundreds place is 7800, and to the thousands place is 8000. 26. Estimate 85,049 to a given place value by rounding it to that place value. 85,049 rounded to the tens place is 85,050, to the hundreds place is 85,000, and to the thousands place is 85,000. 28. To round 41,529 to the nearest thousand, observe that the digit in the hundreds place is 5. Since this digit is at least 5, we add 1 to the digit in the thousands place. Therefore 41,529 miles rounded to the nearest thousand is 42,000 miles. 30. To round 60,149 to the nearest hundred, observe that the digit in the tens place is 4. Since this digit is less than 5, we do not add 1 to the digit in the hundreds place. Therefore, 60,149 days rounded to the nearest hundred is 60,100 days.
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ISM: Prealgebra and Introductory Algebra
32. To round 334,696,947 to the nearest million, observe that the digit in the hundred-thousands place is 6. Since this digit is at least 5, we add 1 to the digit in the millions place. Therefore, 334,696,947 rounded to the nearest million is 335,000,000. 34. To round 74,686 to the nearest hundred, observe that the digit in the tens place is 8. Since this digit is at least 5, we add 1 to the digit in the hundreds place. Therefore, $74,686 rounded to the nearest hundred is $74,700. 36. 2022: To round 13,700,000,000 to the nearest billion, observe that the digit in the hundredmillions place is 7. Since this digit is at least 5, we add 1 to the digit in the billions place. The number 13,700,000,000 rounded to the nearest billion is 14,000,000,000. 2021: To round 15,100,000,000 to the nearest billion, observe that the digit in the hundredmillions place is 1. Since this digit is less than 5, we do not add 1 to the digit in the billions place. The number 15,100,000,000 rounded to the nearest billion is 15,000,000,000. 38.
52 33 15 29
rounds to rounds to rounds to rounds to
50 30 20 + 30 130
555 235
rounds to rounds to
560 240 320
4050 3133 1220
rounds to rounds to rounds to
4100 3100 1200 8400
Chapter 1: The Whole Numbers
50. 542 + 789 + 198 is approximately 540 + 790 + 200 = 1530. The answer of 2139 is incorrect. 52. 5233 + 4988 is approximately 5200 + 5000 = 10,200. The answer of 9011 is incorrect. 54.
89 rounds to 90 97 rounds to 100 100 rounds to 100 79 rounds to 80 75 rounds to 80 82 rounds to + 80 530 The total score is approximately 530.
56.
588 689 277 143 59 802
rounds to rounds to rounds to rounds to rounds to rounds to
600 700 300 100 100 + 800 2600 The total distance is approximately 2600 miles.
58.
1360 1240
rounds to rounds to
1400 1200 200
The difference in price is approximately $200. 40.
42.
44.
1989 1870
rounds to rounds to
2000 1900 100
46.
799 1655 271
rounds to rounds to rounds to
800 1700 + 300 2800
60.
64 41 133
rounds to rounds to rounds to
60 40 + 130 230
The total distance is approximately 230 miles. 62.
6615 1737
rounds to rounds to
6600 1700 8300 The total enrollment is approximately 8300 students.
64. 588 hundred-thousands is 58,800,000 in standard form. 58,800,000 rounded to the nearest million is 59,000,000. 58,800,000 rounded to the nearest ten-million is 60,000,000.
48. 522 + 785 is approximately 520 + 790 = 1310. The answer of 1307 is correct.
66. 433 hundred-thousands is 43,300,000 in standard form. 43,300,000 rounded to the nearest million is 43,000,000. 43,300,000 rounded to the nearest ten-million is 40,000,000.
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
68. 5698, for example, rounded to the nearest ten is 5700. 70. The largest possible number that rounds to 1,500,000 when rounded to the nearest hundredthousand is 1,549,999.
6. Area length width (360 miles)(280 miles) 100,800 square miles The area of Wyoming is 100,800 square miles. 7.
72. answers may vary 74.
5950 7693 8203
rounds to rounds to rounds to
6 000 7 700 8 200 21, 900
The perimeter is approximately 21,900 miles.
16 45 80 640 720 The printer can print 720 pages in 45 minutes.
8. 5 84 420 12 3 36
Section 1.5 Practice Exercises 1. a.
The total cost is $456.
60=0 9.
b. (1)8 = 8 c.
(50)(0) = 0
d. 75 1 = 75 2. a.
6(4 + 5) = 6 4 + 6 5
163 391
rounds to rounds to
200 400 80,000 There are approximately 80,000 words on 391 pages.
Calculator Explorations
b.
30(2 + 3) = 30 2 + 30 3
1. 72 48 = 3456
c.
7(2 + 8) = 7 2 + 7 8
2. 81 92 = 7452
5
3. a.
29 6 174
420 36 456
3. 163 94 = 15,322 4. 285 144 = 41,040 5. 983(277) = 272,291
44
b.
648 5 3240
6. 1562(843) = 1,316,766 Vocabulary, Readiness & Video Check 1.5 1. The product of 0 and any number is 0.
4.
5.
1 0
306 81 306 24 480 24, 786 726 142 1 452 29 040 72 600 103, 092
2. The product of 1 and any number is the number. 3. In 8 12 = 96, the 96 is called the product and 8 and 12 are each called a factor. 4. Since 9 10 = 10 9, we say that changing the order in multiplication does not change the product. This property is called the commutative property of multiplication.
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ISM: Prealgebra and Introductory Algebra 5. Since (3 4) 6 = 3 (4 6), we say that changing the grouping in multiplication does not change the product. This property is called the associative property of multiplication. 6. Area measures the amount of surface of a region.
Chapter 1: The Whole Numbers
22.
9021 3 27, 063
24.
91 72 182 6370 6552
26.
526 23 1 578 10 520 12, 098
7. Area of a rectangle = length width. 8. We know 9(10 + 8) = 9 10 + 9 8 by the distributive property. 9. distributive
10. to show that 8649 is actually multiplied by 70 and not by just 7 11. Area is measured in square units, and here we have meters by meters, or square meters; the answer is 63 square meters, or the correct units are square meters.
28.
12. Multiplication is also an application of addition since it is addition of the same addend. Exercise Set 1.5
30.
2. 55 1 = 55 4. 27 0 = 0 6. 7 6 0 = 0 8. 1 41 = 41
34. (240)(1)(20) = (240)(20) = 4800 36.
12. 6(1 + 4) = 6 1 + 6 4 14. 12(12 + 3) = 12 12 + 12 3 79 3 237
18.
638 5 3190
20.
882 2 1764
720 80 57, 600
32. (593)(47)(0) = 0
10. 5(8 + 2) = 5 8 + 5 2
16.
708 21 708 14 160 14,868
38.
1357 79 12 213 94 990 107, 203
807 127
5 649 16 140 80 700 102, 489 40.
1234 567 8 638 74 040 617 000 699, 678
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
42.
44.
426 110 4 260 42 600 46,860 1876 1407 13 132 750 400 1 876 000 2, 639, 532
62.
3310 3 9930
64.
14 8 112 There are 112 grams of fat in 8 ounces of hulled sunflower seeds.
66.
34 14 136 340 476 There are 476 seats in the room.
46. Area (length)(width) (13 inches)(3 inches) 39 square inches Perimeter length width length width 13 3 13 3 32 inches
68. a. b.
48. Area (length)(width) (25 centimeters)(20 centimeters) 500 square centimeters Perimeter length width length width 25 20 25 20 90 centimeters 50.
52.
982 650
rounds to rounds to
1000 700 700, 000
111 999
rounds to rounds to
100 1000 100, 000
20 3 60 There are 60 apartments in the building.
70. Area (length)(width) (60 feet)(45 feet) 2700 square feet The area is 2700 square feet. 72. Area (length)(width) (776 meters)(639 meters) 495,864 square meters The area is 495,864 square meters. 74.
64 17 448 640 1088 The 17 flash drives hold 1088 GB.
76.
365 3 1095 A cow eats 1095 pounds of grain each year.
54. 2872 12 is approximately 2872 10, which is 28,720. The best estimate is b. 56. 706 409 is approximately 700 400, which is 280,000. The best estimate is d. 58. 70 12 (7 10) 12 7 (10 12) 7 120 840
5 4 = 20 There are 20 apartments on one floor.
60. 9 900 = 8100
1 2
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ISM: Prealgebra and Introductory Algebra
78.
Chapter 1: The Whole Numbers 98. 57 3 = 171 57 6 = 342 The problem is
13 16 78 130 208 There are 208 grams of fat in 16 ounces.
100. answers may vary
Cost per person
Person
Number of persons
Student
24
$5
$120
102. 3 161 = 483 2 479 = 958 483 + 958 + 254 = 1695 LeBron James scored 1695 points during the 20212022 regular season.
Nonstudent
4
$7
$28
Section 1.6 Practice Exercises
Children under 12
5
$2
$10
80.
Total Cost
Cost per Category
1. a.
8 9 72 because 8 9 = 72.
$158 b.
82. 3 26 = 78 There were 78 million Americans age 65+ in 2022. 84.
57 63
126 8 118
40 5 = 8 because 8 5 = 40. 24
c.
4 because 4 6 = 24.
6 2. a.
7
1 because 1 7 = 7.
7 b. 5 1 = 5 because 5 1 = 5.
86. 47 + 26 + 10 + 231 + 50 = 364 88.
90.
c.
19 4 76 The product of 19 and 4 is 76.
d. 4 1 = 4 because 4 1 = 4.
14 9 23 The total of 14 and 9 is 23.
10 10 because 10 1 = 10. 1
f.
21 21 = 1 because 1 21 = 21.
3. a.
4 5 = 5 + 5 + 5 + 5 or 4 + 4 + 4 + 4 + 4
b. answers may vary 96.
e.
0
92. 11 + 11 + 11 + 11 + 11 + 11 = 6 11 or 11 6 94. a.
31 50 1550
11 1 11 because 11 1 = 11.
b. c.
0 because 0 7 = 0.
7 0 8 0 because 0 8 = 0. 7 0 is undefined because if 7 0 is a number, then the number times 0 would be 7.
d. 0 14 = 0 because 0 14 = 0.
Copyright © 2025 Pearson Education, Inc.
13
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
4. a.
b.
818 6 4908 48 10 6 48 48 0 Check: 818 6 4908 553 4 2212 20 21 20 12 12 0 Check: 553
b.
Check: 5100 9 = 45,900
6. a.
5. a.
251 753 6 15 15 03 3 0 Check: 251 3 753
b.
3
7
304 2128 21 02 0 28 28 0
4
234 R 3 939 8 13 12 19 16 3
Check: 234 4 + 3 = 939
4 2212 c.
5 100 9 45, 900 45 09 9 000
657 R 2 5 3287 30 28 25 37 35 2 Check: 657 5 + 2 = 3287
7. a.
9067 R 2 9 81, 605 81 06 0 60 54 65 63 2 Check: 9067 9 + 2 = 81,605
Check: 304 7 = 2128
1 4
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ISM: Prealgebra and Introductory Algebra
b.
4
Chapter 1: The Whole Numbers
12. Find the sum and divide by 7.
5827 R 2 23, 310 20 33 3 2 11 8 30 28 2
18 4 7 126 7 7 35 56 16 56 9 0 3 52 126 The average time is 18 minutes.
Check: 5827 4 + 2 = 23,310
Calculator Explorations 524 R 12 8. 17 8920 85 42 34 80 68 12
1. 848 16 = 53 2. 564 12 = 47 3. 5890 95 = 62 4. 1053 27 = 39
49 R 60 9. 678 33, 282 27 12 6 162 6 102 60
5.
32,886 261 126
6.
143, 088 542 264
7. 0 315 = 0 8. 315 0 is an error.
57 10. 3 171 15 21 21 0 Each student got 57 printer cartridges. 44 532 48 52 48 4 There will be 44 full boxes and 4 printers left over.
11. 12
Vocabulary, Readiness & Video Check 1.6 1. In 90 2 = 45, the answer 45 is called the quotient, 90 is called the dividend, and 2 is called the divisor. 2. The quotient of any number and 1 is the same number. 3. The quotient of any number (except 0) and the same number is 1. 4. The quotient of 0 and any number (except 0) is 0. 5. The quotient of any number and 0 is undefined. 6. The average of a list of numbers is the sum of the numbers divided by the number of numbers. 7. 0 8. zero; this zero becomes a placeholder in the quotient.
Copyright © 2025 Pearson Education, Inc.
15
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers 9. 202 102 + 15 = 20,619 10. This tells us we have a division problem since division may be used to separate a quantity into equal parts. 11. addition and division Exercise Set 1.6
526 26. 4 2104 20 10 8 24 24 0 Check: 526 4 = 2104
2. 72 9 = 8 4. 24 6 = 4 6. 0 4 = 0 8. 38 1 = 38 49 10.
12.
14.
1
49 45 5 9 12 is undefined 0
16. 6 6 = 1 18. 7 0 is undefined
28.
0 30
Check: 0 30 = 0 7 30. 8 56 56 0 Check: 7 8 = 56 11 32. 11 121 11 11 11 0 Check: 11 11 = 121
20. 18 3 = 6 34. 7 22. 5
17 85 5 35 35 0
Check: 17 5 = 85
24. 8
80 640 64 00
Check: 80 8 = 640
1 6
0
60 R 6 426 42 06
Check: 60 7 + 6 = 426 413 R 1 36. 3 1240 12 04 3 10 9 1 Check: 413 3 + 1 = 1240
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ISM: Prealgebra and Introductory Algebra
55 R 2 38. 3 167 15 17 15 2 Check: 55 3 + 2 = 167 833 R 1 40. 4 3333 32 13 12 13 12 1 Check: 833 4 + 1 = 3333 32 42. 23 736 69 46 46 0 Check: 32 23 = 736 48 2016 168 336 336 0 Check: 48 42 = 2016
44. 42
44 R 2 1938 176 178 176 2 Check: 44 44 + 2 = 1938
46. 44
Chapter 1: The Whole Numbers
612 R 10 48. 12 7354 72 15 12 34 24 10 Check: 612 12 + 10 = 7354 405 50. 14 5670 56 07 0 70 70 0 Check: 405 14 = 5670 52.
39 R 9 64 2505 192 585 576 9 Check: 39 64 + 9 = 2505
47 54. 123 5781 492 861 861 0 Check: 47 123 = 5781 96 R 52 56. 240 23, 092 21 60 1 492 1 440 52 Check: 96 240 + 52 = 23,092
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17
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
201 R 50 58. 203 40,853 40 6 25 0 253 203 50
3 040 68. 214 650, 560 642 85 0 8 56 8 56 00 0 0
Check: 201 203 + 50 = 40,853 303 R 63 60. 543 164, 592 162 9 1 69 0 1 692 1 629 63
70.
Check: 303 543 + 63 = 164,592 13 62. 8 104 8 24 24 0 603 R 2 64. 5 3017 30 01 0 17 15 2 1714 R 47 66. 50 85, 747 50 35 7 35 0 74 50 247 200 47
1 8
13 R 3 7 94 7 24 21 3 The quotient is 13 R 3.
3 R 20 72. 32 116 96 20 116 divided by 32 is 3 R 20. 15 R 3 78 5 28 25 3 The quotient is 15 R 3.
74. 5
58 4930 425 680 680 0 There are 58 students in the group.
76. 85
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
252000 78. 21 5292000 42 109 105 42 42 0 Each person received $252,000.
Chapter 1: The Whole Numbers
16 88. 320 5280 320 2080 1920 160 There are 16 whole feet in 1 rod. 30 6 180 18 00
3
90. 412 80. 14 5768 56 16 14 28 28 0 The truck hauls 412 bushels on each trip.
37 26 15 29 51 22 180 Average
180 30 6 169 5 845 5 34 30 45 45
21
82. Lane divider = 25 + 25 = 50 105 50 5280 50 28 0 280 250 30 There are 105 whole lane dividers. 23 R 1 84. 8 185 16 25 24 1 Yes, there is enough for a 22-student class. There is one 8-foot length and 1 additional foot of rope left over. That is, she has 9 feet of extra rope. 16 96 6 36 36 0 The players each scored 16 touchdowns.
92.
121 200 185 176 163 845
0 Average
845 169 5
2
94.
92 96 90 85 92 79
6
89 534 48 54 54 0
534 Average
86. 6
96.
53 40 30 123
534 89 6 41 3 123 12 03 3 0
The average temperature is 41.
Copyright © 2025 Pearson Education, Inc.
19
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
120. answers may vary Possible answer: 2 and 2
11
98.
100.
23 407 92 7011 7533
122.
712 54 2 848 35 600 38, 448
102.
712 54
86 46 10 10 76 36 10 10 66 26 10 10 56 16 10 10 46 6 Therefore, 86 10 = 8 R 6.
Mid-Chapter Review
658
1
1.
9 R 25 106. 31 304 279 25
42 63 89 194
2.
7006 451 6555
108. The quotient of 200 and 20 is 200 20, which is choice b.
3.
87 52 174 4350 4524
0 104.
0 because 0 23 = 0
23
110. 40 divided by 8 is 40 8, which is choice c.
112.
1, 955, 000, 000 7,820, 000, 000 4 38 36 22 20 20 2 0 0 000 000 The advertisers shown spent an average of $1,955,000,000. 2, 700, 000, 000 1, 980, 000, 000 1, 540, 000, 000 1, 600, 000, 000 7,820, 000, 000
4
562 4. 8 4496 40 49 48 16 16 0 5. 1 67 = 67 36 is undefined. 0
114. The average will decrease; answers may vary.
6.
116. No; answers may vary Possible answer: The average cannot be less than each of the four numbers.
7. 16 16 = 1
118. 84 21 = 4 The width is 4 inches. 2 0
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 8. 5 1 = 5 9. 0 21 = 0 10. 7 0 8 = 0 11. 0 7 = 0 12. 12 4 = 3 13. 9 7 = 63
Chapter 1: The Whole Numbers
1099 R 2 21. 7 7695 7 06 0 69 63 65 63 2
14. 45 5 = 9 15.
207 69 138
16.
207 69 276
17.
3718 2549 1169
1
11
1861 18. 7965 9826 182 R 4 19. 7 1278 7 57 56 18 14 4 20.
1259 63 3 777 75 540 79, 317
111 R 1 22. 9 1000 9 10 9 10 9 1 663 R 24 23. 32 21, 240 19 2 2 04 1 92 120 96 24 1 076 R 60 24. 65 70, 000 65 50 0 5 00 4 55 450 390 60
25.
4000 2963 1037
26. 10, 000 101 9 899
Copyright © 2025 Pearson Education, Inc.
21
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers 27.
38. 432,198 rounded to the nearest ten is 432,200. 432,198 rounded to the nearest hundred is 432,200. 432,198 rounded to the nearest thousand is 432,000.
303 101 303 30 300 30, 603
39. 6 + 6 + 6 + 6 = 24 6 6 = 36 The perimeter is 24 feet and the area is 36 square feet.
28. (475)(100) = 47,500 1
29.
62 9
40. 14 + 7 + 14 + 7 = 42 14 7 98 The perimeter is 42 inches and the area is 98 square inches.
71 The total of 62 and 9 is 71. 30.
62 9 558 The product of 62 and 9 is 558.
41.
6 R8 31. 9 62 54 8 The quotient of 62 and 9 is 6 R 8. 32.
62 9 53 The difference of 62 and 9 is 53.
33.
200 17 183 17 subtracted from 200 is 183.
34.
432 201 231 The difference of 432 and 201 is 231.
35. 9735 rounded to the nearest ten is 9740. 9735 rounded to the nearest hundred is 9700. 9735 rounded to the nearest thousand is 10,000.
13 9 6 28 The perimeter is 28 miles.
42. The unknown vertical side has length 4 + 3 = 7 meters. The unknown horizontal side has length 3 + 3 = 6 meters. 3 4 3 7 6 3 26 The perimeter is 26 meters. 3
Average
36. 1429 rounded to the nearest ten is 1430. 1429 rounded to the nearest hundred is 1400. 1429 rounded to the nearest thousand is 1000. 37. 20,801 rounded to the nearest ten is 20,800. 20,801 rounded to the nearest hundred is 20,800. 20,801 rounded to the nearest thousand is 21,000.
2 2
24 5 120 10 20 20 0
19 43. 15 25 37 24 120
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120 5
24
ISM: Prealgebra and Introductory Algebra
124 4 496 4 09 8 16 16 0
12
44.
108 131 98 159 496
Average
496
124
4 45.
28, 547 26, 372 2 175 The Twin Span Bridge is longer by 2175 feet.
46.
329 18 2632 3290 5922 The amount spent on toys is $5922.
Section 1.7 Practice Exercises
Chapter 1: The Whole Numbers
12. 36 [20 (4 2)] 43 6 36 [20 8] 43 6 36 12 43 6 36 12 64 6 3 64 6 61 13.
25 8 2 33 25 8 2 27 2(1) 2(3 2) 25 16 27 2 14 2 7
14. 36 6 3 5 6 3 5 18 5 23 15. Area (side)2 (12 centimeters)2 144 square centimeters The area of the square is 144 square centimeters. Calculator Explorations 1. 46 4096
1. 8 8 8 8 84
2. 56 15, 625
2. 3 3 3 33
3. 55 3125
3. 10 10 10 10 10 105
4. 76 117, 649
4. 5 5 4 4 4 4 4 4 52 46
11
5. 2 5. 4 4 4 16
2048
2
6. 68 1, 679, 616
6. 73 7 7 7 343 7. 74 53 2526 7. 111 11
8. 124 84 16, 640
8. 2 3 2 3 3 18 2
9. 63 75 43 10 = 4295
9. 9 3 8 4 27 8 4 27 2 25
10. 8 22 + 7 16 = 288
10. 48 3 22 48 3 4 16 4 64
11. 4(15 3 + 2) 10 2 = 8
11. (10 7)4 2 32 34 2 32 81 2 9 8118 99
12. 155 2(17 + 3) + 185 = 300
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23
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
Vocabulary, Readiness & Video Check 1.7
24. 81 8
1. In 25 32, the 2 is called the base and the 5 is called the exponent.
26. 54 5 5 5 5 625
2. To simplify 8 + 2 6, which operation should be performed first? multiplication
28. 33 3 3 3 27
3. To simplify (8 + 2) 6, which operation should be performed first? addition 4. To simplify 9(3 2) 3 + 6, which operation should be performed first? subtraction 5. To simplify 8 2 6, which operation should be performed first? division
30. 43 4 4 4 64 32. 83 8 8 8 512 34. 112 1111 121 36. 103 10 10 10 1000 38. 141 14
6. exponent; base 7. 1
40. 45 4 4 4 4 4 1024
8. division, multiplication, addition
42. 5 32 5 3 3 45
9. The area of a rectangle is length width. A square is a special rectangle where length = width. Thus, the area of a square is
44. 2 72 2 7 7 98
side side, or (side)2.
48. 100 10 5 + 4 = 10 5 + 4 = 50 + 4 = 54
Exercise Set 1.7 2. 5 5 5 5 5
46. 24 + 6 3 = 24 + 18 = 42
50. 42 7 6 = 6 6 = 0 4
52. 32
4. 6 6 6 6 6 6 6 67
8
54. 3 4 + 9 1 = 12 + 9 = 21
6. 10 10 10 103
693
8. 4 4 3 3 3 4 3 2
56.
3
10. 7 4 4 4 7 43
32
9
3
16. 62 6 6 36 18. 63 6 6 6 216
9
1
9
2
62.
40 8 2
5 3 20. 35 3 3 3 3 3 243
3
3
2
3
2
48 48 3 25 9 16
64. (9 7) (12 + 18) = 2 30 = 60
111111111111 1
5(12 7) 4 66.
2 4
63
60. 5 (10 15) 9 3 5 25 9 3 125 25 81 27 5 81 27 113
14. 6 6 2 9 9 9 9 62 2 94
22. 1
58. 62 (10 8) 62 2 36 2 72
12. 4 6 6 6 6 4 64
12
32 4 36
2
2
5 18
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5(5) 4 25 18
25 4
21 3 25 18 7
ISM: Prealgebra and Introductory Algebra 68. 18 7 0 = undefined 3
84. [15 (11 6) 22 ] (5 1)2 [15 5 22 ] 42 [15 5 4] 42 [3 4] 42
3
70. 2 3 (100 10) 2 3 10 8 3 10 24 10 14
7 42 7 16 23
72. [40 (8 2)] 25 [40 6] 25 34 25 34 32 2 74. (18 6) [(3 5) 2] (18 6) (8 2) 3 (8 2) 3 16 19 76. 35 [32 (9 7) 22 ] 10 3 35 [32 2 22 ] 10 3 35 [9 2 4] 10 3 35 7 10 3 5 10 3 5 30 35 78.
Chapter 1: The Whole Numbers
86. 29 {5 3[8 (10 8)] 50} 29 {5 3[8 2] 50} 29 {5 3(16) 50} 29 {5 48 50} 29 3 26 88. Area of a square (side)2 (9 centimeters)2 81 square centimeters Perimeter 4(side) 4(9 centimeters) 36 centimeters 90. Area of a square (side)2
52 23 14 25 8 1 18 18 9 10 5 4 1 4 2 4 1 4 8 4 2
3 92 3 812 80. 3(10 6) 22 2 1 1 3(4) 84 3(4) 4 1 84 12 4 1 84 8 1 84 7 12 82. 10 2 33 2 20 10 2 27 2 20 5 27 2 20 5 54 20 39
(41 feet)2 1681 square feet Perimeter = 4(side) = 4(41 feet) = 164 feet 92. The statement is true. 94. 49 4 4 4 4 4 4 4 4 4 The statement is false. 96. (2 + 3) (6 2) = (5) (4) = 20 98. 24 (3 2 2) 5 24 (6 2) 5 24 8 5 3 5 15 100. The total perimeter is 1260 feet. 4 1260 = 5040 The total charge is $5040. 102. 253 (45 7 5) 5 253 (45 35) 5 253 (10) 5 15, 625 10 5 156, 250 5 781, 250 104. answers may vary
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25
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
Section 1.8 Practice Exercises 1. x 2 = 7 2 = 5 2. y(x 3) = 4(8 3) = 4(5) = 20 3.
y 6 18 6 24 4 x 6 6
4. 25 z3 x 25 23 1 25 8 1 18 5.
5(F 32) 5(41 32) 5(9) 45 5 9 9 9 9
6. 3( y 6) 6 3(8 6) 0 6 3(2) 0 6 6 6 True Yes, 8 is a solution. 7. 5n + 4 = 34 Let n be 10. 5(10) 4 0 34 50 4 0 34 54 34 False No, 10 is not a solution. Let n be 6. 5(6) 4 0 34 30 4 0 34 34 34 True Yes, 6 is a solution. Let n be 8. 5(8) 4 0 34 40 4 0 34 44 34 False No, 8 is not a solution. 8. a.
Twice a number is 2x.
b.
8 increased by a number is 8 + x or x + 8.
c.
10 minus a number is 10 x.
d.
10 subtracted from a number is x 10.
e.
The quotient of 6 and a number is 6 x or
6
.
x Vocabulary, Readiness & Video Check 1.8 1. A combination of operations on letters (variables) and numbers is an expression. 2. A letter that represents a number is a variable. 2 6
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ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
3. 3x 2y is called an expression and the letters x and y are variables. 4. Replacing a variable in an expression by a number and then finding the value of the expression is called evaluating the expression. 5. A statement of the form “expression = expression” is called an equation. 6. A value for the variable that makes an equation a true statement is called a solution. 7. When a letter and a variable are next to each other, the operation is an understood multiplication. 8. When first replacing f with 8, we don’t know if the statement is true or false. 9. decreased by Exercise Set 1.8 2.
a
b
a+b
ab
ab
ab
24
6
24 + 6 = 30
24 6 = 18
24 6 = 144
24 6 = 4
a
b
a+b
ab
ab
ab
298
0
298 + 0 = 298
298 0 = 298
298 0 = 0
298 0 is undefined.
4.
6.
a
b
a+b
ab
ab
ab
82
1
82 + 1 = 83
82 1 = 81
82 1 = 82
82 1 = 82
8. 7 + 3z = 7 + 3(3) = 7 + 9 = 16 10. 4yz + 2x = 4(5)(3) + 2(2) = 60 + 4 = 64 12. x + 5y z = 2 + 5(5) 3 = 2 + 25 3 = 24 14. 2y + 5z = 2(5) + 5(3) = 10 + 15 = 25 16. y3 z 53 3 125 3 122 18. 3yz 2 1 3(5)(3)2 1 3 5 9 1 135 1 136 20. 3 (2 y 4) 3 (2 5 4) 3 (10 4) 3 6 9 22. x4 ( y z) 24 (5 3) 24 2 16 2 14
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27
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
24.
8 yz 8 5 3 120 8 15 15 15
26.
6 3x 6 3(2) 6 6 12 4 z 3 3 3
28.
2z 6 2 3 6 6 6 12 4 3 3 3 3
30.
70 15 70 15 70 15 7 5 2 2 y z 2 5 3 10 3
32. 3x2 2x 5 3 22 2 2 5 3 4 2 2 5 12 4 5 11 34. (4 y 3z)2 (4 5 3 3)2 (20 9)2 292 841 36. (xz 5)4 (2 3 5)4 (6 5)4 14 1 38. 3x( y z) 3 2(5 3) 3 2(8) 6(8) 48 40. xz(2 y x z) 2 3(2 5 2 3) 2 3(10 2 3) 2 3(9) 6(9) 54 42.
44.
2 8
6z 2 y 6 3 2 5 4 4 18 10 4 28 4 7 F
50
59
68
77
5(F 32) 9
5(50 32) 5(18) 10 9 9
5(59 32) 5(27) 15 9 9
5(68 32) 5(36) 20 9 9
5(77 32) 5(45) 25 9 9
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ISM: Prealgebra and Introductory Algebra
46. Let n be 9. n2 7 92 0 7 7 7 True Yes, 9 is a solution. 48. Let n be 50. 250 5n 250 0 5(50) 250 250 True Yes, 50 is a solution. 50. Let n be 8. 11n 3 91 11(8) 3 0 91 88 3 0 91 91 91 True Yes, 8 is a solution. 52. Let n be 0. 5(n 9) 40 5(0 9) 0 40 5(9) 0 40 45 40 False No, 0 is not a solution. 54. Let x be 2. 3x 6 5x 10 3(2) 6 0 5(2) 10 6 6 0 10 10 0 0 True Yes, 2 is a solution. 56. Let x be 5. 8x 30 2x 8(5) 30 0 2(5) 40 300 10 10 10 True Yes, 5 is a solution. 58. n + 3 = 16 Let n be 9. 9 3 0 16 12 16 False Let n be 11. 11 3 0 16 14 16 False Let n be 13. 13 3 0 16 16 16 True 13 is a solution.
Chapter 1: The Whole Numbers
60. 3n = 45 Let n be 15. 315 0 45 45 45 True Let n be 30. 3 30 0 45 90 45 False Let n be 45. 3 45 0 45 135 45 False 15 is a solution. 62. 4n + 4 = 24 Let n be 0. 4 0 4 0 24 0 4 0 24 4 24 False Let n be 5. 4 5 40 24 20 4 0 24 24 24 True Let n be 10. 4 10 4 0 24 40 4 0 24 44 24 False 5 is a solution. 64. 6(n + 2) = 23 Let n be 1. 6(1 2) 0 23 6(3) 0 23 18 23 False Let n be 3. 6(3 2) 0 23 6(5) 0 23 30 23 False Let n be 5. 6(5 2) 0 23 6(7) 0 23 42 23 False None are solutions. 66. 9x 15 = 5x + 1 Let x be 2. 9 2 15 0 5 2 1 18 15 0 10 1 3 11 False Let x be 4. 9 4 15 0 5 4 1 36 15 0 20 1 21 21 True Copyright © 2025 Pearson Education, Inc.
29
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
Let x be 11. 9 1115 0 5 11 1 99 15 0 55 1 84 56 False 4 is a solution.
100. As F gets larger,
5(F 32) 9
gets larger.
Chapter 1 Vocabulary Check 1. The whole numbers are 0, 1, 2, 3, ...
68. The sum of three and a number is 3 + x. 70. The difference of a number and five hundred is x 500. 72. A number less thirty is x 30.
76. A number divided by 11 is x 11 or
x
. 11
78. The quotient of twenty and a number, decreased 20 by three is 3. x 80. The difference of twice a number, and four is 2x 4. 82. Twelve subtracted from a number is x 12. 84. The sum of a number and 7 is x + 7. 86. The product of a number and 7 is 7x. 88. Twenty decreased by twice a number is 20 2x. 90. Replace x with 3 and y with 5. 6y + 3x = 6(5) + 3(3) = 30 + 9 = 39
5. To find the area of a rectangle, multiply length times width. 6. The digits used to write numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. 7. A letter used to represent a number is called a variable. 8. An equation can be written in the form “expression = expression.” 9. A combination of operations on variables and numbers is called an expression. 10. A solution of an equation is a value of the variable that makes the equation a true statement. 11. A collection of numbers (or objects) enclosed by braces is called a set. 12. The 21 above is called the sum.
92. Replace x with 3 and y with 5. x3 4 y 33 4(5) 27 20 47
13. The 5 above is called the divisor. 14. The 35 above is called the dividend.
94. 2(x y)2 2(23 72)2 2(95)2 2(9025) 18, 050
15. The 7 above is called the quotient. 16. The 3 above is called a factor.
96. 16 y 20x x3 16 72 20 23 233 1152 460 12,167 12,859 x
3. The position of each digit in a number determines its place value. 4. An exponent is a shorthand notation for repeated multiplication of the same factor.
74. A number times twenty is 20x.
98.
2. The perimeter of a polygon is its distance around or the sum of the lengths of its sides.
is the smallest; answers may vary.
17. The 6 above is called the product. 18. The 20 above is called the minuend. 19. The 9 above is called the subtrahend. 20. The 11 above is called the difference.
3 21. The 4 above is called an addend.
3 0
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ISM: Prealgebra and Introductory Algebra
Chapter 1 Review
Chapter 1: The Whole Numbers
16.
583 279 304
17.
428 21 449
18.
819
1. The place value of 4 in 7640 is tens. 2. The place value of 4 in 46,200,120 is tenmillions. 3. 7640 is written as seven thousand, six hundred forty.
1
4. 46,200,120 is written as forty-six million, two hundred thousand, one hundred twenty.
21 840
5. 3158 = 3000 + 100 + 50 + 8 6. 403,225,000 = 400,000,000 + 3,000,000 + 200,000 + 20,000 + 5000 7. Eighty-one thousand, nine hundred in standard form is 81,900.
19.
10. Locate Middle East in the first column and read across to the number in the 2022 column. There were 212,000,000 Internet users in Middle East in 2022.
18 49 67
28 39 67
15.
462 397 65
91 3623 497 4211 82 1647 238 1967 11
23.
74 342 918 1334 The sum of 74, 342, and 918 is 1334. 2
24.
49 529 308 886 The sum of 49, 529, and 308 is 886.
25.
25,862 7 965 17,897 7965 subtracted from 25,862 is 17,897.
1
14.
8000 92 7908
11
22.
1
13.
12 1
21.
11. Locate the smallest number in the 2016 column. Oceania/Australia had the fewest Internet users in 2016. 12. Locate the largest number in the 2013 column. Asia had the greatest number of Internet users in 2013.
4000 86 3914
20.
8. Six billion, three hundred four million in standard form is 6,304,000,000. 9. Locate Europe in the first column and read across to the number in the 2022 column. There were 750,000,000 Internet users in Europe in 2022.
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31
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
26.
39, 007 4 349 34, 658 4349 subtracted from 39,007 is 34,658.
27.
205 7318 7523 The total distance is 7523 miles.
36.
490 250 240 The balance increased by $240 from June to August.
1
1 11
28.
62, 589 65, 340 69, 770 197, 699 Their total earnings were $197,699.
29. 40 + 52 + 52 + 72 = 216 The perimeter is 216 feet. 30. 11 + 20 + 35 = 66 The perimeter is 66 kilometers. 31.
32.
653 341 312 The number of Internet users in Africa increased by 312 million or 312,000,000. 543 350 193 There were 193 million or 193,000,000 more Internet users in Latin America/Caribbean than in North America in 2022.
33. Find the shortest bar. The balance was the least in May. 34. Find the tallest bar. The balance was the greatest in August. 35.
3 2
280 170 110 The balance decreased by $110 from February to April.
37. To round 43 to the nearest ten, observe that the digit in the ones place is 3. Since this digit is less than 5, we do not add 1 to the digit in the tens place. The number 43 rounded to the nearest ten is 40. 38. To round 45 to the nearest ten, observe that the digit in the ones place is 5. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 45 rounded to the nearest ten is 50. 39. To round 876 to the nearest ten, observe that the digit in the ones place is 6. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 876 rounded to the nearest ten is 880. 40. To round 493 to the nearest hundred, observe that the digit in the tens place is 9. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 493 rounded to the nearest hundred is 500. 41. To round 3829 to the nearest hundred, observe that the digit in the tens place is 2. Since this digit is less than 5, we do not add 1 to the digit in the hundreds place. The number 3829 rounded to the nearest hundred is 3800. 42. To round 57,534 to the nearest thousand, observe that the digit in the hundreds place is 5. Since this digit is at least 5, we add 1 to the digit in the thousands place. The number 57,534 rounded to the nearest thousand is 58,000. 43. To round 39,583,819 to the nearest million, observe that the digit in the hundred-thousands place is 5. Since this digit is at least 5, we add 1 to the digit in the millions place. The number 39,583,819 rounded to the nearest million is 40,000,000. 44. To round 768,542 to the nearest hundredthousand, observe that the digit in the tenthousands place is 6. Since this digit is at least 5, we add 1 to the digit in the hundred-thousands place. The number 768,542 rounded to the nearest hundred-thousand is 800,000.
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ISM: Prealgebra and Introductory Algebra
2
45.
Chapter 1: The Whole Numbers
54. 25(9)(4) = 225(4) = 900 or 25(4)(9) = 100(9) = 900
3785 648 2866
rounds to rounds to rounds to
3800 600 2900 7300
5925 1787
rounds to rounds to
5900 1800 4100
57.
630 192 271 56 703 454 329
rounds to rounds to rounds to rounds to rounds to rounds to rounds to
600 200 300 100 700 500 300 2700
586 29 5 274 11 720 16, 994
58.
242 37 1694 7260 8954
59.
642 177 4 494 44 940 64 200 113, 634
55. 26 34 0 = 0 56. 62 88 0 = 0
46.
47.
They traveled approximately 2700 miles. 48.
837, 427, 045 664, 099,841
rounds to rounds to
The population of Europe was approximately 837,000,000 and the population of Latin America/Caribbean was approximately 664,000,000. The difference in population was about 173,000,000. 49.
837,000,000 664,000,000 173,000,000
60.
347 129 3 123 6 940 34 700 44, 763
61.
1026 401 1 026 410 400 411, 426
62.
2107 302 4 214 632 100 636, 314
276 8 2208
50.
349 4 1396
51.
57 40 2280
52.
69 42 138 2760 2898
53. 20(7)(4) = 140(4) = 560
Copyright © 2025 Pearson Education, Inc.
33
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers 63. “Product” indicates multiplication. 250 6 1500 The product of 6 and 250 is 1500.
4 R2 72. 4 18 16 2 Check: 4 4 + 2 = 18
64. “Product” indicates multiplication. 820 6 4920 The product of 6 and 820 is 4920.
73. 918 0 is undefined.
65.
32 15 160 320 480
38 11 38 380 418
480 418 898 The total cost is $898. 66.
17, 265 20 345, 300 The total cost for 20 students is $345,300.
68. Area (length)(width) (25 centimeters)(20 centimeters) 500 square centimeters 69.
70.
36
Check:
7 7 49
4
Check:
9 4 36
9
19 R 7 76. 8 159 8 79 72 7 Check: 19 8 + 7 = 159 24 R 2 626 52 106 104 2
Check: 24 26 + 2 = 626
78. 19
7
7
35 R 15 680 57 110 95 15
Check: 35 19 + 15 = 680
5 R2 71. 5 27 25 2 Check: 5 5 + 2 = 27 34
Check: 0 668 = 0
33 R 2 75. 5 167 15 17 15 2 Check: 33 5 + 2 = 167
77. 26
67. Area (length)(width) (13 miles)(7 miles) 91 square miles
49
74. 0 668 = 0
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ISM: Prealgebra and Introductory Algebra
18 R 2 83. 5 92 5 42 40 2 The quotient of 92 and 5 is 18 R 2.
506 R 10 79. 47 23, 792 23 5 29 0 292 282 10
21 R 2 86 8 06 4 2 The quotient of 86 and 4 is 21 R 2.
Check: 506 47 + 10 = 23,792
80. 53
Chapter 1: The Whole Numbers
84. 4
907 R 40 48,111 47 7 41 0 411 371 40
27 85. 24 648 48 168 168 0 27 boxes can be filled with cans of corn.
Check: 907 53 + 40 = 48,111 2793 R 140 81. 207 578, 291 414 164 2 144 9 19 39 18 63 761 621 140
13 22,880 17 60 5 280 5 280 0 There are 13 miles in 22,880 yards.
86. 1760
Check: 2793 207 + 140 = 578,291 2012 R 60 82. 306 615, 732 612 37 0 3 73 3 06 672 612
87. Divide the sum by 4. 76 49 32 47 204 The average is 51.
4
51 204 20 04 4 0
88. Divide the sum by 4.
60 Check: 2012 306 + 60 = 615,732
23 85 62 66 236 The average is 59.
Copyright © 2025 Pearson Education, Inc.
4
59 236 20 36 36 0
35
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
105. (6 4)3 [102 (3 17)] (6 4)3 [102 20]
89. 82 8 8 64
(6 4)3 [100 20]
90. 5 5 5 5 125 3
23 5 85 40
91. 5 92 5 9 9 405 92. 4 102 4 10 10 400
106. (7 5)3 [92 (2 7)] (7 5)3 [92 9]
93. 18 2 + 7 = 9 + 7 = 16
(7 5)3 [81 9]
94. 12 8 4 = 12 2 = 10
23 9 89 72
5(62 3) 5(36 3) 5(33) 165 95. 32 2 9 2 11 11 15
5 7 3 5 107.
96. 7(163 8) 7(8) 56 7 8 8 2
2(11 32 ) 4 8 111
108. 97. 48 8 2 = 6 2 = 12
35 15
2(11 9)
3(9 23)
32 11 3(9 8)
20 2(2) 21 3(1)
20
4
21 3
5
7
98. 27 9 3 = 3 3 = 9
109. Area (side)2 (7 meters)2 49 square meters
99. 2 3[15 (20 17) 3] 5 2 2 3[15 3 3] 5 2 2 3[1 3 3] 5 2 2 3[1 9] 5 2 2 3 10 5 2 2 30 10 42
110. Area (side)2 (3 inches)2 9 square inches
100. 21 [2 (7 5) 10] 8 2 4
21[24 2 10] 8 2 21[16 2 10] 8 2 21 4 8 2 21 4 16
111.
2x
z
25 2
10
5
2
112. 4x 3 = 4 5 3 = 20 3 = 17 113. x 7 5 7 is undefined. y 0 114.
y
0
55
5x
0
0
25
3 3 115. x 2z 5 2 2 125 2 2 125 4 121
33 2 2 101. 19 2(3 2 ) 19 2(9 4) 19 2(5) 19 10
9 2 2 102. 16 2(4 3 ) 16 2(16 9) 16 2(7) 16 14 2
103. 4 5 2 7 = 20 14 = 6
116.
7 x 7 5 12 2 3z 3 2 6
117. ( y z)2 (0 2)2 22 4 100 118.
x
y
3
100 5
0
20 0 20
3
119. Five subtracted from a number is x 5. 120. Seven more than a number is x + 7.
104. 8 7 3 9 = 56 27 = 29 36
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ISM: Prealgebra and Introductory Algebra
121. Ten divided by a number is 10 x or
10
Chapter 1: The Whole Numbers 175 25 0 150 150 150 True 175 is a solution.
.
x 122. The product of 5 and a number is 5x. 123. Let n be 5. n 12 20 3 5 12 0 20 3 17 17 True Yes, 5 is a solution. 124. Let n be 23. n 8 10 6 23 8 0 10 6 15 16 False No, 23 is not a solution. 125. Let n = 14. 30 3(n 3) 30 0 3(14 3) 30 0 3(11) 30 33 False No, 14 is not a solution. 126. Let n be 20. 5(n 7) 65 5(20 7) 0 65 5(13) 0 65 65 65 True Yes, 20 is a solution. 127. 7n = 77 Let n be 6. 7 6 0 77 42 77 False Let n be 11. 7 11 0 77 77 77 True Let n be 20. 7 20 0 77 140 77 False 11 is a solution. 128. n 25 = 150 Let n be 125. 125 25 0 150 100 150 False Let n be 145. 145 25 0 150 120 150 False Let n be 175.
129. 5(n + 4) = 90 Let n be 14. 5(14 4) 0 90 5(18) 0 90 90 90 True Let n be 16. 5(16 4) 0 90 5(20) 0 90 100 90 False Let n be 26. 5(26 4) 0 90 5(30) 0 90 150 90 False 14 is a solution. 130. 3n 8 = 28 Let n be 3. 3(3) 8 0 28 9 8 0 28 1 28 False Let n be 7. 3(7) 8 0 28 21 8 0 28 13 28 False Let n be 15. 3(15) 8 0 28 45 80 28 37 28 False None are solutions. 131.
485 68 417
132.
729 47 682
133.
732 3 2196
134.
629 4
2516
Copyright © 2025 Pearson Education, Inc.
37
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
22
135.
374 29 698 1101 21
136.
593 52 766 1411
458 R 8 137. 13 5962 52 76 65 112 104 8 237 R 1 138. 18 4267 36 66 54 127 126 1 139.
1968 36 11 808 59 040 70,848
140.
5324 18 42 592 53 240 95,832
141.
2000 356 1644
142.
9000 519 8481
38
143. To round 842 to the nearest ten, observe that the digit in the ones place is 2. Since this digit is less than 5, we do not add 1 to the digit in the tens place. The number 842 rounded to the nearest ten is 840. 144. To round 258,371 to the nearest hundredthousand, observe that the digit in the tenthousands place is 5. Since this digit is at least 5, we add 1 to the digit in the hundred-thousands place. The number 258,371 rounded to the nearest hundred-thousand is 300,000. 145. 24 4 2 = 6 2 = 12 146. (15 3) (8 5) (18)(3) 54 6 8 1 9 23 1 147. Let n be 9. 5n 6 40 5 9 6 0 40 45 6 0 40 39 40 False No, 9 is not a solution. 148. Let n be 3. 2n 6 5n 15 2(3) 6 0 5(3) 15 6 6 0 15 15 0 0 True Yes, 3 is a solution. 53 1714 160 114 96 18 There are 53 full boxes with 18 left over.
149. 32
150.
27 2 54
8 4 32
54 32 86 The total bill before taxes is $86.
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ISM: Prealgebra and Introductory Algebra
Chapter 1 Getting Ready for the Test 1. In the number 28,690,357,004, the digit 5 is in the ten-thousands place; D. 2. In the number 28,690,357,004, the digit 8 is in the billions place; E. 3. In the number 28,690,357,004, the digit 6 is in the hundred-millions place; F. 4. In the number 28,690,357,004, the digit 0 to the far left is in the millions place; B. 5. To simplify 6 3 2, the first operation to perform is multiplying 3 2; C. 6. To simplify (6 3) 2, the first operation to perform is subtracting 3 from 6; B. 7. To simplify 6 3 2, the first operation to perform is dividing 6 by 3; D. 8. To simplify 6 + 3 2, the first operation to perform is adding 6 and 3; A.
9. 5 23 58 40; C 10. Since 35 5 = 7, the expression is a b; B. 11. since 35 5 = 175, the expression is ab; D. 12. Since 35 5 = 30, the expression is a b; A. 13. Since 35 + 5 = 40, the expression is a + b; C. 1
3. 59 Chapter 1 Test 82 1. 82,426 141 in words is eighty-two thousand, four hundred twenty-six. 2. Four hundred two thousand, five hundred fifty in standard form is 402,550.
4.
600 487 113
Chapter 1: The Whole Numbers
5.
496 30 14,880
6. 69
766 R 42 52,896 48 3 4 59 4 14 456 414 42
7. 23 52 2 2 2 5 5 200 8. 98 1 = 98 9. 0 49 = 0 10. 62 0 is undefined. 11. (24 5) 3 (16 5) 3 11 3 33
12. 16 9 3 4 7 16 3 4 7 16 12 7 28 7 21 13. 61 23 6 2 2 2 48 14. 2[(6 4)2 (22 19)2 ] 10 2[22 32 ] 10 82 2[4 9] 10 5 410 2 2[13] 10 62 40 26 10 79 10 36 15. 5698 1000 = 5,698,000 16. Divide the sum by 5.
84 Copyright © 2025 Pearson Education, 90 Inc. 95 410 The average is 82.
10 0
39
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
17. To round 52,369 to the nearest thousand, observe that the digit in the hundreds place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. The number 52,369 rounded to the nearest thousand is 52,000. 18.
6289 5403 1957
rounds to rounds to rounds to
6 300 5 400 + 2 000 13,700
19.
4267 2738
rounds to rounds to
4300 2700 1600
20.
21.
22.
107 15 92 15 107 122
53 16 318 530 848
129 12 258 1290 1548
848 1548 2396 The total cost is $2396.
Area (side)2 (5 centimeters)2 25 square centimeters
107 15 535 1070 1605
29. Perimeter (20 10 20 10) yards 60 yards Area (length)(width) (20 yards)(10 yards) 200 square yards 30. Replace x with 2. 5(x3 2) 5(23 2) 5(8 2) 5(6) 30 31. Replace x with 7 and y with 8. 3x 5 3(7) 5 21 5 16 1 2y 2 8 16 16
17 493 29 203 203 0 Each can cost $17.
24. 29
40
27.
28. Perimeter (5 5 5 5) centimeters 20 centimeters
7 R2 23. 15 107 105 2
25.
26. 45 8 360 There are 360 calories in 8 tablespoons of white granulated sugar.
32. a.
The quotient of a number and 17 is x 17 or x . 17
b. Twice a number, decreased by 20 is 2x 20.
320 139 91 The higher-priced one is $91 more.
33. Replace n with 6. 5n 11 19 5(6) 110 19 30 110 19 19 19 True 6 is a solution.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 1: The Whole Numbers
34. n + 32 = 4n + 2 Replace n with 0. 0 32 0 4(0) 2 32 0 0 2 32 2 False Replace n with 10. 10 32 0 4(10) 2 42 0 40 2 42 42 True Replace n with 20. 20 32 0 4(20) 2 52 0 80 2 52 82 False 10 is a solution.
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41
Chapter 2 Section 2.1 Practice Exercises 1. a.
If 0 represents the surface of the earth, then 3805 below the surface of the earth is 3805.
b. If zero degrees Fahrenheit is represented by 0F, then 85 degrees below zero, Fahrenheit is represented by 85F. 2. 3. a.
0 > 5 since 0 is to the right of 5 on a number line.
b.
3 < 3 since 3 is to the left of 3 on a number line.
c.
7 > 12 since 7 is to the right of 12 on a number line.
4. a.
|6| = 6 because 6 is 6 units from 0.
b.
|4| = 4 because 4 is 4 units from 0.
c.
|12| = 12 because 12 is 12 units from 0.
5. a. b. 6. a.
The opposite of 14 is 14. The opposite of 9 is (9) or 9. |7| = 7
b. |4| = 4 c.
(12) = 12
7. |x| = |6| = 6 8. The planet with the highest average daytime surface temperature is the one that corresponds to the bar that extends the furthest in the positive direction (upward). Venus has the highest average daytime surface temperature. Vocabulary, Readiness & Video Check 2.1 1. The numbers ...3, 2, 1, 0, 1, 2, 3, ... are called integers. 2. Positive numbers, negative numbers, and zero together are called signed numbers. 3. The symbols “<” and “>” are called inequality symbols. 4. Numbers greater than 0 are called positive numbers while numbers less than 0 are called negative numbers. 5. The sign “<” means is less than and “>” means is greater than. 6. On a number line, the greater number is to the right of the lesser number. 7. A number’s distance from 0 on the number line is the number’s absolute value. 8. The numbers 5 and 5 are called opposites.
42
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
9. number of feet a miner works underground 10. The tick marks are labeled with the integers. 11. 0 will always be greater than any of the negative integers. 12. 8; |8| = 8 13. A negative sign can be translated into the phrase “opposite of.” 14. Lake Eyre Exercise Set 2.1 2. If 0 represents the surface of the water, then 25 feet below the surface of the water is 25. 4. If 0 represents sea level, then 282 feet below sea level is 282. 6. If 0 represents 0 degrees Fahrenheit, then 134 degrees above zero is +134. 8. If 0 represents the surface of the ocean, then 14,040 below the surface of the ocean is 14,040. 10. If 0 represents a loss of $0, then a loss of $400 million is 400 million. 12. If 0 represents 0 Celsius, then 10 below 0 Celsius is 10. Since 5 below 0 Celsius is 5 and 10 is less than 5, 10 (or 10 below 0 Celsius) is cooler. 14. From the blue map, the coldest temperature on record for the state of Arkansas is 29F. 16. 18.
20. 22. 24. 8 < 0 since 8 is to the left of 0 on a number line. 26. 12 < 10 since 12 is to the left of 10 on a number line. 28. 27 > 29 since 27 is to the right of 29 on a number line. 30. 13 > 13 since 13 is to the right of 13 on a number line. 32. |7| = 7 since 7 is 7 units from 0 on a number line. 34. |19| = 19 since 19 is 19 units from 0 on a number line. 36. |100| = 100 since 100 is 100 units from 0 on a number line. 38. |10| = 10 since 10 is 10 units from 0 on a number line.
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
80. |8| = 8 |4| = 4 Since 8 < 4, |8| < |4|.
40. The opposite of 8 is negative 8. (8) = 8 42. The opposite of negative 6 is 6. (6) = 6 44. The opposite of 123 is negative 123. (123) = 123 46. The opposite of negative 13 is 13. (13) = 13 48. |11| = 11
82. (38) = 38 Since 22 < 38, 22 < (38). 84. If the number is 13, then the absolute value of 13 is 13 and the opposite of 13 is 13. 86. If the opposite of a number is 90, then the number is 90 and its absolute value is 90. 88. The ‘bar’ that is equal to 0 corresponds to Lake Maracaibo, so Lake Maracaibo has an elevation at sea level.
50. |43| = 43 52. |18| = 18
90. The bar that extends second to the farthest in the negative direction corresponds to Lake Eyre, so Lake Eyre has the second lowest elevation.
54. (27) = 27 56. (14) = 14
92. The smallest number on the graph is 269C, which corresponds to helium.
58. |29| = 29 60. |x| = |8| = 8
94. The number on the graph closest to +300C is 280C, which corresponds to phosphorus.
62. |x| = |10| = 10
96. 9 + 0 = 9
64. |x| = |32| = 32
98. 66. |x| = |1| = 1 68. 4 > 17 since 4 is to the right of 17 on a number line.
72. |17| = 17 (17) = 17 Since 17 < 17, |17| < (17).
|5|, (4), 23, |10|.
76. 45 < 0 since 45 is to the left of 0 on a number line.
44
362 37 90 489
102. |10| = 10, 23 8, |5| = 5, and (4) = 4, so the numbers in order from least to greatest are
74. |24| = 24 (24) = 24 Since 24 = 24, |24| = (24).
78. |45| = 45 |0| = 0 Since 45 > 0, |45| > |0|.
1
100.
70. |8| = 8 |4| = 4 Since 8 > 4, |8| > |4|.
20 15 35
104. 14 1, (3) = 3, |7| = 7, and |20| = 20, so the numbers in order from least to greatest are |7|, 14, (3), |20|. 106. 33 27, |11| = 11, (10) = 10, 4 = 4, |2| = 2, so the numbers in order from least to greatest are |11|, 4, |2|, (10), and 33.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
|0| = 0; since 0 < 4, then |0| > 4 is false.
9. 54 + 20 = 34
b.
|4| = 4; since 4 = 4, then |4| > 4 is false.
10. 7 + (2) = 5
c.
|5| = 5; since 5 > 4, then |5| > 4 is true.
11. 3 + 0 = 3
d. |100| = 100; since 100 > 4, then |100| > 4 is true.
12. 18 + (18) = 0
108. a.
13. 64 + 64 = 0
110. (|(7)|) = (|7|) = 7 112. False; consider 0, where |0| = 0 and 0 is not positive. 114. True; zero is always less than a positive number since it is to the left of it on a number line. 116. No; b > a because b is to the right of a on the number line. 118. answers may vary
14. 6 + (2) + (15) = 4 + (15) = 11 15. 5 (3) 12 (14) 2 12 (14) 14 (14) 0 16. x + 3y = 6 + 3(2) = 6 + 6 = 0 17. x + y = 13 + (9) = 22 18. Temperature at 8 a.m. 7 (4) (7) 3 (7) 4 The temperature was 4F at 8 a.m.
120. no; answers may vary Section 2.2 Practice Exercises 1.
Calculator Explorations 1. 256 + 97 = 159 2. 811 + (1058) = 247
2. 3. 6(15) + (46) = 44 4. 129 + 10(48) = 351 5. 108,650 + (786,205) = 894,855
3.
6. 196,662 + (129,856) = 326,518 Vocabulary, Readiness & Video Check 2.2 4. |3| + |19| = 3 + 19 = 22 The common sign is negative, so (3) + (19) = 22.
1. If n is a number, then n + n = 0.
5. 12 + (30) = 42
2. Since x + n = n + x, we say that addition is commutative.
6. 9 + 4 = 13
3. If a is a number, then (a) = a.
7. |1| = 1, |26| = 26, and 26 1 = 25 26 > 1, so the answer is positive. 1 + 26 = 25
4. Since n + (x + a) = (n + x) + a, we say that addition is associative.
8. |2| = 2, |18| = 18, and 18 2 = 16 18 > 2, so the answer is negative. 2 + (18) = 16
5. Negative; the numbers have different signs and the sign of the sum is the same as the sign of the number with the larger absolute value, 6.
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45
Chapter 2: Integers and Introduction to Solving Equations
ISM: Prealgebra and Introductory Algebra
6. Negative; the numbers have the same sign (both are negative) and we keep this common sign in the sum. 7. The diver’s current depth is 231 feet below the surface. Exercise Set 2.2 2.
6 + (5) = 11 4.
10 + (3) = 7 6.
9 + (4) = 5 8. 15 + 42 = 57 10. |5| + |4| = 5 + 4 = 9 The common sign is negative, so 5 + (4) = 9. 12. 62 + 62 = 0 14. |8| |3| = 8 3 = 5 8 > 3, so the answer is positive. 8 + (3) = 5 16. 8 + 0 = 8 18. |9| |5| = 9 5 = 4 9 > 5, so the answer is negative. 5 + (9) = 4 20. |6| + |1| = 6 + 1 = 7 The common sign is negative, so 6 + (1) = 7. 22. |23| + |23| = 23 + 23 = 46 The common sign is negative, so 23 + (23) = 46. 24. |400| + |256| = 400 + 256 = 656 The common sign is negative, so 400 + (256) = 656. 26. |24| |10| = 24 10 = 14 24 > 10, so the answer is positive. 24 + (10) = 14
46
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 28. |8| |4| = 8 4 = 4 8 > 4, so the answer is negative. 8 + 4 = 4 30. |89| |37| = 89 37 = 52 89 > 37, so the answer is negative. 89 + 37 = 52 32. |62| |32| = 62 32 = 30 62 > 32, so the answer is positive. 32 + 62 = 30 34. |375| |325| = 375 325 = 50 375 > 325, so the answer is negative. 325 + (375) = 50 36. |56| + |33| = 56 + 33 = 89 The common sign is negative, so 56 + (33) = 89.
Chapter 2: Integers and Introduction to Solving Equations 68. The sum of 49, 2, and 40 is 49 + (2) + 40 = 51 + 40 = 11. 70. 0 + (248) + 8 + (16) + (28) + 32 = 248 + 8 + (16) + (28) + 32 = 240 + (16) + (28) + 32 = 256 + (28) + 32 = 284 + 32 = 252 The diver’s final depth is 252 meters below the surface. 72. Since 6 < 3, Smith won Round 4. 74. The bar for 2021 has a height of 94,680 so the net income in 2021 was $94,680,000,000. 76.
38. 1 + 5 + (8) = 4 + (8) = 4 40. 103 + (32) + (27) = 135 + (27) = 162 42. 18 (9) 5 (2) 9 5 (2) 14 (2) 12 44. 34 (12) (11) 213 22 (11) 213 11 213 224 46. 12 + (3) + (5) = 15 + (5) = 20 48. 35 + (12) = 47
57, 410 94, 680 99,800 251,890 The total net income for 2020, 2021, and 2022 was $251,890,000,000.
78. 14 + (5) + (8) + 7 = 9 + (8) + 7 = 1 + 7 = 8 Their total score was 8. 80. 10,412 + (1786) + 15,395 + 31,418 = 12,198 + 15,395 + 31,418 = 3197 + 31,418 = 34,615 The net income for all the years shown is $34,615. 82. 60 + 43 = 17 Georgia’s all-time record low temperature is 17F.
50. 3 + (23) + 6 = 20 + 6 = 14 52. 100 + 70 = 30 54. (45) + 22 + 20 = 23 + 20 = 3 56. 87 + 0 = 87
84. 10,924 + 3245 = 7679 The depth of the Aleutian Trench is 7679 meters. 86. 91 0 = 91
58. 16 6 (14) (20) 10 (14) (20) 24 (20) 44 60. x + y = 1 + (29) = 30
88.
400 18 382
90. answers may vary
62. 3x + y = 3(7) + (11) = 21 + (11) = 10
92. 4 + 14 = 10
64. 3x + y = 3(13) + (17) = 39 + (17) = 22
94. 15 + (17) = 32
66. The sum of 30 and 15 is 30 + 15 = 15. Copyright © 2025 Pearson Education, Inc.
47
Chapter 2: Integers and Introduction to Solving Equations
6. 3 + 4 + (23) + (10); all the subtraction operations are rewritten as additions in one step rather than changing one operation at a time as you work from left to right.
96. True 98. True 100. answers may vary
7. to follow the order of operations
Section 2.3 Practice Exercises
8. The warmest temperature is 263F warmer than the coldest temperature.
1. 13 4 = 13 + (4) = 9 2. 8 2 = 8 + (2) = 10
Exercise Set 2.3
3. 11 (15) = 11 + 15 = 26
2. 6 (6) = 6 + 6 = 0
4. 9 (1) = 9 + 1 = 8
4. 15 12 = 15 + (12) = 3
5. 6 9 = 6 + (9) = 3
6. 2 5 = 2 + (5) = 3
6. 14 5 = 14 + (5) = 19
8. 12 (12) = 12 + 12 = 24
7. 3 (4) = 3 + 4 = 1
10. 25 (25) = 25 + 25 = 0
8. 15 6 = 15 + (6) = 21
12. 2 42 = 2 + (42) = 44
9. 6 5 2 (3) 6 (5) (2) 3 11 (2) 3 13 3 10
14. 8 9 = 8 + (9) = 1
10. 8 (2) 9 (7) 8 (2) (9) 7 6 (9) 7 3 7 4 11. x y = 5 13 = 5 + (13) = 18 12. 3y z = 3(9) (4) = 27 + 4 = 31 13. 29,028 (1312) = 29,028 + 1312 = 30,340 Mount Everest is 30,340 feet higher than the Dead Sea. Vocabulary, Readiness & Video Check 2.3 1. It is true that a b = a + (b). b
3. To evaluate x y for x = 10 and y = 14, we replace x with 10 and y with 14 and evaluate 10 (14). d 4. The expression 5 10 equals 5 + (10). c 5. additive inverse
16. 17 63 = 17 + (63) = 46 18. 844 (20) = 844 + 20 = 864 20. 5 8 = 5 + (8) = 13 22. 12 (5) = 12 + 5 = 7 24. 16 45 = 16 + (45) = 29 26. 22 10 = 22 + (10) = 32 28. 8 (13) = 8 + 13 = 5 30. 50 (50) = 50 + 50 = 0 32. 35 + (11) = 46 34. 4 21 = 4 + (21) = 17 36. 105 68 = 105 + (68) = 173
2. The opposite of n is n. a
48
ISM: Prealgebra and Introductory Algebra
38. 86 98 = 86 + (98) = 12 40. 8 4 1 = 8 + (4) + (1) = 4 + (1) = 3 42. 30 18 12 30 (18) (12) 12 (12) 0
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
44. 10 6 (9) = 10 + (6) + 9 = 16 + 9 = 7
74. 384 (505) = 384 + 505 = 121 The difference in elevation is 121 feet.
46. 15 (8) 4 15 (8) (4) 23 (4) 27
76. 236 (505) = 236 + 505 = 269 The difference in elevation is 269 feet.
48. 23 (17) (9) 23 17 (9) 40 (9) 31 50. (9) 14 (23) 9 (14) (23) 5 (23) 28
78. 512 (92) = 512 + 92 = 604 The difference in elevation is 604 feet. 80. 52 (92) = 52 + 92 = 40 The difference in elevation is 40 feet. 82. 845 (162) = 845 + 162 = 1007 The difference in temperature is 1007F.
52. 6 (8) (12) 7 6 8 (12) (7) 2 (12) (7) 10 (7) 17
84. 6037 4948 = 6037 + (4948) = 1089 The trade balance was 1089 thousand barrels of crude oil per day that week.
54. 5 (18) (21) 2 5 (18) 21 (2) 13 21 (2) 8 (2) 6
88. Add a number and 36 is x + (36).
86. The difference of 3 and a number is 3 x.
96 90.
56. x y = 7 1 = 7 + (1) = 8
3
58. x y = 9 (2) = 9 + 2 = 11 60. 2x y = 2(8) (10) = 16 + 10 = 26 62. 2x y = 2(14) (12) = 28 + 12 = 40 92. 64. From the graph, the maximum and minimum temperatures for Texas are 120F and 23F. 120 (23) = 120 + 23 = 143 The temperature extreme for Texas is 143F. 66. The state with the hottest maximum is California. The maximum and minimum temperatures for California are 134F and 45F. 134 (45) = 134 + 45 = 179 The temperature extreme for California is 179F. 68. 134 (80) = 134 + 80 = 214 Therefore, 134F is 214F warmer than 80F. 70. 93 18 26 93 (18) (26) 75 (26) 49 $49 is now owed on the account. 72. 13,796 (21,857) = 13,796 + 21,857 = 35,653 The difference in elevation is 35,653 feet.
32
3 32 96 9 06 6 0
51 89 459 4080 4539
94. answers may vary 96. 4 8 = 4 + (8) = 12 98. 3 (10) = 3 + 10 = 7 100. |12| |5| = 12 5 = 12 + (5) = 7 102. |8| |8| = 8 8 = 0 104. |23| |42| = 23 42 = 23 + (42) = 19 106. |2 (6)| = |2 + 6| = |4| = 4 |2| |6| = 2 6 = 2 + (6) = 4 Since 4 4, the statement is false. 108. no; answers may vary
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49
Chapter 2: Integers and Introduction to Solving Equations
ISM: Prealgebra and Introductory Algebra
3. The quotient of two negative numbers is a positive number.
Section 2.4 Practice Exercises 1. 3 8 = 24 2. 5(2) = 10
4. The quotient of a negative number and a positive number is a negative number.
3. 0 (20) = 0
5. The product of a negative number and zero is 0.
4. 10(5) = 50
6. The quotient of 0 and a negative number is 0.
5. 8(6)(2) = 48(2) = 96
7. The quotient of a negative number and 0 is undefined.
6. (9)(2)(1) = 18(1) = 18 7. (3)(4)(5)(1) = 12(5)(1) = 60(1) = 60 4 8. (2) (2)(2)(2)(2) 4(2)(2) 8(2) 16
9. 82 (8 8) 64 42 10.
7
9. We can find out about sign rules for division because we know sign rules for multiplication.
6
10. that ab means a b 11. The phrase “lost four yards” in the example translates to the negative number 4.
11. 16 (2) = 8 12.
80 8 10
13.
6 is undefined. 0
14.
0 7
8. When a negative sign is involved in an expression with an exponent, parentheses tell you whether or not the exponent applies to the negative sign. In Video Notebook Number 3, (3)2, the exponent applies to everything within the parentheses, so 3 is squared; in Video Notebook Number 4, 32, the exponent does not apply to the sign and only 3 is squared.
Exercise Set 2.4 2. 5(3) = 15 4. 7(2) = 14 6. 9(7) = 63
0
8. 6(0) = 0
15. xy = 5 (8) = 40
10. 2(3)(7) = 6(7) = 42
16. x 12 4 y 3
12. 8(3)(3) = 24(3) = 72
17. total score = 4 (13) = 52 The card player’s total score was 52. Vocabulary, Readiness & Video Check 2.4 1. The product of a negative number and a positive number is a negative number.
14. 2(5)(4) = 10(4) = 40 16. 3(0)(4)(8) = 0 18. 2(1)(3)(2) = 2(3)(2) = 6(2) = 12 20. 24 (2)(2)(2)(2) 4(2)(2) 8(2) 16
2. The product of two negative numbers is a positive number.
50
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ISM: Prealgebra and Introductory Algebra
4 22. (1) (1)(1)(1)(1) 1(1)(1) 1(1) 1
Chapter 2: Integers and Introduction to Solving Equations
62.
64.
0 14 63 9
0
7
24. 43 (4 4 4) 64 66. 480 (8) 26. (3) (3)(3) 9 2
28. 90 (9) = 10 30.
32.
56
68.
7
8
60
36 12 3 3
8
70. 2 (2 2 2) 8
32 8 4
72. (11)2 (11)(11) 121 74. 1(2)(7)(3) = 2(7)(3) = 14(3) = 42
34. 13 is undefined. 0 0
36.
38.
480
76. (1)33 1, since there are an odd number of factors.
0
15
78. 2(2)(3)(2) = 4(3)(2) = 12(2) = 24
24 2 12
80.
56 43 168 2240 2408 56 43 = 2408
82.
23 70 1610 70 (23) = 1610
40. 0(100) = 0 42. 6 2 = 12 44. 12(13) = 156 46. 9(5) = 45 48. 7(5)(3) = 35(3) = 105
84. ab = 5(1) = 5
50. (5)2 (5)(5) 25 52.
30 6 5
54.
49 7 7
86. ab = (8)(8) = 64 88. ab = (9)(6) = 54 90.
56. 15 3 = 5
x 9 3 y 3 x
92.
58. 6(5)(2) = 30(2) = 60 60. 20 5 (5) (3) 100 (5) (3) 500 (3) 1500
y
0 5
0
94. x 10 1 y 10
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations 96. xy = 20 (5) = 100 x 20 4 y 5
120. 2015 to 2021 is 6 years. 1 969 6 11,814 6 58 5 4 41 36 54 54 0 1969 fewer birds would be banded per year, or 1969 birds per year.
98. xy = 3 0 = 0 x 3 is undefined. y 0 100. 63 (3) = 21 The quotient of 63 and 3 is 21. 102.
49 5 245 49(5) = 245 The product of 49 and 5 is 245.
104. The quotient of 8 and a number is
122. a. 8
or
x 8 x. 106. The sum of a number and 12 is x + (12). 108. The difference of a number and 10 is x (10). 110. Multiply a number by 17 is x (17) or 17x. 112. A loss of $400 is represented by 400. 7 (400) = 2800 His total loss was $2800. 114. A drop of 5 degrees is represented by 5. 6 (5) = 30 The total drop in temperature was 30 degrees. 116. 1 (39) = 39 The melting point of rubidium is 39C. 118. 11 (70) = 770 The melting point of strontium is 770C.
27,200 12,700 = 14,500 There were about 14,500 more digital non3D movie screens in 2021 than in 2011. This is a change of 14,500 screens.
b. 2011 to 2021 is 10 years. 1 450 10 14, 500 10 45 4 0 50 50 00 The average change was 1450 screens per year. 124. 3(7 4) 2 52 3 3 2 52 3 3 2 25 9 2 25 9 50 59 126. 12 (4 2) + 7 = 12 2 + 7 = 6 + 7 = 13 128. 9(11) = 99 130. 4 + (3) + 21 = 7 + 21 = 14 132. 16 (2) = 16 + 2 = 14 134. The product of an even number of negative numbers is positive, so the product of ten negative numbers is positive.
52
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ISM: Prealgebra and Introductory Algebra
136. (1)50 and (7)20 are positive since there are an even number of factors. Note that (7)20 (1)50 since (1)50 1. (1)55 and (7)23 are negative since there are an odd number of factors. Note that (7)23 (1)55 since (1)55 1. 015 0
The numbers from least to greatest are
Chapter 2: Integers and Introduction to Solving Equations 17. 8(6)(1) = 48(1) = 48 18. 18 2 = 9 19. 65 + (55) = 10 20. 1000 1002 = 1000 + (1002) = 2 21. 53 (53) = 53 + 53 = 106 22. 2 1 = 2 + (1) = 3
(7)23 , (1)55 , 015 , (1)50 , (7)20 . 23.
0
0
47
138. answers may vary Mid-Chapter Review 24. 1. Let 0 represent 0F. Then 50 degrees below zero is represented by 50 and 122 degrees above zero is represented by +122 or 122.
36 4 9
25. 17 (59) = 17 + 59 = 42 26. 8 + (6) + 20 = 14 + 20 = 6
2.
95 19 5
3. 0 > 10 since 0 is to the right of 10 on a number line.
27.
4. 4 < 4 since 4 is to the left of 4 on a number line.
28. 9(100) = 900
5. 15 < 5 since 15 is to the left of 5 on a number line. 6. 2 > 7 since 2 is to the right of 7 on a number line. 7. |3| = 3 because 3 is 3 units from 0. 8. |9| = 9 because 9 is 9 units from 0.
30. 4 (8) 16 (9) 4 (8) (16) 9 12 (16) 9 28 9 19 31.
105 is undefined. 0
32. 7(16)(0)(3) = 0 (since one factor is 0)
9. |4| = 4
33. Subtract 8 from 12 is 12 (8) = 12 + 8 = 4.
10. (5) = 5 11. The opposite of 11 is 11.
34. The sum of 17 and 27 is 17 + (27) = 44.
12. The opposite of 3 is (3) = 3. 13. The opposite of 64 is 64.
29. 12 6 (6) = 12 + (6) + 6 = 18 + 6 = 12
35. The product of 5 and 25 is 5(25) = 125. 36. The quotient of 100 and 5 is
14. The opposite of 0 is 0 = 0. 15. 3 + 15 = 12
37. Divide a number by 17 is
16. 9 + (11) = 20
100 5
20.
x or x (17). 17
38. The sum of 3 and a number is 3 + x. Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations 39. A number decreased by 18 is x (18). 40. The product of 7 and a number is 7 x or 7x. 41. x + y = 3 + 12 = 9 42. x y = 3 12 = 3 + (12) = 15
10. 4[6 5(3 5)] 7 4[6 5(2)] 7 4[6 10] 7 4(4) 7 16 7 23 11. x2 (15)2 (15)(15) 225
43. 2y x = 2(12) (3) = 24 (3) = 24 + 3 = 27 44. 3y + x = 3(12) + (3) = 36 + (3) = 33 45. 5x = 5(3) = 15 y 12 46. 4 x 3
2
14. 4 x2 4 (8)2 4 64 60 15. average sum of numbers number of numbers 17 (1) (11) (13) (16) (13) 2 7 35 7 5 The average of the temperatures is 5F.
2. 24 (2)(2)(2)(2) 16 3. 3 62 3(6 6) 3 36 108 25
25 5 5(1) 5 18 6 12 3 3 1 4
Calculator Explorations
3
6. 30 50 (4) 30 50 (64) 80 (64) 16 7.
1.
120 360 48 10
2.
4750 250 2 (17)
3.
316 (458) 258 28 (25)
4.
234 86 74 18 16
23 (4)2 15 8 16 1 8 1 9
8. 2(2 9) (12) 3 2(7) (12) 3 14 (12) 3 26 3 29 9. (5) 8 (3) 23 (5) 8 (3) 23 (5) 8 (3) 8 40 (3) 8 43 8 35
54
2
13. x y (6) (3) 36 (3) 33
1. (2)4 (2)(2)(2)(2) 16
5.
12. 5 y2 5(4)2 5(16) 80 5 y2 5(4)2 5(16) 80
Section 2.5 Practice Exercises
4.
x2 (15)2 (15)(15) 225
Vocabulary, Readiness & Video Check 2.5 1. To simplify 2 2 (3), which operation should be performed first? division 2. To simplify 9 3 4, which operation should be performed first? multiplication
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
3. The average of a list of numbers is sum of numbers . number of numbers
26. 7 6 6 5 (10) 42 6 5 (10) 42 30 (10) 12 (10) 2
4. To simplify 5[9 + (3)] 4, which operation should be performed first? addition
28. 7 (5)2 7 25 18
5. To simplify 2 + 3(10 12) (8), which operation should be performed first? subtraction
30.
6. To evaluate x 3y for x = 7 and y = 1, replace x with 7 and y with 1 and evaluate 7 3(1). 7. A fraction bar means divided by and it is a grouping symbol. 8. to make sure that the entire value of 2, including the sign, is squared 9. Finding an average is a good application of both order of operations and adding and dividing integers. Exercise Set 2.5 2. 24 (2)(2)(2)(2) 16
3 7 72 4 72 4 72 4 49 196
32. 10 53 7 10 125 7 1250 7 1257 34. 82 (5 2)4 82 34 64 81 17 36. |12 19| 7 = |7| 7 = 7 7 = 1 38. (2)3 (8) 8 40. (2 7)2 (4 3)4 (5)2 14 25 1 25 42. 3 15 (4) (16) 12 (4) (16) 12 (4) (16) 48 (16) 3
4. (2)4 (2)(2)(2)(2) 16
44. (20 5) 5 15 (25) 5 15 5 15 20
6. 5 23 5 8 40
46. 3(8 3) (4) 10 3(5) (4) 10 15 (4) 10 1110 1
8. 10 23 12 = 13 12 = 25 10. 8 + 4(3) = 8 + 12 = 4 12. 7(6) + 3 = 42 + 3 = 39 14. 12 + 6 3 = 12 + 2 = 10 16. 5 9 4 20 5 36 20 41 20 21 18.
20.
20 15 5 5 1 1 88
88 8 8 3 11
22. 7(4) (6) = 28 + 6 = 22 24. [9 (2)]3 [7]3 343
48. (4 12) (8 17) (8) (9) 72 50. (4 4) (8 8) = (1) (1) = 2 52. (11 32 )3 (11 9)3 23 8 54. 3(4 8)2 5(14 16)3 3(4)2 5(2)3 3(16) 5(8) 48 (40) 88 56. 12 [7 (3 6)] (2 3)3 12 [7 (3)] (2 3)3 12 (7 3) (1)3 12 10 (1) 2 (1) 1 Copyright © 2025 Pearson Education, Inc.
55
Chapter 2: Integers and Introduction to Solving Equations
58.
10(1) (2)(3) 10 6 2[8 (2 2)] 2[8 (4)] 16 2(2) 16 4 4
60. 2[6 4(2 8)] 25 2[6 4(6)] 25 2[6 (24)] 25 2(18) 25 36 25 11 62. x y z 2 4 (1) 2 4 1 6 1 5
ISM: Prealgebra and Introductory Algebra
40 (20) (10) (15) (5) 5 90 5 18
82. average
84. The two lowest scores are 20 and 5. 5 (20) = 5 + 20 = 15 The difference between the two lowest scores is 15. 20 (5) (5) 30 10 3 3 The average of the scores is 10.
86. average
88. no; answers may vary 90. 90 45 = 2 92. 45 + 90 = 135
64. 5x y 4z 5(2) 4 4(1) 10 4 (4) 14 (4) 18
94. 3 + 5 + 3 + 5 = 16 The perimeter is 16 centimeters.
66. x2 z (2)2 (1) 4 (1) 3 4x 4(2) 8 68. 2 y 4 4
96. 17 + 23 + 32 = 72 The perimeter is 72 meters. 98. (7 3 4) 2 = (21 4) 2 = 17 2 = 34 100. 2 (8 4 20) = 2 (2 20) = 2 (18) = 36 102. answers may vary
70. z 2 (4)2 16
104. answers may vary
72. x2 (3)2 9
106. (17)6 (17)(17)(17)(17)(17)(17) 24,137, 569
74. 3x2 3(3)2 3(9) 27 2
2
76. 3 z 3 (4) 3 16 13 78. 3z 2 x 3(4)2 (3) 3(16) 3 48 3 51 18 (8) (1) (1) 0 4 6 24 6 4
80. average
56
108. 3x2 2x y 3(18)2 2(18) 2868 3(324) (36) 2868 972 (36) 2868 936 2868 1932 110. 5(ab 3)b 5(2 3 3)3
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5(6 3)3 5(3)3 5(27) 135
ISM: Prealgebra and Introductory Algebra
7.
Section 2.6 Practice Exercises 1.
Chapter 2: Integers and Introduction to Solving Equations
4x 3 5 4(2) 3 0 5 830 5 5 5 True Since 5 = 5 is true, 2 is a solution of the equation.
2.
3.
y 6 2 y 6 6 2 6 y4 Check: y 6 2 46 0 2 2 2 True The solution is 4. 2 z 8 2 8 z 8 8 10 z Check: 2 z 8 2 0 10 8 2 2 True The solution is 10.
4. x 2 90 (100) x 88 (100) x 12 The solution is 12. 5.
3y 18 3y 18 3 3 3 18 y 3 3 y 6 Check: 3y 18 3(6) 0 18 18 18 True The solution is 6.
6.
32 8x 32 8x 8 8 32 8 x 8 8 4 x Check: 32 8x 32 0 8(4) 32 32 True The solution is 4.
3y 27 3y 27 3 3 3 27 y 3 3 y9 Check: 3y 27 3 9 0 27 27 27 True The solution is 9. x
8. 4
4 x
7
4 7 4 4 x 4 7 4 x 28 x 7 Check: 4 28 4 0 7 7 7 True The solution is 28. Vocabulary, Readiness & Video Check 2.6 1. A combination of operations on variables and numbers is called an expression. 2. A statement of the form “expression = expression” is called an equation. 3. An equation contains an equal sign (=) while an expression does not. 4. An expression may be simplified and evaluated while an equation may be solved. 5. A solution of an equation is a number that when substituted for the variable makes the equation a true statement. 6. Equivalent equations have the same solution. 7. By the addition property of equality, the same number may be added to or subtracted from both sides of an equation without changing the solution of the equation. 8. By the multiplication property of equality, both sides of an equation may be multiplied or divided by the same nonzero number without changing the solution of the equation. Copyright © 2025 Pearson Education, Inc.
57
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
9. an equal sign
12.
10. We can add the same number to both sides of an equation and we’ll have an equivalent equation. Also, we can subtract the same number from both sides of an equation and have an equivalent equation. 11. To check a solution, we go back to the original equation, replace the variable with the proposed solution, and see if we get a true statement.
Check:
14.
2. y 16 7 9 16 0 7 7 7 True
4.
a 23 16 7 23 0 16 16 16 False Since 16 = 16 is false, 7 is not a solution of the equation.
6.
3k 12 k 3(6) 0 12 (6) 18 012 6 18 18 True Since 18 = 18 is true, 6 is a solution of the equation.
8.
10.
2(b 3) 10 2(2 3) 010 2(1) 0 10 2 10 False Since 2 = 10 is false, 2 is not a solution of the equation. f 4 6 f 4 4 6 4 f 10 Check: f 4 6 10 0 6 6 6 True The solution is 10.
58
s 7 15 8 7 0 15 15 15 True
The solution is 8.
Exercise Set 2.6
Since 7 = 7 is true, 9 is a solution of the equation.
s 7 15 s 7 7 15 7 s 8
1 y 7 1 7 y 7 7 6 y Check: 1 y 7 10 6 7 1 1 True The solution is 6.
16. 50 40 5 z 10 5 z 15 z Check: 50 40 5 z 50 40 5 0 15 10 5 0 15 15 15 True The solution is 15. 18.
6 y 48 6 y 48 6 6 6 48 y 6 6 y 8 Check: 6 y 48 6(8) 0 48 48 48 True The solution is 8.
20.
2x 26 2x 26 2 2 2 26 x 2 2 x 13 Check: 2x 26 2(13) 0 26 26 26 True The solution is 13.
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ISM: Prealgebra and Introductory Algebra
22.
24.
n 5 11 n 11 11 (5) 11 11 n 11 (5) 11 n 55 n 5 Check: 11 55 0 5 11 5 5 True The solution is 55. 7 y 21 7 y 21 7 7 7 21 y 7 7 y 3 Check: 7 y 21 7 (3) 0 21 21 21 True
Chapter 2: Integers and Introduction to Solving Equations
30.
3y 27 3y 27 3 3 3 27 y 3 3 y 9 The solution is 9.
32.
n 4 48 n 4 4 48 4 n 44 The solution is 44.
34.
The solution is 48. x 36. 9
28.
9x 0 9x 0 9 9 9 0 x 9 9 x0 Check: 9x 0 9 0 0 0 0 0 True The solution is 0. 31x 31 31x 31 31 31 31 31 x 31 31 x 1 Check: 31x 31 311 0 31 31 31 True The solution is 1.
9 x
9
9 (9) 9 x 9 (9) 9 x 81 The solution is 81. 9
The solution is 3. 26.
36 y 12 36 12 y 12 12 48 y
38. z = 28 + 36 z=8 The solution is 8. 40.
11x 121 11x 121 11 11 11 121 x 11 11 x 11 The solution is 11. n
42. 5
20
5n
5 (20) 5 5 n 5 (20) 5 n 100 The solution is 100.
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59
Chapter 2: Integers and Introduction to Solving Equations
ISM: Prealgebra and Introductory Algebra
44. 81 27x 81 27x 27 27 81 27 x 27 27 3 x The solution is 3. 46. A number increased by 5 is x + (5). 48. The quotient of a number and 20 is x (20) or
x . 20
50. 32 multiplied by a number is 32 x or 32x. 52. Subtract a number from 18 is 18 x. 54.
56.
n 961 120 n 961 961 120 961 n 841 The solution is 841. y 1098 18 y 18 18 1098 18 18 y 18 1098 18 y 19, 764 The solution is 19,764.
58. answers may vary 60. answers may vary Chapter 2 Vocabulary Check 1. Two numbers that are the same distance from 0 on a number line but are on opposite sides of 0 are called opposites. 2. The absolute value of a number is that number’s distance from 0 on a number line. 3. The integers are ..., 3, 2, 1, 0, 1, 2, 3, .... 4. The negative numbers are numbers less than zero. 5. The positive numbers are numbers greater than zero. 6. The symbols < and > are called inequality symbols. 7. A solution of an equation is a number that when substituted for the variable makes the equation a true statement. 8. The average of a list of numbers is
60
sum of numbers
. number of numbers
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
9. A combination of operations on variables and numbers is called an expression. 10. A statement of the form “expression = expression” is called an equation. 11. The sign “<” means is less than and “>” means is greater than. 12. By the addition property of equality, the same number may be added to or subtracted from both sides of an equation without changing the solution of the equation. 13. By the multiplication property of equality, both sides of an equation may be multiplied or divided by the same nonzero number without changing the solution of the equation. Chapter 2 Review 1. If 0 represents sea level, then 1572 feet below sea level is 1572. 2. If 0 represents sea level, then an elevation of 11,239 feet is +11,239. 3. 4. 5. |11| = 11 since 11 is 11 units from 0 on a number line. 6. |0| = 0 since 0 is 0 units from 0 on a number line. 7. |8| = 8 8. (9) = 9 9. |16| = 16 10. (2) = 2 11. 18 > 20 since 18 is to the right of 20 on a number line. 12. 5 < 5 since 5 is to the left of 5 on a number line. 13. |123| = 123 |198| = 198 Since 123 > 198, |123| > |198|. 14. |12| = 12 |16| = 16 Since 12 > 16, |12| > |16|. 15. The opposite of 18 is 18. (18) = 18 16. The opposite of 42 is negative 42. (42) = 42 17. False; consider a = 1 and b = 2, then 1 < 2. 18. True Copyright © 2025 Pearson Education, Inc.
61
Chapter 2: Integers and Introduction to Solving Equations
19. True
ISM: Prealgebra and Introductory Algebra
37. |43| + |108| = 43 + 108 = 151 The common sign is negative, so 43 + (108) = 151.
20. True 21. |y| = |2| = 2
38. |100| + |506| = 100 + 506 = 606 The common sign is negative, so 100 + (506) = 606.
22. |x| = |(3)| = |3| = 3 23. |z| = |(5)| = |5| = 5
39. 15 + (5) = 20 The temperature at 6 a.m. is 20C.
24. |n| = |(10)| = |10| = 10 25. The bar that extends the farthest in the negative direction corresponds to elevator D, so elevator D extends the farthest below ground. 26. The bar that extends the farthest in the positive direction corresponds to elevator B, so elevator B extends the highest above ground. 27. |5| |3| = 5 3 = 2 5 > 3, so the answer is positive. 5 + (3) = 2
40. 127 + (23) = 150 The diver’s current depth is 150 feet. 41. 7 (6) (2) (2) 13 (2) (2) 15 (2) 17 Her total score was 17. 42. 19 10 = 19 + (10) = 9 The team’s score was 9.
28. |18| |4| = 18 4 = 14 18 > 4, so the answer is positive. 18 + (4) = 14 29. |16| |12| = 16 12 = 4 16 > 12, so the answer is positive. 12 + 16 = 4 30. |40| |23| = 40 23 = 17 40 > 23, so the answer is positive. 23 + 40 = 17
43. 12 4 = 12 + (4) = 8 44. 12 4 = 12 + (4) = 16 45. 7 17 = 7 + (17) = 24 46. 7 17 = 7 + (17) = 10 47. 7 (13) = 7 + 13 = 20 48. 6 (14) = 6 + 14 = 8 49. 16 16 = 16 + (16) = 0
31. |8| + |15| = 8 + 15 = 23 The common sign is negative, so 8 + (15) = 23.
50. 16 16 = 16 + (16) = 32
32. |5| + |17| = 5 + 17 = 22 The common sign is negative, so 5 + (17) = 22.
52. 5 (12) = 5 + 12 = 7
33. |24| |3| = 24 3 = 21 24 > 3, so the answer is negative. 24 + 3 = 21 34. |89| |19| = 89 19 = 70 89 > 19, so the answer is negative. 89 + 19 = 70 35. 15 + (15) = 0 36. 24 + 24 = 0 62
51. 12 (12) = 12 + 12 = 0
53. (5) 12 (3) 5 (12) (3) 7 (3) 10 54. 8 (12) 10 (3) 8 (12) (10) 3 20 (10) 3 30 3 27 55. 600 (92) = 600 + 92 = 692 The difference in elevations is 692 feet.
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ISM: Prealgebra and Introductory Algebra 56. 142 125 43 85 142 (125) 43 (85) 17 43 (85) 60 (85) 25 The balance in the account is 25.
Chapter 2: Integers and Introduction to Solving Equations
74.
72 9 8
75.
38 38 1
57. 223 245 = 223 + (245) = 22 You are 22 feet or 22 feet below ground at the end of the drop.
76.
58. 66 (16) = 66 + 16 = 82 The total length of the elevator shaft for elevator C is 82 feet.
77. A loss of 5 yards is represented by 5. (5)(2) = 10 The total loss is 10 yards.
59. |5| |6| = 5 6 = 5 + (6) = 1 5 6 = 5 + (6) = 1 |5| |6| = 5 6 is true.
78. A loss of $50 is represented by 50. (50)(4) = 200 The total loss is $200.
60. |5 (6)| = |5 + 6| = |1| = 1 5 + 6 = 11 Since 1 11, the statement is false.
45 9
5
79. A debt of $1024 is represented by 1024. 1024 4 = 256 Each payment is $256. 80. A drop of 45 degrees is represented by 45. 45 5 or 45 9 = 5 9 The average drop each hour is 5F.
61. 3(7) = 21 62. 6(3) = 18 63. 4(16) = 64
81. (7)2 (7)(7) 49
64. 5(12) = 60
82. 72 (7 7) 49
65. (5)2 (5)(5) 25 66. (1)5 (1)(1)(1)(1)(1) 1
83. 5 8 + 3 = 3 + 3 = 0 84. 3 12 (7) 10 9 (7) 10 2 10 8
67. 12(3)(0) = 0 68. 1(6)(2)(2) = 6(2)(2) = 12(2) = 24 69. 15 3 = 5 70.
71.
72.
73.
24 8 0 3
3
0
85. 10 + 3 (2) = 10 + (6) = 16 86. 5 10 (3) = 5 (30) = 5 + 30 = 35 87. 16 (2) 4 = 8 4 = 32 88. 20 5 2 = 4 2 = 8 89. 16 (3) 12 4 16 (36) 4 16 (9) 7
46 is undefined. 0
90. 12 10 (5) 12 (2) 14
100
3 2 3 2 91. 4 (8 3) 4 (5) 64 25 39
5
20
92. (3)3 90 27 90 117 Copyright © 2025 Pearson Education, Inc.
63
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
93.
94.
(4)(3) (2)(1) 12 2 10 5 2 10 5 5 4(12 18)
10 (2 3)
4(6)
24
10 (5)
105.
12
2
18 25 (30) 7 0 (2) 6 18 6
95. average
106.
3 96. average 45 (40) (30) (25) 4 140 4 35 97. 2x y = 2(2) 1 = 4 1 = 5
99.
100. 101.
102.
8x 72 8x 72 8 8 8 72 x 8 8 x 9 The solution is 9.
107. 20 7 y 13 y The solution is 13. 108.
98. y2 x2 12 (2)2 1 4 5
10x 30 10x 30 10 10 10 30 x 10 10 x 3 The solution is 3.
3x 3(2) 6 1 6 6 6
x 31 62 x 31 31 62 31 x 31 The solution is 31. n
5 y x 5(1) (2) 5 2 7 7 y 1 1 1 2n 6 16 2(5) 6 016 10 6 0 16 16 16 False Since 16 = 16 is false, 5 is not a solution of the equation.
109. 4
4 n
11
4 (11) 4 4 n 4 (11) 4 n 44 The solution is 44. x 110. 2
2 x
13
2 13 2 2 x 2 13 2 x 26 The solution is 26.
2(c 8) 20 2(2 8) 0 20 2(10) 0 20 20 20 True Since 20 = 20 is true, 2 is a solution of the equation.
103.
n 7 20 n 7 7 20 7 n 13 The solution is 13.
111.
n 12 7 n 12 12 7 12 n 19 The solution is 19.
104.
5 n 15 5 15 n 15 15 20 n The solution is 20.
112.
n 40 2 n 40 40 2 40 n 38 The solution is 38.
64
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 113. 36 6x 36 6x 6 6 36 6 x 6 6 6 x The solution is 6.
Chapter 2: Integers and Introduction to Solving Equations
129.
3 2 4 3 2 4 130. 5(7 6) 4(2 3) 2 5(1) 4(1) 2 5(1) 4(1) 16
5 4 16 1 16 17
114. 40 8 y 40 8 y 8 8 40 8 y 8 8 5 y The solution is 5.
131.
n 9 30 n 9 9 30 9 n 21 The solution is 21.
132.
n 18 1 n 18 18 118 n 17 The solution is 17.
133.
4x 48 4x 48 4 4 4 48 x 4 4 x 12 The solution is 12.
134.
9x 81 9x 81 9 9 9 81 x 9 9 x 9 The solution is 9.
115. 6 + (9) = 15 116. 16 3 = 16 + (3) = 19 117. 4(12) = 48 118.
84
14 6 14 6 20 20 7 2(3) 7 (6) 1
21
4 119. 76 (97) = 76 + 97 = 21 120. 9 + 4 = 5 121. 18 9 = 27 The temperature on Friday was 27C. 122. 11 + 17 = 6 The temperature at noon on Tuesday was 6C. 123. 12,923 (195) = 12,923 + 195 = 13,118 The difference in elevations is 13,118 feet. 124. 32 + 23 = 9 His financial situation can be represented by 9 or $9. 2 3 2 3 125. (3 7) (6 4) (4) (2) 16 8 2
126. 3(4 2) (6) 32 3(6) (6) 32 3(6) (6) 9 18 (6) 9 12 9 3
135.
n 100 2 n 2 2 100 2 2 n 2 100 2 n 200 The solution is 200.
127. 2 4 3 5 2 12 5 10 5 5 128.
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65
ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
y 136. 1 1 1
1 y
x
3
13.
1(3)
3
y3 The solution is 3.
14.
Chapter 2 Getting Ready For the Test 1. The opposite of 2 is (2) = 2; A.
1. 5 + 8 = 3 2. 18 24 = 18 + (24) = 6
3. The absolute value of 2 is |2| = 2 because 2 is 2 units from 0 on a number line; A.
3. 5 (20) = 100
4. The opposite of 2 is 2; B.
4. 16 (4) = 4
5. For xy, the operation is multiplication; C.
5. 18 + (12) = 30
6. For 12(+3), the operation is multiplication; C.
6. 7 (19) = 7 + 19 = 12
7. For 12 + 3, the operation is addition; A.
7. 5 (13) = 65
12 8. For , the operation is division; D. 3
8.
9. For 4 + 6 2, the operation of multiplication is performed first, since multiplication and division are performed before addition and subtraction; C.
66
5
5
10. 14 |20| = 14 20 = 14 + (20) = 6
x 2 4 x 2 2 4 2 x 2 Choice B is correct.
25
9. |25| + (13) = 25 + (13) = 12
10. For 4 + 6 2, the operation of division is performed first, since multiplication and division are performed before addition and subtraction; D.
12.
x2 4 x22 42 x2 Choice A is correct.
Chapter 2 Test
2. The absolute value of 2 is |2| = 2 because 2 is 2 units from 0 on a number line; A.
3x 6 3 3 x 2 Choice B is correct.
6
3(6) 3 x 18 Choice D is correct.
1 y 1(3)
11. 3x 6
x3
11. |5| |10| = 5 10 = 50 10 10 12. 2 5 5
13. 8 + 9 (3) = 8 + (3) = 11 14. 7 (32) 12 5 7 (32) (12) 5 39 (12) 5 51 5 46 15. (5)3 24 (3) 125 24 (3) 125 (8) 125 8 117
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
17. (7)2 7 (4) 49 7 (4) 7 (4) 28
28. Subtract the elevation of the Romanche Gap from the elevation of Mount Washington. 6288 (25,354) = 6288 + 25,354 = 31,642 The difference in elevations is 31,642 feet.
18. 3 (8 2)3 3 63 3 216 3 (216) 213
29. Subtract the depth of the lake from the elevation of the surface. 1495 5315 = 1495 + (5315) = 3820 The deepest point of the lake is 3820 feet below sea level.
16. (5 9)2 (8 2)3 (4)2 (6)3 16 216 3456
4 19.
20.
21.
82
2 16
4
64 2 4 2 (4) 2 2 16
3(2) 12 6 12 18 2 1(4 5) 1(9) 9 25 30
2
2(6) 7
5
2
12 7
(5)2 5
22. 5(8) [6 (2 4)] (12 16)
30. average
5
5
4
4
9n 45 9 9 9 n 45 9 9 n5 The solution is 5.
5(8) [6 (2)] (12 16)2 5(8) (6 2) (4)2
n 33. 7
7 n
4
7 4 7 7 n 7 4 7 n 28
23. 7x 3y 4z 7(0) 3(3) 4(2) 0 (9) 8 9 8 17
The solution is 28.
24. 10 y2 10 (3)2 10 9 1 25.
16
b. A number subtracted from 20 is 20 x. 32. 9n 45
2
5(8) 8 (4)2 5(8) 8 16 40 8 16 48 16 32
4
The product of a number and 17 is 17 x or 17x.
31. a.
25
12 (13) 0 9
6 1 3(2) 2 y 2(3) 6
3z
26. A descent of 22 feet is represented by 22. 4(22) = 88 Tui is 88 feet below sea level. 27. 129 (79) (40) 35 50 (40) 35 10 35 45 Their new balance can be represented by 45 or $45.
34.
x 16 36 x 16 16 36 16 x 20 The solution is 20.
35. 20 8 8 x 12 8 x 4 x The solution is 4. Cumulative Review Chapters 12 1. The place value of 3 in 396,418 is hundredthousands. 2. The place value of 3 in 4308 is hundreds. 3. The place value of 3 in 93,192 is thousands.
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ISM: Prealgebra and Introductory Algebra
Chapter 2: Integers and Introduction to Solving Equations
4. The place value of 3 is 693,298 is thousands. 5. The place value of 3 in 534,275,866 is tenmillions. 6. The place value of 3 in 267,301,818 is hundredthousands. 7. a.
7 < 7 since 7 is to the left of 7 on a number line.
b.
0 > 4 since 0 is to the right of 4 on a number line.
c.
9 > 11 since 9 is to the right of 11 on a number line.
8. a.
12 > 4 since 12 is to the right of 4 on a number line.
b.
13 > 31 since 13 is to the right of 31 on a number line.
14. Subtract the cost of the camera from the amount in her account. 762 237 525 She will have $525 left in her account after buying the camera. 15. To round 568 to the nearest ten, observe that the digit in the ones place is 8. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 568 rounded to the nearest ten is 570. 16. To round 568 to the nearest hundred, observe that the digit in the tens place is 6. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 568 rounded to the nearest hundred is 600. 17.
4725 2879
rounds to rounds to
4700 2900 1800
9. 13 2 7 8 9 (13 7) (2 8) 9 20 10 9 39
18.
8394 2913
rounds to rounds to
8000 3000 5000
10. 11 3 9 16 (11 9) (3 16) 20 19 39
19. a.
c.
11.
12.
82 < 79 since 82 is to the left of 79 on a number line.
7826 505 7321 Check:
b. 20(4 + 7) = 20 4 + 20 7 c. 7321 505 7826
20. a.
3285 272 3013 Check:
2(7 + 9) = 2 7 + 2 9 5(2 + 12) = 5 2 + 5 12
b.
9(3 + 6) = 9 3 + 9 6
c.
4(8 + 1) = 4 8 + 4 1
21.
631 125 3 155 12 620 63 100 78,875
22.
299 104 1 196 29 900 31, 096
3013 272 3285
13. Subtract 7257 from the radius of Jupiter. 43, 441 7 257 36,184 The radius of Saturn is 36,184 miles.
68
5(6 + 5) = 5 6 + 5 5
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ISM: Prealgebra and Introductory Algebra
23. a. b.
42 7 = 6 because 6 7 = 42. 64
8 because 8 8 = 64.
8
c. 24. a.
7 3 21 because 7 3 = 21. 35
7 because 7 5 = 35.
5 b. 64 8 = 8 because 8 8 = 64.
c.
12 4 48 because 12 4 = 48.
741 25. 5 3705 35 20 20 05 5 0 Check: 741 5 3705 456 26. 8 3648 32 44 40 48 48 0 Check: 456 8 3648
Chapter 2: Integers and Introduction to Solving Equations
27. number of cards number of number of for each person cards employees 238 19 12 R 10 19 238 19 48 38 10 Each employee will receive 12 gift cards. There will be 10 gift cards left over. 28.
cost of each total number of tickets ticket cost 324 36 9 36 324 324 0 Each ticket cost $9.
29. 92 9 9 81 30. 53 5 5 5 125 31. 61 6 32. 41 4 33. 5 62 5 6 6 180 34. 23 7 2 2 2 7 56 35.
7 2 3 32 7 2 3 9 7 6 9 10 2 5(2 1) 5(1) 5 5
36.
62 4 4 23 36 4 4 8 37 25 37 52 36 16 8 12 60 12 5
37. x + 6 = 8 + 6 = 14 38. 5 + x = 5 + 9 = 14
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Chapter 2: Integers and Introduction to Solving Equations
ISM: Prealgebra and Introductory Algebra
|9| = 9 because 9 is 9 units from 0.
45. 7 3 = 21
b.
|8| = 8 because 8 is 8 units from 0.
46. 5(2) = 10
c.
|0| = 0 because 0 is 0 units from 0.
47. 0 (4) = 0
40. a.
|4| = 4 because 4 is 4 units from 0.
48. 6 9 = 54
39. a.
b. |7| = 7 because 7 is 7 units from 0. 41. 2 + 25 = 23 42. 8 + (3) = 5 43. 2a b = 2(8) (6) = 16 (6) = 16 + 6 = 22 44. x y = 2 (7) = 2 + 7 = 5
70
49. 3(4 7) (2) 5 3(3) (2) 5 9 (2) 5 11 5 16 50. 4 8(7 3) (1) 4 8(4) (1) 4 32 (1) 4 32 1 28 1 27
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Chapter 3 14. 4(2x) = (4 2)x = 8x The perimeter is 8x centimeters.
Section 3.1 Practice Exercises 1. a.
8m 14m = (8 14)m = 6m
15. 3(12y + 9) = 3 12y + 3 9 = 36y + 27 The area of the garden is (36y + 27) square yards.
b. 6a + a = 6a + 1a = (6 + 1)a = 7a c.
y2 3y2 7 1y2 3y2 7
Vocabulary, Readiness & Video Check 3.1
(1 3) y2 7 2 y2 7
1.
2. 6z 5 z 4 6z 5 1z (4) 6z 1z 5 (4) (6 1)z 5 (4) 7z 1
14 y2 2x 23 is called an expression while 14 y2 , 2x, and 23 are each called a term. 2. To multiply 3(7x + 1), we use the distributive property.
3. 6y + 12y 6 = 18y 6
3. To simplify an expression like y + 7y, we combine like terms.
4. 7 y 5 y 8 7 y (5) 1y 8 7 y 1y (5) 8 8y 3
4. By the commutative properties, the order of adding or multiplying two numbers can be changed without changing their sum or product.
5. 7 y 2 2 y 9x 12 x 7 y 2 (2 y) (9x) 12 (1x) 7 y (2 y) (9x) (1x) 2 12 9 y 10x 14
5. The term 5x is called a variable term while the term 7 is called a constant term.
6. 6(4a) = (6 4)a = 24a
7. By the associative properties, the grouping of adding or multiplying numbers can be changed without changing their sum or product.
7. 8(9x) = (8 9)x = 72x 8. 8(y + 2) = 8 y + 8 2 = 8y + 16 9. 3(7a 5) = 3 7a 3 5 = 21a 15
6. The term z has an understood numerical coefficient of 1.
8. The terms x and 5x are like terms and the terms 5x and 5y are unlike terms.
10. 6(5 y) = 6 5 6 y = 30 6y
9. For the term 3x2 y, 3 is called the numerical coefficient.
11. 5(2 y 3) 8 5(2 y) 5(3) 8 10 y 15 8 10 y 23
10. because the original expression contains like terms and we can combine like terms 11. distributive property
12. 7(x 1) 5(2x 3) 7(x) (7)(1) 5(2x) 5(3) 7x 7 10x 15 3x 22
12. The 20 is outside the parentheses, so the distributive property does not apply to it. 13. addition; multiplication; P = perimeter; A = area
13. ( y 1) 3y 12 1( y 1) 3y 12 1 y (1)(1) 3y 12 y 1 3y 12 2 y 13
Exercise Set 3.1 2. 8y + 3y = (8 + 3)y = 11y 4. 7z 10z = (7 10)z = 3z
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Chapter 3: Solving Equations and Problem Solving 6. 5b 8b b = (5 8 1)b = 4b 8. 8y + y 2y 8y = (8 + 1 2 8)y = 1y = y 10. 5b 4b + b 15 = (5 4 + 1)b 15 = 2b 15 12. 4(4x) = (4 4)x = 16x 14. 3(21z) = (3 21)z = 63z
ISM: Prealgebra and Introductory Algebra 42. a 4 7a 5 a 7a 4 5 (1 7)a 4 5 6a 1 44. m 8n m 8n m m 8n 8n (1 1)m (8 8)n 2m 0n 2m 46. 5(x 1) 18 5 x 5 118 5x 5 18 5x 13
16. 13(5b) = (13 5)b = 65b 18. 3(x + 1) = 3 x + 3 1 = 3x + 3 20. 4(y 6) = 4 y 4 6 = 4y 24 22. 8(8y + 10) = 8 8y + (8) 10 = 64y 80 24. 5(6 y) 2 5(6) 5( y) 2 30 5 y 2 5 y 30 2 5 y 28
48. 3(x 2) 11x 3 x 3 2 11x 3x 6 11x 3x 11x 6 8x 6 50. 8(1 v) 6v 8 1 (8) v 6v 8 8v 6v 2v 8 52. 5 y 4 9 y y 15 5 y 9 y y 4 15 (5 9 1) y 4 15 13y 11
26. 10 4(6d 2) 10 4 6d 4 2 10 24d 8 24d 10 8 24d 2 28. 3(5 2b) 4b 3(5) (3)(2b) 4b 15 6b 4b 6b 4b 15 2b 15 30. 8z 5(6 z) 20 8z 5 6 5 z 20 8z 30 5z 20 8z 5z 30 20 13z 50 32. 3(5x 2) 2(3x 1) 3 5x 3 2 2 3x 2 1 15x 6 6x 2 15x 6x 6 2 21x 4 34. (5x 1) 10 1(5x 1) 10 1 5x (1)(1) 10 5x 110 5x 9 36. x + 12x = (1 + 12)x = 13x 38. 12x + 8x = (12 + 8)x = 4x
54. 2(x 4) 8(3x 1) 2 x (2) 4 8 3x 8 1 2x 8 24x 8 2x 24x 8 8 22x 16 56. 6x 4 2x x 3 6x 2x x 4 3 (6 2 1)x 4 3 7x 1 58. 5(c 2) 7c 5 c 5 2 7c 5c 10 7c 5c 7c 10 12c 10 60. 8(m 3) 20 m 8 m 8 3 20 m 8m 24 20 m 8m m 24 20 9m 4 62. 5(2a 3) 10a 5 2a 5 3 10a 10a 15 10a 10a 10a 15 15
40. 8r + s 7s = 8r + (1 7)s = 8r 6s 72
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ISM: Prealgebra and Introductory Algebra 64. (7 y 2) 6 1(7 y 2) 6 1 7 y (1)(2) 6 7 y 2 6 7 y 8 66. (12b 20) 5(3b 2) 1(12b 20) 5(3b 2) 1(12b) (1)(20) 5(3b) 5(2) 12b 20 15b 10 12b 15b 20 10 3b 10
Chapter 3: Solving Equations and Problem Solving 84. Area = (length) (width) = 50 40 = 2000 The area is 2000 square feet. 86. Perimeter 2 (length) 2 (width) 2 120 2 80 240 160 400 The perimeter is 400 feet so the rancher needs 400 feet of fencing. 88. x + x + x 14 = (1 + 1 + 1)x 14 = 3x 14 The perimeter is (3x 14) inches.
68. b 2(b 5) b 2 b 2 5 b 2b 10 3b 10
90. 15 + 23 = 8
70. 3x 4(x 2) 1 3x 4 x 4 2 1 3x 4x 8 1 1x 7 x 7
94. 8 + (8) = 0
92. 7 (4) = 7 + 4 = 3
72. 3x + x + 7 + 4x + 12 + 5x = 3x + x + 4x + 5x + 7 + 12 = (3 + 1 + 4 + 5)x + 7 + 12 = 13x + 19 The perimeter is (13x + 19) feet. 74. 3z 1 5z 1 3z 5z 1 1 (3 5)z 11 8z 2 The perimeter is (8z + 2) meters. 76. 6(9y + 1) = 6 9y + 6 1 = 54y + 6 The perimeter is (54y + 6) kilometers. 78. Area (length) (width) (5x) (8) (5 8)x 40x The area is 40x square centimeters. 80. Area (length) (width) 11(z 6) 11 z 11 6 11z 66 The area is (11z 66) square meters. 82. Area (length) (width) 12(2x 3) 12 2x 12 3 24x 36 The area is (24x + 36) square feet.
96. 2(4x 1) 2(4x) (2)(1) 8x (2) 8x 2 The expressions are not equivalent. 98. 8(ab) = 8 ab = 8ab The expressions are not equivalent. 100. 12 y (3y 1) 12 y (1)(3y 1) 12 y (1)(3y) (1)(1) 12 y 3y 1 The expressions are equivalent. 102. 6(x 5) 2 6 x 6 5 2 6x 30 2 6x 32 The expressions are not equivalent. 104. The order of the terms is changed. This is the commutative property of addition. 106. The order of the terms is changed, but the grouping is not. This is the commutative property of addition. 108. Add the areas of each rectangle. 12(3x 5) 4(5x 1) 12 3x 12 5 4 5x 4 1 36x 60 20x 4 36x 20x 60 4 56x 64 The total area is (56x 64) square kilometers.
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving 110. 76(268x + 592) 2960 = 76 268x + 76 592 2960 = 20,368x + 44,992 2960 = 20,368x + 42,032 112. answers may vary 114. answers may vary
5. 4 10 4 y 5 y 14 y 14 1y 1 1 14 y 6.
13x 4(3x 1) 13x 4 3x 4 1 13x 12x 4 13x 12x 12x 4 12x x 4
7.
5 y 2 17 5 y 2 2 17 2 5 y 15 5 y 15 5 5 y3 Check: 5 y 2 17 5(3) 2 017 15 2 0 17 17 17 True The solution is 3.
Section 3.2 Practice Exercises 1.
2.
3.
x 6 1 3 x 6 2 x 6 6 2 6 x 8 Check: x 6 1 3 8 6 0 1 3 2 2 True The solution is 8. 10 2m 4m 10 2m 10 2m 2 2 5 m Check: 10 2m 4m 10 0 2(5) 4(5) 10 0 10 (20) 10 0 10 20 10 10 True The solution is 5. 8 6 2
a 3 a
3 a 3(2) 3 3 3 3(2) a 3 6 a 4. 6 y 1 7 y 17 2 6 y 7 y 1 17 2 y 1 19 y 1 1 19 1 y 20
8. 4(x 2) 60 2 10 4x 8 60 2 10 4x 68 8 4x 68 68 8 68 4x 60 4x 60 4 4 x 15 9. a. b.
5 decreased by a number is 5 x.
c.
Three times a number is 3x.
d.
A number subtracted from 83 is 83 x.
e.
The quotient of a number and 4 is
x 4
or
x . 4 10. a. b.
74
The sum of 3 and a number is 3 + x.
The product of 5 and a number, decreased by 25, is 5x 25. Twice the sum of a number and 3 is 2(x + 3).
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
c.
The quotient of 39 and twice a number is 39 39 (2x) or . 2x
Vocabulary, Readiness & Video Check 3.2 1. The equations 3x = 51 and 3x 51 are called 3 3 equivalent equations. 2. The difference between an equation and an expression is that an equation contains an equal sign, while an expression does not. 3. The process of writing 3x + 10x as 7x is called simplifying the expression.
Chapter 3: Solving Equations and Problem Solving 8. 100 15 y 5 y 100 20 y 100 20 y 20 20 5 y 10. 3x 11 2 3x 9 3x 9 3 3 x 3 12.
2
6. By the multiplication property of equality, y = 8 and 3 y = 3 8 are equivalent equations. 7. Simplify the left side of the equation by combining like terms.
y
16.
9. addition property of equality 10. because order matters with subtraction
2.
4.
x7 23 x7 5 x77 57 x 2 1 8 n 2 7 n 2 7 2 n 2 2 9 n
6. 10 y y 45 9 y 45 9 y 45 9 9 y5
4 z
14. 5 y 9 y 14 (14) 4 y 28 4 y 28 4 4 y7
8. Simplify the left side of the equation by using the distributive property.
Exercise Set 3.2
z
4 z 4 (2) 4 4 8 z
4. For the equation 5x 1 = 21, the process of finding that 4 is the solution is called solving the equation. 5. By the addition property of equality, x = 2 and x + 7 = 2 + 7 are equivalent equations.
20 22
32 52 3 y 20 3 y 3 3(20) 3 y 60
18. 3 5x 4x 13 3 x 13 3 x 3 13 3 x 16 20. 7 10 4x 6 3x 3 x6 3 6 x 6 6 9x 22.
6(3x 1) 19x 6 3x 6 1 19x 18x 6 19x 18x 18x 6 19x 18x 6 x
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving 24.
17x 4(4x 6) 17x 4 4x 4 6 17x 16x 24 17x 16x 16x 16x 24 x 24
26.
28z 9(3z 2) 28z 9 3z 9 2 28z 27z 18 28z 27z 27z 27z 18 z 18
28.
2(1 3y) 7 y 2 (1) (2)(3y) 7 y 2 6y 7y 2 6y 6y 7y 6y 2 y
30.
3y 12 0 3y 12 12 0 12 3y 12 3y 12 3 3 y4
32.
5m 1 46 5m 11 46 1 5m 45 5m 45 5 5 m9
34.
11 3t 2 11 2 3t 2 2 9 3t 9 3t 3 3 3 t
36.
76
4(3y 5) 14 y 4 3y 4 5 14 y 12 y 20 14 y 12 y 12 y 20 14 y 12 y 20 2 y 20 2 y 2 2 10 y
38.
5(x 6) 2 8 5 x 5 6 10 5x 30 10 5x 30 30 10 30 5x 20 5x 20 5 5 x4
40.
2(x 5) 2 8 4 2 x 2 5 2 12 2x 10 2 12 2x 12 12 2x 12 12 12 12 2x 0 2x 0 2 2 x0
42.
x 63 10x x x 63 10x x 63 9x 63 9x 9 9 7 x
44. 35 (3) 3(x 2) 17 35 3 3 x 3 2 17 38 3x 6 17 38 3x 11 38 11 3x 11 11 27 3x 27 3x 3 3 9 x 46. 7 (10) x 5 7 10 x 5 17 x 5 17 5 x 5 5 22 x 48.
y 8 5 1 y 8 6 y 8 8 6 8 y2
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ISM: Prealgebra and Introductory Algebra 50. y 6 y 20 5 y 20 5y 20 5 5 y 4
Chapter 3: Solving Equations and Problem Solving
64.
52. 6x 5 35 6x 30 6x 30 6 6 x 5
66. 10x 11x 5 9 5 x5 4 x55 45 x 1
54.
x 12 x 5 12
9 14
12 (5) 12 60 x 56.
y 68.
6 y
x 12
6 y
8x 8 32 6
8x 8 8 32 8 8x 40 8x 40 8 8 x5 58. 30 t 9t 30 10t 30 10t 10 10 3 t 60. 42 20 2x 13x 22 11x 22 11x 11 11 2 x 62.
10x 6(2x 3) 10x 6 2x 6 3 10x 12x 18 10x 12x 12x 12x 18 2x 18 2x 18 2 2 x9
65 y 8(8 y 9) 65 y 8 8 y 8 9 65 y 64 y 72 65 y 64 y 64 y 64 y 72 y 72
6 y
6 (1) 6 1 7
6 7 6 y 42
70. 8x 4 6x 12 22 8x 6x 4 10 2x 4 10 2x 4 4 10 4 2x 6 2x 6 2 2 x 3 72.
4(x 7) 30 3 37 4 x 4 7 30 3 37 4x 28 30 34 4x 58 34 4x 58 58 34 58 4x 24 4x 24 4 4 x 6
74. Negative eight plus a number is 8 + x. 76. A number subtracted from 12 is 12 x. 78. Twice a number is 2x. 80. The quotient of negative six and a number is 6 6 6 x or or . x x Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
82. Negative four times a number, increased by 18 is 4x + 18.
108.
84. Twice a number, decreased by thirty is 2x 30. 86. The product of 7 and a number, added to 100 is 100 + 7x. 88. The difference of 9 times a number, and 1 is 9x 1. 90. Twice the sum of a number and 5 is 2[x + (5)] or 2(x 5). 92. The quotient of ten times a number and 4 is 10x (4) or
10x 10x . or 4 4
94. The quotient of 20 and a number, decreased by 20 3 or 20 three is (20 x) 3 or 3. x x 96. The bars with length less than $500 correspond to cities with daily costs less than $500. The cities with daily costs less than $500 are Oslo and Riyadh. 98. For a business traveler, the daily cost is $524 for Stavanger, $497 for Oslo, and $528 for Stockholm. 524 + 497 + 528 = 1549 The combined daily cost is $1549. 100. answers may vary 102. no; answers may vary 104. answers may vary x 106.
4 6 57
13 x 4096 78,125 13 x 74, 029 13 x 13 13 (74, 029) 13 x 962, 377
y (8)2 20 (2)2 10 y 64 20 4 10 y 64 20 4 10 y 48 10 y 10 10 48 10 y 480
110. 4(x 11) 90 86 25 5x 4(x 11) 90 86 32 5x 4(x 11) 36 5x 4x 44 36 5x 4x 8 5x 4x 4x 8 5x 4x 8 x Mid-Chapter Review 1. 7x 5y + 14 is an expression because it does not contain an equal sign. 2. 7x = 35 + 14 is an equation because it contains an equal sign. 3. 3(x 2) = 5(x + 1) 17 is an equation because it contains an equal sign. 4. 9(2x + 1) 4(x 2) + 14 is an expression because it does not contain an equal sign. 5. To simplify an expression, we combine any like terms. 6. To solve an equation, we use properties of equality to find any value of the variable that makes the equation a true statement. 7. 7x + x = (7 + 1)x = 8x 8. 6y 10y = (6 10)y = 4y 9. 2a + 5a 9a 2 = (2 + 5 9)a 2 = 2a 2 10.
6a 12 a 14 6a a 12 14 (6 1)a 12 14 5a 26
11. 2(4x + 7) = 2 4x + (2) 7 = 8x 14 12. 3(2x 10) = 3(2x) (3)(10) = 6x + 30 78
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ISM: Prealgebra and Introductory Algebra 13. 5( y 2) 20 5 y 5 2 20 5 y 10 20 5 y 10 14. 12x 3(x 6) 13 12x 3 x 3 6 13 12x 3x 18 13 (12 3)x 18 13 15x 31 15. Area (length) (width) 3(4x 2) 3 4x 3 2
Chapter 3: Solving Equations and Problem Solving 20. 6 (5) x 5 65 x5 11 x 5 11 5 x 5 5 6x Check: 6 (5) x 5 6 (5) 0 6 5 650 65 11 11 True The solution is 6. x
12x 6 The area is (12x 6) square meters.
21.
16. Perimeter = x + x + 2 + 7 = 2x + 9 The perimeter is (2x + 9) feet. 17.
12 11x 14x 12 3x 12 3x 3 3 4 x Check: 12 11x 14x 12 0 11(4) 14(4) 12 0 44 56 12 12 True The solution is 4.
The solution is 15. 22.
18. 8 y 7 y 45 15y 45 15y 45 15 15 y 3 Check: 8 y 7 y 45 8(3) 7(3) 0 45 24 (21) 0 45 45 45 True The solution is 3. 19.
x 12 45 23 x 12 22 x 12 12 22 12 x 10 Check: x 12 45 23 10 12 0 45 23 22 22 True The solution is 10.
14 9 3 x 5 3 x 3 3(5) 3 x 15 x 14 9 Check: 153 0 14 9 3 5 5 True
z 23 7 4 z 30 4 z 4 4 (30) 4 z 120 z 23 7 Check: 4 120 0 23 7 4 30 30 True The solution is 120.
23. 6 2 4x 1 3x 4 x 1 4 1 x 11 5 x Check: 6 2 4x 1 3x 6 2 0 4(5) 1 3(5) 4 0 20 115 4 4 True The solution is 5.
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
24.
5 8 5x 10 4x 3 x 10 3 10 x 10 10 13 x Check: 5 8 5x 10 4x 5 8 0 5(13) 10 4(13) 3 0 65 10 52 3 3 True The solution is 13.
25.
6(3x 4) 19x 6 3x 6 4 19x 18x 24 19x 18x 18x 24 19x 18x 24 x Check: 6(3x 4) 19x 6[3(24) 4] 0 19(24) 6(72 4) 0 456 6(76) 0 456 456 456 True
28. 8 (14) 80 y 20 81y 22 y 20 22 20 y 20 20 42 y Check: 8 (14) 80 y 20 81y 8 (14) 0 80(42) 20 81(42) 22 0 3360 20 3402 22 22 True The solution is 42. 29.
3x 16 10 3x 16 16 10 16 3x 6 3x 6 3 3 x2 Check: 3x 16 10 3(2) 16 0 10 6 16 0 10 10 10 True The solution is 2.
30.
4x 21 13 4x 21 21 13 21 4x 8 4x 8 4 4 x2 Check: 4x 21 13 4 2 21 0 13 8 210 13 13 13 True The solution is 2.
The solution is 24. 26.
25x 6(4x 9) 25x 6 4x 6 9 25x 24x 54 25x 24x 24x 24x 54 x 54 Check: 25x 6(4x 9) 25(54) 0 6[4(54) 9] 1350 0 6(216 9) 1350 0 6(225) 1350 1350 True The solution is 54.
27. 36x 10 37x 12 (14) x 10 12 14 x 10 2 x 10 10 2 10 x 12 36x 10 37x 12 (14) Check: 36(12) 10 37(12) 0 12 (14) 432 10 444 0 12 14 2 2 True The solution is 12.
80
31. 8z 2z 26 (4) 10z 26 4 10z 30 10z 30 10 10 z 3 8z 2z 26 (4) Check: 8(3) 2(3) 0 26 (4) 24 6 0 26 4 30 30 True The solution is 3.
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
32. 12 (13) 5x 10x 25 5x 25 5x 5 5 5 x Check: 12 (13) 5x 10x 12 (13) 0 5(5) 10(5) 25 0 25 50 25 25 True The solution is 5.
38. The quotient of 10 and a number is 10 x or 10 . x
33.
4(x 8) 11 3 26 4 x (4) 8 11 3 26 4x 32 11 23 4x 43 23 4x 43 43 23 43 4x 20 4x 20 4 4 x 5 Check: 4(x 8) 11 3 26 4(5 8) 110 3 26 4(3) 110 23 12 110 23 23 23 True The solution is 5.
34.
6(x 2) 10 4 10 6 x (6)(2) 10 4 10 6x 12 10 14 6x 22 14 6x 22 22 14 22 6x 36 6x 36 6 6 x6 Check: 6(x 2) 10 4 10 6(6 2) 10 0 4 10 6(4) 10 0 14 24 10 0 14 14 14 True The solution is 6.
35. The difference of a number and 10 is x 10. 36. The sum of 20 and a number is 20 + x. 37. The product of 10 and a number is 10x.
39. Five added to the product of 2 and a number is 2x + 5. 40. The product of 4 and the difference of a number and 1 is 4(x 1). Section 3.3 Practice Exercises 1.
7x 12 3x 4 7x 12 12 3x 4 12 7x 3x 16 7x 3x 3x 16 3x 4x 16 4x 16 4 4 x 4 Check: 7x 12 3x 4 7(4) 210 3(4) 4 28 12 0 12 4 16 16 True The solution is 4.
2.
40 5 y 5 2 y 10 4 y 45 5 y 6 y 10 45 5 y 45 6 y 10 45 5 y 6 y 55 5 y 6 y 6 y 55 6 y y 55 Check: 40 5 y 5 2 y 10 4 y 40 5(55) 5 0 2(55) 10 4(55) 40 275 5 0 110 10 220 320 320 True The solution is 55.
3.
6(a 5) 4a 4 6a 30 4a 4 6a 30 4a 4a 4 4a 2a 30 4 2a 30 30 4 30 2a 34 2a 34 2 2 a 17 The solution is 17.
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
4.
4(x 3) 1 13 4x 12 1 13 4x 13 13 4x 13 13 13 13 4x 0 4x 0 4 4 x0 Check: 4(x 3) 1 13 4(0 3) 1 013 12 1 0 13 13 13 True The solution is 0.
5. a.
Vocabulary, Readiness & Video Check 3.3 1. An example of an expression is 3x 9 + x 16, while an example of an equation is 5(2x + 6) 1 = 39. x 2. To solve we use the multiplication 10, 7 property of equality. 3. To solve x 7 = 10, we use the addition property of equality. 4. To solve 9x 6x = 10 + 6, first combine like terms.
The difference of 110 and 80 is 30 translates to 110 80 = 30.
b.
The product of 3 and the sum of 9 and 11 amounts to 6 translates to 3(9 + 11) = 6.
c.
The quotient of 24 and 6 yields 4 24 translates to 4. 6
5. To solve 5(x 1) = 25, first use the distributive property. 6. To solve 4x + 3 = 19, first use the addition property of equality. 7. the addition property of equality; to make sure we get an equivalent equation 8. remove parentheses; the distributive property
Calculator Explorations 9. gives; amounts to 1. Replace x with 12. 76(12 25) = 988 Yes
Exercise Set 3.3 2.
7x 1 8x 4 7x 7x 1 8x 7x 4 1 x 4 1 4 x 4 4 5 x
4.
5x 3 2x 18 5x 3 3 2x 18 3 5x 2x 15 5x 2x 2x 2x 15 3x 15 3x 15 3 3 x 5
6.
4 7m 3m 4 4 7m 7m 3m 7m 4 4 4m 4 4 4 4m 4 4 0 4m 0 4m 4 4 0m
2. Replace x with 35. 47 35 + 862 = 783 Yes 3. Replace x with 170. 170 + 562 = 392 3 (170) + 900 = 390 No 4. Replace x with 18. 55(18 + 10) = 440 75 (18) + 910 = 440 Yes 5. Replace x with 21. 29 (21) 1034 = 1643 61 (21) 362 = 1643 Yes 6. Replace x with 25. 38 25 + 205 = 745 25 25 + 120 = 745 No 82
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ISM: Prealgebra and Introductory Algebra 8.
57 y 140 54 y 100 57 y 140 54 y 54 y 54 y 100 3y 140 100 3y 140 140 100 140 3y 240 3y 240 3 3 y 80
10. 2x 10 3x 12 x 20 2x 3x 10 12 20 x 5x 10 32 x 5x x 10 32 x x 6x 10 32 6x 10 10 32 10 6x 42 6x 42 6 6 x 7 12. 19x 2 7x 31 6x 15 19x 7x 2 3115 6x 12x 2 16 6x 12x 6x 2 16 6x 6x 6x 2 16 6x 2 2 16 2 6x 18 6x 18 6 6 x3 14. 22 42 4(x 1) 20 4x 4 20 4 4x 4 4 16 4x 16 4x 4 4 4 x 16.
Chapter 3: Solving Equations and Problem Solving 18.
3(z 2) 5z 6 3z 6 5z 6 3z 3z 6 5z 3z 6 6 2z 6 6 6 2z 6 6 0 2z 0 2z 2 2 0 z
20.
1( y 3) 10 1y 3 10 y 3 10 y 3 3 10 3 y 13 y 13 1 1 y 13
22.
4 3c 4(c 2) 4 3c 4c 8 4 3c 3c 4c 3c 8 4 c 8 4 8 c 8 8 12 c
24. 4(3t 4) 20 3 5t 12t 16 20 3 5t 12t 4 3 5t 12t 5t 4 3 5t 5t 7t 4 3 7t 4 4 3 4 7t 7 7t 7 7 7 t 1 26.
3x 51 3x 51 3 3 x 17
28.
y 6 11 y 6 6 11 6 y 5
30.
7 z 15 7 7 z 15 7 z 8 1(z) 18 z 8
2(x 5) 8 0 2x 10 8 0 2x 18 0 2x 18 18 0 18 2x 18 2x 18 2 2 x 9
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Chapter 3: Solving Equations and Problem Solving
32.
2 10 12
z
7n 5 12n 10 7n 5 7n 12n 10 7n 5 5n 10 5 10 5n 10 10 15 5n 15 5n 5 5 3n
50.
10w 8 w 10 10w w 8 w w 10 9w 8 10 9w 8 8 10 8 9w 18 9w 18 9 9 w 2
52.
4 7m 3m 4 7m 7m 3m 7m 4 4m 4 4m 4 4 1 m
54.
10 4v 6 0 4v 16 0 4v 16 16 0 16 4v 16 4v 16 4 4 v 4
56.
2(z 2) 5z 17 2z 4 5z 17 2z 4 2z 5z 17 2z 4 3z 17 4 17 3z 17 17 21 3z 21 3z 3 3 7 z
z 10
7 y 3 38 7 y 3 3 38 3 7 y 35 7 y 35 7 7 y5 2b 5 7 2b 5 5 7 5 2b 12 2b 12 2 2 b6
38.
4x 11x 14 14 7x 28 7x 28 7 7 x4
40.
5(3y 2) 16 y 15 y 10 16 y 15 y 15 y 10 16 y 15 y 10 y
42. 9 20 19x 4 18x 11 x 4 11 4 x 4 4 15 x 44. 9(x 2) 25 19 19 9x 18 25 38 9x 7 38 9x 7 7 38 7 9x 45 9x 45 9 9 x5
84
48.
10
120 z
36.
6 y 8 3y 7 6 y 3y 8 3y 3y 7 3y 8 7 3y 8 8 7 8 3y 15 3y 15 3 3 y5
10 z
10 (12) 10
34.
46.
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ISM: Prealgebra and Introductory Algebra
58.
4 3c 2(c 2) 4 3c 2c 4 4 3c 2c 2c 2c 4 4c 4 4c4 44 c0
78. y3 3xyz (1)3 3(3)(1)(0) 1 0 1 80. ( y)3 3xyz [(1)]3 3(3)(1)(0) 13 0 1 0 1
60. 4(2t 5) 21 7t 6
82. The first step in solving 3x + 2x = x 4 is to add 3x and 2x, which is choice c.
8t 20 21 7t 6 8t 1 7t 6 8t 7t 1 7t 7t 6 t 1 6 t 11 6 1 t 5
84. The first step in solving 9 5x = 15 is to subtract 9 from both sides, which is choice c. 86. The error is in the second line. 37x 1 9(4x 7) 37x 1 36x 63 37x 11 36x 63 1 37x 36x 64 37x 36x 36x 64 36x x 64
62. 14 4(w 5) 6 2w 14 4w 20 6 2w 4w 6 6 2w 4w 2w 6 6 2w 2w 6w 6 6 6w 6 6 6 6 6w 12 6w 12 6 6 w2 64.
Chapter 3: Solving Equations and Problem Solving
88. 32 x (9)3 9x 729 9x 729 9 9 x 81
6(5 c) 5(c 4) 30 6c 5c 20 30 6c 5c 5c 5c 20 30 c 20 30 30 c 20 30 c 50
90.
66. The difference of 30 and 10 equals 40 translates to 30 10 = 40. 68. The quotient of 16 and 2 yields 8 translates to 16 8. 2
x 452 542 x 2025 2916 x 2025 2025 2916 2025 x 891
92. no; answers may vary Section 3.4 Practice Exercises 1. a.
“Four times a number is 20” is 4x = 20.
b.
“The sum of a number and 5 yields 32” is x + (5) = 32.
70. Negative 2 times the sum of 3 and 12 is 30 translates to 2(3 + 12) = 30.
c.
“Fifteen subtracted from a number amounts to 23” is x 15 = 23.
72. Seventeen subtracted from 12 equals 29 translates to 12 17 = 29.
d.
“Five times the difference of a number and 7 is equal to 8” is 5(x 7) = 8.
e.
“The quotient of triple a number and 5 gives 3x 1. 1” is 3x 5 = 1 or 5
74. From the height of the 2015 bar, 132 million or 132,000,000 returns were filed electronically in 2015. 76. answers may vary
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Chapter 3: Solving Equations and Problem Solving 2. “The sum of a number and 2 equals 6 added to three times the number” is x 2 6 3x x x 2 6 3x x 2 6 2x 2 6 6 6 2x 4 2x 4 2x 2 2 2 x 3. Let x be the distance from Denver to San Francisco. Since the distance from Cincinnati to Denver is 71 miles less than the distance from Denver to San Francisco, the distance from Cincinnati to Denver is x 71. Since the total of the two distances is 2399, the sum of x and x 71 is 2399. x x 71 2399 2x 71 2399 2x 71 71 2399 71 2x 2470 2x 2470 2 2 x 1235 The distance from Denver to San Francisco is 1235 miles. 4. Let x be the amount her son receives. Since her husband receives twice as much as her son, her husband receives 2x. Since the total estate is $57,000, the sum of x and 2x is 57,000. x 2x 57, 000 3x 57, 000 3x 57, 000 3 3 x 19, 000 2x = 2(19,000) = 38,000 Her husband will receive $38,000 and her son will receive $19,000. Vocabulary, Readiness & Video Check 3.4 1. The phrase is “a number subtracted from 20” so 20 goes first and we subtract the number from that. 2. The phrase is “three times the difference of some number and 5.” The “difference of some number and 5” translates to the expression x 5, and in order to multiply 3 times this expression, we need parentheses around the expression.
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3. The original application asks for the fastest speeds of a pheasant and a falcon. The value of x is the speed in mph for a pheasant, so the falcon’s speed still needs to be found. Exercise Set 3.4 2. “Five subtracted from a number equals 10” is x 5 = 10. 4. “The quotient of 8 and a number is 2” is 8 2. x 6. “Two added to twice a number gives 14” is 2x + 2 = 14. 8. “Five times a number is equal to 75” is 5x = 75. 10. “Twice the sum of 17 and a number is 14” is 2(17 + x) = 14. 12. “Twice a number, subtracted from 60 is 20” is 60 2x 20 60 60 2x 20 60 2x 40 2x 40 2 2 x 20 14. “The sum of 7, 9, and a number is 40” is 7 9 x 40 16 x 40 16 16 x 40 16 x 24 16. “Eight decreased by a number equals the quotient of 15 and 5” is 15 8 x 5 8 x 3 88 x 38 x 5 x 5 1 1 x5 18. “The product of a number and 3 is twice the sum of that number and 5” is 3x 2(x 5) 3x 2x 10 3x 2x 2x 2x 10 x 10
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ISM: Prealgebra and Introductory Algebra 20. “Five times the sum of a number and 2 is 11 less than the number times 8” is 5(x 2) 8x 11 5x 10 8x 11 5x 5x 10 8x 5x 11 10 3x 11 10 11 3x 11 11 21 3x 21 3x 3 3 7x 22. “Seven times the difference of some number and 1 gives the quotient of 70 and 10” is 70 7(x 1) 10 7(x 1) 7 7x 7 7 7x 7 7 7 7 7x 14 7x 14 7 7 x2 24. “Twice a number equals 25 less triple that same number” is 2x 25 3x 2x 3x 25 3x 3x 5x 25 5x 25 5 5 x5 26. The equation is x + 4x = 50. x 4x 50 5x 50 5x 50 5 5 x 10 Maryland has 10 electoral votes, and Texas has 4(10) = 40 electoral votes.
Chapter 3: Solving Equations and Problem Solving
28. Let x be the width of the playing surface for wheelchair pickleball play. Since the length of the playing surface is 14 feet less than twice the width, the length of the playing surface is 2x 14. Since the total of the width and length is 118 feet, the sum of x and 2x 14 is 118. x 2x 14 118 3x 14 118 3x 14 14 118 14 3x 132 3x 132 3 3 x 44 2x 14 = 2(44) 14 = 88 14 = 74 The width of the playing surface for wheelchair pickleball play is 44 feet and the length of the playing surface is 74 feet. 30. Let x be the width of the playing surface for new construction of pickleball courts. Since the length of the playing surface is the width increased by 30 feet, the length of the playing surface is x + 30. Since the total of the width and length is 98 feet, the sum of x and x + 30 is 98. x x 30 98 2x 30 98 2x 30 30 98 30 2x 68 2x 68 2 2 x 34 The width of the playing surface for new construction of a pickleball court is 34 feet and the length of the playing surface is 34 feet + 30 feet = 64 feet. 32. Let x be the number of rulers for Liechtenstein. Since Norway has had three times as many rulers as Liechtenstein, Norway has had 3x rulers. Since the total number of rulers for both countries is 56, the sum of x and 3x is 56. x 3x 56 4x 56 4x 56 4 4 x 14 Thus, Liechtenstein has had 14 rulers and Norway has had 3(14) = 42 rulers.
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34. Let x be the 2021 average daily circulation (in thousands) of The New York Times. Since the average daily circulation of The Wall Street Journal was 367 thousand more than that of The New York Times, the average daily circulation of The Wall Street Journal was x + 367. Since the total circulation of the two papers averaged 1027 thousand per day, the sum of x and x + 367 was 1027. x x 367 1027 2x 367 1027 2x 367 367 1027 367 2x 660 2x 660 2 2 x 330 The average daily circulation of The New York Times in 2021 was 330 thousand and the average daily circulation of The Wall Street Journal was 330 + 367 = 697 thousand.
40. Let x be the distance from Los Angeles to Tokyo. Since the distance from New York to London is 2001 miles less than the distance from Los Angeles to Tokyo, the distance from New York to London is x 2001. Since the total of the two distances is 8939, the sum of x and x 2001 is 8939. x x 2001 8939 2x 2001 8939 2x 2001 2001 8939 2001 2x 10, 940 2x 10, 940 2 2 x 5470 The distance from Los Angeles to Tokyo is 5470 miles.
36. Let x be the recognition value (in billions of dollars) of Google. Since the recognition value of Amazon is $23 billion more than that of Google, the recognition value of Amazon is x + 23. Since the total recognition value of the two brands is $527 billion, the sum of x and x + 23 is 527. x x 23 527 2x 23 527 2x 23 23 527 23 2x 504 x 252 The recognition value of Google is $252 billion, and the recognition value of Amazon is $252 + $23 = $275 billion. 38. Let x be the price of Hasbro Game Night. Since the price of Mario Kart 8 is $44 more than the price of Hasbro Game Night, the price of Mario Kart 8 is x + 44. Since the total of the two prices is $74, the sum of x and x + 44 is 74. x x 44 74 2x 44 74 2x 44 44 74 44 2x 30 x 15 The price of Hasbro Gam Night is $15, and the price of Mario Kart 8 is $15 + $44 = $59.
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42. Let x be the top speed (in miles per hour) of an IndyCar Series car. Since the top speed of an NHRA Top Fuel dragster is 102 mph faster than the top speed of an IndyCar Series car, the top speed of an NHRA Top Fuel dragster is x + 102. Since the combined top speed for these two cars is 574 mph, the sum of x and x + 102 is 574. x x 102 574 2x 102 574 2x 102 102 574 102 2x 472 2x 472 2 2 x 236 The top speed of an IndyCar Series car is 236 mph, and the top speed of an NHRA Top Fuel dragster is 236 + 102 = 338 mph. 44. Let x be the 2022 Vermont population of Indigenous peoples. Since the 2022 California population of Indigenous peoples was 101 times that of Vermont, the 2022 California population of Indigenous peoples was 101x. Since the total of these two populations was 816,000, the sum of x and 101x is 816,000. x 101x 816, 000 102x 816, 000 102x 816, 000 102 102 x 8000 The 2022 Vermont population of Indigenous peoples was 8000, and the 2022 California population of Indigenous peoples was 101(8000) = 808,000.
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ISM: Prealgebra and Introductory Algebra
46. Let x be the speed of the truck. Since the speed of the car is twice that of the truck, the speed of the car is 2x. Since the combined speed is 105 miles per hour, the sum of x and 2x is 105. x 2x 105 3x 105 3x 105 3 3 x 35 The speed of the truck is 35 miles per hour and the speed of the car is 2(35) = 70 miles per hour. 48. Let x be the value of the plow. Since the tractor is worth seven times as much as the plow, the tractor is worth 7x. Since their combined worth is $1200, the sum of x and 7x is 1200. x 7x 1200 8x 1200 8x 1200 8 8 x 150 The plow is worth $150 and the tractor is worth 7($150) = $1050. 50. Let x be the maximum length (in yards) of a soccer field for under-6-year-olds. Since the maximum length of a under-13-year-old soccer field is 80 yards longer than the length for under6-year-olds, the maximum length for a under-13-year-old soccer field is x + 80. Since the total length for these two categories of soccer fields is 140 yards, the sum of x and x + 80 is 140. x x 80 140 2x 80 140 2x 80 80 140 80 2x 60 2x 60 2 2 x 30 The maximum length of a soccer field for under6-year-olds is 30 yards, and the maximum length of a under-13-year-old soccer field is 30 + 80 = 110 yards. 52. Let x be the number of points scored by the University of Connecticut Huskies. Since the South Carolina Gamecocks scored 15 points more than the Huskies, the Gamecocks scored x + 15 points. Since the total number of points scored by the two teams was 113, the sum of x and x + 15 is 113.
Chapter 3: Solving Equations and Problem Solving x x 15 113 2x 15 113 2x 15 15 113 15 2x 98 2x 98 2 2 x 49 x + 15 = 49 + 15 = 64 The South Carolina Gamecocks scored 64 points. 54. Let x be the amount (in millions of dollars) that is estimated to be spent on TV advertising in 2026. Since it is estimated that the amount spent on digital advertising will be $320,830 million more than will be spent on TV advertising, the amount estimated to be spent on digital advertising is x + 320,830. Since the total amount estimated to be spent on digital and TV advertising in 2026 is $450,110 million, the sum and x + 320,830 is 450,110. x x 320,830 450,110 2x 320,830 450,110 2x 320,830 320,830 450,110 320,830 2x 129, 280 2x 129, 280 2 2 x 64, 640 x + 320,830 = 64,640 + 320,830 = 385,470 The amount estimated to be spent on TV advertising in 2026 is $64,640 million, and the amount estimated to be spent on digital advertising is $385,470 million. 56. To round 82 to the nearest ten, observe that the digit in the ones place is 2. Since this digit is less than 5, we do not add 1 to the digit in the tens place. 82 rounded to the nearest ten is 80. 58. To round 52,333 to the nearest thousand, observe that the digit in the hundreds place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. 52,333 rounded to the nearest thousand is 52,000. 60. To round 101,552 to the nearest hundred, observe that the digit in the tens place is 5. Since this digit is at least 5, we add one to the digit in the hundreds place. 101,552 rounded to the nearest hundred is 101,600. 62. yes; answers may vary
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
64. Use P = A + C, where P = 165,000 and A = 156,750. P AC 165, 000 156, 750 C 165, 000 156, 750 156, 750 156, 750 C 8250 C The agent will receive $8250. 66. Use P = C + M where P = 12 and C = 7. PCM 12 7 M 12 7 7 7 M 5M The markup is $5. Chapter 3 Vocabulary Check 1. An algebraic expression is simplified when all like terms have been combined. 2. Terms that are exactly the same, except that they may have different numerical coefficients, are called like terms. 3. A letter used to represent a number is called a variable. 4. A combination of operations on variables and numbers is called an algebraic expression. 5. The addends of an algebraic expression are called the terms of the expression. 6. The number factor of a variable term is called the numerical coefficient. 7. Replacing a variable in an expression by a number and then finding the value of the expression is called evaluating the expression for the variable. 8. A term that is a number only is called a constant. 9. An equation is of the form expression = expression. 10. A solution of an equation is a value for the variable that makes an equation a true statement. 11. To multiply 3(2x + 1), we use the distributive property. 12. By the multiplication property of equality, we may multiply or divide both sides of an equation by any nonzero number without changing the solution of the equation.
90
13. By the addition property of equality, the same number may be added to or subtracted from both sides of an equation without changing the solution of the equation. Chapter 3 Review 1. 3y + 7y 15 = (3 + 7)y 15 = 10y 15 2.
2 y 10 8 y 2 y 8 y 10 (2 8) y 10 6 y 10
3. 8a a 7 15a 8a a 15a 7 (8 1 15)a 7 6a 7 4. y 3 9 y 1 y 9 y 3 1 (1 9) y 3 1 8 y 2 5. 2(x + 5) = 2 x + 2 5 = 2x + 10 6. 3(y + 8) = 3 y + (3) 8 = 3y 24 7. 7x 3(x 4) x 7x 3 x 3 4 x 7x 3x 12 x 7x 3x x 12 (7 3 1)x 12 11x 12 8. (3m 2) m 10 1(3m 2) m 10 1 3m (1) 2 m 10 3m 2 m 10 3m m 2 10 (3 1)m 2 10 4m 12 9. 3(5a 2) 20a 10 3 5a 3 2 20a 10 15a 6 20a 10 15a 20a 6 10 (15 20)a 6 10 5a 4 10. 6 y 3 2(3y 6) 6 y 3 2 3y 2 6 6 y 3 6 y 12 6 y 6 y 3 12 (6 6) y 3 12 12 y 9
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ISM: Prealgebra and Introductory Algebra 11. 6 y 7 11y y 2 6 y 11y y 7 2 (6 11 1) y 7 2 16 y 5 12. 10 x 5x 12 3x x 5x 3x 10 12 (1 5 3)x 10 12 1x 2 x2
Chapter 3: Solving Equations and Problem Solving
22.
23.
n 18 10 (2) n 18 10 2 n 18 12 n 18 18 12 18 n 6
24.
c 5 13 7 c 5 6 c 5 5 6 5 c 1
25.
7x 5 6x 20 x 5 20 x 5 5 20 5 x 25
26.
17x 2(8x 4) 17x 2 8x 2 4 17x 16x 8 17x 16x 16x 16x 8 x 8
27.
5x 7 3 5x 7 7 3 7 5x 10 5x 10 5 5 x 2
28.
14 9 y 4 14 4 9 y 4 4 18 9 y 18 9 y 9 9 2 y
13. Perimeter = 2(2x + 3) = 2 2x + 2 3 = 4x + 6 The perimeter is (4x + 6) yards. 14. Perimeter = 4 5y = 20y The perimeter is 20y meters. 15. Area (length) (width) 3(2x 1) 3 2x 31 6x 3 The area is (6x 3) square yards. 16. Add the areas of the two rectangles. 10(x 2) 7(5x 4) 10 x 10 2 7 5x 7 4 10x 20 35x 28 10x 35x 20 28 (10 35)x 20 28 45x 8 The area is (45x + 8) square centimeters. 17.
z 5 7 z 5 5 7 5 z 2
18.
3x 10 4x 3x 3x 10 4x 3x 10 x
19. 3y 21 3y 21 3 3 y 7 20. 3a 15 3a 15 3 3 a5 21.
x 2 6 x 6 6 2 6 x 12
y 3 15 y 15 15 (3) 15 y 45
29.
z
8 (6) 4 z 8 6 4 z 2 4 z 4 4 (2) 4 z 8
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
30. 1 (8) 9
x
37.
2(x 4) 10 2(7) 2 x 2 4 10 14 2x 8 10 14 2x 2 14 2x 2 2 14 2 2x 12 2x 12 2 2 x 6
38.
3(x 6) 13 20 1 3 x (3) 6 13 19 3x 18 13 19 3x 31 19 3x 31 31 19 31 3x 12 3x 12 3 3 x4
5 x
5 x 5 (9) 5 5 45 x 31. 6 y 7 y 100 105 y 5 y 5 1 1 y5 32. 19x 16x 45 60 3x 15 3x 15 3 3 x 5 33.
34.
35.
36.
92
9(2x 7) 19x 9 2x 9 7 19x 18x 63 19x 18x 18x 63 19x 18x 63 x
39. The product of 5 and a number is 5x.
5(3x 3) 14x 5 3x 5 3 14x 15x 15 14x 15x 15x 15 14x 15x 15 x
42. The quotient of 2 and a number is 2 x or
3x 4 11 3x 4 4 11 4 3x 15 3x 15 3 3 x5 6 y 1 73 6 y 1 1 73 1 6 y 72 6 y 72 6 6 y 12
40. Three subtracted from a number is x 3. 41. The sum of 5 and a number is 5 + x. 2 x 2
or . x 43. The product of 5 and a number, decreased by 50 is 5x 50. 44. Eleven added to twice a number is 2x + 11. 45. The quotient of 70 and the sum of a number and 70 6 is 70 (x + 6) or . x6 46. Twice the difference of a number and 13 is 2(x 13). 47.
2x 5 7x 100 2x 2x 5 7x 2x 100 5 5x 100 5 100 5x 100 100 105 5x 105 5x 5 5 21 x
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ISM: Prealgebra and Introductory Algebra
48.
6x 4 x 66 6x 6x 4 x 6x 66 4 7x 66 4 66 7x 66 66 70 7x 70 7x 7 7 10 x
49.
2x 7 6x 1 2x 2x 7 6x 2x 1 7 4x 1 7 1 4x 11 8 4x 8 4x 4 4 2 x
50.
5x 18 4x 5x 5x 18 4x 5x 18 9x 18 9x 9 9 2x
51.
5(n 3) 7 3n 5n 15 7 3n 5n 3n 15 7 3n 3n 2n 15 7 2n 15 15 7 15 2n 22 2n 22 2 2 n 11
52.
7(2 x) 4x 1 14 7x 4x 1 14 7x 4x 4x 4x 1 14 3x 1 14 14 3x 114 3x 15 3x 15 3 3 x 5
Chapter 3: Solving Equations and Problem Solving 53. 6x 3 (x) 20 5x 7 6x 3 x 20 5x 7 7x 3 27 5x 7x 5x 3 27 5x 5x 2x 3 27 2x 3 3 27 3 2x 30 2x 30 2 2 x 15 54.
x 25 2x 5 2x 10 25 3x 15 2x 25 3x 2x 15 2x 2x 25 x 15 25 25 x 15 25 x 10
55.
3(x 4) 5x 8 3x 12 5x 8 3x 3x 12 5x 3x 8 12 2x 8 12 8 2x 8 8 4 2x 4 2x 2 2 2 x
56.
4(x 3) 2x 48 4x 12 2x 48 4x 2x 12 2x 2x 48 6x 12 48 6x 12 12 48 12 6x 36 6x 36 6 6 x 6
57. 6(2n 1) 18 0 12n 6 18 0 12n 12 0 12n 12 12 0 12 12n 12 12n 12 12 12 n 1
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93
Chapter 3: Solving Equations and Problem Solving
68. “The difference of a number and 3 is the quotient 8 of 8 and 4” is x 3 . 4
58. 7(3y 2) 7 0 21y 14 7 0 21y 21 0 21y 21 21 0 21 21y 21 21y 21 21 21 y 1 59.
95x 14 20x 10 10x 4 95x 14 30x 14 95x 14 14 30x 14 14 95x 30x 95x 30x 30x 30x 65x 0 65x 0 65 65 x0
60. 32z 11 28z 50 2z (1) 4z 11 50 2z 1 4z 11 51 2z 4z 2z 11 51 2z 2z 2z 11 51 2z 1111 5111 2z 40 2z 40 2 2 z 20 61. The difference of 20 and 8 is 28 translates to 20 (8) = 28. 62. Nineteen subtracted from 2 amounts to 21 translates to 2 19 = 21. 63. The quotient of 75 and the sum of 5 and 20 is 75 equal to 3 translates to 3. 5 20 64. Five times the sum of 2 and 6 yields 20 translates to 5[2 + (6)] = 20. 65. “Twice a number minus 8 is 40” is 2x 8 = 40. 66. “The product of a number and 6 is equal to the sum of the number and 2a” is 6x = x + 2a. 67. “Twelve subtracted from the quotient of a x number and 2 is 10” is 12 10. 2 94
ISM: Prealgebra and Introductory Algebra
69. “Five times a number subtracted from 40 is the same as three times the number” is 40 5x 3x 40 5x 5x 3x 5x 40 8x 40 8x 8 8 5 x The number is 5. 70. “The product of a number and 3 is twice the difference of that number and 8” is 3x 2(x 8) 3x 2x 16 3x 2x 2x 2x 16 x 16 The number is 16. 71. Let x be the number of votes for the Independent candidate. Since the Democratic candidate received 272 more votes than the Independent candidate, the Democratic candidate received x + 272 votes. The total number of votes is 18,500. x x 272 14, 000 18, 500 2x 14, 272 18, 500 2x 14, 272 14, 272 18, 500 14, 272 2x 4228 2x 4228 2 2 x 2114 The Democratic candidate received 2114 + 272 = 2386 votes. 72. Let x be the number of shows he watches on Hulu each month. Since he watches twice as many shows each month on Netflix, the number of shows he watches on Netflix each month is 2x. Since the total number of shows he watches each month is 126, the sum of x and 2x is 126. x 2x 126 3x 126 3x 126 3 3 x 42 He watches 2(42) = 84 shows each month on Netflix. 73. 9x 20x = (9 20)x = 11x
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 74. 5(7x) = (5 7)x = 35x
Chapter 3: Solving Equations and Problem Solving
y 84.
3 y
75. 12x 5(2x 3) 4 12x 10x 15 4 22x 19 3
76. 7(x 6) 2(x 5) 7x 42 2x 10 7x 2x 42 10 9x 32 77.
c 5 13 7 c 5 6 c 5 5 6 5 c 1
78. 7 x 5 6x 20 x 5 20 x 5 5 20 5 x 25
81. 9x 12 8x 6 (4) x 12 10 x 12 12 10 12 x 22 82. 17x 14 20x 2x 5 (3) 17x 20x 2x 14 5 3 x 14 8 x 14 14 8 14 x 6 83.
5(2x 3) 11x 10x 15 11x 10x 10x 15 11x 10x 15 x
6
3(6) 3 y 18
85.
12 y 10 70 12 y 10 10 70 10 12 y 60 12 y 60 12 12 y 5
86.
4n 8 2n 14 4n 2n 8 2n 2n 14 2n 8 14 2n 8 8 14 8 2n 22 2n 22 2 2 n 11
87.
6(x 3) x 4 6x 18 x 4 6x 6x 18 x 6x 4 18 7x 4 18 4 7x 4 4 14 7x
79. 7x 3x 50 2 4x 52 4x 52 4 4 x 13 80. x 8x 38 4 7 x 42 7x 42 7 7 x 6
3 y
1 5
14 7x 7 7 2 x 88. 9(3x 4) 63 0 27x 36 63 0 27x 27 0 27x 27 27 0 27 27x 27 27x 27 27 27 x 1
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95
ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving 89. 5z 3z 7 8z 1 6 2z 7 8z 7 2z 8z 7 8z 8z 7 10z 7 7 10z 7 7 7 7 10z 0 10z 0 10 10 z0 90.
The amount of energy consumption for Vermont was 126 trillion BTU, and the amount of energy consumption for the District of Columbia was 126 + 18 = 144 trillion BTU.
4x 3 6x 5x 3 30 10x 3 5x 33 10x 5x 3 5x 5x 33 5x 3 33 5x 3 3 33 3 5x 30 5x 30 5 5 x 6
91. “Three times a number added to twelve is 27” is 12 3x 27 12 12 3x 27 12 3x 15 3x 15 3 3 x5 The number is 5. 92. “Twice the sum of a number and four is ten” is 2(x 4) 10 2x 8 10 2x 8 8 10 8 2x 2 2x 2 2 2 x 1 The number is 1. 93. Let x be the amount of energy consumption (in trillions of BTU) for Vermont. Since the District of Columbia used 18 trillion BTU more than Vermont, the amount of energy consumption for the District of Columbia was x + 18. Since the total number of BTU used by both entities was 270 trillion BTU, the sum of x and x + 18 is 270. x x 18 270 2x 18 270 2x 18 18 270 18 2x 252 2x 252 2 2 x 126 96
94. Let x be the number of roadway miles in South Dakota. Since North Dakota has 4489 more roadway miles than South Dakota, North Dakota has x + 4489 roadway miles. Since the total number of roadway miles is 169,197, the sum of x and x + 4489 is 169,197. x x 4489 169,197 2x 4489 169,197 2x 4489 4489 169,197 4489 2x 164, 708 2x 164, 708 2 2 x 82, 354 South Dakota has 82,354 roadway miles and North Dakota has 82,354 + 4489 = 86,843 roadway miles. Chapter 3 Getting Ready for the Test 1. x 7x = 1x 7x = (1 7)x = 6x Choice A 2.
2x 3 8x 3 2x 8x 3 3 (2 8)x (3 3) 6x 0 6x Choice B
3. 3(2x) = (3 2)x = 6x Choice A 4. 2(2x + 1) = 2 2x + 2 1 = 4x + 2 Choice C 5. (x 2) 5x 4 1(x 2) 5x 4 1 x (1)(2) 5x 4 x 2 5x 4 x 5x 2 4 1x 5x 2 4 (1 5)x (2 4) 4x 2 Choice C 6. 3(1 2x) 3 31 3 2x 3 3 6x 3 6x 3 3 6x 0 6x Choice B
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ISM: Prealgebra and Introductory Algebra
7.
5x 10 0 5x 10 10 0 10 5x 10 Choice B
8.
4(2x 1) 8 14x 10 8x 4 8 14x 10 8x 4 14x 10 Choice A
Chapter 3: Solving Equations and Problem Solving
6.
5 (2) 2 x 5 2 2 x 3 2 x 2 2 (3) 2 x 6
8.
5x 12 4x 14 22 x 2 22 x 2 2 22 2 x 24
9.
4x 7 15 4x 7 7 15 7 4x 8
10. The product of 3 and a number is 3 x or 3x; choice A. 11. 5 subtracted from a number is x 5; choice D.
13. Twice a number gives 14 is 2x = 14; choice B. x 14. A number divided by 6 yields 14 is
14;
6
choice C. 15. Two subtracted from a number equals 14 is x 2 = 14; choice E. 16. Six times a number, added to 2 is 14 is 2 + 6x = 14; choice F. Chapter 3 Test 1.
7x 5 12x 10 7x 12x 5 10 (7 12)x 5 10 5x 5
2. 2(3y + 7) = 2 3y + (2) 7 = 6y 14 3. (3z 2) 5z 18 1(3z 2) 5z 18 1 3z (1) 2 5z 18 3z 2 5z 18 3z 5z 2 18 8z 20 4. perimeter = 3(5x + 5) = 3 5x + 3 5 = 15x + 15 The perimeter is (15x + 15) inches.
x
7.
9. The sum of 3 and a number is 3 + x; choice C.
12. 5 minus a number is 5 x; choice E.
12 y 3y 12 2 y 12 2 y 2 2 6 y
4x 8 4 4 x 2 10.
2(x 6) 0 2x 12 0 2x 12 12 0 12 2x 12 2x 12 2 2 x6
11. 4(x 11) 34 10 12 4x 44 34 10 12 4x 10 2 4x 10 10 2 10 4x 12 4x 12 4 4 x3
5. Area (length) (width) 4 (3x 1) 4 3x 4 1 12x 4 The area is (12x 4) square meters.
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97
ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
12.
5x 2 x 10 5x x 2 x x 10 4x 2 10 4x 2 2 10 2 4x 8
20. Let x be the number of free throws Jo made. Since Thanh made twice as many free throws as Jo, Thanh made 2x free throws. Since the total number of free throws was 12, the sum of x and 2x is 12. x 2x 12 3x 12 3x 12 3 3 x4 Thanh made 2(4) = 8 free throws.
4x 8 4 4 x 2 13.
4(5x 3) 2(7x 6) 20x 12 14x 12 20x 12 14x 14x 12 14x 6x 12 12 6x 12 12 12 12 6x 0 6x 0 6 6 x0
21. Let x be the number children participating in the event. Since the number of adults participating in the event is 112 more than the number of children, the number of adults is x + 112. Since the total number of participants in the event is 600, the sum of x and x + 112 is 600. x x 112 600 2x 112 600 2x 112 112 600 112 2x 488 2x 488 2 2 x 244 244 children participated in the event.
14. 6 2(3n 1) 28 6 6n 2 28 6n 4 28 6n 4 4 28 4 6n 24 6n 24 6 6 n4
Cumulative Review Chapters 13
15. The sum of 23 and a number translates to 23 + x.
1. 308,063,557 in words is three hundred eight million, sixty-three thousand, five hundred fiftyseven.
16. Three times a number, subtracted from 2 translates to 2 3x.
2. 276,004 in words is two hundred seventy-six thousand, four.
17. The sum of twice 5 and 15 is 5 translates to 2 5 + (15) = 5.
3. 2 + 3 + 1 + 3 + 4 = 13 The perimeter is 13 inches.
18. Six added to three times a number equals 30 translates to 3x + 6 = 30.
4. 6 + 3 + 6 + 3 = 18 The perimeter is 18 inches.
19. The difference of three times a number and five times the same number is 4 translates to 3x 5x 4 2x 4 2x 4 2 2 x 2 The number is 2.
5.
900 174 726
Check:
726 174 900
6.
17,801 8216 9585
Check:
9585 8216 17,801
7. To round 248,982 to the nearest hundred, observe that the digit in the tens place is 8. Since this digit is at least 5, we add 1 to the digit in the hundreds place. The number 248,982 rounded to the nearest hundred is 249,000. 98
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
8. To round 844,497 to the nearest thousand, observe that the digit in the hundreds place is 4. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. The number 844,497 rounded to the nearest thousand is 844,000.
16. 2a2 5 c 2 22 5 3 24 53 8 53 13 3 10
9.
17. 2n 30 = 10 Let n = 26. 2 26 30 010 52 30 0 10 22 10 False 26 is not a solution. Let n = 40. 2 40 30 010 80 30 0 10 50 10 False 40 is not a solution. Let n = 20. 2 20 30 010 40 30 0 10 10 10 True 20 is a solution.
25 8 200
10.
395 74 1 580 27 650 29, 230
208 11. 9 1872 18 07 0 72 72 0 Check: 208 9 1872
18. a.
14 < 0 because 14 is to the left of 0 on the number line.
b. (7) = 7, so (7) > 8 because 7 is to the right of 8 on the number line. 19. 5 + (2) = 3
86 12. 46 3956 368 276 276 0 Check: 86 46 516 3440 3956
20. 3 + (4) = 7 21. 15 + (10) = 25 22. 3 + (7) = 4 23. 2 + 25 = 23 24. 21 + 15 + (19) = 36 + (19) = 17 25. 4 10 = 4 + (10) = 14
13. 2 4 3 3 = 8 3 3 = 8 1 = 7
26. 2 3 = 2 + (3) = 5
14. 8 4 + 9 3 = 32 + 9 3 = 32 + 3 = 35
27. 6 (5) = 6 + 5 = 11
15. x2 z 3 52 4 3 25 4 3 29 3 26
28. 19 (10) = 19 + 10 = 29 29. 11 (7) = 11 + 7 = 4 30. 16 (13) = 16 + 13 = 3
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ISM: Prealgebra and Introductory Algebra
Chapter 3: Solving Equations and Problem Solving
31.
12 2 6
32.
30 6 5
46. 3y 15 3y 15 3 3 y 5 47.
2x 6 18 2x 6 6 18 6 2x 24 2x 24 2 2 x 12
48.
3a 5 1 3a 5 5 1 5 3a 6 3a 6 3 3 a 2
33. 20 (4) = 5 34. 26 (2) = 13 35.
36.
48 3
16
120 10 12
2 37. (3) (3)(3) 9 5 38. 2 (2 2 2 2 2) 32 2 39. 3 (3 3) 9 2 40. (5) (5)(5) 25
41. 2 y 6 4 y 8 2 y 4 y 6 8 (2 4) y 6 8 6y 2 42. 6x 2 3x 7 6x 3x 2 7 (6 3)x 2 7 3x 9 43.
3y 1 3 3(1) 1 0 3 3 1 0 3 2 3 False Since 2 = 3 is false, 1 is not a solution of the equation.
44.
5x 3 7 5(2) 3 0 7 10 3 0 7 7 7 True Since 7 = 7 is true, 2 is a solution of the equation.
45.
12x 36 12x 36 12 12 x3
100
49. Let x be the price of the software. Since the price of the computer system is four times the price of the software, the price of the computer system is 4x. Since the combined price is $2100, the sum of x and 4x is 2100. x 4x 2100 5x 2100 5x 2100 5 5 x 420 The price of the software is $420 and the price of the computer system is 4($420) = $1680. 50. Let x be the number. “Two times the number plus four is the same amount as three times the number minus seven” translates to 2x 4 3x 7 2x 2x 4 3x 2x 7 4 x7 47 x77 11 x The number is 11.
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Chapter 4 b.
Section 4.1 Practice Exercises 1. In the fraction
11
, the numerator is 11 and the
2 denominator is 2. 2. In the fraction
c. 13. a.
10 y
, the numerator is 10y and 17 the denominator is 17.
3. 3 out of 8 equal parts are shaded:
b. c.
3 14.
8
9
1
9 4. 1 out of 6 equal parts is shaded:
1 6
5. 7 out of 10 equal parts are shaded:
15. 7 10
6. 9 out of 16 equal parts are shaded:
16.
9 16
17.
9. number of planets farther 5 number of planets in our solar system 8
4
0
4
13 is undefined 0
19.
13 13 1 2
20. a.
5
of the planets in our solar system are farther 8 from the Sun than Earth is.
75 2
7
7
6
c.
10
9 10
d.
21. a.
4
35 2 7
37 7
2 3 6 2 18 2 20 3 3 3 3
b.
1
of a whole and there are 5 parts 4 shaded, or 1 whole and 1 more part. 5 1 ;1 4 4
5
1
of a whole and there are 8 parts 3 shaded, or 2 wholes and 2 more parts. 8 2 ;2 3 3
12. a.
1
18. 8. answers may vary; for example,
11. Each part is
0
1
7. answers may vary; for example,
10. Each part is
6 1 6
10 10 9 100 9 109 10 10 10
1 5 4 1 20 1 21 5 5 5 5
1 59 5 4 9 4 1 5 5
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
b.
2 9 23 18 5
3. The fraction
3 3 the fraction
23 5 2 9 9
c.
d.
10
6. Each shape is divided into 3 equal parts. 7. how many equal parts to divide each whole number into 8. The fraction is equal to 1. 9. The operation of addition is understood in mixed 1 1 number notation; for example, 1 means 1 . _ 3 3 10.
7 7 51 49 2
division Exercise Set 4.1 2. In the fraction
1
, the numerator is 1 and the 4 denominator is 4. Since 1 < 4, the fraction is proper.
1 20 21 20 1
53 4. In the fraction 21 , the numerator is 53 and the denominator is 21. Since 53 > 21, the fraction is improper.
21 1 1 20 20
6. In the fraction
1. The number
17
is called a fraction. The number 31 31 is called its denominator and 17 is called its numerator.
2. If we simplify each fraction, 4 0
9
26
, the numerator is 26 and the 26 denominator is 26. Since 26 26, the fraction is improper.
Vocabulary, Readiness & Video Check 4.1
102
is called a mixed number.
5. When the numerator is greater than or equal to the denominator, you have an improper fraction.
4 13 62 52 10
and we say
is called a proper fraction, and
4. The value of an improper fraction is always 1 and the value of a proper fraction is always <1.
12 4 48 4 8 8 0 48 12 4
51 2 7 7 7
f.
8
is called an improper fraction,
3 8
62 10 4 13 13
e.
8
0
1, 0, 9 4
8. 5 out of 6 equal parts are shaded:
5 6
10. Each part is
1
of a whole and there are 10 parts 4 shaded, or 2 wholes and 2 more parts.
is undefined. a.
10 4
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ISM: Prealgebra and Introductory Algebra
b.
28. answers may vary; for example,
2 4
2
12. Each part is
Chapter 4: Fractions and Mixed Numbers
1
of a whole and there are 11 parts 3 shaded, or 3 wholes and 2 more parts. a.
30. answers may vary; for example,
11 3
b.
32. answers may vary; for example,
2 3
3
14. 5 out of 8 equal parts are shaded:
5 8
12 7
34. medical degrees 22 employees 63 22 of the employees have medical degrees. 63
8
36. a.
5
16. 5 out of 12 equal parts are shaded:
18. 7 out of 8 equal parts are shaded:
20. Each part is
1
of a whole and there are 6 parts 5 shaded, or 1 whole and 1 more part. a.
b.
6 5
b.
1
1 5
22. Each part is
no medical degree 41 employees 63 41 of the employees do not have medical 63 degrees.
38. planets with longer days
1
of a whole and there are 23 parts 5 shaded, or 4 wholes and 3 more parts. a.
number of employees who do not have medical degrees = 63 22 = 41
23 5
number of planets in solar system 8 4 of the planets in our solar system have longer 8 days than Earth has. 40. 5 of 12 inches is
b.
4
5
of a foot.
12
3 4 5
42. 37 of 60 minutes is 24. 3 out of 8 equal parts are shaded:
of an hour.
60
3 8
26. 13 out of 16 equal parts are shaded:
37
13 16
44. number with white lettering 9 number of teams in league 20 9 of the teams in the league have white 20 lettering on their shirts.
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
46. There are 50 states total. Consumer fireworks are legal in 46 states. a.
Consumer fireworks are legal in
46
of the states.
50 b. 50 46 = 4 Consumer fireworks are illegal in 4 states. c.
Consumer fireworks are illegal in
4 50
48. a.
of the states.
number of watercolors 15 total number of pieces 37 15 of the inventory are watercolor paintings. 37
b.
number of oil paintings 17 total number of pieces 37 17 of the inventory are oil paintings. 37
c.
37 15 17 = 5 There are 5 sculptures.
d.
numbers of sculptures 5 total number of pieces 37 5 of the inventory are sculptures. 37
50.
52.
54.
56.
58.
3 1 3
60.
20 20 1
62.
104
0 8
0
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ISM: Prealgebra and Introductory Algebra
64.
14 1 14
66.
7 is undefined 0
68.
5
Chapter 4: Fractions and Mixed Numbers
14 88. 14 196 14 56 56 0
1
196 14 14
5 13 17 1 13 17 13 30 70. 1 17 17 17 17 72. 2
1 90. 143 149 143 6
5 9 2 5 18 5 23 9 9 9 9
149 6 1 143 143
3 8 7 3 56 3 59 74. 7 8 8 8 8 76. 12
2 5 12 2 60 2 62 5 5 5 5 14
78. 10
7 92. 123 901 861 40
27 10 14
27
27
270 14 27
27
2 7 114 2 798 2 800 80. 114 7 7 7 7 1 82. 7 13 7 6 13 6 1 7 7 7 84. 9 64 63 1 64 1 7 9 9 5 86. 12 65 60 5
901 40 7 123 123
284
94. 43 4 4 4 64 96. 34 3 3 3 3 81 98. 21 21 21 4 4 4 100.
45 57
45 45 57 57
102. answers may vary 104. 106. 11 is close to 12, so
11
is close to 1. 5
11
12 12 rounded to the nearest whole number is 6. 108. 5 + 4 + 9 + 8 = 26 9 of the 26 national parks in these states are in 9 California: 26
65 5 5 12 12
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105
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
Section 4.2 Practice Exercises
1
64 4 16 4 16 16 8. 20 45 4 5 5
30 2 15 23 5
1. a.
1
1
1
is in simplest form. 9 21 3 7 3 7 7 27 3 9 3 9 9 1
7 Since
21 and
9 equivalent.
2 2 2 3 3 23 3 2
27
7 both simplify to
, they are 9
11. Not equivalent, since the cross products are not equal: 13 5 = 65 and 4 18 = 72
2. 60 2 30 2 2 15
12.
2 2 3 5 22 3 5 11 3. 3 33 3 99 3 297 The prime factorization of 297 is 33 11.
7 mountains over 8000 meters in Nepal 14 mountains over 8000 meters 7 14 7 1 72 1
30 15 2 15 2 2 2 1 45 15 3 15 3 3 3
5.
39x 313 x 3 13x 13x 13x 1 51 317 3 17 17 17 9 3 3 50 255 9 Since 9 and 50 have no common factors, is 50 already in simplest form.
6.
7 7 7 7 7 7 1 112 7 16 7 16 16 16
7 1 7 2
1
4.
106
1
1
72 2 36
1 2
1 of the mountains over 8000 meters tall are 2 located in Nepal. Calculator Explorations 1.
128 4 224 7
2.
231 7 396 12
3.
340 20 459 27
49
1
7 10.
2 2 18 2 2 29
7.
1
7aaa 7 a a a a 9. 2 7 8 a a 8 7 8 a a 56a
2 2 2 7 23 7 c.
1
7a3
56 2 28 2 2 14
b.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
4.
Exercise Set 4.2
999 37 1350 50
5.
432 8 810 15
6.
225 5 315 7
7.
54 2 243 9
2. 12 2 6 2 2 3 22 3 4. 75 3 25 3 5 5 3 52 6. 64
Vocabulary, Readiness & Video Check 4.2 1. The number 40 equals 2 2 2 5. Since each factor is prime, we call 2 2 2 5 the prime factorization of 40. 2. A natural number, other than 1, that is not prime is called a composite number. 3. A natural number that has exactly two different factors, 1 and itself, is called a prime number. Since 11 and 48 have no common factors other 11 is in simplest form. than 1, the fraction 48
5. Fractions that represent the same portion of a whole are called equivalent fractions. 5 6. In the statement
15
12 36 called cross products.
, 5 36 and 12 15 are
7. Check that every factor is a prime number and check that the product of the factors is the original number.
24
8
is not in simplest form;
8. 128 2 64 2 2 32 2 2 2 16 2 2 2 2 8 2 22 224 2 2 2 2 2 2 2 27 10. 130 2 65 2 5 13 2 5 13 12. 93 = 3 31 14. 836 2 418 2 2 209 2 2 1119 22 1119
5 12
9. You can simplify the two fractions and then 3 6 1 compare them. and both simplify to , 9 18 3 so the original fractions are equivalent. 10
16. 504 2 252 2 2 126 2 2 2 63 2 2 2 3 21
8. an equivalent form of 1 or a factor of 1
10.
8
2 4 2 4 2 2 2 2 2 2 26
455 35 8. 689 53
4.
Chapter 4: Fractions and Mixed Numbers
2 2 2 3 3 7 23 32 7
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107
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
18.
5 30
20.
9y
5 1 1 5 6 6
3 3 y 3 16
48 22.
22
34
2 11 2 17
48. 3 y
16
144 y 18 8 y 8 162 y 18 9 y 9 270x4 y3 z3
3y 50.
16
15x3 y3 z3
11 17
18 15 x x x x y y y z z z
115 x x x y y y z z z 18 x 1 18x
24.
70 7 10 7 80 8 10 8
52. Equivalent, since the cross products are equal: 3 10 = 30 and 6 5 = 30
26.
25z 5 5 z 5 55z 5 11 z 11
54. Not equivalent, since the cross products are not equal: 2 11 = 22 and 5 4 = 20
21 73 3 49 77 7
56. Equivalent, since the cross products are equal: 4 15 = 60 and 10 6 = 60
28.
58. Equivalent, since the cross products are equal: 2 28 = 56 and 8 7 = 56
59b 9 9 30. 2 16b 5 16 b b 16 b 80b 45b
32.
60. Not equivalent, since the cross products are not equal: 16 16 = 256 and 20 9 = 180
32 2 2 2 2 2 63 3 3 7 Since 32 and 63 have no common factors,
32
is
62. Not equivalent, since the cross products are not equal: 6 35 = 210 and 14 21 = 294
63 already in simplest form. 34.
64.
36 y 66 y 6 6 42 yz 6 7 y z 7 z 7z
36. 28 4 7 7 60 4 15 15 38.
66.
5 a 5a 3 3b2 b b b b b
28abc
44.
60ac
74abc 15 4 a c
1 20
1
5 20 5 1 20 centimeters is 5 of a meter.
68. a.
7b 15
7b 15
65
5 13 5 234 18 13 18
46.
108
20 centimeters 100 centimeters
60a2b 5 12 a a b 36ab3 312 a
60 30 2 2 40. 150 30 5 5 42.
200 caps 200 1 1 2000 caps 200 10 10 1 200 caps represents of the total caps. 10
15 3 5 5 423 3141 141 5 of the National Park Service areas can 141 be found in New Mexico.
b. 423 15 = 408 408 National Park Service areas are found outside New Mexico.
78 6 13 13 13 90x 6 15 x 15 x 15x
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ISM: Prealgebra and Introductory Algebra
c.
70.
10 5 2 2 35 5 7 7 2 of the students made an A on the first test. 7
72. a.
b.
74.
76.
408 3136 136 423 3141 141 136 of the National Park Service areas are 141 found outside New Mexico.
34,200 15,200 = 19,000 $19,000 was not covered by the trade-in. $19, 000 3800 5 5 $34, 200 3800 9 9 5 of the purchase price was not covered by 9 the trade-in.
4300 employees 100 43 43 20, 000 employees 100 200 200 43 of the employees work at the Hallmark 200 headquarters in Kansas City, Missouri.
88. 131, 625 3 43,875 3 314, 625 3 3 3 4875 3 3 3 31625 3 3 3 3 5 325 3 3 3 3 5 5 65 3 3 3 3 5 5 5 13 131, 625 34 53 13 90. answers may vary 92. yes; answers may vary 94. The piece representing Nursing is labeled
8
. 100
8 42 2 100 4 25 25 2 of entering college freshmen plan to major 25 in Nursing. 96. answers may vary
y3 (5)3 125 25 5 5 5
98. The piece representing Memorials is labeled 76 . 1000 76 4 19 19 1000 4 250 250 19 of National Park Service areas are 250 Memorials.
78. 5a = 5(4) = 20 80. answers may vary 82.
Chapter 4: Fractions and Mixed Numbers
9506 7 1358 7 12, 222 9 1358 9
84. 37 + 7 = 44 blood donors have an O blood type. 44 donors 4 11 11 100 donors 4 25 25 11 of blood donors have an O blood type. 25 86. 9 + 1 = 10 blood donors have a B blood type. 10 donors 110 1 100 donors 10 10 10 1 of blood donors have a B blood type. 10
100. answers may vary 102. 1235, 2235, 85, 105, 900, and 1470 are divisible by 5 because each number ends with a 0 or 5. 8691, 786, 2235, 105, 222, 900, and 1470 are divisible by 3 because the sum of each number’s digits is divisible by 3. 2235, 105, 900, and 1470 are divisible by both 3 and 5. 104. 15; answers may vary
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
Section 4.3 Practice Exercises
13.
3 5 3 5 15 1. 7 11 7 11 77 2.
1 1 11 1 3 9 3 9 27
6 7 6 7 2 3 7 3 3 77 8 77 8 7 11 2 4 11 4 44 4 3 4. 4 3 1 4 3 1 1 27 8 27 8 3 9 4 2 9 2 18 3.
5.
1 11 111 11 2 28 2 28 56
10 2 10 9 4 9 4 2 10 9 42 259 42 5 9 4 45 4
5 y3 5y 4 4 1 3y 1 4 5 y3
3y 14.
3y 1 4 5 y3 3 y 1 45 y y y 31 45 y y 3 20 y2
4 33 4 33 4 311 3 6. 11 16 1116 11 4 4 4 7.
1 3 25 1 3 25 6 10 16 6 10 16 1 3 5 5 2 3 2 5 16 1 5 2 2 16 5 64
8.
9.
2 3y 23 y y y 3 2 3 2 1 1 a3
b
a3 b b2 a2
b2 a2
aaab bb a a
a
b
3
10. a.
3 3 3 3 3 3 3 27 4 4 4 4 4 4 4 64 2
b.
4 4 4 4 4 16 5 5 25 5 5 5
11.
8 2 8 9 8 9 2 4 9 4 9 36 7 9 7 2 72 72 7 7
12.
4 1 4 2 42 8 9 2 9 1 9 1 9
110
3y
3
15. 2 9 7 2 9 7 3 14 15 314 15 2 3 3 7 3 7 2 15 3 7 7 15 3 15 7 7 315 77 45 49 3 9 3 9 27 16. a. xy 4 2 42 8
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ISM: Prealgebra and Introductory Algebra
b.
Chapter 4: Fractions and Mixed Numbers 6. The word “of” indicates multiplication.
3 9 x y 4 2 3 2 4 9 3 2 49 1 3 2 2 2 3 3 1 23 1 6
7. We have a negative fraction times a positive fraction, and a negative number times a positive number is a negative number. 8. Yes; the negative sign is inside the parentheses and the exponent applies to everything in the parentheses. 9. When writing the reciprocal of a fraction, the denominator becomes the numerator, and the numerator becomes the denominator. 10. There are only prime numbers in the denominator, and neither 4 nor 9 has a factor that includes these prime numbers. 11. radius 1 diameter 2
9 2x 17. 49 9 2 0 4 8 2 90 9 1 8 4 29 9 18 0 4 29 9 0 1 2 49 94 True 4 4 9 Yes, is a solution of the equation. 8 18.
Exercise Set 4.3 2. 4.
5 7 5 7 35 9 4 9 4 36 4
1
15
20
3
1 1 90 1 90 1 5 18 90 18 5 5 1 5 1 5 1 There are 18 roller coasters in Cedar Point 6.
Vocabulary, Readiness & Video Check 4.3
1. To multiply two fractions, we write a c a c . b d bd 2. Two numbers are reciprocals of each other if their product is 1. 3. The expression
23 7
222
11 8.
4 1
15 20 4 1 15 4 5 1 15 5 1 75
11
12
311 3111 1 1112 11 3 4 4
5 64 y 5 64 y 5 32 2 y y 32 100 32 100 32 2 5 10 10
2 2 6 y3 3 10. 6 y 3 3 1 2 6 y3 31 2 3 2 y3
while
7
3
2 2 2. 2 7 7 7 7
4. Every number has a reciprocal except 0. a 5. To divide two fractions, we write
b
c d
ad bc
.
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31 2 2 y3 1 4 y3
111
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
12.
14.
a b a b b 3 a 3 b3 a 3 1 a b bbb a a a 1 a a bb 1 2 a b2
28.
11 0 0 12
27 10 16 27 10 16 32 13 30 32 13 30 3 3 310 16 16 2 1310 3 3 3 2 13 9 26
8 4 3 15 3 15 8 43 15 8 4 31 5 3 4 2 1 5 2 1 10
9z 32.
2
4
1 1 1 1 1 22. 2 2 2 2 2 1111 2222 1 16
20
2
9
9z 9 9z 9 81z 20 2 20 2 40
5 5 10 34. 10 6 65 11 6 10 5 1 6 10 5 1 65 2 1 6 2 1 12
3
3 1 3 3 3 1 24. 4 3 4 4 4 3 3 3 31 4 4 43 3 31 444 9 64
112
4
30. 10 4 10 5x 4 11 5x 11 10 5x 11 4 2 55 x 11 2 2 5 5 x 11 2 25x 22
8 8 8 8 64 8 20. 9 9 9 9 9 81
3
16. 13x 5 13x 5 20 6x 20 6x 13 x 5 456 x 13 46 13 24 18.
5 4 5 4 26. 5 4 5 5 8 4 8 3 83 4 2 3 23 6 5
36.
y3 3 y3 3 y y y y y2 9 y y 3 3 3 y2 y3 y2 9
38.
8 5 8 5 40 11 9 11 9 99
40.
2 5y 2 11 2 11 22 5 y 11 5 y 5 y 5 y 5 y 25 y 2
3
9
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3
ISM: Prealgebra and Introductory Algebra
42.
Chapter 4: Fractions and Mixed Numbers 1 5 1 1 5 12 54. 2 6 12 2 6 1 1 5 12 2 6 1 1 5 6 2 2 6 1 1 5 1 5
12 21y 47y 12 21y 47y 12 y 7 21 4 y 4 3 y 7 1 7 3 4 y 1 1
44.
24 5 24 5 8 3 5 1 1 45 8 45 8 5 3 38 3
2 31 56. 16a 2 31 16a 24a 1 24a 2 16a 31 1 24a
5
1 1 1 1 1 1 46. 2 2 2 2 2 2 11111 21 2 2 2 2
32 b
48.
a
ba
3
baaa
3
a
1 3 62a 3
a
a2 b 3 a2 b 3 a a b b b b b b2
50. 100
1
1 2 1 2 100 2 1 1 100 2 11 200
100
58. 1 6 1 6 6 5 7 5 7 35 60.
62.
9 16 9 15 9 15 135 2 15 2 16 2 16 32 17 y2 24x
2 52. 7x
2 8 a a 31 3 a 2 18 a 31
13y
18x
2
17 y2 18x 24x 13y
17 y2 18x 24x 13y 17 y y 6 3 x 6 4 x 13 y 17 y 3 4 13 51y 52
14x 14x 7x 3 3 1 7x2 3 1 14x 7x2 3 114x 7 x x 3 1 2 7 x x 3 1 2 3x 2
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 64. 33 2 5 33 2 5 11 9 1 11 9 33 11 5 1 2 9 3311 5 1 2 9
31111 5 2 3 3 1111 5 23 605 6
66. 15c3
3c2 15c3 3c2 5 1 5 15c3 5 1 3c2
b.
3
15c 5
76. a. b.
xy
xy
78.
8 1 8 1 2 4 1 2 1 2 9 4 94 94 9 9
x y 7
8 1 8 4 8 4 32 9 4 9 1 9 1 9
6
1
2
7 1 62
7 12
1 6 2 7 2 6 1 2 7 6 1 72 2 31 7 3
x y
ac b23 a c b b b b b b2 ac b3 baac a a 2 b a c ba c
3 8 2 38 2 70. 4 9 5 4 9 5 3 4 2 2 4 3 3 5 2 2 3 5 2 5 3 2 2 5 3 2 5 3
114
74. a.
1 3c2 5 3 c c c 5 1 3 c c 55c 1 25c 68.
3311 5 29
5 4 5 5 4 5 72. 8 7 16 8 7 16 5 4 5 8 744 5 5 8 28 5 28 8 5 5 28 8 5 57 4 4 2 5 7 2
2
y
7
6
3 11 2 9 6 0 3 11 11 29 6 0 311 11 2 6 0 11 23 6 0 11 11 6 6 True 11 11 9 Yes, is a solution of the equation. 11
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ISM: Prealgebra and Introductory Algebra
5x
80.
1
Chapter 4: Fractions and Mixed Numbers
88.
3 1
3 5 0 5 3 5 3 1 0 1 5 3 53 0 1 15 3 3 1 False 1 3 3 No, is not a solution of the equation. 5
82.
84.
86.
1
1 of 200 5 5 1 1 40 1 40
1
of 44 million
1
44, 000, 000 11 1 44, 000, 000 11 1 111 4, 000, 000 111 4, 000, 000 The increase in the number of people who 11
subscribed to Hulu was approximately 4 million. 90.
3 3 of 8 16 16 3 3 3 2 3 2 3 The screw sinks inches deep after eight turns. 2
92. d 2 r 7 2 20 2 7 1 20 27 1 20 27 1 2 10 7 10 7 The diameter is foot. 10
5
5 of 24 8 8 5 5 5 15 1 15
1
1 of 3000 5 5 1 1 600 The diet can contain 600 calories from fat per day.
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
94.
3
of 355,000
3
102.
50
50 3 3 3 21, 300 The family pays $21,300 to the real estate companies. 96.
3
of 662
1 120 The family drove 120 miles for medical.
100
104.
811 42 69 922
106.
882 773 109
3
662 331 3 662 331 1 3 662 3311 3 2 331 3311 3 2 1 6 In the first 44 seasons of Survivor, 6 contestants have appeared four times.
1
331
1 1 3 98. area length width 2 16 3 The area is square mile. 16 3 100.
25
of 12,000
3
25
3 3 3 1440 The family drove 1440 miles for shopping.
116
108. answers may vary 2 2 1 8 39 8 1 8 39 8 110. 2 13 16 9 2 13 16 9 2 1 8 13 3 2 4 138 2 3 3 2 2 1 4 3 2 4 4 1 3 3 2 44 1 3 3 2 16 1 9 2 16 2 9 1 16 2 9 1 32 9
13
112.
of 3720
13
3720 40 13 3720 40 1 13 3720 40 1 13 40 93 40 1 13 93 1 1209 There were approximately 1209 two-year colleges in the U.S. in 2020. 40
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ISM: Prealgebra and Introductory Algebra
114.
3
of 36 on the first bus and
1
of 30 on the 4 3 second bus are wearing orange polo shirts. 3 1 4 3 3 1 3 1 3 10 1 27 10 37 There are 37 students on the two buses wearing orange polo shirts.
Section 4.4 Practice Exercises 1.
6 2 62 8 13 13 13 13 5
2.
8x 3.
1 8x
11
6
5 1 6 2 3 3 8x 8x 2 4x 4x
11 6
5 12 12
7 8
and
11
is
16
16.
7.
2 7y 2 7y 5 5 5
8.
4 6 3 4 6 3 5 5 or 11 11 11 11 11 11 10
5
12 12
13. 30 1 = 30 30 2 = 60 30 3 = 90 30 4 = 120 30 5 = 150
Not a multiple of 25. Not a multiple of 25. Not a multiple of 25. Not a multiple of 25. A multiple of 25. 23 1 and is 150. 25 30
14. 40 =
8 4 8 4 4 4 or 17 17 17 17 17
9. x y
13 11 13 11 2 1 2 1 4 4 4 4 2 2 2 1 The jogger ran mile farther on Monday. 2
The LCD of
7 2 7 2 5 1 5 1 15 15 15 15 3 5 3
6.
11.
12 12 5.
10. perimeter 3 3 3 3 20 20 20 20 3 333 20 12 20 3 4 5 4 3 5 3 The perimeter is mile. 5
12. 16 is a multiple of 8, so the LCD of
20 6 7 20 6 7 33 or 3 11 11 11 11 11
4.
Chapter 4: Fractions and Mixed Numbers
10 5 5 5 or 12 12 12
2225 108 = 2 2 3 3 3 LCD = 2 2 2 3 3 3 5 = 1080 3 11 The LCD of and is 1080. 40 108
15. 20 = 2 2 5 24 = 2 2 2 3 45 = 3 3 5 LCD = 2 2 2 3 3 5 = 360 7 1 13 The LCD of , , and is 360. 20 24 45 16. y = y 11 = 11 The LCD of
7 y
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and
6 11
is 11y.
117
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
17.
7 7 7 7 7 49 8 8 7 8 7 56 1
18.
19.
4
Exercise Set 4.4
1 5 1 5 5 4 5 4 5 20
2.
9
4.
9 9 9 9 81 4x 9 4x 9 36x
Vocabulary, Readiness & Video Check 4.4 9 13 1. The fractions and are called like 11 11 3 1 fractions while and are called unlike 3 4 fractions. 2. a b
c b
a c and a b
b
3. As long as b is not 0,
2
3
2
c
9 2 11 17 17
3 2 5 1 5 1 10 10 2 5 2
3 1 3 1 2 1 2 1 6. 8 8 8 8 42 4
4x
10 10
4 6 4 6 24 1 6 1 6 6
9
17 17
3x 3x 6 3x 6 18x 7 7 6 76 42
20. 4 21.
11. Multiplying by 1 does not change the value of a fraction.
ac
5 7 5 (7) 12 112 1 24 2 12 2 8. 24 24 24 10.
3 2 3 2 5 1 5 1 10 y 10 y 10 y 10 y 2 5 y 2 y
12.
7z 15
3z
z
15 15
.
7z 3z 1z
3z 15
b
b
a
a
a . b b b
15
3 z z3 5 5
4. The distance around a figure is called its perimeter.
14.
9 5 95 4 13 13 13 13
5. The smallest positive number divisible by all the denominators of a list of fractions is called the least common denominator (LCD).
16.
5 1 5 1 4 2 2 2 6 6 6 6 3 2 3
18.
4 7 4 7 3 3 z z z z z
6. Fractions that represent the same portion of a whole are called equivalent fractions. 7. To add like fractions, we add the numerators and keep the same denominator. 8. The y-value is negative and follows the minus sign, so parentheses are used around the y-value to separate the two signs. 5 7 5 7 9. P ; 2 meters 12 12 12 12
20. 37 18 37 18 45 45 45 45 37 18 45 19 45 19 45
10. 45 is the smallest number that both denominators, 15 and 9, divide into evenly.
118
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ISM: Prealgebra and Introductory Algebra
22.
24.
27 5 28 27 5 28 28 28 28 28 6 28 3 2 14 2 3 14
42.
44.
18b 3 18b 3 5 5 5
26.
15 85 15 85 70 7 10 7 200 200 200 200 20 10 20
28. 15 15 15 15 26 y 26 y 26 y 30 26 y 2 15 2 13 y 15 13y 30.
2x 15
7 15
2x 7 15
32.
3 15 3 15 12 3 4 3 16z 16z 16z 16z 44 z 4z
34.
1 15 2 115 2 12 43 3 8 8 8 8 8 42 2
36.
9y 2y 5y 4y 9y 2y 5y 4y 8 8 8 8 8 12 y 8 y 4 3 34 2 y 2
38. x y
Chapter 4: Fractions and Mixed Numbers
7 9 7 9 2 1 2 1 8 8 8 8 42 4
1 5 1 5 4 2 2 2 40. x y 6 6 6 6 23 3
3 6 4 3 2 3 6 4 3 2 18 13 13 13 13 13 13 13 18 The perimeter is feet. 13 1 1 1 1 1111 4 2 2 2 6 6 6 6 6 6 23 3 2 centimeter. The perimeter is 3
46. To find the remaining distance, subtract the 10 16 miles already run from the total miles. 8 8 16 10 16 10 6 2 3 3 8 8 8 8 2 4 4 3 He must run mile more. 4 48. To find the fraction that had speed limits greater 21 than 70 mph, subtract the that have 70 mph 50 37 speed limits from the that have speed limits 50 70 mph or greater. 37 21 37 21 16 2 8 8 50 50 50 50 2 25 25 8 of the states had speed limits that were 25 greater than 70 mph. 50. Asia makes up
30 100
of the world’s land area,
while Africa makes up
20 100
of the world’s land
area. 30 20 30 20 50 1 50 1 100 100 100 100 2 50 2 1 of the world’s land area is within Asia and 2 Africa. 52. Asia makes up
30
of the world’s land area,
100 while Australia makes up
6 100
of the world’s
land area.
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 30 6 30 6 24 6 4 6 100 100 100 100 25 4 25 6 Asia’s land area is greater than that of 25 Australia’s.
78.
80.
54. Multiples of 20: 20 1 = 20, not a multiple of 12 20 2 = 40, not a multiple of 12 20 3 = 60, a multiple of 12 LCD: 60 56. Multiples of 90: 90 1 = 90, a multiple of 15 LCD: 90 58. 4 = 2 2 14 = 2 7 20 = 2 2 5 LCD = 2 2 5 7 = 140 60. y = y 70 = 2 5 7 LCD = 2 5 7 y = 70y 62. 24 = 2 2 2 3 45 = 3 3 5 LCD = 2 2 2 3 3 5 = 360 64. 11 = 11 33 = 3 11 121 = 11 11 LCD = 3 11 11 = 363 66.
68.
70.
2 20 40 5 5 20 100 Find those goods whose fraction is over
82. 43 4 4 4 64 84. 34 3 3 3 3 81 86. 54 5 5 5 5 625 88. 42 5 4 4 5 80 90.
3 1 3 1 2 1 2 1 4 4 4 4 22 2
92. answers may vary 94.
96.
3 3 4 3 3 4 2 1 2 1 8 8 8 8 8 4 2 4 1 He is mile from home. 4 108 108 19 108 19 2052 215 y 215 y 19 215 y 19 4085 y
3
100. False;
3 20 3 20 60 5 5 20 5 20 100 1 1 10 110 10 5 5 10 5 10 50
21
19z 6 19z 6 114z 21 6 21 6 126
74.
76.
7 7 6a 7 6a 42a 6 6 6a 6 6a 36a
40
. 100 Books and magazines, clothing, and shoes have 2 over of the goods bought online. 5
98. answers may vary
3x 3x 6 3x 6 18x 2 2 6 26 12
120
is the largest fraction, so clothing has the 100 largest fraction bought online.
5 5 4 5 4 20 6 6 4 6 4 24
19z 72.
60
7 12
7 4 7 4 28 12 4 12 4 48
Section 4.5 Practice Exercises 1. The LCD is 21. 2 8 23 8 6 8 14 2 7 2 7 21 7 3 21 21 21 21 3 7 3 2. The LCD is 18. 5 y 2 y 5 y 3 2 y 2 15 y 4 y 19 y 6 9 63 92 18 18 18
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ISM: Prealgebra and Introductory Algebra
3. The LCD is 20. 1 4 9 1 9 5 20 5 4 20 4 9 20 20 5 20 1 5 45 1 4
Chapter 4: Fractions and Mixed Numbers
10. The LCD is 30. 3 3 1 3 6 3 3 1 2 5 10 15 5 6 10 3 15 2 18 9 2 30 30 30 29 30 29 The homeowners need cubic yard of cement. 30
4. The LCD is 70. 5 9 5 10 9 7 50 63 13 70 7 10 7 10 10 7 70 70 5. The LCD is 24. 5 1 1 5 3 18 1 2 8 3 12 8 3 38 12 2 15 8 2 24 24 24 5 24
11. The LCD is 12. 3 2 3 3 2 4 9 8 1 4 3 4 3 3 4 12 12 12 1 The difference in length is foot. 12 Calculator Explorations
5 6. Recall that 5 . The LCD is 4. 1 5 y 5 4 y 20 y 20 y 1 4 1 4 4 4 4 4 7. The LCD is 40. Write each fraction as an equivalent fraction with a denominator of 40. 5 5 5 25 8 8 5 40 11 11 2 22 20 20 2 40 25 22 5 11 Since 25 > 22, , so . 40 40 8 20 8. The LCD is 20. Write each fraction as an equivalent fraction with a denominator of 20. 17 20 4 44 16 5 5 4 20 17 16 17 4 Since 17 < 16, , so . 20 20 20 5
1.
1 2 37 16 5 80
2.
3 2 23 20 25 100
3.
4
9
7
8
95 72
4.
9 5 163 11 12 132
5.
10 12 394 17 19 323
6.
14 15 253 31 21 217
Vocabulary, Readiness & Video Check 4.5 1. To add or subtract unlike fractions, we first write the fractions as equivalent fractions with a common denominator. The common denominator we use is called the least common denominator. 2. The LCD for
3.
1
5
1 5 and is 24. 6 8
1 4 5 3 4 15 19. 6 4 8 3 24 24 24
9. The LCD is 99. 5 4 5 9 4 11 45 44 1 x y 11 9 11 9 9 11 99 99 99
6
8
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
4.
1
6
1 4 5 3 11 4 15 . 8 6 4 8 3 24 24 24 5
5. x y is an expression while 3x
1 is an 5
equation. 6. Since 10 < 1, we know that
10
13
1
. 13
7. They are unlike terms and so cannot be combined. 8. Once the fractions have the same denominator, we then just compare numerators. 3
9.
1 ; 6 2 3 34
10.
15
4 cm and 2
15
cm
Exercise Set 4.5
12. The LCD is 20. y 5 y 5 20 1 20 5 20 y 1 20 20 100 y 20 20 100 y 20 14. The LCD is 36. 7 5 7 3 5 2 21 10 11 12 18 12 3 18 2 36 36 36 16. The LCD is 10. 7 10 7 10 10 1 10 10 10 7 110 10 100 7 10 10 93 10
2. The LCD is 12. 5 1 5 2 1 10 1 11 6 12 6 2 12 12 12 12
18. The LCD is 18. 7 x 2x 7x 2x 2 7x 4x 11x 18 9 18 9 2 18 18 18
4. The LCD is 12. 2 1 2 4 1 3 8 3 5 3 4 3 4 4 3 12 12 12
20. The LCD is 12. 5z 1 5z 2 1 10z 1 10z 1 12 6 12 6 2 12 12 12
6. The LCD is 9. 5 1 5 1 3 5 3 2 9 3 9 3 3 9 9 9
22. The LCD is 5x. 2 3 2 x 3 5 2x 15 2x 15 5x 5 x 5 x x 5 5x 5x
8. The LCD is 15. 4 2 2 2 23 2 6 15 5 15 5 3 15 15 15
24. The LCD is 9. 5 1 5 1 3 5 3 8 9 3 9 3 3 9 9 9
10. The LCD is 25. 2 y 3 y 2 y 5 3 y 10 y 3 y 13 y 5 25 5 5 25 25 25 25
26. The LCD is 15. 4 2 4 3 2 12 2 10 2 5 2 5 15 5 3 15 15 15 15 3 5 3 28. The LCD is 25. 2b 3 2b 5 3 10b 3 10b 3 5 25 5 5 25 25 25 25 30. The LCD is 36. 5 7 5 2 7 3 10 21 11 18 12 18 2 12 3 36 36 36
122
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ISM: Prealgebra and Introductory Algebra
32. The LCD is 40. 5 3 5 5 3 2 25 6 19 8 20 8 5 20 2 40 40 40 34. The LCD is 130. 10 7 10 10 7 13 100 91 9 13 10 13 10 10 13 130 130 130 36. The LCD is 18. 7 2 7 2 2 7 4 11 18 9 18 9 2 18 18 18 38.
3
4
13 13
3 4 1 13 13
40. The LCD is 60. 1 1 2 1 20 115 2 12 3 4 5 3 20 4 15 5 12 20 15 24 60 60 60 11 60 42. The LCD is 26. 5z 3 5z 2 3 10z 3 10z 3 13 26 13 2 26 26 26 26 44. The LCD is 58. 1 3 1 29 3 2 29 6 35 2 29 2 29 29 2 58 58 58 46. The LCD is 16. z z 2z z 4 z 2 2z 4 8 16 4 4 8 2 16 4z 2z 2z 16 16 16 8z 16 8 z 8 2 z 2
Chapter 4: Fractions and Mixed Numbers
52. The LCD is 60. 7 3 72 3 3 30 20 30 2 20 3 14 9 60 60 5 60 1 5 12 5 1 12 54. The LCD is 12x. 1 5 1 x 5 12 12 x 12 x x 12 x 60 12x 12x x 60 12x 56. The LCD is 20. 6 3 1 6 4 3 5 110 5 4 2 5 4 4 5 2 10 24 15 10 20 20 20 29 20 58. The LCD is 24. 11 7 11 2 7 22 7 15 3 5 5 12 24 12 2 24 24 24 24 3 8 8 60. The LCD is 24x. 3 5 3 3x 5 2 8 12x 8 3x 12x 2 9x 10 24x 24x 9x 10 24x
48. The LCD is 16. 9 3 9 3 2 9 6 3 16 8 16 8 2 16 16 16 50. The LCD is 28. 3y y 3y 7 y 4 21y 4 y 25 y 4 7 47 7 4 28 28 28
62. The LCD is 14. 5 3 1 5 3 2 1 7 14 7 2 14 5 76 2 72 7 14 14 14 6 14 3 2 72 3 7
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Chapter 4: Fractions and Mixed Numbers
66. The LCD is 99. Write each fraction as an equivalent fraction with a denominator of 99. 5 5 11 55 9 9 11 99 6 6 9 54 11 11 9 99 55 54 5 6 Since 55 > 54, , so . 99 99 9 11 68. The LCD is 48. Write each fraction as an equivalent fraction with a denominator of 48. 7 76 42 8 8 6 48 5 5 8 40 6 68 48 42 40 7 5 Since 42 < 40, , so . 48 48 8 6 70. The LCD is 117. Write each fraction as an equivalent fraction with a denominator of 117. 2 2 13 26 9 9 13 117 3 3 9 27 13 13 9 117 26 27 2 3 Since 26 > 27, , so . 117 117 9 13
3
3
1 4 1 4 4 4 3 3 3 3 9
80. The LCD is 21. 10 1 10 1 10 1 3 10 1 3 21 7 21 7 21 7 3 21 7 3 10 3 10 3 21 21 21 21 26 5 or 1 21 21 26 5 yards or 1 yards. The perimeter is 21 21
84. The difference of a number and x
7
.
20
124
2 5 translates as
2 x . 5
1
78. The LCD is 8. 3 5 1 3 5 1 4 8 8 2 8 8 24 3 5 4 8 8 8 12 8 43 42 3 1 or 1 2 2 3 1 The perimeter is miles or 1 miles. 2 2
82. A number increased by
72. The LCD is 12. 5 1 3 1 4 3 3 4 9 x y 3 4 3 4 4 3 12 12 12 74. x y
1 3 3 4 2 3 3 4 2 4 3 3 3 4 4 3 8 9 12 12 17 12
76. 2x y 2
64. The LCD is 10. 9x 1 x 9x 1 5 x 2 10 2 5 10 2 5 5 2 9x 5 2x 10 10 10 11x 5 10
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7 20
translates as
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
86. The LCD is 264. 1 5 1 66 5 66 5 61 4 264 4 66 264 264 264 264 61 A killer bee will chase a person mile 264 farther than a domestic honeybee. 88.
96. The piece representing historical parks is labeled 8 and the piece representing memorials is 25 19 labeled . The LCD is 250. 250 8 19 8 10 19 25 250 25 10 250 80 19 25 250 99 250 99 of the areas maintained by the National 250 Park Service are designated as historical parks or memorials.
3 1 1 3 2 1 1 4 8 8 42 8 8 6 1 1 8 8 8 4 8 1 4 24 1 2 1 The inner diameter is inch. 2
90. The LCD is 100. 67 1 67 1 20 67 20 47 100 5 100 5 20 100 100 100 47 The eSIM (MFF2) card is mm thicker than 100 the eSIM (WLCSP) card.
100. 8 6 4 7 = 8 24 7 = 16 7 = 23 102. 50 (5 2) = 50 10 = 5 104. a.
92. The LCD is 16. 11 5 11 11 5 2 11 16 8 16 16 8 2 16 11 10 11 16 16 16 32 16 2 The total width is 2 inches. 94. The LCD is 25. 1 1 1 1 5 1 5 6 25 5 25 5 5 25 25 25 The Arctic and Indian Oceans account for
6 25
the world’s water surface.
98. 1 17 1000 17 983 1000 1000 1000 1000 983 of the areas maintained by the National 1000 Park Service are not national parkways or scenic trails.
of
b.
There seems to be an error.
c.
The LCD is 8. 2 5 3 5 3 2 5 6 1 not 8 4 8 4 2 8 8 8 4
106. The LCD is 600. 1 9 7 9 60 7 3 1 200 10 200 3 10 60 200 3 3 200 540 21 200 600 600 600 519 200 600 600 319 600
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
11. 315 3105 3 3 35 3 3 5 7
108. The LCD is 1352. 19 968 19 52 968 26 1352 26 52 1352 988 968 1352 1352 20 1352 54 338 4 5 338
315 32 5 7
12.
441 3 147 3 3 49 3 3 7 7
110. 1; answers may vary
441 32 72
Mid-Chapter Review 13.
2 2 1 1 14 2 7 7
2. Each part is 1 of a whole and there are 5 parts 4 5 1 shaded, or 1 whole and 1 more part: or 1 4 4
14.
24 6 4 6 20 5 4 5
3. number that get fewer than 8 73 total number of people 85 73 of people get fewer than 8 hours of sleep 85 each night.
16.
1. 3 out of 7 equal parts are shaded:
3 7
4. 5.
7.
8.
3
9. 65 = 5 13 10. 70 2 35 2 5 7 70 = 2 5 7
126
165z3
15 11 z z z 15 14 z
210z 20.
7 is undefined 0
9 8 10 10
90 30 3 3 240 y 30 8 y 8 y
17 0
89
18.
1 0
80
54x 27 2 x 2x 135 27 5 5
11 1 11 17
72
17.
19. 6.
56 14 4 14 60 15 4 15
15.
11 z z 14
11z2 14
35 7 a b 35 11 a a b b b 385a b 7 11 a b b 7 11ab2 245ab
2 3
21. Not equivalent, since the cross products are not equal: 7 10 = 70, 8 9 = 72 22. Equivalent, since the cross products are equal: 10 18 = 180, 12 15 = 180
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
number not adjacent 2 total number 50
31.
2 1 2 1 of the states are not 50 25 2 25 adjacent to any other states.
32.
50 2 = 48; 48 states are adjacent to other states.
33.
1 3 1 3 3 5 5 5 5 25
48 2 24 24 of the states are adjacent 50 2 25 25 to other states.
34.
1 3 1 5 1 5 1 5 5 5 3 5 3 3
24. a. digital 3-D screens in U.S./Canada 14, 000
35.
2 5 2 6 26 2 23 2 2 4 3 6 3 5 3 5 3 5 5 5
36.
2a 5 2a 5 2 a 5 5 5 3 6a 3 6a 3 2 3 a 3 3 9
23. a.
b.
c.
14, 000 7 2000 7 of the screens in 44, 000 22 2000 22 the U.S./Canada are digital 3-D screens.
b. 44,000 14,000 = 30,000 30,000 of the screens in the U.S./Canada are not digital 3-D. 30, 000 c.
15 2000
15
of the screens in 44, 000 22 2000 22 the U.S./Canada are not digital 3-D screens.
25. 5 = 5 6=23 LCM = 2 3 5 = 30
1
44, 000
screens in U.S./Canada
1 3 1 3 4 5 5 5 5 3
5
1 3
5
2
2 5 5
5
37. The LCD is 6y. 2 5 2 2 5 3y 6 y 3y 2 6 y 22 5 3y 2 6 y 4 5 6y 6y 1 6y
28.
7 7 4 7 4 28 9 9 4 9 4 36
38. The LCD is 6. 2x 5x 2x 2 5x 3 6 3 2 6 2x 2 5x 3 2 6 4x 5x 6 6 9x 6 3 3 x 3 2 3x 2
29.
11 11 5 11 5 55 15 15 5 15 5 75
1 7 1 7 1 39. 7 18 7 18 18
30.
5 5 8 5 8 40 6 6 8 6 8 48
40. 4 3 4 3 4 3 4 4 9 7 9 7 3 3 7 3 7 21
26. 2 = 2 14 = 2 7 LCM = 2 7 = 14 27. 6 = 2 3 18 = 2 3 3 30 = 2 3 5 LCM = 2 3 3 5 = 90
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
7z 2 7z 6z 2 6z 8 8 1 7z 1 2 8 6z 7z 1 8 6z2 7 z 1 86 z z 7 48z 9 9 5 9 1 9 1 9 42. 5 10 10 1 10 5 10 5 50
41.
43. The LCD is 40. 7 1 7 5 1 2 35 2 37 8 20 8 5 20 2 40 40 40
1 x . 5
128
3 1 5
2
2 1530 3 1 2 1530 31 2 3 510 31 2 510 1 1020
1530
3
of 1530 is 1020.
52. 18
46. The LCD is 50. 3y y 6 3y 5 y 10 6 2 10 5 25 10 5 5 10 25 2 15 y 10 y 12 50 50 50 25 y 12 50 50 25 y 12 50 2 2 2 47. of a number translates as x or x.
48. The quotient of a number and
51.
translates as
translates as
6 increased by a number translates as 6 x. 11 11
3
45. The LCD is 18. 2 1 1 2 2 1 1 6 9 18 3 9 2 18 3 6 4 1 6 18 18 18 11 18
3
50.
8 9
8 x. 9
2
44. The LCD is 36. 5 1 5 3 1 4 15 4 11 12 9 12 3 9 4 36 36 36
3
49. A number subtracted from
3
18
3
4
1 4 18 4 1 3 18 4 1 3 3 6 4 1 3 64 1 24 They can sell 24 lots.
53. The LCD is 16. 7 1 1 72 1 1 8 16 16 8 2 16 16 14 1 1 16 16 16 12 16 43 44 3 4 3 The inner diameter is foot. 4
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
54. The LCD is 21. 17 5 17 5 3 17 15 2 21 7 21 7 3 21 21 21 2
of the 21 2022 FIA Formula One World Championship races.
Red Bull driver Sergio Perez won
Section 4.6 Practice Exercises 7y 7y 1 7y 5 7 y 5 7 y 5 7 y 1. 10 1 10 5 10 1 10 1 2 5 1 2 5 13 1 11 23 6 2. 23 26 33 24 4
3
4 1 6 12 4 12 6 1 426 6 1 8 or 8 1
1 2 7 2. To simplify , which operation do we 2 3 8 perform first? multiplication 1 2 7 3. To simplify , which operation do we 2 3 8 perform first? division
3
12 1 12 1
2 6 12 43 12 32
62 89 8 or 8 1
4. The LCD is 20. 3 20 3
3
7 1 2 8 2 3 , which operation do we perform first? subtraction
4. To simplify
3
1 1 9 3 5. To simplify , which operation 3 4 11 8 do we perform first? addition
20 3
15
20 5x 20 1 4x 20
4 x4 5 1 20 x5 1
1. A fraction whose numerator or denominator or both numerator and denominator contain fractions is called a complex fraction.
4
Vocabulary, Readiness & Video Check 4.6
3. The LCD is 12. 1 1 12 1 1 2 6 2 6 32 12 3 2 3
4
3 3 2 5 5 23 3 1 5 5 4 5
43 34 3 1 96 68 12 12 4 16 12
1 1 5 1 1 5 1 2 5 1 6. 2 5 5 2 8 8 2 5 8 8 5 2 5 1 10 10 8 8 3 6 10 8 3 2 3 2 5 8 3 3 5 8 9 40 3 3 3 2 7. xy 5 5 10 3
4
8 8 54 46 2 2 5. 2 27 27 27 27 3
2
6. To simplify 9 3 , which operation do we 4 perform first? evaluate the exponential expression 7. distributive property
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
8. They have the same denominator so they are like fractions.
12.
9. Since x is squared and the replacement value is negative, we use parentheses to make sure the whole value of x is squared. Without parentheses, the exponent would not apply to the negative sign. Exercise Set 4.6 2.
5 6 5 15 5 12 5 12 1 5 6 2 2 15 6 12 6 15 6 15 653 3 12 25
4. 11 25 6
25
25 11 6
14.
25 6 11 25 25 6 11 25 6 11 11 3y 1 3y 2 3y 2 6 y 1 11 2 11 1 11 1 11 2
8. The LCD is 18. 7 2 18 7 2 6 3 6 3 38 18 3 8
2
9
2 9 18 7 18 2 6 3 18 23 18 98
2112 27 16 33 11 3
10. The LCD is 10y. 3 2 3 2 10 y 10 10 2 10 y 2
5y
7 1 1 7 4 1 7 4 1 1 8 4 7 8 1 7 4 2 1 7 2 2
3y
6.
1 1 1 1 11 2 6 3 2 63 1 1 2 18 1 9 1 2 9 18 9 1 18 18 10 18 25 29 5 9
1 2 1 16. 3 9 4 2 9 1 1 4 94 1 1 4 4 36 1 4 4 35 4
18. 1 1 1 1 1 2 1 1 2 1 5 10 5 10 5 2 10 5 2 10 2 1 2 1 10 10 1 3 1 01010 1 3 1 10 10 0 3 100
5y
10 y 3 10 y 2 10
4 3y 20 y 4 23y 4
130
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ISM: Prealgebra and Introductory Algebra
3
20. 2 7 4 9 4 3 9 3 3 9 3 4 1 9 3 9 1 4 3 9 1 4 27 4 22.
2 1 2 5 2 1 5 1 1 5 2 5 1 2 2 2 10 1 5 2 2 1 2 9 1 5 2 29 1 5 2 9 5 55 4 5 2
Chapter 4: Fractions and Mixed Numbers
2
8 2 8 2 3 2 24. 9 2 3 9 1 3 3 2
8 6 2 9 38 2 34 9 3 64 4 81 3 64 3 81 4 16 4 3 27 3 4 16 27
3 4 3 3 4 2 3 26. 2 3 2 3 3 3 2 9 8 6 6 3 1 6 1 1 1 6 6 6 1 216 3
2
3
2
1 1 2 3 1 1 2 2 3 28. 6 3 5 4 6 3 2 5 4 2 3 6 1 2 6 6 20
3
2
3 6 6 3 2 1 20 3 2 10 1 9 8 100 1 25 9 2 8 25 100 2 25 18 200 200 43 200 5 1 5 1 6 2 30. 2z x 2 3 3 3 6 3
1
3 32. y x 5 2
z
5 6 2 30 5
13 30 56
12 (10) 25 2 25
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 2 2 2 2 5 34. x z 1 3 6 1 1 5 5 3 3 6 6 1 25 9 36 1 4 25 9 4 36 4 25 36 36 21 36 7 3 12 3 7 12
1 5 36. (1 x)(1 z) 1 1 3 6 1 1 56 1 3 3 1 6 5 3 3 6 6 4 1 3 6 2 2 1 3 2 23
9 7
38.
7
14z
10 25 14z 10 25
5 1 1 1 5 2 1 1 40. 21 2 7 3 21 1 7 3 5 2 11 211 7 3 10 1 21 21 11 21
2
3 3 9 3 9 3 2 9 6 15 42. 4 8 16 8 16 8 2 16 16 16
10 3 10 1 30 5 25 44. 2 2 2 4 51 10 4 15 10 4 10 15 40 2 42
10 3 1
3 1
2
3 1 5 1 1 2 48. 1 3 10 20 5 1 10 2 20 5 5 3 1 5 2 20 20 55 5 4 20 5
54 5 4 5 1 5
7 25 10 14z 755 5 27 2 z 5 2 2 z 5 4z
132
2
46. 1 3 1 1 3 3 2 4 2 2 4 4 1 9 4 4 16 9 1 4 4 16 4 9 165 16 16
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ISM: Prealgebra and Introductory Algebra 2
2 5 2 3 2 50. 9 3 9 3 32 5 6 9 9 2 1 9 1 1 9 9 1 81 5
7
52.
7
Chapter 4: Fractions and Mixed Numbers 60. 4
1 5
64.
3 9 10 3 9 5 10 5 10
2
1
66.
1 2 1 2 1 2 1 7 54. 2 7 2 7 2 7 2 2 1 2 1 7 27 22 2 7 14 4 22 77 14 2 4 7 4 49 28 28 45 28
x 2 38 8 2 38
8 1 x 4 56. 1 4
5 7 18 5 7 6 9 6 9
2
18(2) 18 5 18 7 6
15 18 2 14 36 29 36
9
68. no; answers may vary 70. False; the average cannot be less than the smallest number. 72. False, if the absolute value of the negative fraction is greater than the absolute value of the positive fraction, the sum is negative. 74. True 76. No; answers may vary.
8 1 8 x 4 8 2 8 83
78. Addition, multiplication, subtraction, division 80. Subtraction, multiplication, division, addition
8 2x 16 3 8 2x 19
3
10(2) 10 53 10 109 10 2 69 20 15 20 5 3 5 4 3 4
14 10 16 15 24 31
2
4 5 1 20 1 19 4 or 3 1 5 5 5 5 5 5
62. yes; answers may vary
1 20 10 2 10 2 20 4 3 43 5 4 5 4 20 7 20 1 10 2 20 54 20 43
58. 2
2 3 2 6 2 8 2 or 2 1 3 3 3 3 3 3
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 3 4 82. 4x y 4 4 7 4 3 4 14 7 12 4 4 7 12 7 4 4 47 74 84 16 28 28 68 28 4 17 47 17 7
84.
9 14
x y
7 3 97 3 97 97 2 15 1 15 1 5 3 5 or 19 5 7 Estimate: 6 rounds to 6 and 3 6 = 18, so the 15
5. 3 6
answer is reasonable. 6.
7
9 7.
9 14
47 28 9 14 28 34 47
3 4
1. 11 2 2 11 5 11 5 11 or 1 2. 1 3 15 3 15 3 5 3 9 9 5 5 18 5 18 5 6 3 15 18 or 15 6 6 1 6 1 6 1 1
1 3 16 11 16 11 4 4 11 44 4 4. 3 2 or 8 5 4 5 4 5 4 5 4 5 5 3 1 Estimate: 3 5 rounds to 3, 2 4 rounds to 3.
10.
23 31 14 7 14 23 14 7 31 23 7 2 7 31 46 15 or 1 31 31
1
5 30 6 2 12 4 4 5 30 17 6 30 1 2 Estimate: 2 rounds to 2, 4 rounds to 4, and 6 5 2 + 4 = 6, so the answer is reasonable. 2
3 2
5 14 6 7
3 3 = 9, so the answer is reasonable.
134
4 1 4 1 4 9 7 9 7 63
18
Section 4.7 Practice Exercises
1
4 3 8 19 15 5 15 5 8 5 15 19 85 15 19 85 5 319 8 57
28 3 28 4 4 7
9.
7
8
7
18 2116 18 5
9
3.
4
2 3 8. 3 2
4
Copyright © 2025 Pearson Education, Inc.
2
3
5
14 12 2 14 3 3 17 5 1 6 5 14 14 14
ISM: Prealgebra and Introductory Algebra
11.
12.
12
12
6 3 7 1 2 5
30 3 35 7 2 35 2 2 37 17 1 18 17 35 35 35 7
32
32
9 5
16
16
18
14 18 5
18 9 1 16 18 16 2
Chapter 4: Fractions and Mixed Numbers
1 22 16. 44 3 44 7 7 44 7 1 22 2 22 7 1 22 14 or 14 1 Therefore, 14 dresses can be made from 44 yards. 17. 9
18. 5 13.
9 4
7
9
15 3 5
4
7 8
15 9
4
15
22 15 9
19.
25 10
24 2 9
10 14
20.
9 2 97
15.
23 19
1 4 5 12
23 19
12
22 19
15 12 5
12 10 3 5 3 12 6 The girth of the largest known American beech 5 tree is 3 feet larger than the girth of the largest 6 known sugar maple tree.
66 7
10 11 5 10 65 11 11 11 4
5 8
46 1 9 5 5
4 8 37 32 5 9 5 46 45 1
21. 2 3 3 3 11 18 4 5 4 5 1118 45 11 2 9 225 99 9 or 9 10 10
3 12 5
79 3 7
8
9
7
37
15 13 4 15
9 14.
3
22. 4
2
1 30 5 1 7 4 7 4 30 4 7 5 30 4 75 56 4 75 24 3 or 3 7 7
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
23.
12 6
3
12
4 2
6
3 2
3
9
the 5 is called the whole number part 5 , 4 3 and is called the fraction part. 4
12 8
12 1 6 12
1
3. To estimate operations on mixed numbers, we round mixed numbers to the nearest whole number.
6 12 6 3 4 12 2 9 24. 7 11 30 14
9
2
4 9 14 11 30 14 15 40 1 39 14 14
30
7
11
40
14
1
5
365
31 3776 35 35
26
14 6.
290 13
1 Exercise Set 4.7 2.
4 13
3941 14 231 17 17
6. 5
Vocabulary, Readiness & Video Check 4.7 1. The number 5
3 4
136
1 3
4.
2769 3 7. 92 30 10 8.
5. The denominator of the mixed number we’re 4 graphing, 3 , is 5. 5
9. We’re adding two mixed numbers with unlike signs, so the answer has the sign of the mixed number with the larger absolute value, which in this case is negative.
14 22
.
8. To find how much more 11 1 in. is than 3 3 in., 4 5 we subtract the mixed numbers, which includes subtracting the fractional parts.
17 3923 4. 186 21 21 5.
8
written as an improper
8
21 fraction is
5
7. The fractional part of a mixed number should always be a proper fraction.
11
14 1019 2. 67 15 15 3. 107
The mixed number 2
6. To multiply mixed numbers, we first write them as equivalent improper fractions and then multiply as we multiply for fractions.
280
11
4.
14
Calculator Explorations 1. 25
3
2. For
is called a mixed number.
9 10
1
rounds to 5. 6 5 3 rounds to 4. 7 5 4 = 20 The best estimate is d.
Copyright © 2025 Pearson Education, Inc.
3 8
ISM: Prealgebra and Introductory Algebra
8. 20
3
Chapter 4: Fractions and Mixed Numbers
28. Exact:
rounds to 20.
14 8 4 rounds to 5. 11 20 5 = 4 The best estimate is c. 10.
5
4
9
1 5
3 10
30. Exact: 12 4
1 1 9 57 513 1 14. Exact: 2 7 or 16 4 8 4 8 32 32 1 1 Estimate: 2 rounds to 2, 7 rounds to 7. 4 8 2 7 = 14 16. Exact: 5 3 35 38 7 5 2 19 133 1 5 7 or 44 6 5 6 5 235 3 3 5 3 Estimate: 5 rounds to 6, 7 rounds to 8. 6 5 6 8 = 48 1 6 10 2 310 20 18. 6 3 20 3 1 3 1 3 1
5
21
26.
81 1
3 1
rounds to 2. 2 82=6 The best estimate is c.
5
5
12
12 1
4
12 2 12 7 12
Estimate: 12
5
rounds to 12, 4
1
rounds to 4. 12 6 12 + 4 = 16 so the answer is reasonable. 32.
2
6 8
34.
13 7
8
8
9 32 10 9 21
36.
3
23
5 8
8
15
4
1
3 2 9 5 1 3 6
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26 7
26 11 14 26
2
8
4
6
26
38. rounds to 8.
11 6
16
5 5 25 22. 4 5 5 9 5 9 5 9 21 9 21 189 7 24. 3 rounds to 4. 8 1 2 rounds to 2. 5 42=2 The best estimate is d.
11 2
6
4 5 14 21 2 7 21 147 7 or 7 20. 2 2 5 8 5 8 5 2 4 20 20 1
4
11 4 2 Estimate: 7 rounds to 7, 3 rounds to 3. 11 11 7 + 3 = 10 so the answer is reasonable.
5 21 5 3 7 7 1 or 2 9 5 3 3 5 3 3
2 9 5 9 3 9 3 27 2 or 5 12. 9 1 3 1 3 1 5 1 5 5 5
5
7
14 63
32 30 63 44 49 63
9
23 8
9 15 8
15 17 2 2 31 311 32 15 15 15 10 30 12 9 30 5 3 30 27 9 16 16 30 10 4
137
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
7
40. Exact:
3
4 9 2
52.
8
9 2 4 9 4 2 Estimate: 7 rounds to 7, 3 rounds to 3. 9 9 7 3 = 4 so the answer is reasonable. 12
42. Exact:
4
5
12
12 1
4
5 12 2
6
12 3 1 8 8 12 4 5 1 Estimate: 12 rounds to 12, 4 rounds to 4. 12 6 12 4 = 8 so the answer is reasonable. 44.
2
5
13 7 4 26
46.
5
4
26 7 4 26
7 10
48.
86 27
2 15 3
86 27
10
30 9
85 27
30 58
30 25
138
58
22
15 9
8
8
15
23 15 9
15 14 14 15
1 1 26 13 56. 5 3 5 4 5 4 26 4 5 13 2 13 4 5 13 2 4 5 8 5 3 1 5 3
5
25
2 5
15
3
5
25
10
20
25
13 25
3 3 6 3 3 60. 5 2 5 2 3 8 16 16 16 16 7 16 1 6 2 3 9 8
7 16 8 6 16 6 9 16 21 5 5 18 18 1 19 16 16 16
3
3
34 30 9
30 5 5 50. 83 2 8 8 7 7 1 8 8
15 3
1 9 3 27 1 13 54. 4 3 2 2 1 2 2
62.
4
8
23
5
58. 15
10 7 10 7 1 10 3 6 10
8 1
30 26 7 4 26 23 26 4
8
23
5 6
64.
47
5
12 19 23 24
47 23
10 24 19
46 23
24 23
34 24 19 24 15 24
23
5 8
2 3 20 11 20 11 4 5 11 55 1 66. 6 2 18 3 4 3 4 3 4 3 4 3 3
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ISM: Prealgebra and Introductory Algebra
68.
7
3
7
7
14
15 20
6
15 1
20
2
70. The sum of 8
7
14 13 42 14 3 4
and a number translates as
3 8 x. 4
Chapter 4: Fractions and Mixed Numbers
1 1 9 19 3 319 57 1 or 28 82. 4 6 2 3 2 3 23 2 2 57 1 The area is or 28 square yards. 2 2 1 13 5 65 1 16 84. 3 5 4 4 1 4 4 1 The perimeter is 16 yards. 4 86.
3
1
72. Divide a number by 6 translates as 11 1 x 6 . 11
3 8
3
8 8 3
8 5 28 The remaining piece is 2
5
feet.
8
74. 27 3 1 111 1 111 4 111 111 4 4 4 4 4 1 1 This will make 111 quarter-pound hamburgers. 76. Subtract the standard track gauge in Portugal from the standard gauge in Spain. 9 18 65 65 20 10 11 11 65 65 20 20 7 20 7 Spain’s standard track gauge is inch wider 20 than that of Portugal. 78.
5
6
42
43
88. 23 1 4 47 4 47 1 47 1 47 5 7 2 2 1 2 4 24 8 8 7 The length of each side is 5 feet. 8 90.
2
8 5 41 8 3 1 8 The short crutch should be lengthened 3 1 inches. 8
7 15 1
15 8 6 15 The total duration for the eclipses in even8 numbered years is 6 minutes. 15
8
5 41 8
4
92.
7 15 37 2 60
28 60 37 2 60
4
It will be 1
4
17 20
88 60 37 2 60 51 17 1 1 60 20 3
minutes longer.
5 7 5 7 35 1 or 17 2 1 2 1 2 2 2 35 1 The area is or 17 square inches. 2 2
80. 5 3
1
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139
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 94. 3 5 3 2 23 11 6 3 6 3 23 3 6 11 23 3 2 311 23 22 1 1 22 5 2 5 2 96. 17 14 17 14 9 3 9 3 5 6 17 14 9 9 14 6 16 14 9 9 8 2 9 7 5 7 5 98. 31 26 31 26 8 12 8 12 21 10 31 26 24 24 11 5 24 100. 1 3 3 1 7 7 4 2 4 2 7 2 4 7 72 42 7 4 1 2 2 3 4 3 7 102. 20 30 20 30 50 5 10 10 10 10
1 3 17 19 104. 4 2 4 8 4 8 17 8 4 19 17 4 2 4 19 34 19 15 1 19 1 1 106. (5 y) 1 y y 5 5 y 5 9 10 9 10 m 1 m m 108. m 10 9 10 9 7 3 3 110. a. 6 6 1 7 4 4 4 11 3 3 b. 5 5 2 7
4 c.
7
12 16
7
4
3 4
The answer is d, all of them. 112. Incorrect; to multiply mixed numbers, first write
each mixed number as improper fraction. 114. answers may vary 19 1 9 12 9 21 20 10 The best sum estimate is d.
116. 11
118. answers may vary Section 4.8 Practice Exercises 2 1.
y
5
12 2 5 2 y 3 3 12 3 5 24 y 12 3 4 5 8 y 12 12 13 y 12 3 2
140
4
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Check:
y
2
5
3 12 13 2 5 0 12 3 12 13 2 4 5 0 12 3 4 12 13 8 5 0 12 12 12 5 5 True 12 12 13 The solution is . 12 2.
1
y2
5 1
5 y 5 2 5 y 10 1 Check: y2 5 1 10 0 2 5 2 2 True The solution is 10. 3.
4.
Chapter 4: Fractions and Mixed Numbers
5
b 25 7 7 5 7 b 25 5 7 5 7 25 1b 5 b 35 5 Check: b 25 7 5 35 0 25 7 25 25 True The solution is 35. 7 2 x 10 5 10 7 10 2 x 7 10 7 5 10 2 x 7 5 4 x 7
Check:
7
x
2
10 5 4 2 0 10 7 5 28 2 0 70 5 2 2 True 5 5 4 The solution is . 7 7
5x
5.
3
4 1 3 5x 5 5 4 1 3 x 5 4 3 x 20
1
Check: 5
5x
3
3
4 3
0
20 4 15 3 0 20 4 3 3 True 4 4 3 The solution is . 20 6. The LCD is 15. 11 3 x 15 5 11 3 15 x 15 15 5 11x 9 11x 9 11 11 9 x 11 The solution is
9
.
11
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Chapter 4: Fractions and Mixed Numbers
7. Solve by multiplying by the LCD: The LCD is 8. y 3 2 8 4 y 3 8 8(2) 8 y 3 8 4 8 8(2) 8 4 y 6 16 y 6 6 16 6 y 10 Solve with fractions: y 3 2 8 4 y 3 3 3 2 8 4 4 4 y 8 3 8 4 4 y 5 8y 4 5 8 8 8 4 1 245 8 y 8 1 4 y 10 y
8.
x
3
2 8 4 10 3 02 8 4 10 6 02 8 8 16 02 8 2 2 True The solution is 10. Check:
x
1
5 5 x 1 5 5 x 5 5 x 1 5 5(x) 5 5 5 x 5x 1 4x 1 4x 1 4 4 1 x 4
142
ISM: Prealgebra and Introductory Algebra
To check, replace x with
1
in the original 4 equation to see that a true statement results. The 1 solution is . 4 9. The LCD is 10. y y 3 2 5 2 y y 3 10 10 2 5y 2 y 3 10 10 10 2 55 y 2 y 15 2 5 y 2 y 2 y 2 y 15 3y 15 3y 15 3 3 y5 y y 3 Check: 2 5 2 5 5 3 0 12 5 2 1 2 0 11 2 2 1 1 2 2 True 2 2 The solution is 5. 9 y 9 3 y 10 10. 10 3 10 3 3 10 9 3 y 10 10 3 310 27 10 y 30 30 27 10 y 30 Vocabulary, Readiness & Video Check 4.8 1. The LCD of 3 and 6 is 6. 2. The LCD of 21 and 7 is 21. 3. The LCD of 5 and 3 is 15. 4. The LCD of 11 and 2 is 22. 5. addition property of equality
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
6. We want z with an understood coefficient of 1. 4 The coefficient of z is . Two numbers are 9 reciprocals if their product is 1, so we use the multiplication property of equality to multiply 9 both sides of the equation by , the reciprocal 4 4 of . 9 7. We multiply by 12 because it is the LCD of all fractions in the equation. The equation no longer contains fractions.
4.
5
4
14 z 4 5 14 14 14 14 45 z 14 9 z 14 5
14 5
4
1
1 2 5 5x 11 4x 11 11 5 4 1 4 2 5 4 0 11 11 11 11 11 20 1 16 2 5 0 11 11 11 11 11 3 3 True 11 11 4 The solution is . 11
x
z
2 5 4x 11 11 11 1 3 x 11 11 1 1 3 1 x 11 11 11 11 4 x 11
6. 5x
Exercise Set 4.8 1 7 9 9 1 1 7 1 x 9 9 9 9 8 x 9 1 7 Check: x 9 9 8 1 7 0 9 9 9 8 1 7 0 9 9 7 7 True 9 9 8 The solution is . 9
5
14 14 9 5 4 0 14 14 14 95 4 0 14 14 4 4 True 14 14 9 The solution is . 14
8. Multiplying through by the LCD is a step in solving an equation, but we don’t have an equationwe have an expression.
2.
z
Check:
Check:
8.
y
8
1
9 3 8 8 1 8 y 9 9 3 9 1 3 8 y 3 3 9 3 8 y 9 9 11 y 9
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143
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
Check:
y
8
1
9 3 11 8 1 0 9 9 3 3 1 0 9 3 1 1 True 3 3 11 The solution is . 9 1
10.
a
11 8x 7 2 14 11 x 7 14 2 2 11 2 x 7 7 14 7 11 2 2 x 14 7 2 11 4 x 14 14 15 x 14
12. 9x
144
2
8x
11
7 14 15 2 15 11 9 8 0 14 7 14 14 135 4 120 11 0 14 14 14 14 11 11 True 14 14 15 The solution is . 14
3
2 8 1 1 1 3 a 2 2 2 8 1 4 3 a 24 8 4 3 a 8 8 7 a 8 1 3 Check: a 2 8 1 7 3 0 2 8 8 4 7 3 0 8 8 8 3 3 True 8 8 7 The solution is . 8
9x
Check:
14.
1
y
7
4 10 1 7 y 7 7 4 10 10 10 1 5 7 2 y 4 5 10 2 5 14 y 20 20 9 y 20 1 7 y Check: 4 10 1 9 7 4 0 20 10 1 5 9 72 4 5 0 20 10 2 5 0 9 14 20 205 20 5 True 20 20 9 The solution is . 20
2
5x 4
16.
1 1 (5x) 4 5 5 4 x 5 18.
1
x6
3 1
3 x 3 6 3 x 18
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
20.
Chapter 4: Fractions and Mixed Numbers
4
x 8 7 7 4 7 x 8 4 7 4 x
32.
56
4 x 14 22.
11 2 x 10 7 10 11 10 2 x 11 10 11 7 20 x 77
1 2
26.
28.
12
36.
25 1 1 12 4z 4 4 25 12 z 4 25 3 z 25 3
7 y 20 35 7 20 y 20 5 20 12 y 7 12 y 7 12 12 7 y 12 x
30.
1
4z
7 1 x5 5 7 5 1 5 5 5 x 5 5 1 7 5 x5 7 x55 75 x 12
1
x
3 4 12 2 1 x 12 3 4 12 12 2 1 12 12 x 3 4 83 x 5 x
12 5 2 12 5 z 24
2z
2
34.
2z 5
24.
x x 6 3 x 3 x 3(6) 3 x 3 3(x) 18 3 x 3x 18 2x 18 2x 18 2 2 x9
a a 5 2 7 2 a a 5 14 14 2 a 5 7 7a 14 2 14 7 2 7a 2a 35 7a 2a 2a 2a 35 5a 35 5a 35 5 5 a7
5 y 5 8 y 9 40 9 y 40 9 y 38. 9 8 9 8 8 9 72 72 72 40. 2
42.
7x
3
2 3 7x 6 7x 6 7x 1 3 3 3 3 3
9x 5x 9x 3 5x 4 27x 20x 7x 8 6 8 3 6 4 24 24 24
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145
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 2
44.
y
3 54.
5 10 5 2 5 3 y 2 5 2 10 15 y 20 3 y 4
56.
x 2
4 x 4 (2) 4 x 8
x 21 45 x4 20 21 20 20 20 5 4 4 x 20 20 21 5 4 16 5x 21 16 5x 16 2116 5x 5 5x 5 5 5 x 1
58.
3 9 x 24 9 4 43 x 3 4 3 2 49 x 3 2 2 2 3 3 x 3 2 x 6 y
60.
5 5 a 16 6 16 5 16 5 a 5 16 5 6 2 8 5 a 5 8 23 a
48.
1 4 1
4
46.
8 1 8 2 1 11 16 11 5 11 2 11 2 2 11 22 22 22
4 y 3 y 3 3(4 y) 3 y 3(4) 3 y y 12 3y y 3y 12 3y 3y 2 y 12 2 y 12 2 2 y6
3 30n 34n
50.
4n
3 20 3
20 1 1 3 4n 4 4 20 3 n 80 y
52.
146
1
2 7 7 y 1 7 2 7 7 7 y 7 72 1 7 y 14 1 y 14 14 114 y 15
62.
7 5 4 x 9 18 18 7 9 x 9 18 9 7 9 9 x 7 9 7 18 99 x 7 9 2 9 x 14
64. 27x 30x
3x
4 9 4
9 1 4 3x 3 3 9 4 x 27 1
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ISM: Prealgebra and Introductory Algebra
66.
68.
70.
5
1 7 y 1 7 5 4 2 10 20 y 20 4 2 10 1 7 25 y 20 20 2 10 25 y 10 14 25 y 4 25 y 4 25 25 4 y 25
Chapter 4: Fractions and Mixed Numbers
x
76. a.
b.
a a 1 6 3 2 a a 1 6 6 6 3 2 a 1 a 6 6 3 2 a 2a 3 a 2a 2a 3 2a a 3 a 3 1 1 a 3
78.
y y 2 4 5 3 y y 15 2 15 4 5 3 y y 15 15 2 15 15 4 5 3 3y 30 5 y 60 3y 5 y 30 5 y 5 y 60 2 y 30 60 2 y 30 30 60 30 2 y 30 2 y 30 2 2 y 15
80.
5 2 x6 53 6 6(2) 6 3 x 5 6 6 12 6 3 x 10 12 x 10 10 12 10 x 22 x 5 x 5 2 x 10 x 10 6 3 6 3 2 6 6 6
19 353x 23 53 1431 27 19 353x 23 1431 1431 1431 27 53 353x 23 27 19 1431 1431 1431 27 513 353x 1219 513 1219 353x 1219 1219 706 353x 706 353x 353 353 2 x 5 3 5 4 5 4 5 4 5 12 4 12 3 12 3 4 3 3 9 5 The length is inch. 9
Chapter 4 Vocabulary Check 1. Two numbers are reciprocals of each other if their product is 1. 2. A composite number is a natural number greater than 1 that is not prime. 3. Fractions that represent the same portion of a whole are called equivalent fractions.
72. To round 576 to the nearest hundred, observe that the digit in the tens place is 7. Since this digit is at least 5, we add 1 to the digit in the hundreds place. 576 rounded to the nearest hundred is 600. 74. To round 2333 to the nearest ten, observe that the digit in the ones place is 3. Since this digit is less than 5, we do not add 1 to the digit in the tens place. 2333 rounded to the nearest ten is 2330.
4. An improper fraction is a fraction whose numerator is greater than or equal to its denominator. 5. A prime number is a natural number greater than 1 whose only factors are 1 and itself. 6. A fraction is in simplest form when the numerator and the denominator have no factors in common other than 1.
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Chapter 4: Fractions and Mixed Numbers
ISM: Prealgebra and Introductory Algebra
7. A proper fraction is one whose numerator is less than its denominator. 8. A mixed number contains a whole number part and a fraction part. 9. In the fraction
7
, the 7 is called the numerator and the 9 is called the denominator.
9 10. The prime factorization of a number is the factorization in which all the factors are prime numbers. 11. The fraction
3
is undefined.
0 12. The fraction
0
0.
5 13. Fractions that have the same denominator are called like fractions. 14. The LCM of the denominators in a list of fractions is called the least common denominator. 15. A fraction whose numerator or denominator or both numerator and denominator contain fractions is called a complex fraction. a 16. In
b
c
, a d and b c are called cross products.
d
Chapter 4 Review 1. 2 out of 6 equal parts are shaded:
2 6
2. 4 out of 7 equal parts are shaded:
4 7
3. Each part is
1
of a whole and there are 7 parts shaded, or 2 wholes and 1 more part:
3 4. Each part is
1
of a whole and there are 13 parts shaded, or 3 wholes and 1 more part:
148
131 23 = 108 108 cars are not blue. not blue 108 total 131 108 of the cars on the lot are not blue. 131 Copyright © 2025 Pearson Education, Inc.
1 3
13 4
5. successful free throws 11 total free throws 12 11 of the free throws were made. 12
b.
or 2
3
4
6. a.
7
or 3
1 4
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
3 7. 1 3 8.
9.
10.
20 20 0 1
1
0
4 is undefined 0
11.
12.
13.
14. 3 15. 4 15 12 3 15 3 3 4 4 3 16. 13 39 39 0 39 3 13 17. 2
1 5 2 1 10 1 11 5 5 5 5
8 9 3 8 27 8 35 18. 3 9 9 9 9 19.
12 4 3 3 28 7 4 7
20.
15 3 5 5 27 3 9 9 Copyright © 2025 Pearson Education, Inc.
149
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
33. 24x 15 24x 15 3 5 8x3 5 8x 8 3 x 5 3 5 8 x x x 3 3 x x 9 2 x
29ab
29 a b
32 a b c
32abc 18xyz 24.
3 2 22. 36 y 1 36 y y y y 72 y 2 36 y 2
23.
1 21. 25x 1 25 x 3 25 x x 3x 75x2
29 32c
27 y3
18 x y z
18z 23 x y 23
7
27 y3 7
3 9 y y y 7
y
34.
45x2 y 9 5 x x y 5x 3 2 25. 9 3 x y y y 27xy 3y
21 18 y2 2118 y2 3 7 9 2 y y 2 3 111 1 1 1 1 1 35. 3 3 3 3 27 3 3 3
42ab2c 6 7 a b b c 7b 3 2 65 a b c c c 26. 30abc 5c
5 5 5 5 5 25 12 12 12 36. 12 12 144
27.
28.
23xy
3 3 3 8 3 8 3 4 2 2 2 37. 4 8 4 3 43 1 4 3 1
8 42 2 12 inches 12 4 3 3 2 8 inches represents of a foot. 3 8 inches
2
38.
15 6 = 9 cars are not white. 9 non-white cars 9 3 3 3 15 total cars 15 3 5 5 3 of the cars are not white. 5
29.
Not equivalent, since the cross products are not equal: 34 4 = 136 and 10 14 = 140
30.
Equivalent, since the cross products are equal: 30 15 = 450 and 50 9 = 450
21a 7a 21a 5 4 5 4 7a 21a 5 4 7a 7 3 a 5 47a 3 5 4 15 4
9 1 9 3 9 3 27 39. 2 3 2 1 2 1 2 5 5 2y 5 1 5 1 5 40. 2 y 3 3 1 3 2y 3 2 y 6y
3 1 31 3 31. 5 2 5 2 10 65 5 6 5 5 32. 6 5 7 12 7 12 762 72 14
150
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41. x y
9
52. 4 = 2 2 8=222 12 = 2 2 3 LCD = 2 2 2 3 = 24
3
7 4 9 4 7 3 94 73 3 3 4 73 3 4 7 12 7
42. ab 7
Chapter 4: Fractions and Mixed Numbers
7 9 79 63 10 1 10 110 10
53.
2 2 10 2 10 20 3 3 10 310 30
54.
5 5 7 5 7 35 8 8 7 8 7 56
55.
7a 7a 7 7a 7 49a 6 6 7 67 42
56.
9b 9b 5 9b 5 45b 4 4 5 45 20
9
43. area length width
11 7 11 7 77 6 8 6 8 48
57.
4
The area is
2 2 22 4 3 3 3 3 9
square meter.
58.
59.
7 3 7 3 10 11 11 11 11
47.
4 2 4 2 6 23 2 9 9 9 9 3 3 3 1
5
12 12 48. 49.
11x x 11x x 12x 3 4 x 4x 15 15 15 15 3 5 5 4y
21 50.
1 5 4 1 4 1 12 12 3 4 3
4
3
4y 3 21
2
21
3
15 15 15
4 3 2 1 1 15 15 15
51. 3 = 3 x=x LCD = 3 x = 3x
5
5 2 5 2 10 9 y 2 9 y 2 18 y
60. 46.
9y
9 45.
4 10 4 10 40 5x 5x 10 5x 10 50x
77 The area is square feet. 48 44. area side side
4
3 2 1 3 2 1 6 2 3 3 8 8 8 8 8 2 4 4 3 He did of his homework that evening. 4 9
3
9
3
9 393 16 16 16 16 16 24 16 38 2 8 3 2 3 The perimeter is miles. 2
61. The LCD is 18. 7 2 7 2 2 7 4 11 18 9 18 9 2 18 18 18 62. The LCD is 26. 4 1 4 2 1 8 1 7 13 26 13 2 26 26 26 26 63. The LCD is 12. 1 1 1 4 1 3 4 3 1 3 4 3 4 4 3 12 12 12
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Chapter 4: Fractions and Mixed Numbers
64. The LCD is 12. 2 1 2 4 1 3 8 3 5 3 4 3 4 4 3 12 12 12 65. The LCD is 55. 5x 2 5x 5 2 25x 2 25x 2 11 55 11 5 55 55 55 55 66. The LCD is 15. 4 b 4 b 3 4 3b 4 3b 15 5 15 5 3 15 15 15 67. The LCD is 36. 5 y 2 y 5 y 3 2 y 4 15 y 8 y 7 y 12 9 12 3 9 4 36 36 36 68. The LCD is 18. 7 x 2x 7x 2x 2 7x 4x 11x 18 9 18 9 2 18 18 18 69. The LCD is 9y. 4 5 4 y 5 9 4 y 45 4 y 45 9 y 9 y 9 y y 9 9y 9 y 70. The LCD is 11y. 9 3 9 y 311 9 y 33 9 y 33 11 y 11 y y 11 11y 11y 11y 71. The LCD is 150. 4 23 7 4 6 23 2 7 3 25 75 50 25 6 75 2 50 3 24 46 21 150 150 150 91 150 72. The LCD is 18. 2 2 1 2 6 2 2 1 3 3 9 6 3 6 9 2 6 3 12 4 3 18 18 18 5 18
152
ISM: Prealgebra and Introductory Algebra
73. The LCD is 18. 2 5 2 5 2 2 53 2 2 53 9 6 9 6 92 63 9 2 63 4 15 4 15 18 18 18 18 38 18 2 19 29 19 9 19 The perimeter is 9 meters. 74. The LCD is 10. 1 3 7 1 2 3 2 7 5 5 10 5 2 5 2 10 2 6 7 10 10 10 15 10 3 5 25 3 2 The perimeter is
3 feet. 2
75. The LCD is 50. 9 3 92 3 18 3 21 25 50 25 2 50 50 50 50 21 of the donors have type A blood. 50 76. The LCD is 12. 2 5 24 5 8 5 3 1 3 1 3 12 3 4 12 12 12 12 4 3 4 1 The difference in length is yard. 4
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ISM: Prealgebra and Introductory Algebra 2x
77.
2x
5 7 10
7 5 10
Chapter 4: Fractions and Mixed Numbers 1
x 81.
yz
2x 10 5 7 2x 10 57 2 x 5 2 5 7 4x 7
78.
15 20 24 15 4
4 10
20 2 20 1 5
2
20 43 20 107 8 10 15 14 2 1 2
5 1 12 y 56 14 80. 6 4 1 12 y 1
12 y
12 y
1 10 y 3y 1 7y 1 7 y
84.
12 y 5 12 y 1 6
x y 12 23 4 z 5 1 30 2 2 3 4 30 5 1 30 30 2 2 3 6 4 15 20 24 5 24 5 1 4 83. 5 2 4 5 2 4 8 13 2 5 13 1 5 131 5 13
82.
20 3 7
4 3 5
3y 7 3y 11 3y 7 3y 7 3y 11 7 7 7 11 7 11 11 7 20 2 1 21 5 2 5 2 3 7 4 10
2
2
30 3 5 15 30 2 30 4
79.
2
23 45 30 1
4
85.
2
1 2 1 27 3 27 9 2 1 3 27 9 3 2 3 27 27 1 27 2
9 1 2 1 9 1 2 1 10 3 5 11 10 3 5 11 3 2 10 55 311 2 2 10 11 55 2 33 4 110 110 29 110
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
86. 2 1 3 2 1 2 3 7 5 10 7 5 2 10 2 2 3 72 510 10 7 10 25 7 25 1 7
87.
7
3
8 5 9 6 1 3 12
3 hamburger patty.
9 24 20 9 24 2 3 24 31 7 7 19 19 1 20 24 24 24 7
89. 1
66 55 15 5 55 51 2 55 7
5
rounds to 2. 8 1 3 5 rounds to 3.
91. 34115 1 341 31 341 2 11 31 2 22 2 1 2 1 31 1 31 We would expect 22 miles on one gallon.
154
3 3 13 3 13 3 2 13 6 7 10 5 10 5 10 5 2 10 10 10 7 The unknown measurement is yard. 10
94. 1
1 3 2 3 5 12 15 12 15 27 7 14 14 14 14
7 7 7 7 96. 23 24 24 23 8 10 8 10 7 28 68 24 24 23 10 40 40 7 35 35 23 23 23 8 40 40 33 40 7 7 33 23 24 8 10 40 1 7 16 27 97. 3 2 5 10 5 10 16 10 5 27 16 5 2 5 27 32 27 5 1 27
An estimate is 2 3 = 6. 5 1 13 16 13 16 138 2 26 1 1 3 5 8 5 8 5 8 5 5 5 85 3 2 27 9 27 7 9 3 7 21 1 90. 6 1 5 49 4 7 4 7 4 9 4 4
3
7 3 4 1 93. 18 10 8 8 8 8 8 2 1 17 2 17 1 17 1 8 2 4 2 2 1 2 2 4 4 1 Each measurement is 4 inches. 4
95.
1 88. 8 5 rounds to 8. 3 5 rounds to 5. 11 An estimate is 8 5 = 3. 1 11 8 8 5 55 3 15 5 5 55 11
92. 7 1 5 22 5 22 5 110 36 2 3 3 1 31 3 3 110 2 There are or 36 grams of fat in a 5-ounce
1 3 9 7 63 15 98. 2 1 3 4 4 4 4 16 16
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ISM: Prealgebra and Introductory Algebra
2 1 a 2 32 61 2 a 3 3 6 3
99.
Chapter 4: Fractions and Mixed Numbers
104.
1 2 2 6 3 2 1 4 a 6 6 5 a 6 a
7 8x 5 1 10 7 x 5 10 1 1 7 1 x 5 5 10 5 7 2 x 10 10 9 x 10
100. 9x
1
3 x6 5 5 5 3 x 6 3 5 3 x 10
101.
2
102.
y
105.
y 11 2 5 5 y 11 5 2 5 5 5 y 5 5 2 11 5 y 10 11 y 10 10 1110 y 1 1
x 17 1 6 x4 12 17 12 12 12 6 4 1 x 12 12 17 6 4 2 3x 17 2 3x 2 17 2 3x 15 3x 15 3 3 x5
106.
4
9 3 9 2 9 4 y 2 9 2 3 9 4 y 23 3 3 2 2 y 6 23 y 1 y 6 107.
x 5 x 1 5 4 2 20 x 5 x 1 20 20 5 4 x 5 x 1 20 2 20 20 20 20 5 4 2 20 4x 25 10x 1 4x 25 10x 10x 110x 25 6x 1 25 6x 25 1 25 6x 24 6x 24 6 6 x 4 6 5 65 3 2 5 1 15 8 15 8 5 3 2 4 4
x 103.
3
6
7 7 x 6 7 3 7 7 7 x 7 7 3 6 7 x 21 6 x 21 21 6 21 x 15
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 5x2 108.
10x3
5x2
y3
y
y
y3
115. 12
10x3
5x2 y3
9
1
rounds to 12.
7 3
rounds to 10. 5 An estimate is 12 10 = 2. 1 5 12 12 7 35 3 21 9 9 5 35
y 10x3 5 x x y y y y 5 2 x x x y2 2x
3 1 3 1 2 1 2 1 10 10 10 10 5 2 5 7 2 72 72 7 110. 8x 3 8x 3 24 x3 12x 109.
2x 111.
x
5 5 2 2
112. 5 11 55
11 5
55
8 2 8 3
4 1 8 1 8 8
16 6 8 1 22 1 or 3 7 7
117. 3 2 4 3 2 3 4 8 3 9 8 3 3 9 3 6 4 8 9 9 3 2 8 9 3 2 89 3 2 2 4 3 3 1 12
25 55
2
55
rounds to 3.
8
1 9 rounds to 10. 2 An estimate is 3 + 10 = 13. 7 7 2 2 8 8 1 4 9 9 2 8 11 3 3 11 111 12 8 8 8
156
4 81 1 8
8 2 3
3 1 8 1 1 113. 5 4 5 4 8 4 5 1 8 4 5 1 32 2 or 6 5 5 7
4 1 1 8
2x 4 x 3 3 4 3 4 43 8x 3x 12 12 8x 3x 12 11x 12
114. 2
2 3
116.
23
55
118. 11x 2 10x 13 7 14 2 13 x 7 14 2 2 13 2 x 7 7 14 7 13 2 2 x 14 7 2 13 4 x 14 14 9 x 14
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11 9
40 35 21
35 19 2 35
ISM: Prealgebra and Introductory Algebra
119.
120.
Chapter 4: Fractions and Mixed Numbers
3 4 x 5 15 5 3 5 4 x 3 5 3 15 5 4 x 315 5 4 x 3 5 3 4 x 9
5. 4
2 23 16 7 23 7 2 ; B 8 8
8 2 8 2 16 , the operation is 11 11 1111 121 multiplication; C.
7. Since
2 8 11 8 11 8 2 4 4, the 11 11 11 2 11 2 2 2 operation is division; D.
8. Since
4 1 81 11 8111 81 1 122. 7 11 5 2 11 2 11 2 2 or 40 2 81 1 or 40 square feet. The area is 2 2 Chapter 4 Getting Ready for the Test
2.
2 2 2
3.
2
1 2
1; B
2
8
2 82 6 , the operation is 11 11 11 11 subtraction; B.
9. Since
8
2 8 2 10 , the operation is 11 11 11 11 addition; A.
10. Since
11.
5
7 12.
13.
1
8
5 1
7
7
6
; F 7
5 1 5 1 5 ; H 7 7 7 7 49 5
7 5 14.
1; A
3 5 4 3 20 3 23 ; C 5 5 5 5
6. 8
x 5 3 12 6 4 3 x 5 12 12 12 6 4 x 5 12 12 9 12 6 x 10 9 x 10 10 9 10 x 19
2
0; C
2
1 50 11 121. 50 5 2 1 2 50 2 11 1 2 2 100 11 2 2 89 2 1 44 2 1 The length of the remaining piece is 44 yards. 2
1.
0
4.
7
1
7
1
5 7 57 5 5; B 7 1 7 1 1 5 1
7
4 ; D 7 7
15. Since multiplication and division are performed before addition and subtraction, the first operation that should be performed to simplify 1 1 1 1 1 is the multiplication of and ; C. 2 5 4 5 4
is undefined; D
0
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
16. Since operations in parentheses (grouping symbols) are performed first, the first operation 1 1 1 that should be performed to simplify 2 5 4 1 1 is the addition of and ; A. 2 5
18 3. 4 75 4 35 32 3 75 3 18 4 4
17. Since multiplication and division are performed in order from left to right, the first operation that 1 1 1 should be performed to simplify is the 2 5 4 1 1 division of by ; D. 2 5
4.
18. Since operations in parentheses (grouping symbols) are performed first, the first 1operation 1 1 that should be performed to simplify 2 5 4 1 1 is the multiplication of and ; C. 5 4
6. Check the cross products. 5 11 = 55 7 8 = 56 Since 55 56, the fractions are not equivalent.
x
19.
3
2 10 5 x 3 10 10 5 10 2 x 3 10 10 10 2 10 5 x 6 20 The correct equivalent equation is choice D.
20. Since
2
1
1
does not contain an equal sign, it is 3 9 an expression; A.
21. Since
2
3 9 equation; B.
contains an equal sign, it is an
Chapter 4 Test 1. 7 of the 16 equal parts are shaded:
2. 7
5. 42x 3 14 x 3x 70 5 14 5
7. Check the cross products. 6 63 = 378 27 14 = 378 Since 378 = 378, the fractions are equivalent. 2
8. 84 2 42 2 2 21 2 2 3 7 2 3 7 9. 495 3 165 3 3 55 3 3 5 11 32 5 11 10.
2 3 7 2 21 2 23 3 3 3 3
7
. 16
12.
7x x 7x x 8x 9 9 9 9
13. The LCD is 7x. 1 3 1 x 3 7 x 21 x 21 7 x 7 x x 7 7x 7 x 7x 3 3 14. xy z xy z x y y y z y y y2 z xy z xy x y z 1
2 8 2 8 16 15. 3 15 315 45 16.
158
4 3 4 4 44 4 4 4 4 3 43 3
11. 4 4 4 4 4 3 4 3 4 3
x 9
24 64 4 210 6 7 5 35
9a 2 9a 2 2 9a 4 9a 4 10 5 10 5 2 10 10 10
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers
2 17. 8 2 8 2 10 2 5 15 y 15 y 15 y 15 y 3 5 y 3y 18. 3a 16 3a 16 3 a 8 2 1 1 8 6a3 8 6a3 8 2 3 a a a a a a2
19.
11 3 5 11 2 3 3 5 12 8 24 12 2 8 3 24 22 9 5 24 24 24 22 9 5 24 18 24
3 6 46
20.
7
3
35 40 16 7 40 30 2 40 81 1 1 12 12 2 14 40 40 40 3
8 2 7 5 3 2 4
21.
19 2
11 11 3 2 11 8 16 18
3 11
2 5 3 9 3 2 2 59 2 45 1 or 22 2 2 2 1 2 6 1 24. 6 7 6 7 1 6 2 6 6 1 7 1 6 6 2 36 1 6 6 7 2 35 7 6 2 35 76 275 7 23 5 2 or 1 3 3
3 4
23. 3 1 6 3 10 27 3 4 3 4 10 27 3 4
25.
1 2 3 1 3 3 1 3 3 9 2 3 4 2 2 4 2 2 4 16 2
4 6 6 2 3 9 4 6 9 9 16 6 9 6 16 9 9 23 2 8 9 3 8
11 16 22.
3
3
2
3 2 5 3 2 2 5 26. 4 3 6 4 2 3 2 6 3 4 5
16
12 12 3 3 16 12 3 3 16 3 4 3 3 64 1 or 21 3 3
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ISM: Prealgebra and Introductory Algebra
Chapter 4: Fractions and Mixed Numbers 5 4 7 52 44 7 3 27. 3 6 3 12 6 10 16 7 3 2 3 4 12 12 12 12 33 3 12 1 33 1 12 3 331 12 3 11 31 12 3 11 12 28.
5x 7
2
20x2 21
5x 7
20x 21
5x 21 7 20x2 5 x 3 7 7 45 x x 3 4x
14 5 14 3 70 6 76 29. 2 71 14 2 17 14 2 14 17 28 7 21 2 2 2 5 3
160
x 24 x 5 5 x 24 5 x 5 5 5 x 5 5 x 24 5 x 5x 24 6x 24 6x 24 6 6 x 4
3 3 x 8 4 8 3 8 3 x 3 8 3 4 83 x 3 4 423 x 3 4 x 2
2 32.
x
5
x
3 4 12 2 2 x 5 x 12 3 4 12 12 2 2 x 5 x 12 12 12 12 3 4 12 2 8 3x 5 6x 8 3x 6x 5 6x 6x 8 3x 5 8 8 3x 5 8 3x 3 3x 3 3 3 x 1
14 5 3
13 or 3 21 30.
31.
1 1 5 1 5 1 5 2 33. 5x 5 2 2 1 2 1 2 2 34. x y
1
3
7
2 8 1 31 2 8 1 8 2 31 18 2 31 1 2 4 2 31 4 31
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
35.
6 2
1
6
2 3
2
4
2
5
4 3
2
4
The remaining piece is 3
39. perimeter 1
4 3
3 feet.
17 50
Food:
3
25 17 3 17 3 2 17 6 23 50 25 50 25 2 50 50 50 23 of spending goes for housing and food 50 combined. 37. Education:
50
3 258 43 258 4 43 6 4 24 4 1 4 1 43 1 43 Expect to travel 24 miles on 1 gallon of gas.
40. 258 10
2. 115 in words is one hundred fifteen.
3
100 1 4 3 1 2 44 3 50 25 100 50 2 25 4 100 2 16 3 100 100 100 21 100 21 of spending goes for education, 100 transportation, and clothing. 38.
3. 27,034 in words is twenty-seven thousand, thirty-four. 4. 6573 in words is six thousand, five hundred seventy-three. 5.
46 713 759
6.
587 44 631
7.
543 29 514 Check:
2
47, 000 2 47, 000 25 1 2 25 1880 25 1 2 1880 1 3760 Expect $3760 was spent on health care. 25
2
1. 546 in words is five hundred forty-six.
4 25
Clothing:
1
Cumulative Review Chapters 14
1
Transportation:
2
3 3 3 2 3 2 3 3 3 3 10 3 1 3 3 2 2 area length width 1 3 3 1 The perimeter is 3 feet and the area is 3 2 square foot. 3
4 3 3 4 4
36. Housing:
6
Chapter 4: Fractions and Mixed Numbers
8.
514 29 543
995 62 933
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Chapter 4: Fractions and Mixed Numbers
Check:
933 62 995
15. 7 7 7 73 16. 7 7 72
9. To round 278,362 to the nearest thousand, observe that the digit in the hundreds place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. 278,362 rounded to the nearest thousand is 278,000. 10. To round 1436 to the nearest ten, observe that the digit in the ones place is 6. Since this digit is at least 5, we add 1 to the digit in the tens place. 1436 rounded to the nearest ten is 1440. 11.
12.
96 12 192 960 1152 Twelve boxes of single-serve coffee maker pods 22. hold 1152 pods. 435 3 1305 The family travels 1305 miles in 3 days.
7089 13. 8 56, 717 56 07 0 71 64 77 72 5
17. 3 3 3 3 9 9 9 34 93 18. 9 9 9 9 5 5 94 52 19. 2(x y) = 2(6 3) = 2(3) = 6 20. 8a 3(b 5) 8 5 3(9 5) 8 5 3 4 40 12 52 21. Let 0 represent the surface of Earth. Then 7257 feet below the surface is represented as 7257. Let 0 represent a temperature of 0F. Then 21F below zero is represented as 21. 23. 7 + 3 = 4 24. 3 + 8 = 5 25. 7 8 (5) 1 7 8 5 1 7 (8) 5 (1) 1 5 (1) 4 (1) 3 26. 6 (8) (9) 3 6 (8) 9 3 2 9 3 73 10 27. (5)2 (5)(5) 25
56,717 8 = 7089 R 5 Check: 7089 8 + 5 = 56,712 + 5 = 56,717
28. 24 2 2 2 2 16 29. 3(4 7) (2) 5 3(3) (2) 5 9 (2) (5) 11 (5) 16
379 14. 12 4558 36 95 84 118 108 10
2 2 2 2 30. (20 5 ) (20 25) (5) 25
31. 2 y 6 4 y 8 (2 y 4 y) (6 8) 6 y 2
4558 12 = 379 R 10 Check: 379 12 + 10 = 4548 + 10 = 4558 162
ISM: Prealgebra and Introductory Algebra
32. 5x 1 + x + 10 = (5x + x) + (1 + 10) = 6x + 9
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ISM: Prealgebra and Introductory Algebra 33. 5x 2 4x 7 19 x 2 12 x 2 2 12 2 x 14
40.
41. 42x 6 7 x 7x 66 6 11 11
17 7x 3 3x 21 3x 20 7x 21 6x 20 7x 7x 21 6x 7x 20 21 x 20 21 21 21 x 1 x or x 1
36.
7 5 39 35 4 39 4 7 5 5
34. 9 y 1 8 y 3 20 y 1 17 y 11 17 1 y 18 35.
Chapter 4: Fractions and Mixed Numbers
70
42.
105 y
35 2 2 35 3 y 3y
1 7 10 7 43. 3 3 8 3 8 10 7 38 25 7 3 2 4 57 3 4 35 11 or 2 12 12
9x 2 7x 24 9x 7x 2 7x 7x 24 2x 2 24 2x 2 2 24 2 2x 22 2x 22 2 2 x 11
37. Two of five equal parts are shaded:
2
44.
2 3
4
2 4 24 8 2 or 2 3 1 3 1 3 3
5 38. 156 2 78 2 2 39 2 2 3 13
45.
5 3 5 4 54 5 4 5 5 16 4 16 3 16 3 4 4 3 4 3 12
46. 1
1 3 11 28 11 5 11 5 11 5 10 5 10 5 10 28 5 2 28 56
156 22 3 13
39. a.
4
b.
1
2 9 4 2 36 2 38 9 9 9 9
8 111 8 11 8 19 11 11 11 11
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163
Chapter 5 Section 5.1 Practice Exercises 1. a.
11. 209.986 209
1000 2 493 209 2 500 493 209 500
0.06 in words is six hundredths.
b.
200.073 in words is negative two hundred and seventy-three thousandths.
c.
0.0829 in words in eight hundred twentynine ten-thousandths.
2. 87.31 in words is eighty-seven and thirty-one hundredths. 3. 52.1085 in words is fifty-two and one thousand eighty-five ten-thousandths. 4. The check should be paid to “CLECO,” for the amount of “207.40,” which is written in words as 40 “Two hundred seven and .” 100
5. Five hundred and ninety-six hundredths is 500.96. 6. Thirty-nine and forty-two thousandths is 39.042. 7. 0.051
1000 97 100
9. 0.12
12 3 4 3 100 25 4 25
10. 64.8 64
8 10
64
12. 29.208 26.28 0 < 8 so 26.208 < 26.28 13. 0.12 0.026 1 > 0 so 0.12 > 0.026 14. 0.039 0.0309 9 > 0 so 0.039 > 0.0309 Thus, 0.039 < 0.0309. 15. To round 482.7817 to the nearest thousandth, observe that the digit in the ten-thousandths place is 7. Since this digit is at least 5, we add 1 to the digit in the thousandths place. The number 482.7817 rounded to the nearest thousandth is 482.782. 16. To round 0.032 to the nearest hundredth, observe that the digit in the thousandths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the hundredths place. The number 0.032 rounded to the nearest hundredth is 0.03.
51
8. 29.97 29
986
24 25
64
4 5
17. To round 3.14159265 to the nearest tenthousandth, observe that the digit in the hundredthousandths place is 9. Since this digit is at least 5, we add 1 to the digit in the ten-thousandths place. The number 3.14159265 rounded to the nearest the ten-thousandth is 3.1416, or 3.1416. 18. $24.62 rounded to the nearest dollar is $25, since 6 5.
164
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
Vocabulary, Readiness & Video Check 5.1 1. The number “twenty and eight hundredths” is written in words and “20.08” is written in standard form. 2. Another name for the distance around a circle is its circumference. 3. Like fractions, decimals are used to denote part of a whole. 4. When writing a decimal number in words, the decimal point is written as and. 5. The place value tenths is to the right of the decimal point while tens is to the left of the decimal point. 6. The decimal point in a whole number is after the last digit. 7. as “and” 8. 9.8 is nine and eight tenthsthe 8 should be in the hundredths place; 9.08 9. Reading a decimal correctly gives you the correct place value, which tells you the denominator of your equivalent fraction. 10. left to right 11. When rounding, we look to the digit to the right of the place value we’re rounding to. In this case we look to the hundredths-place digit, which is 7. Exercise Set 5.1 2. 9.57 in words is nine and fifty-seven hundredths. 4. 47.65 in words is forty-seven and sixty-five hundredths. 6. 0.495 in words is negative four hundred ninety-five thousandths. 8. 233.056 in words is two hundred thirty-three and fifty-six thousandths. 10. 5000.02 in words is five thousand and two hundredths. 12. 410.3 in words is four hundred ten and three tenths. 14. 31.04 in words is thirty-one and four hundredths. 16. The check should be paid to “Amani Dupre,” for the amount “513.70,” which is written in words as “Five hundred 70 thirteen and .” 100
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
18. The check should be paid to “Kroger,” for the amount of “387.49,” which is written in words as “Three hundred 49 eighty-seven and .” 100
20. Five and one tenth is 5.1. 22. Twelve and six hundredths is 12.06. 24. Negative eight hundred four and three hundred ninety-nine thousandths is 804.399. 26. Eighty-three ten-thousandths is 0.0083. 28. 0.9
9 10
30. 0.39 32. 0.8
39 100
8 10
34. 5.4 5
24 4 25 5
4 10
5
36. 0.024
22 2 5 25 5
24 1000 5
38. 9.005 9
1000 40. 11.406 11
9
406 1000
38 3 125 8 125
1 5 9 1 200 5 200
11
2 203 203 11 2 500 500
42. 0.2006 2006 2 1003 1003 10, 000 2 5000 5000 44. Five tenths is 0.5 and as a fraction is
5
1 . 10 2
46. In words, 0.019 is nineteen thousandths. As a fraction, 0.019
166
19 . 1000
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ISM: Prealgebra and Introductory Algebra
68. To round 0.892 to the nearest hundredth, observe that the digit in the thousandths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the hundredths place. The number 0.892 rounded to the nearest hundredth is 0.89.
48. 0.12 0.15 2 < 5 so 0.12 < 0.15 50. 0.59 0.52 9 > 2 so 0.59 > 0.52 Thus, 0.59 < 0.52.
70. To round 63.4523 to the nearest thousandth, observe that the digit in the ten-thousandths place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousandths place. The number 63.4523 rounded to the nearest thousandth is 63.452.
52. 0.0756 0.2 0 < 2 so 0.0756 < 0.2
72. To round 3.14159265 to the nearest one, observe that the digit in the tenths place is 1. Since this digit is less than 5, we do not add 1 to the digit in the ones place. The number 3.14159265 rounded to the nearest ones is 3.
54. 0.98400 0.984 4 = 4 so 0.98400 = 0.984
74. To round 3.14159265 to the nearest hundredthousandth, observe that the digit in the millionths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the hundredthousandths place. The number 3.14159265 rounded to the nearest hundred-thousandth is 3.14159.
56. 519.3405 519.3054 4 > 0 so 519.3405 > 519.3054 58. 18.1 18.01 1 > 0 so 18.1 > 18.01 Thus, 18.1 < 18.01. 60.
Chapter 5: Decimals
76. To round 14,769.52 to the nearest one, observe that the digit in the tenths place is 5. Since this digit is at least 5, we add 1 to the digit in the ones place. The number 14,769.52 rounded to the nearest one is 14,770. The amount is $14,770.
0.01 0.1 + > so 0.01 > 0.1
78. To round 0.7633 to the nearest hundredth, observe that the digit in the thousandths place is 3. Since this digit is less than 5, we do not add 1 to the digit in the hundredths place. The number 0.7633 rounded to the nearest hundredth is 0.76. The amount is $0.76.
62. 0.562 0.652 5 < 6 so 0.562 < 0.652 Thus, 0.562 > 0.652. 64. To round 0.64 to the nearest tenth, observe that the digit in the hundredths place is 4. Since this digit is less than 5, we do not add 1 to the digit in the tenths place. The number 0.64 rounded to the nearest tenth is 0.6.
80. To round 0.112 to the nearest hundredth, observe that the digit in the thousandths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the hundredths place. The number 0.112 rounded to the nearest hundredth is 0.11. Norway won 0.11 of the medals awarded.
66. To round 68,934.543 to the nearest ten, observe that the digit in the ones place is 4. Since this digit is less than 5, we do not add 1 to the digit in the tens place. The number 68,934.543 rounded to the nearest ten is 68,930.
82. To round 3.36 to the nearest tenth, observe that the digit in the hundredths place is 6. Since this digit is at least 5, we add 1 to the digit in the tenths place. The number 3.36 rounded to the nearest tenth is 3.4. The speed is 3.4 mph.
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
84. To round 89.61 to the nearest one, observe that the digit in the tenths place is 6. Since this digit is at least 5, we add 1 to the digit in the ones place. The number 89.61 rounded to the nearest one is 90. The amount is $90 billion.
110. a. b.
90.
8 945 4 536 13, 481
c.
Moving the decimal point in 100,800. two places to the left gives 1008. The approximate number of games is 1008 thousand.
Section 5.2 Practice Exercises
4002 3897 105
1. a.
19.520 5.371 24.891
b.
40.080 17.612 57.692
c.
0.125 422.800 422.925
2. a.
34.5670 129.4300 2.8903 166.8873
b.
11.210 46.013 362.526 419.749
92. To round 146.059 to the nearest ten, observe that the digit in the ones place is 6. Since this digit is at least 5, we add 1 to the digit in the tens place. The number 146.059 rounded to the nearest ten is 150, which is choice d. 94. To round 146.059 to the nearest tenth, observe that the digit in the hundredths place is 5. Since this digit is at least 5, we add 1 to the digit in the tenths place. The number 146.059 rounded to the nearest tenth is 146.1, which is choice b. 96. answers may vary 98. 17 268 17.268 1000 100. 0.00026849576 in words is twenty-six million, eight hundred forty-nine thousand, five hundred seventy-six hundred-billionths. 3. 102. answers may vary 104. answers may vary 106. 0.0612 and 0.0586 rounded to the nearest hundredth are 0.06. 0.066 rounds to 0.07. 0.0506 rounds to 0.05. 108. From smallest to largest, 0.01, 0.0839, 0.09, 0.1
168
4800 21 4 800 96 000 100,800 4800 thousand 21 = 100,800 thousand
86. To round 24.6229 to the nearest thousandth, observe that the digit in the ten-thousandths place is 9. Since this digit is at least 5, we add 1 to the digit in the thousandths place. The number 24.6229 rounded to the nearest thousandth is 24.623. The day length is 24.623 hours. 88.
4820 thousand rounded to the nearest hundred thousand is 4800 thousand.
19.000 26.072 45.072
4. 7.12 + (9.92) Subtract the absolute values. 9.92 7.12 2.80 Attach the sign of the larger absolute value. 7.12 + (9.92) = 2.8
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
5. a.
6.70 3.92 2.78
Check:
2.78 3.92 6.70
13. 4.3y 7.8 20.1y 14.6 4.3y 20.1y 7.8 14.6 (4.3 20.1) y (7.8 14.6) 24.4 y 22.4
b.
9.720 4.068 5.652
Check:
5.652 4.068 9.720
14.
6. a.
73.00 29.31 43.69
Check:
43.69 29.31 73.00
721.38 148.29 152.50 1022.17 The total monthly cost is $1022.17.
b.
210.00 68.22 141.78
Check:
141.78 68.22 210.00
15.
71.3 70.8 0.5 The average height in Germany is 0.5 inch greater than the average height in Luxembourg.
7.
25.91 19.00 6.91
Calculator Explorations 1. 315.782 + 12.96 = 328.742
8. 5.4 9.6 = 5.4 + (9.6) Add the absolute values. 5.4 9.6 15.0 Attach the common sign. 5.4 9.6 = 15
2. 29.68 + 85.902 = 115.582 3. 6.249 1.0076 = 5.2414 4. 5.238 0.682 = 4.556
9. 1.05 (7.23) = 1.05 + 7.23 Subtract the absolute values. 7.23 1.05 6.18 Attach the sign of the larger absolute value. 1.05 (7.23) = 6.18 10. a.
Exact 58.10 326.97 385.07
Estimate 1 60 300 360
Estimate 2 60 330 390
b. Exact 16.080 0.925 15.155
Estimate 1 16 1 15
Estimate 2 20 1 19
11. y z = 11.6 10.8 = 0.8 12.
y 4.3 7.8 12.1 4.3 0 7.8 7.8 7.8 True Yes, 12.1 is a solution.
5.
12.555 224.987 5.2 622.65 865.392
6.
47.006 0.17 313.259 139.088 499.523
Vocabulary, Readiness & Video Check 5.2 1. The decimal point in a whole number is positioned after the last digit. 2. In 89.2 14.9 = 74.3, the number 74.3 is called the difference, 89.2 is the minuend, and 14.9 is the subtrahend. 3. To simplify an expression, we combine any like terms. 4. To add or subtract decimals, we line up the decimal points vertically.
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
5. True or false: If we replace x with 11.2 and y with 8.6 in the expression x y, we have 11.2 8.6. false
12. Exact:
734.89 640.56 1375.45 Estimate: 730 640 1370
6. True or false: If we replace x with 9.8 and y with 3.7 in the expression x + y we have 9.8 + 3.7. false 7. Lining up the decimal points also lines up place values, so we only add or subtract digits in the same place values.
14. Exact:
200.89 7.49 62.8 271.18 Estimate: 200 7 60 267
8. check subtraction by addition 9. so the subtraction can be written vertically with decimal points lined up 10. Two: There are 2 x-terms and there are 2 constants.
16.
65.0000 5.0903 6.9003 76.9906
18.
Check:
3.6 4.1 7.7
8.9 3.1 5.8
5.8 3.1 8.9
20.
Check:
5.17 3.70 8.87
28.0 3.3 24.7
24.7 3.3 28.0
22.
863.230 39.453 823.777
Check:
823.777 39.453 863.230
11. To calculate the amount of border material needed, we are actually calculating the perimeter of the triangle. Exercise Set 5.2 2.
4.
6.
32.4000 1.5800 0.0934 34.0734
24. Exact: 6.4 3.04 = 3.36 Estimate: 6 3 = 3 Check: 3.36 + 3.04 = 6.40
8. 18.2 + (10.8) Add the absolute values. 18.2 10.8 29.0 Attach the common sign. 18.2 + (10.8) = 29 10. 4.38 + (6.05) Subtract the absolute values. 6.05 4.38 1.67 Attach the sign of the larger absolute value. 4.38 + (6.05) = 1.67 170
26. Exact:
2000.00 327.47 1672.53 Estimate: 2000 300 1700
Check:
28.
800.0 8.9 791.1
Check: 791.1 8.9 800.0
Copyright © 2025 Pearson Education, Inc.
1672.53 327.47 2000.00
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
30. 8.63 5.6 = 8.63 + (5.6) Add the absolute values. 8.63 Check: 14.23 5.60 5.60 14.23 8.63 Attach the common sign. 8.63 5.6 = 14.23
46. 40.3 700 = 40.3 + (700) Subtract the absolute values. 700.0 40.3 659.7 Attach the sign of the larger absolute value. 40.3 700 = 659.7
32. 8.53 17.84 = 8.53 + (17.84) Subtract the absolute values. 17.84 Check: 9.31 8.53 8.53 9.31 17.84 Attach the sign of the larger absolute value. 8.53 17.84 = 9.31
48.
845.9 45.8 800.1
50.
100.007 2.080 5.014 107.101
34. 9.4 (10.4) = 9.4 + 10.4 Subtract the absolute values. 10.4 Check: 1.0 9.4 9.4 10.4 1.0 Attach the sign of the larger absolute value. 9.4 (10.4) = 1 36.
38.
40.
7.000 0.097 6.903
Check:
6.903 0.097 7.000
45.0 9.2 35.8
Check:
35.8 9.2 45.0
0.7 3.4 4.1
42. 5.05 + 0.88 Subtract the absolute values. 5.05 0.88 4.17 Attach the sign of the larger absolute value. 5.05 + 0.88 = 4.17 44.
600.47 254.68
52. 0.004 + 0.085 Subtract the absolute values. 0.085 0.004 0.081 Attach the sign of the larger absolute value. 0.004 + 0.085 = 0.081 54. 36.2 10.02 = 36.2 + (10.02) Add the absolute values. 36.20 10.02 46.22 Attach the common sign. 36.2 10.02 = 46.22 56. 6.5 (3.3) = 6.5 + 3.3 Subtract the absolute values. 6.5 3.3 3.2 Attach the sign of the larger absolute value. 6.5 (3.3) = 3.2 58. y + x = 5 + 3.6 = 8.6 60. y z = 5 0.21 = 4.79 62. x + y + z = 3.6 + 5 + 0.21 = 8.6 + 0.21 = 8.81
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345.79
64.
x 5.9 8.6
171
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
66.
45.9 z 23 45.9 (22.9) 0 23 23 23 True Yes, 22.9 is a solution.
68.
1.9 x x 0.1 1.9 0.9 0 0.9 0.1 1 1 True Yes, 0.9 is a solution.
86. To find the late fee, add. 23.15 6.85 30.00 The average late fee was $30.00. 88. To find the amount of rainfall in Yuma, add 3.006 to the amount in the Atacama Desert. 3.006 0.004 3.010 Yuma receives an average of 3.01 inches of rainfall each year.
70. 14.2z 11.9 9.6z 15.2 14.2z 9.6z 11.9 15.2 (14.2 9.6)z (11.9 15.2) 4.6z 3.3 72. 8.96x 2.31 4.08x 9.68 8.96x 4.08x 2.31 9.68 (8.96 4.08)x (2.31 9.68) 13.04x 7.37 74. Change = 20 18.26 20.00 18.26 1.74 If Bai paid with two $10 bills, his change was $1.74. 76. Subtract the cost of the frames from the cost of the glasses. 347.89 97.23 250.66 The lenses cost $250.66. 78. Perimeter = 4.2 + 5.78 + 7.8 = 17.78 The perimeter is 17.78 inches. 80. Perimeter 6.43 3.07 6.43 3.07 19.0 The perimeter is 19.0 inches. 82. The phrase “how much more” indicates that we should subtract the average rainfall in Omaha from that in New Orleans. 64.16 30.08 34.08 Therefore, on average New Orleans receives 34.08 inches more rain than Omaha. 84. To find the increase, subtract. 6.97 6.77 0.20 The increase was 0.20 hour per day. 172
90. Add the lengths of the sides to get the perimeter. 15.7 10.6 15.7 10.6 52.6 Therefore, 52.6 feet of railing was purchased. 92.
145.290 144.416 0.874 The difference is 0.874 mile per hour.
94. The shortest bar indicates the least consumption per person, so Sweden has the least chocolate consumption per person. 96.
17.8 17.4 0.4 The difference in consumption is 0.4 pound per year.
98. 23 2 = 46 100. 39 3 = 117 3
1 1 1 1 111 1 102. 5 5 5 5 5 5 125 5 104. It is incorrect. After borrowing, there should be 10 tenths. 8 9 910
900.0 96.4 803.6 106. 17.67 (5.26 + 7.82) = 17.67 13.08 = 4.59 The unknown length is 4.59 meters.
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ISM: Prealgebra and Introductory Algebra
108. 2 quarters, 3 dimes, 4 nickels, and 2 pennies: 0.25 + 0.25 + 0.10 + 0.10 + 0.10 + 0.05 + 0.05 + 0.05 + 0.05 + 0.01 + 0.01 = 1.02 The value of the coins shown is $1.02. 110. 5 nickels: 0.05 + 0.05 + 0.05 + 0.05 + 0.05 = 0.25 1 dime and 3 nickels: 0.10 + 0.05 + 0.05 + 0.05 = 0.25 2 dimes and 1 nickel: 0.10 + 0.10 + 0.05 = 0.25 1 quarter: 0.25 112.
256, 436.012 256, 435.235 0.777 The difference in measurements is 0.777 mile.
Chapter 5: Decimals 6. 203.004 100 = 20,300.4 7. (2.33)(1000) = 2330 8. 6.94 0.1 = 0.694 9. 3.9 0.01 = 0.039 10. (7682)(0.001) = 7.682 11. 61.45 million 61.451 million 61.451, 000, 000 61, 450, 000 12. 7y = 7(0.028) = 0.196 13.
114. no; answers may vary
Yes, 5.5 is a solution.
116. 14.271 8.968x + 1.333 201.815x + 101.239x = 8.968x 201.815x + 101.239x + 14.271 + 1.333 = (8.968 201.815 + 101.239)x + (14.271 + 1.333) = 109.544x + 15.604 Section 5.3 Practice Exercises 1.
34.8 0.62 696 20 880 21.576
1 decimal place 2 decimal places
2. 0.0641 27 4487 1 2820 1.7307
4 decimal places 0 decimal places
14. C = 2r = 2 11 = 22 22(3.14) = 69.08 The circumference is 22 meters 69.08 meters. 15.
60.5 5.6 36 30 302 50 338.80 338.8 ounces of fertilizer are needed.
Vocabulary, Readiness & Video Check 5.3 1 2 3 decimal places
1. When multiplying decimals, the number of decimal places in the product is equal to the sum of the number of decimal place in the factors. 2. In 8.6 5 = 43, the number 43 is called the product while 8.6 and 5 are each called a factor.
4 0 4 decimal places
3. (7.3)(0.9) = 6.57 (Be sure to include the negative sign.) 4. Exact:
6x 33 6(5.5) 0 33 33 33 True
30.26 2.89 2 7234 24 2080 60 5200 87.4514
5. 46.8 10 = 468
Estimate:
30 3 90
3. When multiplying a decimal number by powers of 10 such as 10, 100, 1000, and so on, we move the decimal point in the number to the right the same number of places as there are zeros in the power of 10. 4. When multiplying a decimal number by powers of 10 such as 0.1, 0.01, and so on, we move the decimal point in the number to the left the same number of places as there are decimal places in the power of 10. 5. The distance around a circle is called its circumference.
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6. When multiplying decimals, we need to learn where to place the decimal point in the product.
14. Exact:
7. knowing whether we placed the decimal point correctly in our product 8. We just need to know how to move the decimal point. 100 has two zeros, so we move the decimal point two places to the right.
16.
9. 3(5.7) (0.2) 10. We used an approximation for . The exact answer is 10 cm. 11. This is an application problem and needs units attached. The complete answer is 24.8 grams.
4.
0.23 9 2.07
2 decimal places 0 decimal places 2 0 2 decimal places
6.8 0.3 2.04
1 decimal place 1 decimal place 11 2 decimal places
174
8.3 2.7 5 81 16 60 22.41
0.864 0.4 0.3456
1 decimal place 3 decimal places
1 3 4 decimal places
18. 7.2 100 = 720 20. 23.4 0.1 = 2.34
26. (4.72)(0.01) = 0.0472 28. 36.41 0.001 = 0.03641 30.
8. The product (7.84)(3.5) is positive. 7.84 2 decimal places 3.5 1 decimal place 3 920 23 520 27.440 2 1 3 decimal places
12.
2 6 12
24. 0.5 100 = 50
6. The product (4.7)(9.02) is negative. 9.02 2 decimal places 4.7 1 decimal place 6 314 36 080 42.394 2 1 3 decimal places and include the negative sign.
10. Exact:
300.9 0.032 6018 9 0270 9.6288
Estimate:
22. (1.123)(1000) = 1123
Exercise Set 5.3 2.
2.0005 5.5 1 00025 10 00250 11.00275
Estimate:
3 decimal places 1 decimal place 3 1 4 decimal places
8 3 24
0.216 0.3 0.0648
32. (345.2)(100) = 34,520 34.
0.42 5.7 294 2 100 2.394
36. (0.001)(562.01) = 0.56201 38. 993.5 0.001 = 0.9935 40.
9.21 3.8 7 368 27 630 34.998
42. 48.3 million 48.31 million 48.31, 000, 000 48, 300, 000 olds own at About 48,300,000 American househ least one dog.
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ISM: Prealgebra and Introductory Algebra 44. 343.8 billion 343.81 billion 343.81, 000, 000, 000 343,800, 000, 000 The restaurant industry had projected online delivery sales of $343,800,000,000 in 2022. 46. yz = (0.2)(5.7) = 1.14 48. 5y + z = 5(0.2) + 5.7 = 1.0 + 5.7 = 6.7 50.
52.
100z 14.14 100(1414) 0 14.14 141, 400 14.14 False No, 1414 is not a solution.
Chapter 5: Decimals 64. Circumference = 2 radius C = 2 3950 = 7900 7900 3.14 316 00 790 00 23700 00 24806.00 The circumference is 7900 miles, which is approximately 24,806 miles. 66. C = d C = 80 = 80 3.14 80 251.20 The circumference is 80 meters or approximately 251.2 meters.
0.7x 2.52 0.7(3.6) 0 2.52 2.52 2.52 True Yes, 3.6 is a solution.
54. C = d is (22 inches) = 22 inches C 22(3.14) inches = 69.08 inches 56. C = 2r is 2 5.9 kilometers = 11.8 kilometers C 11.8 (11.8)(3.14) kilometers 37.052 kilometers
68. Volume length width height 16 7.6 2 243.2 The volume of the box is 243.2 cubic inches. 70. a.
58. pay before taxes = 19.52 20 19.52 20 390.40 The worker’s pay for last week was $390.40. 60. Multiply the number of servings by the number of grams of saturated fat in 3.5 ounces. 3 0.1 = 0.3 There is 0.3 gram of saturated fat in 3 servings. 62. Area = length width 6.43 3.1 643 19 290 19.933 The area is 19.933 square inches.
Circumference = diameter Smaller circle: C = 16 = 16 C 16(3.14) = 50.24 The circumference of the smaller circle is approximately 50.24 inches. Larger circle: C = 32 = 32 C 32(3.14) = 100.48 The circumference of the larger circle is approximately 100.48 inches.
b. Yes, the circumference gets doubled when the diameter is doubled. 72. 6.7775 10,000 = 67,775 They would pay $67,775 for 10,000 bushels of corn. 74.
6.7918 300 2037.5400 You would receive 2037.54 Chinese yuan for 300 American dollars.
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Chapter 5: Decimals
76.
92. answers may vary
1.0722 130 32 1660 107 2200 139.3860 The tourist spent 139.39 euros.
Section 5.4 Practice Exercises 1. 8
78. Multiply the doorway height in meters by the number of inches in 1 meter. 39.37 2.15 1 9685 3 9370 78 7400 84.6455 The doorway is approximately 84.6455 inches.
2. 48
8 80. 365 2920 2920 0 162 82.
84.
9
3. a.
162
75 25 75 25 9 9 18 25 3 25 9 18 3 1 54 7.20 0.14 98.60 105.94
86.
100.0 48.6 51.4
88.
3.6 0.04
Check: 46.3 8 370.4
0.71 34.08 33 6 48 48 0
Check:
1.135 14 15.890 14 18 14 49 42 70 70 0
0.71 48 5 68 28 40 34.08
Check: 1.135 14 4 540 11 350 15.890
Thus, 15.89 14 = 1.135.
b.
0.027 104 2.808 2 08 728 728
Check: 0.027 104 108 2 700
0
2.808
Thus, 2.808 (104) = 0.027
0.144 90. 8.3 minutes = 8.3 60 seconds = 498 seconds 498 1.86 29 88 398 40 498 00 926.28 926.28 100,000 = 92,628,000 The radio wave travels 92,628,000 miles. 176
46.3 370.4 32 50 48 24 2 4 0
29.8 4. 5.6 166.88 becomes 56 1668.8 112 548 504 44 8 44 8 0
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ISM: Prealgebra and Introductory Algebra
12.35 5. 0.16 1.976 becomes 16 197.60 16 37 32 56 48 80 80 0 41.052 41.05 6. 0.57 23.4 becomes 57 2340.000 228 60 57 3 00 2 85 150 114 36 7.9 7. 91.5 722.85 becomes 915 7228.5 6405 823 5 823 5 0 8 Estimate: 90 720 8.
362.1
0.49
0.3621
x
3.9
100 39
0 3.9 100 0.39 3.9 False No, 39 is not a solution. 11.84 12. 1250 14800.00 1250 2300 1250 10500 10000 5000 5000 0 He needs 11.84 bags or 12 whole bags. Calculator Explorations 1. 102.62 41.8 100 40 = 4000 Since 4000 is not close to 428.9516, it is not reasonable. 2. 174.835 47.9 200 50 = 4 Since 4 is close to 3.65, it is reasonable. 3. 1025.68 125.42 1000 100 = 900 Since 900 is close to 900.26, it is reasonable.
Vocabulary, Readiness & Video Check 5.4 1. In 6.5 5 = 1.3, the number 1.3 is called the quotient, 5 is the divisor, and 6.5 is the dividend.
0.049
10 10. x y = 0.035 0.02 0.02 0.035 becomes 2
11.
4. 562.781 + 2.96 563 + 3 = 566 Since 566 is not close to 858.781, it is not reasonable.
1000 9.
Chapter 5: Decimals
1.75 3.50 2 15 14 10 10 0
2. To check a division exercise, we can perform the following multiplication: quotient divisor = dividend. 3. To divide a decimal number by a power of 10 such as 10, 100, 1000, or so on, we move the decimal point in the number to the left the same number of places as there are zeros in the power of 10. 4. True or false: If we replace x with 12.6 and y with 0.3 in the expression y x, we have 0.3 (12.6) true Copyright © 2025 Pearson Education, Inc.
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Chapter 5: Decimals
5. a whole number 6. deciding if our decimal point is in the correct place in the quotient 7. We just need to know how to move the decimal point. 1000 has three zeros, so we move the decimal point in the decimal number three places to the left. 8. This actually means 4 divides into 1 zero times, so we place a zero in the quotient. 9. We want the answer rounded to the nearest tenth, so we go to one extra place value, to the hundredths place, in order to round. Exercise Set 5.4 5.9 2. 4 23.6 20 36 3 6 0
9.9 10. Exact: 2.2 21.78 becomes 22 217.8 198 19 8 19 8 0 11 Estimate: 2 22 0.519 12. 16 8.304 8 0 30 16 144 144 0 14. A positive number divided by a negative number is a negative number. 900 0.04 36 becomes 4 3600 36 000
0.085 4. 6 0.510 48 30 30 0
36 (0.04) = 900 16. A negative number divided by a negative number is a positive number. 4
0.9 3.6 becomes 9 36 36 0
500 6. 0.04 20 becomes 4 2000 20 000
(3.6) (0.9) = 4 18. 0.34 2.176 becomes 34
8. 0.36 1.764 becomes 36
178
4.9 176.4 144 32 4 22 4 0
6.4 217.6 204 13 6 13 6 0
800 20. 0.03 24 becomes 3 2400 24 000
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ISM: Prealgebra and Introductory Algebra
22. 0.92 3.312 becomes 92
3.6 331.2 276 55 2 55 2 0
24. A negative number divided by a negative number is a positive number. 2.2 9.9 21.78 becomes 99 217.8 198 19 8 19 8 0 21.78 (9.9) = 2.2 8.6 26. Exact: 6.3 54.18 becomes 63 541.8 504 37 8 37 8 0 9 Estimate: 6 54 4.4 28. 7.7 33.88 becomes 77 338.8 308 30 8 30 8 0 63 30. 0.051 3.213 becomes 51 3213 306 153 153 0 3.213 63 0.051
Chapter 5: Decimals
66.488 32. 0.75 49.866 becomes 75 4986.600 450 486 450 36 6 30 0 6 60 6 00 600 600 0 0.0045 34. 2.96 0.01332 becomes 296 1.3320 1 184 1480 1480 0 0.0462 0.046 36. 0.98 0.0453 becomes 98 4.5300 3 92 610 588 220 196 24 28.23 28.2 38. 3.5 98.83 becomes 35 988.30 70 288 280 83 70 1 30 1 05 25 40.
64.423
0.64423
100 42.
13.49
1.349
10 44. 13.49 (10,000) = 0.001349
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Chapter 5: Decimals
14.5 46. 9 130.5 9 40 36 45 4 5
50, 000 62. 0.0007 35 becomes 7 350, 000 35 00 000 35 (0.0007) = 50,000 2.2 64. 0.98 2.156 becomes 98 215.6 196 19 6 19 6 0
0 17.7 48.
50.
1.77
10 986.11
2.156 0.98 = 2.2
0.098611
10, 000
66. 86.79 (1000)
0.045 52. 12 0.540 48 60 60 0
54. 0.7 4.9 becomes 7
7 49 49 0
0.08679
23.8 680 0.035 70. z x = 4.52 5.65
3.2 56. 0.42 1.344 becomes 42 134.4 126 84 8 4 0 1.344 0.42 = 3.2 900 58. 0.03 27 becomes 3 2700 27 000
0.8 5.65 4.52 becomes 565 452.0 452 0 0 z x = 4.52 5.65 = 0.8 72. y 2 = 0.8 2 0.4 2 0.8 8 0 y 2 = 0.8 2 = 0.4 y
27 0.03 = 900
74. 50 200 20 00
7.128
0.89 0 0.89
8 0.89 0.89 True Yes, 7.12 is a solution.
20 0.4 = 50
180
1000
680 68. 0.035 23.8 becomes 35 23800 210 280 280 00
4.9 (0.7) = 7
60. 0.4 20 becomes 4
86.79
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ISM: Prealgebra and Introductory Algebra
x 76.
0.23
10 23
0 0.23 10 2.3 0.23 False No, 23 is not a solution. 78. Number of boxes = 486 36 13.5 36 486.0 36 126 108 18 0 18 0 0 Since whole boxes must be used, 14 boxes are needed. 19.68 19.7 80. 2.54 50 becomes 254 5000.00 254 2460 2286 1740 1524 21 60 20 32 1 28 There are approximately 19.7 inches in 50 centimeters. 82. Divide the price per hundred pounds by 100. 121.20 1.2120 1.21 100 The average price per pound was $1.21. 84. Each dose is 0.5 teaspoon. 48 0.5 24 becomes 5 240 20 40 40 0 There are 48 doses in the bottle.
Chapter 5: Decimals
86. From Exercise 83, we know there are 24 teaspoons in 4 fluid ounces. Thus, there are 48 half teaspoons (0.5 tsp) or doses in 4 fluid ounces. To see how long the medicine will last, if a dose is taken every 6 hours, there are 24 6 = 4 doses taken per day. 48 (doses) 4 (per day) = 12 days. The medicine will last 12 days. 21.90 21.9 88. 19.35 423.8 becomes 1935 42380.00 3870 3680 1935 17450 17415 350 Drake’s car averages about 21.9 miles per gallon. 345.5 90. 24 8292.0 72 109 96 132 120 12 0 12 0 0 There were 345.5 thousand books sold each hour. 92.
3
7
5 10
3 10 310 3 5 2 6 5 7 5 7 5 7 7
94. 3 1 3 7 1 2 4 14 4 7 14 2 3 7 1 2 4 7 14 2 21 2 28 28 23 28 96.
1.278 0.3 0.3834
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3 decimal places 1 decimal place 3 1 4 decimal places
181
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
98.
1.278 0.300 0.978
100.
7.20 0.05 49.10 56.35
102.
87.2 10, 000
134.78 134.8 1.15 155 becomes 115 15500.00 115 400 345 550 460 90 0 805 950 9 20 30
0.00872
104. 1.437 + 20.69 is approximately 1 + 21 = 22, which is choice b. 106. 302.729 28.697 is approximately 300 30 = 270, which is choice a. 108.
56 75 80 211 70.3 3 3 45.2 180.8 16 20 20 08 8 0 The length of a side is 45.2 centimeters.
The range of wind speeds is 113.9 134.8 knots. 13.6 116. 185.35 2523.86 becomes 18535 252386.0 18535 67036 55605 11431 0 11121 0
110. 4
310 0 No; it will take the student more than 13 months to pay it off at the minimum rate. Mid-Chapter Review 1.
1.60 0.97 2.57
2.
3.20 0.85 4.05
3.
9.8 0.9 8.9
4.
10.2 6.7 3.5
5.
0.8 0.2 0.16
112. answers may vary 113.91 113.9 114. 1.15 131 becomes 115 13100.00 115 160 115 450 345 1050 1035 150 115 35
182
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ISM: Prealgebra and Introductory Algebra
6.
Chapter 5: Decimals
0.6 0.4 0.24
15. 3.4 29.24 becomes 34
0.27 7. 8 2.16 16 56 56
29.24 (3.4) = 8.6 5.4 16. 1.9 10.26 becomes 19 102.6 95 76 7 6 0 10.26 (1.9) = 5.4
0
8. 6
0.52 3.12 3 0 12 12
17. 2.8 100 = 280
0 9.
10.
11.
12.
8.6 292.4 272 20 4 20 4 0
18. 1.6 1000 = 1600
9.6 0.5 4.80 (9.6)(0.5) = 4.8 8.7 0.7 6.09 (8.7)(0.7) = 6.09 123.60 48.04 75.56
19.
96.210 7.028 121.700 224.938
20.
0.268 1.930 142.881 145.079 0.56 25.76 23 0 2 76 2 76 0 25.76 (46) = 0.56
21. 46
325.20 36.08 289.12
13. Subtract absolute values. 25.000 0.026 24.974 Attach the sign of the larger absolute value. 25 + 0.026 = 24.974 14. Subtract absolute values. 44.000 0.125 43.875 Attach the sign of the larger absolute value. 0.125 + (44) = 43.875
22. 43
0.63 27.09 25 8 1 29 1 29 0
27.09 43 = 0.63
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
23.
24.
12.004 2.3 3 6012 24 0080 27.6092
3 decimal places 1 decimal place
28.006 5.2 5 6012 140 0300 145.6312
3 decimal places 1 decimal place
3 1 4 decimal places
25.
28 30. 0.061 1.708 becomes 61 1708 122 488 488 0 1.708 28 0.061
3 1 4 decimal places
10.0 4.6 5.4
26. Subtract absolute values. 18.00 0.26 17.74 Attach the sign of the greater absolute value. 0.26 18 = 17.74 27. 268.19 146.25 = 268.19 + (146.25) Add absolute values. 268.19 146.25 414.44 Attach the common sign. 268.19 146.25 = 414.44
31.
160.00 43.19 116.81
32.
120.00 101.21 18.79
33. 15.62 10 = 156.2 34. 15.62 10 = 1.562 35.
15.62 10.00 25.62
36.
15.62 10.00 5.62
37.
53.7 79.2 71.2 204.1 The exact distance is 204.1 miles. 50 80 70 200 The estimated distance is 200 miles.
38.
16.75 13.85 2.90 It costs $2.90 more to send the package as Priority Mail.
39.
18.8 21.8 40.6 The total number of vinyl albums shipped was 40.6 million.
28. 860.18 434.85 = 860.18 + (434.85) Add absolute values. 860.18 434.85 1295.03 Attach the common sign. 860.18 434.85 = 1295.03 34 29. 0.087 2.958 becomes 87 2958 261 348 348 0 2.958 34 0.087
184
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ISM: Prealgebra and Introductory Algebra
2.1538 2.154 4. 13 28.0000 26
Section 5.5 Practice Exercises 1. a.
0.4 5 2.0 2.0 0
2
3. a.
b.
9
40 9.000 8 0 1 00 80 200 200 0
0.375 2. 8 3.000 2 4 60 56 40 40 0
0.4
20 13 70 65 50 39 110 104 6
5
0.225 b.
Chapter 5: Decimals
0.225
40
3 0.375 8
0.833... 6 5.000 4 8 20 18 20 18 2
5
0.22... 9 2.00 18 20 18 2
2
0.83 6
6.
9
0.2
5
53 16 16 3.3125 16 53.0000 48 50 4 8 20 16 40 32 80 80 0 5 Thus, 3 3.3125. 16
5. 3
7.
3
3 2 6 0.6 5 5 2 10 3
50
3 2 6 0.06 50 2 100
0.2 8. 5 1.0 10 0 Since 0.2 < 0.25, then
1
0.25.
5
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
9. a.
1
0.5 and 0.5 < 0.54, so
2 b.
9
1
0.54.
2 0.55... 5.00 4 5 50 45 5
12. 8.69(3.2 1.8) = 8.69(1.4) = 12.166 13. (0.7)2 2.1 0.49 2.10 2.59 20.06 (1.2)2 10 14.
c.
15. Area
base height
2 1
7 2.1 2 0.5 7 2.1 7.35 The area of the triangle is 7.35 square meters. 16. 1.7y 2 = 1.7(2.3) 2 = 3.91 2 = 1.91
2 0.714 < 0.72, so
1
20.06 1.44 10
0.02 20.06 0.144 0.02 19.916 0.02 995.8
0.02
5 0.5 0.55..., so 0.5 . 9 0.714 7 5.000 4 9 10 7 30 28
5
0.72.
Vocabulary, Readiness & Video Check 5.5
7 10. a.
1. The number 0.5 means 0.555. false
1
0.333... 3 0.302 = 0.302 3 0.375 8 1 3 0.302, , 3 8
2. To write
19 19 9. true
4. To simplify 8.6(4.8 9.6), we first subtract. true 5. We place a bar over just the repeating digits and only 6 repeats in our decimal answer. 6. It is easier to compare decimal numbers. 7. The fraction bar serves as a grouping symbol. 8. A = l w; 0.248 sq yd
0.4 = 0.40 0.41 = 0.41 3 0.43 7 3 0.4, 0.41, 7
9. 4(0.3) (2.4) Exercise Set 5.5
11. 897.8 100 10 = 8.978 10 = 89.78
186
as a decimal, perform the division
3. (1.2)2 means (1.2)(1.2) or 1.44. false
b. 1.26 = 1.26 1 1 1.25 4 2 1 1.40 5 1 2 1 , 1.26, 1 4 5 c.
9
0.05 2. 20 1.00 1 00 0
Copyright © 2025 Pearson Education, Inc.
1 20
0.05
ISM: Prealgebra and Introductory Algebra
4. 25
0.52 13.00 12 5 50 50 0
0.125 6. 8 1.000 8 20 16 40 40 0
8. 25
0.12 3.00 2 5 50 50 0
13 25
1
8
0.4166... 12. 12 5.0000 4 8 20 12 80 72 80 72 8
5
0.76 19.00 17 5 1 50 1 50 0
0.125
8
1.6 10. 5 8.0 5 30 3 0 0
14. 25
0.52
3 25
0.12
1.6
5
0.416 12
19
Chapter 5: Decimals
16. 40
0.775 31.000 28 0 3 00 2 80 200 200 0
0.777... 18. 9 7.000 6 3 70 63 70 63 7 0.5625 20. 16 9.0000 8 0 1 00 96 40 32 80 80 0 0.8181... 22. 11 9.0000 88 20 11 90 88 20 11 9
31
0.775
40
7 0.7 9
9
0.5625
16
9
0.81 11
0.76
25
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Chapter 5: Decimals
0.875 24. 8 7.000 6 4 60 56 40 40 0 0.424 26. 375 159.000 150 0 9 00 7 50 1 500 1 500 0 28.
30.
7 0.777 0.78 9
9
0.5625 0.56
16 32.
9
0.8181 0.8
11 0.45 34. 20 9.00 8 0 1 00 1 00 0 0.054 36. 125 8.000 7 50 500 500 0
188
4
7
4.875
8
159 0.424 375
0.00000125 38. 800, 000 1.00000000 800000 2000000 1600000 4000000 4000000 0 40. 0.983 0.988 3 < 8 so 0.983 < 0.988 0.725 42. 40 29.000 28 0 1 00 80 200 200 0 29 0.725 40 44. 0.00563 0.00536 6 > 3 so, 0.00563 > 0.00536 Thus, 0.00563 < 0.00536. 0.117 0.12 46. 17 2.000 17 30 17 130 119 11 2 2 0.12 and 0. 1 0.12, so 0. 1 17 17 .
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ISM: Prealgebra and Introductory Algebra
0.5454... 48. 11 6.0000 5 5 50 44
Chapter 5: Decimals 12.7142 12.714 89.0000 7 19 14 50 4 9 10 7
54. 7
60 55 50 44
30 28 20 14 6
6 6 6 0.54 and 0.583 0.54, so 0.583 . 11 11 0.555... 50. 9 5.000 4 5 50 45 50 45 5 5 0.555 and 0.555 0.557, so 5 0.557. 9 9
52. 59
22
89
12.714 and 12.713 < 12.714, so
7 12.713
89 . 7
56. 0.40, 0.42, 0.47 58. 0.72 = 0.720 0.72, 0.727, 0.728 60. 7.56 = 7.560 67 7.444... 9 7.562 = 7.562 67 , 7.56, 7.562 9
0.37288 0.3729 22.00000 17 7 4 30 4 13 170 118 520 472 480 472 8
62.
5
0.833
6 5 0.821, , 0.849 6 64. (2.5)(3) 4.7 = 7.5 4.7 = 12.2
0.3729 and 0.372 < 0.3729, so
66. (0.05)2 3.13 0.0025 3.13 3.1325
59 0.372
22
.
68. (8.2)(100) (8.2)(10) = 820 82 = 738
59 70.
0.222 2.13 1.908 0.159 12 12
72. 0.9(5.6 6.5) = 0.9(0.9) = 0.81 74. (1.5)2 0.5 2.25 0.5 2.75
Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
76.
78.
3
3 (9.6)(5) 48 0.75 48 47.25 4 4 3
(4.7 5.9)
8
3
(1.2) 0.375(1.2) 0.45
8
80. Area
1
bh
2 1
17 4.4 2 0.5 17 4.4 37.4 square feet
0.0817 0.082 100. 15, 371 1257.0000 1229 68 27 320 15 371 11 9490 10 7597 1 1893 Approximately 0.08 of radio stations had a Spanish format. 0.650 0.65 102. 15, 371 10000.000 9222 6 777 40 768 55 8 550 The top six formats accounted for about 0.65 of radio stations.
82. Area l w
7 8 (1.2)(0.875) 1.05 square miles (1.2)
84. y2 (0.3)2 0.09
104. answers may vary 86. x z = 6 (2.4) = 6 + 2.4 = 8.4 x 88.
90.
2z 6 2(2.4) 20 4.8 15.2 0.3 y 4
19
Section 5.6 Practice Exercises 1.
z 0.9 1.3 z 0.9 0.9 1.3 0.9 z 0.4 Check: z 0.9 1.3 0.4 0.9 0 1.3 1.3 1.3 True The solution is 0.4.
2.
0.17x 0.34
4 2 19 8 19 11 1 11 2 22 22 22 22 2
11 22 2
3
2 3 2 2 3 3 3 92. 3 2 3 3 2 2 2 2 2 3 3 3 3 3 2 2 2 3 2
0.17x 0.34 0.17 0.17 x 2 Check: 0.17x 0.34 0.17(2) 0 0.34 0.34 0.34 True
94. 1.0000 = 1 96. 101 > 99, so
101 99
98.
99 99
190
1
The solution is 2.
1 3.
2.9 1.7 0.3x 2.9 1.7 1.7 0.3x 1.7 1.2 0.3x 1.2 0.3x 0.3 0.3 4x The solution is 4.
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ISM: Prealgebra and Introductory Algebra
4.
5.
6.
8x 4.2 10x 11.6 8x 4.2 4.2 10x 11.6 4.2 8x 10x 7.4 8x 10x 10x 10x 7.4 2x 7.4 2x 7.4 2 2 x 3.7 The solution is 3.7.
Chapter 5: Decimals
6.
8. 3y 25.8 3y 25.8 3 3 y 8.6
6.3 5x 3(x 2.9) 6.3 5x 3x 8.7 6.3 5x 6.3 3x 8.7 6.3 5x 3x 2.4 5x 3x 3x 2.4 3x 8x 2.4 8x 2.4 8 8 x 0.3 The solution is 0.3. 0.2 y 2.6 4 10(0.2 y 2.6) 10(4) 10(0.2 y) 10(2.6) 10(4) 2 y 26 40 2 y 26 26 40 26 2 y 14 2 y 14 2 2 y7 The solution is 7.
Vocabulary, Readiness & Video Check 5.6
10.
7.1 0.2x 6.1 7.1 0.2x 7.1 6.1 7.1 0.2x 1 0.2x 1 0.2 0.2 x5
12.
7x 9.64 5x 2.32 7x 9.64 9.64 5x 2.32 9.64 7x 5x 11.96 7x 5x 5x 5x 11.96 2x 11.96 2x 11.96 2 2 x 5.98
14.
5(x 2.3) 19.5 5x 11.5 19.5 5x 11.5 11.5 19.5 11.5 5x 8 5x 8 5 5 x 1.6
16.
0.7x 0.1 1.5 10(0.7x 0.1) 10(1.5) 7x 1 15 7x 11 15 1 7x 14 7x 14 7 7 x2
18.
3y 7 y 24.4 10 3y 10(7 y 24.4) 30 y 70 y 244 30 y 70 y 70 y 70 y 244 40 y 244 40 y 244 40 40 y 6.1
1. so that we are no longer working with decimals 2. We would have avoided working with a negative coefficientsubtracting 2x would have given us a positive coefficient for x. Exercise Set 5.6 2.
y 0.5 9 y 0.5 0.5 9 0.5 y 9.5
4. 0.4x 50 0.4x 50 0.4 0.4 x 125
9.7 x 11.6 9.7 11.6 x 11.6 11.6 1.9 x
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
20.
1.5x 2 1.2x 12.2 10(1.5x 2 1.2x) 10 12.2 15x 20 12x 122 3x 20 122 3x 20 20 122 20 3x 102 3x 102 3 3 x 34
22.
x 5.7 8.4 x 5.7 5.7 8.4 5.7 x 2.7
24.
9 y 0.162 9 y 0.162 9 9 y 0.018
26.
34.
24 y 10 20 y 17 24 y 10 10 20 y 17 10 24 y 20 y 7 24 y 20 y 20 y 20 y 7 4 y 7 4 y 7 4 4 y 1.75
36.
1.5 0.4x 0.5 1.5 0.5 0.4x 0.5 0.5 1 0.4x 1 0.4x 0.4 0.4 2.5 x
38.
50x 0.81 40x 0.48 100(50x 0.81) 100(40x 0.48) 5000x 81 4000x 48 5000x 81 48 4000x 48 48 5000x 129 4000x 5000x 5000x 129 4000x 5000x 129 1000x 129 1000x 1000 1000 0.129 x
40.
4(2x 1.6) 5x 6.4 8x 6.4 5x 6.4 8x 6.4 6.4 5x 6.4 6.4 8x 5x 8x 5x 5x 5x 3x 0 3x 0 3 3 x0
42.
y 5 0.3y 4.1 10( y 5) 10(0.3y 4.1) 10 y 50 3y 41 10 y 50 50 3y 41 50 10 y 3y 91 10 y 3y 3y 3y 91 7 y 91 7 y 91 7 7 y 13
2x 4.2 8.6 2x 4.2 4.2 8.6 4.2 2x 12.8 2x 12.8 2 2 x 6.4
28.
30.
32.
192
9 y 6.9 6 y 11.1 9 y 6.9 6.9 6 y 11.1 6.9 9 y 6 y 4.2 9 y 6 y 6 y 6 y 4.2 3y 4.2 3y 4.2 3 3 y 1.4 2.3 500x 600x 0.2 2.3 2.3 500x 600x 0.2 2.3 500x 600x 2.5 500x 600x 600x 600x 2.5 100x 2.5 100x 2.5 100 100 x 0.025 7(x 8.6) 6x 7x 60.2 6x 7x 7x 60.2 6x 7x 60.2 x 60.2 x 1 1 60.2 x
44. x + 14 5x 17 = (x 5x) + (14 17) = 4x 3
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ISM: Prealgebra and Introductory Algebra
1 1 16 55 16 6 16 2 3 32 46. 5 9 3 6 3 6 3 55 3 55 55 48. 50 14 50.
9 13
49
13
9 4 14 35 13 13 13
y 15.9 3.8 y 15.9 15.9 3.8 15.9 y 12.1
Chapter 5: Decimals
68. 100, although answers may vary 70. answers may vary 72. 7.68 y 114.98496 7.68 y 114.98496 7.68 7.68 y 14.972 74.
52. x 4.1 x 4.02 x x 4.1 4.02 2x 8.12 54. 9a 5.6 3a 6 9a 3a 5.6 6 6a 0.4 56.
58.
9.7 x 4.3 9.7 4.3 x 4.3 4.3 5.4 x
Section 5.7 Practice Exercises
2.8x 3.4 13.4 2.8x 3.4 3.4 13.4 3.4 2.8x 16.8 2.8x 16.8 2.8 2.8 x 6
1. Mean
9.21 x 4x 11.33 x 4x 9.21 11.33 3x 20.54
62.
6(x 1.43) 5x 6x 8.58 5x 6x 6x 8.58 5x 6x 8.58 x 8.58 x 1 1 8.58 x 8.4z 2.6 5.4z 10.3 10(8.4z 2.6) 10(5.4z 10.3) 84z 26 54z 103 84z 26 26 54z 103 26 84z 54z 129 84z 54z 54z 54z 129 30z 129 30z 129 30 30 z 4.3
66. 3.2x 12.6 8.9x 15.2 3.2x 8.9x 12.6 15.2 12.1x 2.6
87 75 96 91 78 427 85.4 5 5
2.
60.
64.
6.11x 4.683 7.51x 18.235 6.11x 4.683 4.683 7.51x 18.235 4.683 6.11x 7.51x 13.552 6.11x 7.51x 7.51x 7.51x 13.552 1.4x 13.552 1.4x 13.552 1.4 1.4 x 9.68
32 34 46 38 42 95 50 42 41 420 9 9 46.67 The mean salary of all staff members except G is about $46.67 thousand or $46,670. 3. gpa
4 2 3 4 2 5 1 2 4 2 2 4522
40
2.67
15
4. Because the numbers are in numerical order, and there are an odd number of items, the median is the middle number, 24. 5. Write the numbers in numerical order: 36, 65, 71, 78, 88, 91, 95, 95 Since there are an even number of scores, the median is the mean of the two middle numbers. 78 88 median 83 2 6. Mode: 15 because it occurs most often, 3 times. 7. Median: Write the numbers in order. 15, 15, 15, 16, 18, 26, 26, 30, 31, 35 Median is mean of middle two numbers, 18 26 22. 2 Mode: 15 because it occurs most often, 3 times. Copyright © 2025 Pearson Education, Inc.
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Chapter 5: Decimals
Vocabulary, Readiness & Video Check 5.7 1. Another word for “mean” is average. 2. The number that occurs most often in a set of numbers is called the mode. 3. The mean (or average) of a set of number items sum of items is . number of items 4. The median of a set of numbers is the middle number. If the number of numbers is even, it is the mean (or average) of the two middle numbers. 5. An example of weighted mean is a calculation of grade point average. 6. The answer is not exact; it’s an approximation since we rounded to the nearest tenth. 7. Place the data numbers in numerical order (or verify that they already are). 8. All occurrences of a number will be grouped together, making it easier to locate and count numbers that occur more than once.
Median: Write the numbers in order: 0.1, 0.3, 0.4, 0.5, 0.6, 0.6, 0.7, 0.8, 0.8 The middle number is 0.6. Mode: Since 0.6 and 0.8 occur twice, there are two modes, 0.6 and 0.8. 8. Mean: 451 356 478 776 892 500 467 780 8 4700 8 587.5 Median: Write the numbers in order: 356, 451, 467, 478, 500, 776, 780, 892 478 500 The mean of the middle two: 489 2 Mode: There is no mode, since each number occurs once. 10. Because the numbers are in numerical order, the median height is the middle number (of the top 5), 1972 feet. 12. Mean:
Exercise Set 5.7 45 36 28 46 52 207 41.4 5 5 Median: Write the numbers in order: 28, 36, 45, 46, 52 The middle number is 45. Mode: There is no mode, since each number occurs once.
14. answer may vary
2. Mean:
4. Mean: 4.9 7.1 6.8 6.8 5.3 4.9 35.8 6.0 6 6 Median: Write the numbers in order: 4.9, 4.9, 5.3, 6.8, 6.8, 7.1 5.3 6.8 Median is mean of middle two: 6.05 2 Mode: Since 6.8 and 4.9 occur twice, there are two modes, 6.8 and 4.9. 6. Mean: 0.6 0.6 0.8 0.4 0.5 0.3 0.7 0.8 0.1 9 4.8 9 0.5 194
2717 2073 1972 1965 1819 1776 6 12, 322 6 2053.7 feet
16. gpa
18. gpa
11 0 1 2 4 3 5 11 4 5
24
2.18
11
3 2 3 2 2 3 4 3 3 3 39 3.0 2 2333 13
20. Median: Write the numbers in order. 4.7, 6.9, 6.9, 7.0, 7.5, 7.8 6.9 7.0 The mean of the middle two: 6.95 2 22. Mean:
93 85 89 79 88 91 525 87.5 6 6
24. Mode: There is no mode, since each number occurs only once. 26. Median: Write the numbers in order. 64, 65, 66, 68, 70, 70, 71, 71, 72, 75, 77, 78, 80, 82, 86 The middle number is 71.
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ISM: Prealgebra and Introductory Algebra
28. The mean is 73 (from Exercise 25). There were 6 pulse rates higher than the mean. They are 75, 77, 78, 80, 82, and 86. 30. answers may vary 32. Mean: 4160 4000 3915 12, 075 4025 miles 3 3 Median: 4000 miles 34.
Chapter 5: Decimals
9. The mean of a list of items with number values is sum of items . number of items 10. When 2 million is written as 2,000,000, we say it is written in standard form. Chapter 5 Review 1. In 23.45, the 4 is in the tenths place.
4x 4 x x 36 4 9 9
2. In 0.000345, the 4 is in the hundred-thousandths place.
35a3
3. 23.45 in words is negative twenty-three and forty-five hundredths.
5 7 a a a 7a 36. 2 5 20 a a 20 100a
38. Since the mode is 21, 21 must occur at least twice in the set. Since there are an odd number of numbers in the set, the median, 20 is in the set. Therefore the missing numbers are 20, 21, 21.
4. 0.00345 in words is three hundred forty-five hundred-thousandths. 5. 109.23 in words is one hundred nine and twentythree hundredths. 6. 200.000032 in words is two hundred and thirtytwo millionths.
40. answers may vary Chapter 5 Vocabulary Check
7. Eight and six hundredths is 8.06.
1. Like fractional notation, decimal notation is used to denote a part of a whole. 2. To write fractions as decimals, divide the numerator by the denominator. 3. To add or subtract decimals, write the decimals so that the decimal points line up vertically.
8. Negative five hundred three and one hundred two thousandths is 503.102. 9. Sixteen thousand twenty-five and fourteen tenthousandths is 16,025.0014. 10. Fourteen and eleven thousandths is 14.011. 16
4 4 4 100 25 4 25
4. When writing decimals in words, write “and” for the decimal point.
11. 0.16
5. When multiplying decimals, the decimal point in the product is placed so that the number of decimal places in the product is equal to the sum of the number of decimal places in the factors.
12. 12.023 12
6. The mode of a set of numbers is the number that occurs most often.
13.
7. The distance around a circle is called the circumference.
14. 25
8. The median of a set of numbers in numerical order is the middle number. If there are an even number of numbers, the mode is the mean of the two middle numbers.
15. 0.49 0.43 9 > 3 so 0.49 > 0.43
23 1000
231 0.00231 100, 000 1
25
4
25 100
25.25
16. 0.973 = 0.9730
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Chapter 5: Decimals
27. Add the absolute values. 6.40 0.88 7.28 Attach the common sign. 6.4 + (0.88) = 7.28
17. 38.0027 38.00056 2 > 0 so 38.0027 > 38.00056 Thus, 38.0027 < 38.00056. 18. 0.230505 0.23505 0 < 5 so 0.230505 < 0.23505 Thus, 0.230505 > 0.23505. 19. To round 0.623 to the nearest tenth, observe that the digit in the hundredths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the tenths place. The number 0.623 rounded to the nearest tenth is 0.6.
28. Subtract the absolute values. 19.02 6.98 12.04 Attach the sign of the larger absolute value. 19.02 + 6.98 = 12.04 29.
20. To round 0.9384 to the nearest hundredth, observe that the digit in the thousandths place is 8. Since this digit is at least 5, we add 1 to the digit in the hundredths place. The number 0.9384 rounded to the nearest hundredth is 0.94.
200.490 16.820 103.002 320.312
30.
21. To round 42.895 to the nearest hundredth, observe that the digit in the thousandths place is 5. Since this digit is at least 5, we add 1 to the digit in the hundredths place. The number 42.895 rounded to the nearest hundredth is 42.90.
0.00236 100.45000 48.29000 148.74236
31.
4.9 3.2 1.7
32.
5.23 2.74 2.49
22. To round 16.34925 to the nearest thousandth, observe that the digit in the ten-thousandths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the thousandths place. The number 16.34925 rounded to the nearest thousandth is 16.349. 23. 890.5 million 890.51 million 890.51, 000, 000 890, 500, 000 24. 600 thousand 600 1 thousand 600 1000 600, 000 25.
8.6 9.5 18.1
26.
3.9 1.2 5.1
196
33. 892.1 432.4 = 892.1 + (432.4) Add the absolute values. 892.1 432.4 1324.5 Attach the common sign. 892.1 432.4 = 1324.5 34. 0.064 10.2 = 0.064 + (10.2) Subtract the absolute values. 10.200 0.064 10.136 Attach the sign of the larger absolute value. 0.064 10.2 = 10.136 35.
100.00 34.98 65.02
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ISM: Prealgebra and Introductory Algebra
36.
200.00000 0.00198 199.99802
37.
19.9 15.1 10.9 6.7 52.6 The total distance is 52.6 miles.
Chapter 5: Decimals
0.0877 47. 3 0.2631 24 23 21 21 21 0
38. x y = 1.2 6.9 = 5.7 39. Perimeter = 6.2 + 4.9 + 6.2 + 4.9 = 22.2 The perimeter is 22.2 inches. 40. Perimeter = 11.8 + 12.9 + 14.2 = 38.9 The perimeter is 38.9 feet. 41. 7.2 10 = 72 42. 9.345 1000 = 9345 43. A negative number multiplied by a positive number is a negative number. 34.02 2 decimal places 1 decimal place 2.3 10 206 68 040 2 1 3 decimal places 78.246 34.02 2.3 = 78.246
15.825 48. 20 316.500 20 116 100 16 5 16 0 50 40 100 100 0 49. A negative number divided by a negative number is a positive number. 70 0.3 21 becomes 3 210 21 00 21 (0.3) = 70
44. A negative number multiplied by a negative number is a positive number. 839.02 2 decimal places 87.3 1 decimal place 251 706 5873 140 67121 600 73246.446 2 1 3 decimal places 839.02 (87.3) = 73,246.446
50. A negative number divided by a positive number is a negative number. 0.21 0.03 0.0063 becomes 3 0.63 6 03 3 0 0.0063 0.03 = 0.21
45. C = 2r = 2 7 = 14 meters C 14 3.14 = 43.96 meters 46. C = d = 20 = 20 inches C 20 3.14 = 62.8 inches
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals 8.0588 8.059 51. 0.34 2.74 becomes 34 274.0000 272 20 0 2 00 1 70 300 272 280 272 8 52. 19.8 601.92 becomes 198
30.4 6019.2 594 79 0 79 2 79 2 0
53.
23.65
0.02365
1000 54.
93 10
9.3 7.31 7.3
55. 3.28 24 becomes 328 2400.00 2296 104 0 98 4 5 60 3 28 2 32 There are approximately 7.3 meters in 24 feet. 45 56. 69.71 3136.95 becomes 6971 313695 27884 34855 34855 0 The loan will be paid off in 45 months.
0.8 57. 5 4.0 4 0 0 0.9230 0.923 58. 13 12.0000 11 7 30 26 40 39 10 0 10 12 0.923 13 0.3333... 59. 3 1.0000 9 10 9 10 9 10 9 1 2
1 3
60. 60
13 16
2.3 or 2.333 0.2166... 13.0000 12 0 1 00 60 400 360 400 360 40 0.216 or 0.217
61. 0.392 = 0.39200
198
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4 5
0.8
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
69. 7.6 1.9 + 2.5 = 14.44 + 2.5 = 11.94
62. 0.0231 0.0221 3 > 2 so 0.0231 > 0.0221, thus 0.0231 < 0.0221.
70. (2.3)2 1.4 5.29 1.4 3.89 71. 0.0726 10 1000 = 0.00726 1000 = 7.26
0.571 0.57 63. 7 4.000 3 5 50
72. 0.9(6.5 5.6) = 0.9(0.9) = 0.81 (1.5)2 0.5 73.
49 10 7 3 4
0.57 and 0.57 < 0.625, so
7
74. 4
0.625.
64. 17
5
5
76. Area
8 5
5 17
3
.
1
b h
2 1
1
b h
2 1
77.
x 3.9 4.2 x 3.9 3.9 4.2 3.9 x 0.3
78.
70 y 22.81 70 22.81 y 22.81 22.81 92.81 y
79.
2x 17.2 2x 17.2 2 2 x 8.6
0.428
7 3 0.42, , 0.43 7
68.
55
0.05
, 0.626, 0.685
8 67.
0.05
2.75
(5.2)(2.1) 2 0.5(5.2)(2.1) 5.46 square inches
0.294 and 0.293 < 0.294, so 0.293
0.625
(4.6)(3) 2 0.5(4.6)(3) 6.9 square feet
65. 0.832, 0.837, 0.839 66.
2.25 0.5
7 0.74 7.74 129 0.06 0.06
0.2941 0.294 5.0000 3 4 1 60 1 53 70 68 20 17 3
17
0.05
75. Area
7
18
1.6363
11 1.63 = 1.63 19 1.583 12 19 18 , 1.63, 12 11
80. 1.1y 88 1.1y 88 1.1 1.1 y 80 81.
3x 0.78 1.2 2x 3x 0.78 0.78 1.2 2x 0.78 3x 1.98 2x 3x 2x 1.98 2x 2x x 1.98
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
82.
x 0.6 2x 4x 0.9 3x 0.6 4x 0.9 3x 0.6 0.6 4x 0.9 0.6 3x 4x 1.5 3x 4x 4x 4x 1.5 x 1.5
83.
1.3x 9.4 0.4x 8.6 10(1.3x 9.4) 10(0.4x 8.6) 13x 94 4x 86 13x 94 94 4x 86 94 13x 4x 180 13x 4x 4x 4x 180 9x 180 9x 180 9 9 x 20
84.
3(x 1.1) 5x 5.3 3x 3.3 5x 5.3 3x 3.3 3.3 5x 5.3 3.3 3x 5x 2 3x 5x 5x 5x 2 2x 2 2x 2 2 2 x 1 13 23 33 14 6 89 17.8 5 5 Median: Write the numbers in order. 6, 13, 14, 23, 33 The middle number is 14. Mode: There is no mode, since each number occurs once.
85. Mean:
45 86 21 60 86 64 45 7 407 7 58.1 Median: Write the numbers in order. 21, 45, 45, 60, 64, 86, 86 The middle number is 60. Mode: There are 2 numbers that occur twice, so there are two modes, 45 and 86.
86. Mean
14, 000 20, 000 12, 000 20, 000 36, 000 45, 000 147, 000 24, 500 6 6 Median: Write the numbers in order. 12,000, 14,000, 20,000, 20,000, 36,000, 45,000 20, 000 20, 000 The mean of the middle two: 20, 000 2 Mode: 20,000 is the mode because it occurs twice.
87. Mean
200
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
560 620 123 400 410 300 400 780 430 450 4473 447.3 10 10 Median: Write the numbers in order. 123, 300, 400, 400, 410, 430, 450, 560, 620, 780 410 430 The mean of the two middle numbers: 420 2 Mode: 400 is the mode because it occurs twice.
88. Mean
89. gpa
4 3 4 3 2 2 3 3 2 1 39 3.25 3 3 2 3 1 12
90. gpa
3 3 3 4 2 2 1 2 3 3 36 2.57 3 4 2 23 14
91. 200.0032 in words is two hundred and thirty-two ten-thousandths. 92. Negative sixteen and nine hundredths is 16.09 in standard form. 93. 0.0847
94.
6
847 10, 000
0.857
7 8
0.889 9 0.75 = 0.750 6 8 0.75, , 7 9 95.
7 0.07 100
0.1125 96. 80 9.0000 8 0 1 00 80 200 160 400 400 0
9
0.1125
80
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201
ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
51.0571 51.057 97. 175 8935.0000 875 185 175 10 0 0 10 00 875 1 250 1 225 250 175 75 8935 51.057 175
104. Subtract absolute values. 4.9 3.2 1.7 Attach the sign of the larger absolute value. 3.2 4.9 = 1.7 105.
106. Subtract absolute values. 102.06 89.30 12.76 Attach the sign of the larger absolute value. 102.06 + 89.3 = 12.76
98. 402.000032 402.00032 0 < 3 so 402.000032 < 402.00032 Thus, 402.000032 > 402.00032. 99.
6
0.54
11 0.54 0.55, so
9.12 3.86 5.26
6
0.55 11
100. To round 86.905 to the nearest hundredth, observe that the digit in the thousandths place is 5. Since this digit is at least 5, we add 1 to the digit in the hundredths place. The number 86.905 rounded to the nearest hundredth is 86.91. 101. To round 3.11526 to the nearest thousandth, observe that the digit in the ten-thousandths place is 2. Since this digit is less than 5, we do not add 1 to the digit in the thousandths place. The number 3.11526 rounded to the nearest thousandth is 3.115. 102. To round 123.46 to the nearest one, observe that the digit in the tenths place is 4. Since this digit is less than 5, we do not add 1 to the digit in the ones place. The number $123.46 rounded to the nearest dollar (or one) is $123.00.
107.
4.021 10.830 () 0.056 14.907
108.
2.54 2 decimal places .2 1 decimal place 508 7 620 8.128 2 1 3 decimal places
109. The product of a negative number and a positive number is a negative number. 3.45 2 decimal places 2.1 1 decimal place 345 6 900 7.245 2 1 3 decimal places The product is 7.245. 4900 110. 0.005 24.5 becomes 5 24500 20 45 45 000
103. To round 3645.52 to the nearest one, observe that the digit in the tenths place is 5. Since this digit is at least 5, we add 1 to the digit in the ones place. The number $3645.52 rounded to the nearest dollar (or one) is $3646.00.
202
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
23.9043 23.904 111. 2.3 54.98 becomes 23 549.8000 46 89 69 20 8 20 7 10 0 100 92 80 69 11
Chapter 5 Getting Ready for the Test
112. length = 115.9 120 width = 77.3 80 Area length width 120 80
B. x y = 3 0.2 = 3.0 0.2 = 2.8
rounds to rounds to rounds to
2. In the number 1026.89704, the digit 2 is in the tens place; A. 3. In the number 1026.89704, the digit 0 to the left of the decimal point is in the hundreds place; B. 4. In the number 1026.89704, the digit 0 to the right of the decimal point is in the tenthousandths place; E. Replace x with 3 and y with 0.2 in each expression. A. x + y = 3 + 0.2 = 3.0 + 0.2 = 3.2
C. xy = (3)(0.2) = 0.6 D.
9600 square feet 113. $3.79 $2.49 $1.99
1. In the number 1026.89704, the digit 8 is in the tenths place: C.
x y 3 0.2
3 0.2
$4 $2 $2 $8
310 0.2 10
30
15
2
5. Since xy = 0.6, choice C is the correct expression.
Yes, the items can be purchased with a $10 bill.
6. Since x + y = 3.2, choice A is the correct expression. 7. Since x y = 2.8, choice B is the correct expression.
2 10.24 0.1024 114. (3.2) 100 100
115. (2.6 + 1.4)(4.5 3.6) = (4)(0.9) = 3.6 116. Mean: 73 82 95 68 54 372 74.4 5 5 Median: Write the numbers in order. 54, 68, 73, 82, 95 The median is the middle value: 73. Mode: There is no mode, since each number occurs once. 117. Mean: 952 327 566 814 327 729 3715 6 6 619.17 Median: Write the numbers in order. 327, 327, 566, 729, 814, 952 The mean of the two middle values: 566 729 647.5 2 Mode: 327, since this number appears twice.
8. since x y = 15, choice D is the correct expression. 9. 2(0.5)2 0.3 2(0.25) 0.3 0.50 0.3 0.8; B 10.
5(x 3.2) 9.5 5 x 5 3.2 9.5 5x 16 9.5 Choice B is the correct equation.
11.
0.3x 1.8 2.16 100(0.3x 1.8) 100(2.16) 100 0.3x 100 1.8 100 2.16 30x 180 216 Choice C is the correct equation.
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Chapter 5: Decimals 7 9 10 13 13 52 10.4 5 5 Median: Since the numbers are in order, the median is the middle number, 10. Mode: 13 occurs more often than any other value. Mean:
12. The median is 10; B. 13. The mode is 13; C.
0.00843 (0.23) 0.037
14. The mean is 10.4; A. Chapter 5 Test 1. 45.092 in words is forty-five and ninety-two thousandths. 2. Three thousand and fifty-nine thousandths in standard form is 3000.059. 3.
2.893 4.210 10.492 17.595
5. Subtract the absolute values. 30.25 9.83 20.42 Attach the sign of the larger absolute value. 9.83 30.25 = 20.42 10.2 4.01 102 40 800 40.902
8. To round 34.8923 to the nearest tenth, observe that the digit in the hundredths place is 9. Since this digit is at least 5, we add 1 to the digit in the tenths place. 34.8923 rounded to the nearest tenth is 34.9. 9. To round 0.8623 to the nearest thousandth, observe that the digit in the ten-thousandths place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousandths place. 0.8623 rounded to the nearest thousandth is 0.862. 10. 25.0909 25.9090 0 < 9 so 25.0909 < 25.9090
4. Add the absolute values. 47.92 3.28 51.20 Attach the common sign. 47.92 3.28 = 51.20
6.
0.0366 0.037 7. 0.23 0.00843 becomes 23 0.8430 69 153 138 150 138 12
0.444... 11. 9 4.000 3 6 40 36 40 36 4 0.444 0.445, so
1 decimal place 2 decimal places
0.445.
9
1 2 3 decimal places
12. 0.345 345 5 69 69 1000 5 200 200 13. 24.73 24
14.
204
4
73 100
13 113 1 1 5 5 0.5 26 2 13 2 25 10
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 0.9411 0.941 17 16.0000 15. 15 3 70 68 20 17 30 17 13
22. Mean: 8 10 16 16 14 12 12 13 101 12.625 8 8 Median: List the numbers in order. 8, 10, 12, 12, 13, 14, 16, 16 12 13 The mean of the middle two: 12.5 2 Mode: 12 and 16 each occur twice, so the modes are 12 and 16. 23. gpa
16 0.941 17 16. (0.6) 1.57 0.36 1.57 1.93 2
17.
0.23 1.63 1.86 6.2 0.3 0.3
20.
4 3 3 3 2 3 3 4 4 1 43 3.07 3 3 3 4 1 14
24. 4, 583 million 45831 million 45831, 000, 000 4, 583, 000, 000 1 (4.2 miles)(1.1 miles) 2 0.5(4.2)(1.1) square miles 2.31 square miles
25. Area
18. 2.4x 3.6 1.9x 9.8 (2.4x 1.9x) (3.6 9.8) 0.5x 13.4 19.
Chapter 5: Decimals
26. C = 2r = 2 9 = 18 miles C 18 3.14 = 56.52 miles
0.2x 1.3 0.7 0.2x 1.3 1.3 0.7 1.3 0.2x 0.6 0.2x 0.6 0.2 0.2 x 3
27. a.
2(x 5.7) 6x 3.4 2x 11.4 6x 3.4 2x 11.4 11.4 6x 3.4 11.4 2x 6x 14.8 2x 6x 6x 6x 14.8 4x 14.8 4x 14.8 4 4 x 3.7 26 32 42 43 49 192 38.4 5 5 Median: The numbers are listed in order. The middle number is 42. Mode: There is no mode since each number occurs once.
Area length width (123.8) (80) 9904 The area is 9904 square feet.
b. 9904 0.02 = 198.08 Teru needs to purchase 198.08 ounces. 28.
14.2 16.1 23.7 54.0 The total distance is 54 miles.
Cumulative Review Chapters 15 1. 72 in words is seventy-two.
21. Mean:
2. 107 in words is one hundred seven. 3. 546 in words is five hundred forty-six. 4. 5026 in words is five thousand twenty-six. 5.
46 713 759
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Chapter 5: Decimals
6. 3 + 7 + 9 = 19 The perimeter is 19 inches. 7.
543 29 514
14.
Check:
514 29 543
25 17 19 39 100 5 4 4 The average is 25.
15. 2 4 3 3 = 8 3 3 = 8 1 = 7 16. 77 11 7 = 7 7 = 49
121 R 1 8. 27 3268 27 56 54 28 27 1
17. 92 9 9 81 18. 53 5 5 5 125 19. 34 3 3 3 3 81 20. 103 10 10 10 1000
9. To round 278,362 to the nearest thousand, observe that the digit in the hundreds place is 3. Since this digit is less than 5, we do not add 1 to the digit in the thousands place. The number 278,362 rounded to the nearest thousand is 278,000. 10. 30 = 2 15 = 2 3 5 11.
c.
e.
b.
The opposite of 2 is (2) = 2.
c.
The opposite of 0 is 0. The opposite of 7 is (7) = 7.
b.
The opposite of 4 is 4.
c.
The opposite of 1 is (1) = 1.
Check: 7 1 = 7
25. 2 + (21) = 23
25 1 = 25
Check: 25 1 = 25
26. 7 + (15) = 22
6
1
12 12 = 1 20
Check: 1 6 = 6
27. 5 62 5 36 180 28. 4 23 4 8 32
20
1
f.
The opposite of 13 is 13.
7 17
6 d.
22. 2a 4 2 7 4 14 4 18 6 c 3 3 3
24. a.
12. 236 86 0 = 0
b.
x 5 y 35 5 5 35 25 10 2 y 5 5 5
23. a.
236 86 1 416 18 880 20, 296
13. a.
21.
1 18 18
Check: 1 12 = 12 Check: 20 1 = 20
Check: 1 18 = 18
29. 72 (7 7) 49 5 30. (2) (2)(2)(2)(2)(2) 32
31. (5)2 (5)(5) 25 32. 32 (3 3) 9
206
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ISM: Prealgebra and Introductory Algebra
33. Each part represents
1
of a whole. Four parts
Chapter 5: Decimals
42.
7 5
3 are shaded, or 1 whole and 1 part. 4 1 or 1 3 3
36. Each part represents
37. 252 2 126 2 2 63 2 2 3 21 2 2 3 3 7 252 22 32 7
38.
87 25 62
72 36 2 36 39. 26 13 2 13 40. 9 7 8 9 7 72 7 79 8 8 8 8 41.
16 2 8 2 40 5 8 5 10 2 5 2 25 5 5 5 The fractions are equivalent.
4
5 . 7 9
or 4 9 36 and 5 7 35 7 9 63 9 7 63 36 35 4 5 Since 36 > 35, then and . 63 63 7 9 43.
2 5 2 5 10 3 11 311 33
5 4 21 4 21 4 7 3 4 3 1 or 1 44. 2 8 7 8 7 8 7 427 2 2
1
of a whole. Fourteen 3 parts are shaded, or 4 wholes and 2 parts. 14 2 or 4 3 3
0.556
Since 0.571 > 0.556, then
1 of a whole. Seven parts 4 are shaded or 1 whole and 3 parts. 7 3 or 1 4 4 1 of a whole. Eleven parts 4 are shaded or 2 wholes and 3 parts. 11 3 or 2 4 4
0.571
9
34. Each part represents
35. Each part represents
4
45.
1 1 11 1 4 2 42 8
46. 7 5
2 7 z
47.
7 37 7 37 37 37 1 7 1 7 1
11 5
4 z 6 4 z 4 4 6 4 z 24
48. 6x 12 5x 20 x 12 20 x 12 12 20 12 x 8 49.
763.7651 22.0010 43.8900 829.6561
50.
89.2700 14.3610 127.2318 230.8628
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ISM: Prealgebra and Introductory Algebra
Chapter 5: Decimals
51.
52.
208
23.6 0.78 1 888 16 520 18.408 43.8 0.645 2190 1 7520 26 2800 28.2510
1 decimal place 2 decimal places
1 2 3 decimal places 1 decimal place 3 decimal places
1 3 4 decimal places
Copyright © 2025 Pearson Education, Inc.
Chapter 6 Section 6.1 Practice Exercises 1. a.
The ratio of 3 parts oil to 7 parts gasoline is 3. 7
b. Convert 3 hours to minutes. 3 hours = 3 60 minutes = 180 minutes The ratio of 40 minutes to 3 hours is 40 2 . 180 9 2.
1.68 1.68 100 168 7 4.8 4.8 100 480 20
3. a.
3
63
8 x 3 x 8 63 3x 504 3x 504 3 3 x 168 5.
2x 1 x 3 7 5 5(2x 1) 7(x 3) 10x 5 7x 21 10x 7x 26 3x 26 3x 26 3 3 26 x 3
The wall is 17 feet long. 7. ounces 5 8 ounces gallons 16 x gallons 5 x 16 8 5x 128 5x 128 5 5 3 x 25.6 or 25 5 3 Therefore, 25.6 or 25 gallons of gas can be 5 treated with 8 ounces of alcohol.
The ratio of the length of the shortest side to the length of the longest side is 3 6 meters 6 . 10 meters 10 5
b. Perimeter = 6 + 8 + 10 = 24 meters The ratio of the length of the longest side to 10 meters 10 5 the perimeter is . 24 meters 24 12 4.
6. Let x be the length of the wall. feet 4 x feet inches 1 4 14 inches 1 4 4 1 x 4 4 17 x 1 4 17 x
Vocabulary, Readiness & Video Check 6.1 1.
4.2
8.4
1 is called a proportion, while the 2
fraction
7 8
2. In
a
c
may also be called a ratio.
, a d and b c are called cross
b d products.
3. In order to simplify the ratio, quantities must have the same units of measurement to divide out. 4. No; proportions are actually equations containing rational expressions, so they can also be solved by using the steps to solve those equations. 5. There are also many ways to set up an incorrect proportion, so just checking your solution in your proportion isn’t enough. You need to determine if your solution is reasonable from the relationships given in the stated application.
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Chapter 6: Ratio, Proportion, and Percent
22. 32,056 31,623 = 433
Exercise Set 6.1 2. The ratio of 18 disks to 41 disks is
4. The ratio of 15 miles to 40 miles is
36.38 36.4 11.9 433 becomes 119 4330.00 357 760 714 46 0 35 7 10 30 9 52 78 433 miles were driven and the ratio is 36.4 miles to 1 gallon.
18
. 41
15
3 . 40 8
6. 2 dollars = 2 4 quarters = 8 quarters The ratio of 12 quarters to 2 dollars is
12
3 . 8 2
8. 2 days = 2 24 hours = 48 hours The ratio of 60 hours to 2 days is 60 5 . 48 4 24. 10. The ratio of 8.1 to 10 is
8.1 8.110 81 . 10 10 10 100
12. The ratio of 1.5 to 9.75 is 1.5 1.5 100 150 2 75 2 . 9.75 9.75 100 975 13 75 13 14. The ratio of the land area of Tuvalu to the land 10 sq mi 2 5 5 area of San Marino is . 24 sq mi 2 12 12
26.
28.
20. 16,895 16,543 = 352
22.27 22.3 15.8 352 becomes 158 3520.00 316 360 316 440
316 1240 1106 134 352 miles were driven and the ratio is 22.3 miles to 1 gallon.
210
9 6 4x 2 9 2 4x 6 18 24x 18 24x 24 24 3 x 4
16. Perimeter = 94 + 50 + 94 + 50 = 288 feet The ratio of width to perimeter is 50 feet 2 25 25 . 288 feet 2 144 144 18. The ratio of white blood cells to red blood cells 1 is . 600
x 16 2 6 x 6 2 16 6x 32 6x 32 6 6 16 x 3
30.
a 3 5 2 a 2 53 2a 15 2a 15 2 2 15 a 2 y 5 y 16 3 3y 5( y 16) 3y 5 y 80 2 y 80 2 y 80 2 2 y 40
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ISM: Prealgebra and Introductory Algebra
32.
34.
36.
38.
x 1 5 x2 3 3(x 1) 5(x 2) 3x 3 5x 10 3 2x 10 7 2x 7 2x 2 2 7 x 2
42. Let x be the number of reams. reams 5 x reams weeks 3 15 weeks 5 15 3 x 75 3x 75 3x 3 3 25 x They will need 25 reams or 25 3.125 4 cases. 8
6 27 11 3x 2 6(3x 2) 11 27 18x 12 297 18x 309 18x 309 18 18 103 x 6
44. Let x be the the time to reach the restaurant (in seconds). distance 800 500 distance x time time 60 800 x 60 500 800x 30, 000 800x 30, 000 800 800 x 37.5 It takes 37.5 seconds to reach the restaurant.
7
8 x 3 2x 7 2x 8(x 3) 14x 8x 24 6x 24 6x 24 6 6 x 4 x
Chapter 6: Ratio, Proportion, and Percent
46. a.
Let x be the number of ounces. x ounces ounces 3 feet 150 5280 3 5280 150 x 15,840 150x 15,840 150x 150 150 105.6 x 105.6 ounces of pesticide are needed. feet
2x 5
3 7 7 x 3(2x 5) 7x 6x 15 x 15 40. Let x be the number of ounces. applications 4 35 applications ounces 3 x ounces 4 x 3 35 4x 105 4x 105 4 4 x 27 They should purchase 27 ounces.
b. Let x be the number of cups. ounces 8 105.6 ounces x cups cups 1 8 x 1105.6 8x 105.6 8 8 x 13.2 x 13 This is about 13 cups of pesticide.
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211
Chapter 6: Ratio, Proportion, and Percent
48. Let x be the amount spent on pet food at PetSmart. x pet food 50 pet food total sales 125 6, 676, 700, 000 total sales 50 6, 676, 700, 000 125 x 333,835, 000, 000 125x 333,835, 000, 000 125x 125 125 2, 670, 680, 000 x We expect that $2,670,680,000 was spent on pet food. 50. a.
Let x be the length of your index finger. x index finger index finger 8 height height 111 5 8 5 111 x 40 111x 40 111x 111 111 40 x 0.36 111 40 The proposed length is foot or about 111 0.36 foot.
b. answers may vary 52. Let x be the number of milligrams. milligrams 76 x milligrams ounces 3 8 ounces 76 8 3 x 608 3x 608 3x 3 3 2 202 x 3 2 There are 202 milligrams of cholesterol in 3 8 ounces of chicken. 54. Let x be the number of men that blame. blame 2 x blame total 5 40 total 2 40 5 x 80 5x 80 5x 5 5 16 x You would expect 16 men to blame their not eating well on fast food.
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56. Let x be the number of people who are retired. x retired retired 13 not retired 5 280 not retired 13 280 5 x 3640 5x 3640 5x 5 5 728 x You would expect 728 people age 65 and older who are retired. 58. Let x be the number of pounds. pounds 11 x pounds cups 2 4 6 cups 1 1 6 2 x 4 9 6 x 4 4 6 4 9 x 9 1 9 4 8 x 3 2 x2 3 2 The recipe requires 2 pounds of firmly packed 3 brown sugar. 60. Let x be the cups of salt. ice 5 18 34 ice salt 1 x salt 3 5 x 118 4 75 5x 4 1 1 75 5x 5 5 4 15 x 4 3 x3 4 3 Mix 3 cups of salt with the ice. 4
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
62. a.
b.
64. a.
Let x be the number of gallons of oil needed. 1 oil oil x gas 10 20 gas x 20 10 1 20x 10 20 20 x 0.5 0.5 gallon of oil is needed. Let x be the number of quarts. gallons 1 0.5 gallons x quarts quarts 4 1 x 4 0.5 x2 0.5 gallon is 2 quarts. Let x be the milligrams of medicine. 80 milligrams milligrams x pounds
66.
608
24
608 4 152 6 She should be given 152 milligrams every 6 hours. 300 2 150 2 2 75 2 2 3 25 22 3 5 5 22 3 52
68.
81 3 27 3 3 9 3 3 3 3 34
72. Let x be the number of ml. mg 15 33 mg x ml ml 1 15 x 1 33 15x 33 15x 33 15 15 x 2.2 2.2 ml of the medicine should be administered. 74. Let x be the number of ml. mg 8 6 mg ml 1 x ml 8 x 1 6 8x 6 8x 6 8 8 x 0.75 0.75 ml of the medicine should be administered.
pounds
190 25 25 x 190 80 25x 15, 200 25x 15, 200 25 25 x 608 The daily dose is 608 milligrams. b.
Chapter 6: Ratio, Proportion, and Percent
Section 6.2 Practice Exercises 1. Since 27 students out of 100 students in the club 27 . Then are freshmen, the fraction is 100 27 27%. 100 2.
31
31%
100 3. 49% = 49(0.01) = 0.49 4. 3.1% = 3.1(0.01) = 0.031 5. 175% 175(0.01) = 1.75 6. 0.46% = 0.46(0.01) = 0.0046 7. 600% = 600(0.01) = 6.00 or 6 8. 50% 50
1 100
9. 2.3% 2.3
50 1 50 1 100 2 50 2
2.3
1 100
10. 150% 150
1 100
70. answers may vary
Copyright © 2025 Pearson Education, Inc.
2.3 10
100
100 10
150
100
3 50 2 50
23 1000
3 2
or 1
1 2
213
ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent 1 11. 66 2 % 66 2 3 3 100 200 1 3 100 2 100 1 3 2 3 12. 12% 12
1
100
20.
3 4 3 100 25 4 25
12
13. 0.14 = 0.14(100%) = 14.% or 14% 14. 1.75 = 1.75(100%) = 175.% or 175% 15. 0.057 = 0.057(100%) = 05.7% or 5.7%
3
3 100% 300 % 17.65% 17 17 17 1 17 17.647 17.65 17 300.000 17 130 119 11 0 10 2 80 68 120 119 1 3 Thus, is approximately 17.65%. 17
3
100%
16. 0.5 = 0.5(100%) = 050.% or 50% 3
3 100 300 17. 100% % % 12% 25 25 25 1 25 18.
9 40
19. 5
3
9
100% 40 9 100 % 40 1 900 % 40 20 22 % 40 1 22 % 2
1 11 2 2 11 100% 2 11 100 % 2 1 1100 % 2 550%
21. As a decimal 27.5% = 27.5(0.01) = 0.275. 1 As a fraction 27.5% 27.5 100 27.5 100 27.5 10 100 10 275 1000 11 25 40 25 11 . 40 Thus, 27.5% written as a decimal is 0.275, and 11 written as a fraction is . 40 3 7 7 7 100% 700 % 175% 22. 1 100% 4 4 4 4 1 4 3 Thus, a “1 times” increase is the same as a 4 “175% increase.” Vocabulary, Readiness & Video Check 6.2 1. Percent means “per hundred.” 2. 100% = 1. 3. The % symbol is read as percent. 4. To write a decimal or a fraction as a percent, multiply by 1 in the form of 100%.
214
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ISM: Prealgebra and Introductory Algebra
5. To write a percent as a decimal, drop the % symbol and multiply by 0.01. 6. To write a percent as a fraction, drop the % 1 symbol and multiply by . 100 7. Percent means “per 100.” 8. 100% 9. Multiplying by 100% is the same as multiplying by 1. 10. The difference is how the percent symbol is replacedfor a decimal, replace % with 0.01 and for a fraction, replace % with
1 . 100
Chapter 6: Ratio, Proportion, and Percent 1
26. 2% 2
81% of this player’s attempts were made. 4. Only 3 out of 100 adults preferred volleyball. 3 3% 100 6. 20 of the adults preferred basketball, while 15 preferred baseball. Thus, 20 + 15 = 35 preferred basketball or baseball. 35 35% 100
10. 3% = 3(0.01) = 0.03
14. 45.7% = 45.7(0.01) = 0.457 16. 1.4% = 1.4(0.01) = 0.014 18. 0.9% = 0.9(0.01) = 0.009 20. 500% = 500(0.01) = 5.00 or 5 22. 72.18% = 72.18(0.01) = 0.7218
100
22 100
1 50
1 28. 7.5% 7.5 100 7.5 100 7.5 10 100 10 75 1000 3 25 40 25 3 40 30. 275% 275
1
100
275
100
11 25 4 25
11 4
or 2
3 4
1
100 8.75 100 8.75 100 100 100 875 10, 000 7 125 80 125 7 80
7 7 1 36. 21 % 21 8 8 100 175 1 8 100 175 800 7 25 32 25 7 32
12. 136% = 136(0.01) = 1.36
1
2 50
3 3 1 31 1 31 34. 7 % 7 4 4 100 4 100 400
8. 62% = 62(0.01) = 0.62
24. 22% 22
100
32. 8.75% 8.75
81 81% 100
2 1
100
Exercise Set 6.2 2.
2
2 11 2 50
11
38. 0.44 = 0.44(100%) = 44%
50
40. 0.008 = 0.008(100%) = 0.8% 42. 2.7 = 2.7(100%) = 270%
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ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent
44. 0.019 = 0.019(100%) = 1.9%
68.
50. 0.8 = 0.8(100%) = 80%
54.
56.
58.
60.
9
100%
900
10
10
10
2
2
200
5
5
41
100%
% 90%
100%
% 40%
4100
50
50
50
5
5
500
16
16
5
5
6
100%
%
16
100%
6
4
70. 1
% 31 % 4
500
1 % 88 % 6 3
7
2
10 66.
216
11
7 10
27
100%
10
1100 100% % 12 12 12 91.666 91.67 12 1100.000 108 20 12 80 7 2 80 72 80 72 8 11 is approximately 91.67%. 12
11
100%
%
Percent
Decimal
Fraction
52.5%
0.525
21 40
75%
0.75
3 4
66 1 %
0.665
133 200
3
0.8333
5 6
100%
1
1
14%
0.14
7 50
Percent
Decimal
Fraction
800%
8
8
320%
3.2
3 15
608%
6.08
2 6 25
926%
9.26
9 13
2
83 1 %
1 1 31 3100 62. 6 6 100% 100% % 620% 5 5 5 5 64. 2
1000
10 is approximately 90.91%. 11
% 82% 125
100%
99 1
5 41
10
11 11 90.909 90.91 11 1000.000 99 10 0 9 9 100
48. 9.00 = 9.00(100%) = 900%
9
11
46. 0.1115 = 0.1115(100%) = 11.15%
52.
10
2700
270%
10 72.
74. 38% = 38(0.01) = 0.38 1 38 19 38% 38 100 100 50 76. 70% = 70(0.01) = 0.70 1 70 7 70% 70 100 100 10
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ISM: Prealgebra and Introductory Algebra
78. 20.2% = 20.2(0.01) = 0.202 1 20.2% 20.2 100 20.2 100 20.2 10 100 10 202 1000 101 500
Chapter 6: Ratio, Proportion, and Percent
92. 6 2 4 5 20 29 3 6 3 6 20 6 3 29 20 2 3 3 29 40 29 11 1 29 94. a.
80. 9.6% = 9.6(0.01) = 0.096 1 9.6 10 96 12 9.6% 9.6 100 100 10 1000 125 82. 59.5% = 59.5(0.01) = 0.595 1 59.5% 59.5 100 59.5 100 59.5 10 100 10 595 1000 119 200
0.5269% rounded to the nearest tenth percent is 0.5%.
b. 0.5269% rounded to the nearest hundredth percent is 0.53%. 96. a.
0.231 = 0.231(100%) = 23.1% CORRECT
b. 5.12 = 5.12(100%) = 512% INCORRECT c.
3.2 = 3.2(100%) = 320% CORRECT
d. 0.0175 = 0.0175(100%) = 1.75% INCORRECT a and c are correct.
84. 0.1144 = 0.1144(100%) = 11.44%
98. 45% + 30% + 20% = 95% 100% 95% = 5% Other is 5% of the top four components of bone.
1 100% 100 % 25% 86. 4 4 4
100. 5 of 8 equal parts are shaded so 5 5 500 100% % 62.5% 8 8 8
88. 18.6% 18.6 1 100 18.6 100 18.6 10 100 10 186 1000 93 500
102. A decimal written as a percent is less than 100% when the decimal is less than 1.
2 5 2 5 7 3 90. 11 11 11 11 11 11 73 1111 21 121
108. The percent change for statistician is 33%, so 33% = 33(0.01) = 0.33.
1
104.
56 0.5490 0.549 102 0.549(100%) = 54.9%
106. The second longest bar corresponds to data scientist, so that is predicted to be the second fastest growing occupation.
110. answers may vary
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Chapter 6: Ratio, Proportion, and Percent
Section 6.3 Practice Exercises
9. 15% of wh– a– t nu– mb– er is 2.4?
1. 8 is wha–t perce – nt – of 48? 8
x
15% x 2.4 0.15 x 2.4 0.15x 2.4 0.15x 0.15 x 16 Then 15% of 16 is 2.4. Is this result reasonable? To see, round 15% to 20%. Then 20% of 16 or 0.20(16) = 3.2, which is close to 2.4.
48
2. 2.6 is 40% of wh– a– t nu– mb– er ? 2.6 40%
x
3. Wh– atnu– m– ber is 90% of 0.045? x
90% 0.045
4. 56% of 180 is wh– a– t nu– mb– er ? 56% 180
x
5. 12% of wh– a– t nu– mb– er is 21? 12%
x
21
11. Wh atpe – rc – e–nt of 90 is 27?
6. Wh– atperc ent of 95 is 76? –– x
1 10. 18 is 4 % of wha–– t nu– mb– er ? 2 1 x 18 4 % 2 18 0.045x 18 0.045x 0.045 0.045 400 x 1 Then 18 is 4 2 % of 400.
95 76
7. Wh– atnu– m– ber is 25% of 90? x 25% 90 x = 25% 90 x = 0.25 90 x = 22.5 Then 22.5 is 25% of 90. Is this reasonable? To see, round 25% to 30%. Then 30% or 0.30(90) is 27. Our result is reasonable since 22.5 is close to 27.
x 90 27 90x 27 90x 27 90 390 x 10 or x 0.30 Since we are looking for percent, we can write 3 or 0.30 as a percent. x = 30% 10 Then 30% of 90 is 27. To check, see that 30% 90 = 27.
8. 95% of 400 is wh– a– t nu– mb– er ? 95% 400 x 0.95 400 x 380 x Then 95% of 400 is 380. Is this result reasonable? To see, round 95% to 100%. Then 100% of 400 or 1.0(400) = 400, which is close to 380. 218
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12. 63 is wh– at perce nt of 45? ––
Chapter 6: Ratio, Proportion, and Percent
63 x 45 63 45x 63 45x 45 45 7 x 5 1.4 x 140% x Then 63 is 140% of 45.
9.2 =
x
2. The word of usually translates to “multiplication.” 3. In the statement “10% of 90 is 9,” the number 9 is called the amount, 90 is called the base, and 10 is called the percent. 4. 100% of a number = the number 5. Any “percent greater than 100%” of “a number” = “a number greater than the original number.” 6. Any “percent less than 100%” of “a number” = “a number less than the original number.” 7. “of” translates to multiplication; “is” (or something equivalent) translates to an equal sign; “what” or “unknown” translates to our variable. 8. It is already solved for x; we just need to simplify the left side.
2. 36% of 72 is wh– a– t nu– mb– er ? x
x
40%
x
=6
= 25% 55
375 = 300
14. 25% 68 x 0.25 68 x 17 x 25% of 68 is 17. 16. x = 18% 425 x = 0.18 425 x = 76.5 76.5 is 18% of 425. 18. 0.22 44% x 0.22 0.44x 0.22 0.44x 0.44 0.44 0.5 x 0.22 is 44% of 0.5 1 20. 4 % x 45 2 0.045x 45 0.045x 45 0.045 0.045 x 1000 1 4 % of 1000 is 45. 2 22.
4. 40% of wh– a– t nu– mb– er is 6?
92
12. Watpe h – rc – e–nt of 375 is 300?
1. The word is translates to “=.”
Exercise Set 6.3
x
10. Wh–at nu– m– ber is 25% of 55?
Vocabulary, Readiness & Video Check 6.3
36% 72 =
of 92?
8. 9.2 is wha–t perce – nt –
x 40 60 40 x 60 40 40 x 1.5 x 150% 60 is 150% of 40.
6. 0.7 is 20% of wh– at–nu– mb– er ? 0.7 = 20%
x Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
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24.
26.
48 x 50 48 x 50 50 50 0.96 x 96% x 48 is 96% of 50.
40.
4.8% 32 x 0.048 32 x 1.536 x 4.8% of 32 is 1.536.
42.
x 500 2 x 500 2 500 500 x 0.004 x 0.4% 0.4% of 500 is 2.
44.
9.75 7.5% x 9.75 0.075x 9.75 0.075x 0.075 0.075 130 x 9.75 is 7.5% of 130.
0.5 5% x 0.5 0.05x 0.5 0.05x 0.05 0.05 10 x 0.5 is 5% of 10.
28. 300% 56 x 3.00 56 x 168 x 300% of 56 is 168.
30.
32.
1 7.2 6 % x 4 7.2 0.0625x 7.2 0.0625x 0.0625 0.0625 115.2 x 1 7.2 is 6 % of 115.2. 4 2.64 x 25 2.64 x 25 25 25 0.1056 x 10.56% x 2.64 is 10.56% of 25.
46. 2520 x 3500 2520 x 3500 3500 3500 0.72 x 72% x 2520 is 72% of 3500. 48.
50.
x 120 76.8 x 120 76.8 120 120 x 0.64 x 64% 64% of 120 is 76.8.
52.
38. 160% x 40 1.6x 40 1.6 x 40 1.6 1.6 x 25 160% of 25 is 40. 220
35
7
x 5 35 5 x 7 175 7x 175 7x 7 7 25 x
34. x = 36% 80 x = 0.36 80 x = 28.8 28.8 is 36% of 80. 36.
2 x 40 2 x 40 40 40 0.05 x 5% x 2 is 5% of 40.
54.
x 6 3 13 x 13 3 6 13x 18 13x 18 13 13 5 x 1 13 20 x 25 10
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ISM: Prealgebra and Introductory Algebra
56.
Chapter 6: Ratio, Proportion, and Percent
2.
5 15 6 x
amount 30 p 90 100
58. 68 8 n 68 8 n 8 8 68 n 8 Choice b is correct.
4.
64. Since 100% of 98 is 98, the percent is 100%; a.
68. Since 230% is greater than 100%, which is 1, 230% of 45 is greater than 45; b.
5. Whatpe ent of 40 – rc ––
6.
x 75, 528 27, 945.36 x 75, 528 27, 945.36 75, 528 75, 528 x 0.37 x 37% 37% of 75,528 is 27,945.36.
wh– at–nu– mb– er
percent
base It appears after the word of .
54 27 b 100
is 75?
46% of 80 is wh– a– t nu– mb– er ? percent base amount a 46 80 100
7. Wh– at nu– m– ber is 8% of 120?
Section 6.4 Practice Exercises of
percent base amount 75 p 40 100
72. Since 180% is greater than 100%, which is 1, 180% of y is greater than y, so y is less than 45; c. 74. answers may vary
65% of wh– at–nu– mb– er ?
680 65 100 b
70. Since 30% is less than 100%, which is 1, 30% of y is less than y, so y is greater than 45; b.
680 is
amount percent base
66. Since 100% of 35 is 35, and 50 is greater than 35, x must be greater than 100%; b.
27%
base
amount percent base a 25 116 100
62. answers may vary
1.
percent
3. Wh– at nu– m– ber is 25% of 116?
60. n = 0.7 12 n = 8.4 Choice a is correct.
76.
30 is wha–t perce nt of 90? ––
is
54? amount It is the part compared to the whole
amount percent base a 8 120 100 a 100 120 8 100a 960 100a 960 100 100 a 9.6 9.6 is 8% of 120.
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8.
percent base amount 52 65 b 100 52 100 b 65 5200 65b 5200 65b 65 65 80 b Thus, 65% of 80 is 52. 9.
Vocabulary, Readiness & Video Check 6.4
65% of wh– at–nu– mb– er is 52?
15.4 is
5% of wh– a– t nu– mb– er ?
amount percent base 15.4 1 15.4 5 or b b 20 100 15.4 20 1 b 308 b So, 154 is 5% of 308. 10. Whatpe – rc – e–nt of 40
414 is wha–t perce nt of 180? –– base amount percent 414 p 180 100 414 100 180 p 41, 400 180 p 41, 400 180 p 180 180 230 p Then 414 is 230% of 180.
222
2. In the question “50% of what number is 28?,” which part of the percent proportion is unknown? base 3. In the question “What number is 25% of 200?,” which part of the percent proportion is unknown? amount 4. In the question “38 is what percent of 380?,” which part of the percent proportion is unknown? percent 5. 45 follows the word “of,” so it is the base. 6. 100 Exercise Set 6.4 2.
is 8?
percent base amount 8 p 1 p or 40 100 5 100 1100 5 p 100 5 p 100 5 p 5 5 20 p So, 20% of 40 is 8. 11.
1. When translating the statement “20% of 15 is 3” to a proportion, the number 3 is called the amount, 15 is the base, and 20 is the percent.
92% of 30 is wh– at–nu– mb– er ? percent base a 92 30 100
amount = a
4. Wh– at nu– m– ber is 7% of 175? amount = a percent base a 7 175 100 6.
1.2 is 47% of wh– a– t nu– mb– er ? amount percent 1.2 47 b 100
8.
base = b
85% of wh– a– t nu– mb– er is 520? percent base = b 520 85 b 100
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amount
ISM: Prealgebra and Introductory Algebra
10. Whatpe – rc – e–nt of 900 is 216? percent = p 216 p 900 100 12.
22.
base amount
9.6 is wha–t perce nt of 96? –– amount percent = p 9.6 p 96 100
14.
Chapter 6: Ratio, Proportion, and Percent
a
25
84 100 a 4 84 1 4a 84 4a 84 4 4
or
a 84
base
24.
1 4
26.
18.
20.
a 60 a 3 or 29 100 29 5 a 5 29.3 5a 87 5a 87 5 5 a 17.4 17.4 is 60% of 29. 55 11 55 55 or b 100 b 20 55 20 b 11 1100 11b 1100 11b 11 11 100 b 55% of 100 is 55. 1.1 11 1.1 44 or b 100 b 25 1.1 25 b 11 27.5 11b 27.5 11b 11 11 2.5 b 1.1 is 44% of 2.5.
24 p 12 p or 50 100 25 100 12 100 25 p 1200 25 p 1200 25 p 25 25 48 p 24 is 48% of 50. 7.4 5 7.4 1 b 100 or b 20
7.4 20 b 1 148 b 7.4 is 5% of 148.
a 21 25% of 84 is 21. 16.
147 p or 3 p 98 100 2 100 3 100 2 p 300 2 p 300 2 p 2 2 150 p 147 is 150% of 98.
a
28.
30.
2.5 90 100 a 100 90 2.5 100a 225 100a 225 100 100 a 2.25 2.25 is 2.5% of 90. 30
6
or
30
b 100 b 30 50 b 3 1500 3b 1500 3b 3 3 500 b 30 is 6% of 500. 32.
550.4
3 50
p
172 100 550.4 100 172 p 55, 040 172 p 55, 040 172 p 172 172 320 p 550.4 is 320% of 172.
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34.
36.
a 53 130 100 a 100 130 53 100a 6890 100a 6890 100 100 a 68.9 68.9 is 53% of 130. 1.6 p 5 100 1.6 100 5 p 160 5 p 160 5 p 5 5 32 p 32% of 5 is 1.6.
44.
46.
38.
221 170 221 17 or b 100 b 10 22110 b 17 2210 17b 2210 17b 17 17 130 b 170% of 130 is 221.
48.
40.
a 7.8 24 100 a 100 24 7.8 100a 187.2 100a 187.2 100 100 a 1.872 7.8% of 24 is 1.872.
50.
42.
224
3 p 500 100 3100 500 p 300 500 p 300 500 p 500 500 0.6 p 0.6% of 500 is 3.
9.18 6.8 b 100 9.18 100 b 6.8 918 6.8b 918 6.8b 6.8 6.8 135 b 9.18 is 6.8% of 135. a 75 14 100 a 100 14 75 100a 1050 100a 1050 100 100 a 10.50 75% of 14 is 10.5. 9310 p 3800 100 9310 100 3800 p 931, 000 3800 p 931, 000 3800 p 3800 3800 245 p 9310 is 245% of 3800. 7 5 7 2 5 3 14 15 14 15 1 12 8 12 2 8 3 24 24 24 24
2 1 8 9 52. 2 4 3 2 3 2 82 93 32 2 3 16 27 6 6 43 6 1 7 6 1
54.
10.78 4.30 0.21 15.29
56.
16.37 2.61 13.76
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ISM: Prealgebra and Introductory Algebra
58. answers may vary
4. The ratio of 1.6 to 4.6 is 1.6 1.6 10 16 8 . 4.6 4.6 10 46 23
520
60.
62.
65 b 100 520 65 0 800 100 520 100 0 800 65 52, 000 52, 000 True Yes the base is 800. 36
5. The ratio of the width to the length is 12 inches 12 2 . 18 inches 18 3 6. a.
7.
8.
64. answers may vary
8,854, 200 110, 736 p 110, 736 110, 736 79.958 p 80.0% of 110,736 is 88,542. Mid-Chapter Review 18
20 2. The ratio of 36 to 100 is
36
9
.
10
9 . 100 25
3. The ratio of 8.6 to 10 is 8.6 8.6 10 86 43 . 10 10 10 100 50
3.5 7 12.5 z 3.5 z 12.5 7 3.5z 87.5 3.5z 87.5 3.5 3.5 z 25 x 7 2x 3 5 5(x 7) 3 2x 5x 35 6x 35 x
9. Let x be the number of weeks that a gross of boxes is likely to last. boxes 5 144 boxes x weeks weeks 3 5 x 3144 5x 432 5x 432 5 5 x 86.4 A gross of boxes of envelopes is likely to last about 86 weeks.
88, 542 p 110, 736 100 88, 542 100 110, 736 p 8,854, 200 110, 736 p
1. The ratio of 18 to 20 is
67 films were rated R.
b. 40 + 61 + 67 + 32 = 200 The ratio of PG-13 films to total films was 61 films 61 . 200 films 200
p
12 100 36 50 0 12 100 1 3 False 2 The percent is not 50. 36 p 12 100 36 100 12 p 3600 12 p 3600 12 p 12 12 300 p The percent is 300 (300%).
66.
Chapter 6: Ratio, Proportion, and Percent
10. Use 4.3 for the number of weeks in a month. Let x be the number of boxes that will be needed in a month. x boxes boxes 5 weeks 3 4.3 weeks 5 4.3 3 x 21.5 3x 21.5 3x 3 3 7.2 x Seven boxes should be purchased to last a month.
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Chapter 6: Ratio, Proportion, and Percent
11. 0.94 = 0.94(100%) = 94%
33. 6.8% = 6.8(0.01) = 0.068 1 6.8 6.8 10 68 17 6.8% 6.8 100 100 100 10 1000 250
12. 0.17 = 0.17(100%) = 17% 3 13.
8
300
100%
8
7 14.
3
2
7 2
% 37.5%
34. 11.25% = 11.25(0.01) = 0.1125 1 11.25% 11.25 11.25 100 100 11.25 100 100 100 1125 10, 000 9 80
8 100%
700% 350%
2
15. 4.7 = 4.7(100%) = 470% 16. 8 = 8(100%) = 800% 17.
9
20 18.
53
20. 3
3 4 1 4
100%
20
50 19. 6
9
53
900 20
100%
5300
50
27
13
4
4
% 45% % 106%
50
27
13
4
100%
100%
2700 4
1300
4
21. 0.02 = 0.02(100%) = 2% 22. 0.06 = 0.06(100%) = 6%
4
% 675%
% 325%
35. 74% = 74(0.01) = 0.74 1 74 37 74% 74 100 100 50 36. 45% = 45(0.01) = 0.45 1 45 9 45% 45 100 100 20 37. 16 1 % 16.33(0.01) 0.163 3 1 49 49 1 49 16 % % 3 3 3 100 300
23. 71% = 71(0.01) = 0.71 24. 31% = 31(0.01) = 0.31 25. 3% = 3(0.01) = 0.03 26. 4% = 4(0.01) = 0.04 27. 224% = 224(0.01) = 2.24 28. 700% = 700(0.01) = 7.0 29. 2.9% = 2.9(0.01) = 0.029 30. 6.6% = 6.6(0.01) = 0.066 31. 7% = 7(0.01) = 0.07 1 7 7% 7 100 100
2 38. 12 % 12.66(0.01) 0.127 3 2 38 38 1 38 19 12 % % 3 3 3 100 300 150 39.
a 15 90 100 a 100 15 90 100a 1350 100a 1350 100 100 a 13.5 15% of 90 is 13.5.
32. 5% = 5(0.01) = 0.05 1 5 1 5% 5 100 100 20
226
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Chapter 6: Ratio, Proportion, and Percent
40.
78 78 b 100 78 100 78 b 7800 78b 7800 78b 78 78 100 b 78% of 100 is 78.
45.
a 40 85 100 a 100 40 85 100a 3400 100a 3400 100 100 a 34 34 is 40% of 85.
41.
297.5 85 b 100 297.5 100 85 b 29, 750 85b 29, 750 85b 85 85 350 b 297.5 is 85% of 350.
46.
128.7 p 99 100 128.7 100 p 99 12,870 99 p 12,870 99 p 99 99 130 p 130% of 99 is 128.7.
42.
78 p 65 100 78 100 65 p 7800 65 p 7800 65 p 65 65 120 p 78 is 120% of 65.
47.
115 p 250 100 115 100 p 250 11, 500 250 p 11, 500 250 p 250 250 46 p 46% of 250 is 115.
43.
44.
23.8 p 85 100 23.8 100 85 p 2380 85 p 2380 85 p 85 85 28 p 23.8 is 28% of 85. a 38 200 100 a 100 38 200 100a 7600 100a 7600 100 100 a 76 38% of 200 is 76.
48.
49.
a 45 84 100 a 100 45 84 100a 3780 100a 3780 100 100 a 37.8 37.8 is 45% of 84. 63 42 b 100 63 100 42 b 6300 42b 6300 42b 42 42 150 b 42% of 150 is 63.
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50.
195 p 3326 100 195 100 3326 p 19, 500 3326 p 19, 500 3326 p 3326 19, 500 6 p It is estimated that the number of nurses employed will increase by 6%.
58.9 95 b 100 58.9 100 95 b 5890 95b 5890 95b 95 95 62 b 95% of 62 is 58.9.
Section 6.5 Practice Exercises
3. Method 1: 864 is 32% of wh– a– t nu– mb– er ?
1. Method 1: Wh– at nu– m– ber is 25% of 2174? x = 25% 2174 x = 0.25 2174 x = 543.5 We predict 543.5 miles of the trail resides in the state of Virginia. Method 2: Wh– at nu– m– ber is 25% of 2174? amount percent base a 25 2174 100 a 100 2174 25 100a 54, 350 100a 54, 350 100 100 a 543.5 We predict 543.5 miles of the trail resides in the state of Virginia. 2. Method 1: 195 is wha–t perce nt of 3326? ––
228
percent
base
amount percent base 864 32 b 100 864 100 b 32 86, 400 32b 86, 400 32b 32 32 2700 b There are 2700 students at Euclid University. 4. Method 1: Wh tanu– m– ber is 1.4% of 287 million? –
195 = x 3326 195 3326x 195 3326x 3326 3326 0.06 x 6% x It is estimated that the number of nurses employed will increase by 6%. Method 2: 195 is wha–t perce – nt – of 3326? amount
864 = 32% x 864 0.32 x 864 0.32x 0.32 0.32 2700 x There are 2700 students at Euclid University. Method 2: 864 is 32% of wh– a– t nu– mb– er ?
x = 1.4% 287 x = 0.014 287 x = 4.018 a.
The increase in the number of registered vehicles on the road in 2022 was 4.018 million.
b.
The total number of registered vehicles on the road in 2022 was 287 million + 4.018 million = 291.018 million
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ISM: Prealgebra and Introductory Algebra
Method 2: h at nu– m– ber is 1.4% of 287 million? – amount percent a 1.4 287 100 a 100 287 1.4 100a 401.8 100a 401.8 100 100 a 4.018
base
Chapter 6: Ratio, Proportion, and Percent
Vocabulary, Readiness & Video Check 6.5 1. The price of the home is $375,000. 2. An improper fraction is greater than 1, so our percent increase is greater than 100%. Exercise Set 6.5
a.
The increase in the number of registered vehicles on the road in 2022 was 4.018 million.
b.
The total number of registered vehicles on the road in 2022 was 287 million + 4.018 million = 291.018 million
5. Find the amount of increase by subtracting the original number of attendants from the new number of attendants. amount of increase = 333 285 = 48 The amount of increase is 48 attendants. amount of increase percent of increase original amount 48 285 0.168 16.8% The number of attendants to the local play, Peter Pan, increased by about 16.8%. 6. Find the amount of decrease by subtracting 18,483 from 20,200. Amount of decrease = 20,200 18,483 = 1717 The amount of decrease is 1717. amount of decrease percent of decrease original amount 1717 20, 200 0.085 8.5% The population decreased by 8.5%.
2. 28 is 35% of what number? Method 1: 28 35% x 28 0.35x 28 0.35x 0.35 0.35 80 x 80 children attend this day care center. Method 2: 28 35 b 100 28 100 b 35 2800 35b 2800 35b 35 35 80 b 80 children attend this day care center. 4. 60% of 220 is what number? Method 1: 60% 220 x 0.60 220 x 132 x The maximum weight resistance is 132 pounds. Method 2: a 60 220 100 100 a 220 60 100a 13, 200 100a 13, 200 100 100 a 132 The maximum weight resistance is 132 pounds. 6. 154 is what percent of 3850? Method 1: 154 x 3850 154 3850 x 3850 3850 0.04 x 4% x Pierre received 4% of his total spending at the food cooperative as a dividend.
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Method 2: 154 p 3850 100 154 100 p 3850 15, 400 3850 p 15, 400 3850 p 3850 3850 4 p Pierre received 4% of his total spending at the food cooperative as a dividend. 8. 62% of 43,600 is what number? Method 1: 62% 43, 600 x 0.62 43, 600 x 27, 032 x 27,032 screens were digital non-3-D. Method 2: a 62 43, 600 100 a 100 43, 600 2 100a 2, 703, 200 100a 2, 703, 200 100 100 a 27, 032 27,032 screens were digital non-3-D. 10. What percent of 121,461 is 73,900? Method 1: x 121, 461 73, 900 121, 461x 73, 900 121, 461 121, 461 x 0.61 x 61% Of the 121,461 veterinarians in the United States in 2021, 61% practiced medicine in a clinic. Method 2: 73, 900 p 121, 461 100 73, 900 100 121, 461 p 7, 390, 000 121, 461p 7, 390, 000 121, 461p 121, 461 121, 461 61 p Of the 121,461 veterinarians in the United States in 2021, 61% practiced medicine in a clinic. 12. 5% of 7640 is what number? Method 1: 5% 7640 x 0.05 7640 x 382 x 230
There are 382 fewer students this year. The current enrollment is 7640 382 = 7258 students. Method 2: a 5 7640 100 a 100 7640 5 100a 38, 200 100a 38, 200 100 100 a 382 There are 382 fewer students this year. The current enrollment is 7640 382 = 7258 students. 14. 22% of 467 million is what number? Method 1: 22% 467 x 0.22 467 x 102.74 x a.
The increase in the revenue from vinyl albums was $102.74 million.
b.
The 2022 revenue from vinyl albums was $467 million + $102.74 million = $569.74 million.
Method 2: a 22 467 100 a 100 467 22 100a 10, 274 100a 10, 274 100 100 a 102.74 a.
The increase in the revenue from vinyl albums was $102.74 million.
b.
The 2022 revenue from vinyl albums was $467 million + $102.74 million = $569.74 million.
16. 1.1% of 2,960,000 is what number? Method 1: 1.1% 2, 960, 000 x 0.011 2, 960, 000 x 32, 560 x The population of Mississippi in 2023 was 2,960,000 32,560 = 2,927,440.
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ISM: Prealgebra and Introductory Algebra
Method 2: a 1.1 2, 960, 000 100 a 100 2, 960, 000 1.1 100a 3, 256, 000 100a 3, 256, 000 100 100 a 32, 560 The population of Mississippi in 2023 was 2,960,000 32,560 = 2,927,440. 18. 28 is what percent of 115? Method 1: 28 x 115 28 115x 115 115 0.24 x 24% x 24% of the runs are easy. Method 2: 28 p 115 100 28 100 115 p 2800 115 p 2800 115 p 115 115 24 p 24% of the runs are easy.
Chapter 6: Ratio, Proportion, and Percent
22. 130 is what percent of 190? Method 1: 130 x 190 130 190 x 190 190 0.684 x 68.4% x 68.4% of the total calories come from fat. Method 2: 130 p 190 100 130 100 190 p 13, 000 190 p 13, 000 190 p 190 190 68.4 p 68.4% of the total calories come from fat. 24. 35 is what percent of 130? Method 1: 35 x 130 35 130 x
20. 5 is what percent of 20? Method 1: 5 x 20 5 20x 20 20 0.25 x 25% x 25% of the total calories come from fat. Method 2: 5 p 20 100 5 100 20 p 500 20 p 500 20 p 20 20 25 p 25% of the total calories come from fat.
35 130 x 130 130 0.269 x 26.9% x 26.9% of the total calories come from fat. Method 2: 35 p 130 100 35 100 130 p 3500 130 p 3500 130 p 130 130 26.9 p 26.9% of the total calories come from fat. 26. 842.40 is 18% of what number? Method 1: 842.40 18% x 842.40 0.18x 842.40 0.18x 0.18 0.18 4680 x The banker’s total pay is $4680.
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Chapter 6: Ratio, Proportion, and Percent
Method 2: 842.40 18 b 100 842.40 100 b 18 84, 240 18b 84, 240 18b 18 18 4680 b The banker’s total pay is $4680. 28. What number Method 1: is 1.04% of 28,350? x = 1.04% 28,350 x = 0.0104 28,350 x = 294.84 There are expected to be 295 defective components in the batch. Method 2: a 1.04 28, 350 100 a 100 28, 350 1.04 100a 29, 484 100a 29, 484 100 100 a 294.84 There are expected to be 295 defective components in the batch. 30. What number is 6.5% of 58,500? Method 1: x = 6.5% 58,500 x = 0.065 58,500 x = 3802.5 The union worker can expect an increase of $3802.50 in salary. The new salary will be $58,500 + $3802.50 = $62,302.50. Method 2: a 6.5 58, 500 100 a 100 58, 500 6.5 100a 380, 250 100a 380, 250 100 100 a 3802.50 The union worker can expect an increase of $3802.50 in salary. The new salary will be $58,500 + $3802.50 = $62,302.50.
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32. 126 is 70% of what number? Method 1: 126 70% x 126 0.70x 126 0.70x 0.70 0.70 180 x $126 is 70% of the total price of $180. Method 2: 126 70 100 126 b 100 b 70 12, 600 70b 12, 600 70b 70 70 180 b $126 is 70% of the total price of $180. 34. 43.8% of 56 million is what number? Method 1: 43.8% 56 million x 0.438 56 million x 24.528 million x 56 million + 24.528 million = 80.528 million The increase is 24.528 million and the projected population in 2040 is 80.528 million. Method 2: a 43.8 56 million 100 a 100 56 million 43.8 100a 2452.8 million 100a 2452.8 million 100 100 a 24.528 million 56 million + 24.528 million = 80.528 million The increase is 24.528 million and the projected population in 2040 is 80.528 million. 36. What number is 3% of 1960 thousand? Method 1: x = 3% 1960 x = 0.03 1960 x = 58.8 The increase is projected to be 58.8 thousand. The number of bachelor degrees awarded in 20282029 is projected to be 1960 thousand + 58.8 thousand = 2018.8 thousand.
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ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent
Method 2: a 3 1960 100 a 100 1960 3 100a 5880 100a 5880 100 100 a 58.8 The increase is projected to be 58.8 thousand. The number of bachelor degrees awarded in 20282029 is projected to be 1960 thousand + 58.8 thousand = 2018.8 thousand. Original Amount
New Amount
Amount of Increase
38.
8
12
12 8 = 4
40.
68
170 68 = 102
170
Original Amount
New Amount
Amount of Decrease
42.
25
20
25 20 = 5
44.
200
162
200 162 = 38
46. percent of decrease
amount of decrease
48. percent of decrease
amount of decrease
Percent of Increase 4
0.5 50%
8 102
1.5 150%
68
Percent of Decrease 5
0.20 20%
25 38
0.19 19%
200
original amount 530 477 530 53 530 0.10 The decrease in the number of employees was 10%.
original amount 10,845 10, 700 10,845 145 10,845 0.013 The decrease in the number of cable TV systems was 1.3%.
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50. percent of decrease
amount of decrease
original amount 125, 536 105,804 125, 536 19, 732 125, 536 0.157 The decrease in population is expected to be 15.7%.
amount of decrease 52. percent of decrease original amount 419, 000 377, 000 419, 000 42, 000 419, 000 0.100 The decrease in correctional officers is expected to be 10.0%.
54. percent of increase
amount of increase
original amount 3.3 3.1 3.1 0.2 3.1 0.065 The increase in registered nurses is expected to be 6.5% 56. percent of decrease
amount of decrease
original amount 2045 1906 2045 139 2045 0.068 The decrease in Ford vehicle sales in the United States from 2020 to 2021 was 6.8%.
58. percent of decrease
amount of decrease
original amount 767 658 767 109 767 0.142 The decrease in boating accidents was 14.2%.
234
60. percent of increase
amount of increase
original amount 6.5 4.7 4.7 1.8 4.7 0.383 The increase in revenue from streaming music was 38.3%.
62. 0.7 29.4 becomes 7
64.
42 294 28 14 14 0
78.00 19.46 58.54
35 5 3 5 3 5 66. 8 12 8 12 8 3 4 32 4 9 4 9 68. 2 3 2 3 5 10 5 10 Subtract the absolute values. 9 9 3 3 10 10 4 8 2 2 5 10 1 1 10
Attach the sign of the larger absolute value. 4 9 1 2 3 1 5 10 10 70. answers may vary 72. What number is 62.1% of 273,800,000? Method 1: x = 62.1% 273,800,000 x = 0.621 273,800,000 x = 170,029,800 The number of Internet users in Indonesia was 170,029,800 in 2021.
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ISM: Prealgebra and Introductory Algebra
Method 2: a 62.1 273,800, 000 100 a 100 273,800, 000 62.1 100a 17, 002, 980, 000 100a 17, 002, 980, 000 100 100 a 170, 029,800 The number of Internet users in Indonesia was 170,029,800 in 2021. 74. 9,200,860 is what percent of 10,420,000? Method 1: 9, 200,860 x 10, 420, 000 9, 200,860 10, 420, 000x 10, 420, 000 10, 420, 000 0.883 x 88.3% of the population of Sweden were Internet users in 2021. Method 2: 9, 200,860 p 10, 420, 000 100 9, 200,860 100 10, 420, 000 p 920, 086, 000 10, 420, 000 p 920, 086, 000 p 10, 420, 000 88.3 p 88.3% of the population of Sweden were Internet users in 2021. 76. 20,090,100 is 16.7% of what number? Method 1: 20, 090,100 16.7% x 20.090,100 0.167 x 20, 090,100 x 0.167 120, 300, 000 x The population of Ethiopia was about 120,300,000 in 2021. Method 2: 20, 090,100 16.7 b 100 20, 090,100 100 b 16.7 2, 009, 010, 000 16.7b 2, 009, 010, 000 16.7b 16.7 16.7 120, 300, 000 b The population of Ethiopia was about 120,300,000 in 2021.
Chapter 6: Ratio, Proportion, and Percent
80. To find the percent of decrease, Payton should have divided by the original amount, which is 180. 180 150 30 percent of decrease 16 2% 180 180 3 Section 6.6 Practice Exercises 1. sales tax tax rate purchase price 8.5% $59.90 0.085 $59.90 $5.09 The sales tax is $5.09. Total Price purchase price sales tax $59.90 $5.09 $64.99 The sales tax on $59.90 is $5.09, and the total price is $64.99. 2. sales tax tax rate purchase price $1665 r $18, 500 1665 r 18, 500 18, 500 18, 500 0.09 r The sales tax rate is 9%. 3. commission commission rate sales 6.6% $47, 632 0.066 $47, 632 $3143.71 The commission for the month is $3143.71. 4. commission commission rate sales $645 r $4300 645 r 4300 0.15 r 15% r The commission rate is 15%.
78. answers may vary
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Chapter 6: Ratio, Proportion, and Percent
5. amount of discount discount rate original price 35% $700 0.35 $700 $245 The discount is $245. sale price original price discount $700 $245 $455 The sale price is $455. Vocabulary, Readiness & Video Check 6.6 1. sales tax = tax rate purchase price 2. total price = purchase price + sales tax 3. commission = commission rate sales 4. amount of discount = discount rate original price 5. sale price = original price amount of discount 6. We rewrite the percent as an equivalent decimal. 7. We write the commission rate as a percent. 8. Replace “amount of discount” in the second equation with “discount rate original price”: sale price = original price (discount rate original price). Exercise Set 6.6 2.
sales tax 6% $188 0.06 $188 $11.28 The sales tax is $11.28.
4. sales tax = 8% $699 = 0.08 $699 = $55.92 Total price = $699 + 55.92 = $754.92 The total price of the stereo system is $754.92. 6.
236
$374 r $6800 374 6800 r 0.055 r The sales tax rate is 5.5%.
8. a.
$76 9.5% p $76 0.095 p $76 p 0.095 $800 p The purchase price of the earrings is $800.
b. total price = $800 + $76 = $876 The total price of the earrings is $876. 10. sales tax = 8% $1890 = 0.08 $1890 = $151.20 total price = $1890 + $151.20 = $2041.20 The sales tax is $151.20 and the total price is $2041.20. 12. $32.85 9% p $32.85 0.09 p $32.85 p 0.09 $365 p The purchase price of the LED TV with DVD player is $365. 14. $103.50 r $1150 103.50 r 1150 0.09 r The sales tax rate is 9%. 16. Purchase price = $35 + $55 + $95 = $185 Sales tax = 6.5% $185 = 0.065 $185 $12.03 Total price = $185 + $12.03 = $197.03 The sales tax is $12.03 and the total price of the clothing is $197.03. 18. commission 12.8% $1638 0.128 $1638 $209.66 Her commission was $209.66. 20.
$3575 r $32, 500 $3575 r $32, 500 0.11 r The commission rate is 11%.
22. commission 5.6% $9638 0.056 $9638 $539.73 Her commission will be $539.73.
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ISM: Prealgebra and Introductory Algebra
24.
Chapter 6: Ratio, Proportion, and Percent
$1750 7% sales $1750 0.07 s $1750 s 0.07 $25, 000 s The selling price of the fertilizer was $25,000. Original Price
Discount Rate
Amount of Discount
Sale Price
26.
$74
20%
20% $74 = $14.80
$78 $14.80 = $59.20
28.
$110.60
40%
40% $110.60 = $44.24
$110.60 $44.24 = $66.36
30.
$370
25%
25% $370 = $92.50
$370 $92.50 = $277.50
32.
$17,800
12%
12% $17,800 = $2136
$17,800 $2136 = $15,664
34. discount = 30% $4295 = 0.30 $4295 = $1288.50 sale price = $4295 $1288.50 = $3006.50 The discount is $1288.50 and the sale price is $3006.50. Purchase Price
Tax Rate
Sales Tax
Total Price
36.
$243
8%
8% $243 = $19.44
$243 + $19.44 = $262.44
38.
$65
8.4%
8.4% $65 = $5.46
$65 + $5.46 = $70.46
Sale
Commission Rate
Commission
40.
$195,450
5%
$195,450 5% = $9772.50
42.
$25,600
$2304
0.09 9%
$2304
$25, 600 44. 500
2
3 40 3 120
25
46. 1000
1
5 50 5 250 20
48. 6000 0.06
3 4
360
3
270
4
50. 10% $200 = 0.10 $200 = $20 $200 + $20 = $220 The best estimate is c; $220.
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Chapter 6: Ratio, Proportion, and Percent
Bill Amount
10%
15%
20%
52.
$15.89 $16
$1.60
1 $1.60 ($1.60) $1.60 $0.80 $2.40 2
2($1.60) = $3.20
54.
$9.33 $9
$0.90
1 $0.90 ($0.90) $0.90 $0.45 $1.35 2
2($0.90) = $1.80
56. A discount of 20% plus an additional 40% is better. Answers may vary. 58.
commission 5.5% $562, 560 0.055 $562, 560 $30, 940.80 agent's amount 60% $30, 940.80 0.60 $30, 940.80 $18, 564.48 The real estate agent will receive $18,564.48.
Section 6.7 Practice Exercises 1. I P R T I $875 7% 5 $875 (0.07) 5 $306.25 The simple interest is $306.25. 2. I P R T I $1500 20%
9
12 9 $1500 (0.20) 12 $225 She paid $225 in interest. 3. I P R T $2100 13%
6
12 6 $2100 (0.13) 12 $136.50 The interest is $136.50. total amount principal interest $2100 $136.50 $2236.50 After 6 months, the total amount paid will be $2236.50.
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4. P = $3000, r = 4% = 0.04, n = 1, t = 6 years nt
r A P 1 n 16 0.04 $3000 1 1 $3000(1.04)6 $3795.96 The total amount after 6 years is $3795.96.
1 5. P = $5500, r 6 % 0.0625, n = 365, t = 5 4 r nt A P 1 n 3655 0.0625 $5500 1 365 $7517.41 The total amount after 5 years is $7517.41.
Calculator Explorations 45
0.09 1. A $600 1 4
$936.31
36515
0.04 2. A $10, 000 1 365
120
0.11 3. A $1200 1 1
$18, 220.59
4.
A
$9674.77
21 0.07 $5800 1 $6213.11 2 44
0.06 5. A $500 1 4
$634.49
Chapter 6: Ratio, Proportion, and Percent
3. Compound interest is computed on not only the original principal, but on interest already earned in previous compounding periods. 4. When interest is computed on the original principal only, it is called simple interest. 5. Total amount (paid or received) = principal + interest 6. The principal amount is the money borrowed, loaned, or invested. 7. Simple interest is calculated on the principal only. 8. semiannually; 2 Exercise Set 6.7 2. simple interest principal rate time $800 9% 3 $800 0.09 3 $216 4. simple interest principal rate time $950 12.5% 5 $950 0.125 5 $593.75 6. simple interest principal rate time 1 $1500 14% 2 4 $1500 0.14 2.25 $472.50 8. simple interest principal rate time 8 $775 15% 12 $775 0.15 2 3 $77.50 10. simple interest principal rate time $1000 10%
36519
0.05 6. A $2500 1 365
1. To calculate simple interest, use I = P R T. 2. To calculate compound interest, use r nt A P 1 . n
12 $1000 0.10 1.5 $150
$6463.85
Vocabulary, Readiness & Video Check 6.7
18
12. simple interest principal rate time $265, 000 8.25% 30 $265, 000 0.0825 30 $655,875 $265,000 + $655,875 = $920,875 The amount of interest is $655,875. The total amount paid is $920,875.
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ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent 14. simple interest principal rate time 9 $4500 14% 12 $4500 0.14 0.75 $472.50 Total amount $4500 $472.50 $4972.50 16. Simple interest principal rate time $2000 8% 5 $2000 0.08 5 $800 Total amount $2000 $800 $2800 r 18. A P 1 nt n 110 0.15 2060 1 1 2060(1.15)10 8333.85 The total amount is $8333.85.
nt
r 20. A P 1 n 415 0.10 1450 1 4 1450(1.025)60 6379.70 The total amount is $6379.70.
nt
r 22. A P 1 n 0.08 36510 3500 1 3650 365 0.08
3500 1 365 $7788.71 The total amount is $7788.71. nt
r 24. A P 1 n 21 0.10 63751 2 2 6375(1.05) 7028.44 The total amount is $7028.44.
nt
r 26. A P 1 n 25 0.08 2000 1 2 10 2000(1.04) 2960.49 The total amount is $2960.49.
nt
r 28. A P 1 n 0.08 3655 2000 1 365 1825 0.08 2000 1 365 2983.52
The total amount is $2983.52. 30. Perimeter = 18 + 16 + 12 = 46 The perimeter is 46 centimeters. 32. Perimeter = 21 + 21 + 21 + 21 = 84 The perimeter is 84 miles. 34. x x x 5 x 5 5 4 x 4 4 5 4 x 36.
3 9 1 3 22 1 3 2 111 2 11 22 3 11 9 3 11 9 3 9
38. answers may vary Chapter 6 Vocabulary Check 1. In a mathematical statement, of usually means “multiplication.” 2. In a mathematical statement, is means “equals.” 3. Percent means “per hundred.” 4. Compound interest is computed not only on the principal, but also on interest already earned in previous compounding periods. amount 5. In the percent proportion percent base = 100 . 6. To write a decimal or fraction as a percent, multiply by 100%. 7. The decimal equivalent of the % symbol is 0.01.
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8. The fraction equivalent of the % symbol is
Chapter 6: Ratio, Proportion, and Percent
1
. 100
5.
9. The percent equation is base percent = amount. 10. percent of decrease =
amount of decrease original amount
11. percent of increase = amount of increase original amount
6.
12. sales tax = tax rate purchase price 13. total price = purchase price + sales tax 14. commission = commission rate sales 15. amount of discount = discount rate original price
7.
16. sale price = original price amount of discount 17. A proportion is a mathematical statement that two ratios are equal. 8. 18. A ratio is the quotient of two numbers or two quantities. Chapter 6 Review 1. 1 dollar = 1 100 cents = 100 cents The ratio of 20 cents to 1 dollar is 1 20 cents 20 . 100 cents 100 5
9.
4.
x 12 2 4 x 4 2 12 4x 24 4x 24 4 4 x6
20 c 2 5 20 5 2 c 100 2c 100 2c 2 2 50 c 2
3 x 1 x 3 2(x 3) 3(x 1) 2x 6 3x 3 6 x3 9 x 4
3
y3 y2 4( y 2) 3( y 3) 4 y 8 3y 9 y 8 9 y 17 y2 5 y 3 3( y 2) 5 y 3y 6 5 y 6 2y 6 2y 2 2 3 y
2. The ratio of four parts red to six parts white is 4 2 . 6 3 3.
32 100 x 100 32 x 100 100 32x 10, 000 32x 10, 000 32 32 x 312.5
10.
x3 2 3x 2 5 5(x 3) 2(3x 2) 5x 15 6x 4 15 x 4 19 x
20 x 1 25 20 25 1 x 500 x
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Chapter 6: Ratio, Proportion, and Percent
11. Let x be the number of parts that can be processed in 45 minutes. parts 300 x parts minutes 20 45 minutes 300 45 20 x 13, 500 20x 13, 500 20x 20 20 675 x The machine can process 675 parts in 45 minutes. 12. Let x be the amount charged for 3 hours of consulting. dollars 90 x dollars hours 8 3 hours 90 3 8 x 270 8x 270 8x 8 8 33.75 x The charge is $33.75 for 3 hours of consulting. 13.
14.
37 37% 100 37% of adults preferred pepperoni. 77
77% 100 77% of free throws were made.
15. 26% = 26(0.01) = 0.26 16. 75% = 75(0.01) = 0.75 17. 3.5% = 3.5(0.01) = 0.035 18. 1.5% = 1.5(0.01) = 0.015 19. 275% = 275(0.01) = 2.75 20. 400% = 400(0.01) = 4.00 or 4 21. 47.85% = 47.85(0.01) = 0.4785 22. 85.34% = 85.34(0.01) = 0.8534 23. 1.6 = 1.6(100%) = 160% 24. 5.2 = 5.2(100%) = 520%
27. 0.71 = 0.71(100%) = 71% 28. 0.65 = 0.65(100%) = 65% 29. 6 = 6(100)% = 600% 30. 9 = 9(100%) = 900% 7 1 31. 7% 7 100 100 1 3 15 32. 15% 15 100 100 20 33. 25% 25
1
25
100 100 1 34. 8.5% 8.5 100 8.5 100 8.510 100 10 85 1000 17 200 1 35. 10.2% 10.2 100 10.2 100 10.2 10 100 10 102 1000 51 500
1 4
36. 16 2 % 50 % 50 1 50 1 3 3 300 6 3 100 10 0 1 37. 33 1 % 100 % 100 1 3 3 3 100 300 3 1 1 10 1 110 38. 110% 110 1 100 10 100 100
25. 0.076 = 0.076(100%) = 7.6%
2 26. 0.085 = 0.085(100%) = 8.5%
242
39.
5
2 100 200 % % 40% 5 1 5
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ISM: Prealgebra and Introductory Algebra
40.
7 7 100 % 700 % 70% 10 10 1 10
41.
7 7 100 % 700 % 175 % 58 1 % 12 12 1 12 3 3
Chapter 6: Ratio, Proportion, and Percent 51. x = 17% 640 x = 0.17 640 x = 108.8 108.8 is 17% of 640. 52.
2 5 100 500 2 42. 1 % % 166 % 3 3 1 3 3 1 5 100 500 % % 125% 43. 1 4 4 1 4 44.
3
5 45.
1
16 5 46. 47.
8
3 100 300 % % 60% 5 1 5
53.
1 100 100 % % 6.25% 16 1 16
5 100 500 % % 62.5% 8 1 8
1250 1.25% x
54.
1250 0.0125x 1250 0.0125x 0.0125 0.0125 100, 000 x 1250 is 1.25% of 100,000. 1 48. x 33 % 24, 000 3 100 1 x 24, 000 3 100 1 x 24, 000 3 x 8000 1 8000 is 33 % of 24,000. 3 49. 124.2 x 540 124.2 540 x 540 540 0.23 x 23% x 124.2 is 23% of 540. 50.
22.9 20% x 22.9 0.20 x 22.9 0.20 x 0.20 0.20 114.50 x 22.9 is 20% of 114.50.
55.
693 x 462 693 462x 462 462 1.5 x 150% x 693 is 150% of 462. 104.5 25 b 100 104.5 100 25 b 10, 450 25b 10, 450 25b 25 25 418 b 104.5 is 25% of 418. 16.5 5.5 b 100 16.5 100 5.5 b 1650 5.5b 1650 5.5b 5.5 5.5 300 b 16.5 is 5.5% of 300. a
30
532 100 a 100 30 532 100a 15, 960 100a 15, 960 100 100 a 159.6 159.6 is 30% of 532. 56.
63 p 35 100 63 100 p 35 6300 35 p 6300 35 p 35 35 180 p 63 is 180% of 35.
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Chapter 6: Ratio, Proportion, and Percent
57.
93.5 p 85 100 93.5 100 p 85 9350 85 p 9350 85 p 85 85 110 p 93.5 is 110% of 85.
58.
a 33 500 100 a 100 33 500 100a 16, 500 100a 16, 500 100 100 a 165 165 is 33% of 500.
Method 2: 200 p 12, 360 100 2000 100 12, 360 p 200, 000 12, 360 p 200, 000 12, 360 p 12, 360 12, 360 16 p 16% of freshmen are enrolled in prealgebra.
244
amount of decrease
61. percent of decrease
59. 1320 is what percent of 2000? Method 1: 1320 x 2000 1320 2000 x 2000 2000 0.66 x 66% x 66% of people own microwaves. Method 2: 1320 p 2000 100 1320 100 2000 p 132, 000 2000 p 132, 000 2000 p 2000 2000 66 p 66% of people own microwaves. 60.
ISM: Prealgebra and Introductory Algebra
2000 is what percent of 12,360? Method 1: 2000 x 12, 360 2000 12, 360 x 12, 360 12, 360 0.16 x 16% x 16% of freshmen are enrolled in prealgebra.
original amount 675 534 675 141 675 0.209 20.9% Violent crime decreased 20.9%.
62. percent of increase
amount of increase
original amount 33 16 16 17 16 1.0625 106.25% The charge will increase 106.25%.
63. Amount of increase = percent increase original amount = 15% $11.50 = 0.15 $11.50 $1.73 Total amount = $11.50 + $1.73 = $13.23 The new hourly rate is $13.23. 64. Amount of decrease percent decrease original amount 4% $215, 000 0.04 $215, 000 $8600 Total amount = $215,000 $8600 = $206,400 $206,400 is expected to be collected next year.
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ISM: Prealgebra and Introductory Algebra 65. sales tax tax rate purchase price 9.5% $250 0.095 $250 $23.75 Total amount = $250 + $23.75 = $273.75 The total price for the coat is $273.75. 66. sales tax tax rate purchase price 8.5% $25.50 0.085 $25.50 $2.17 The sales tax is $2.17. 67. commission commission rate sales 5% $100, 000 0.05 $100, 000 $5000 The sales representative’s commission is $5000. 68. commission commission rate sales 7.5% $4005 0.075 $4005 $300.38 The sales clerk’s commission is $300.38. 69. Amount of discount discount original price 30% $3000 0.30 $3000 $900 Sale price = $3000 $900 = $2100 The amount of discount is $900; the sale price is $2100. 70. Amount of discount discount original price 10% $90 0.10 $90 $9.00 Sale price = $90 $9 = $81 The amount of discount is $9; the sale price is $81. 71. I P R T $4000 12% $4000 0.12
4 12 1 3
$160 The simple interest is $160.
Chapter 6: Ratio, Proportion, and Percent 72. I P R T $650 20%
3
12 $6500 0.20 0.25 $325 The simple interest is $325. nt
r 73. A P 1 n 115 0.12 5500 1 1 5500(1.12)15 30,104.61 The total amount is $30,104.61.
r nt 74. A P 1 n 210 0.11 6000 1 2 6000(1.055)20 17, 506.54 The total amount is $17,506.54. nt
r 75. A P 1 n 45 0.12 100 1 4 100(1.03)20 180.61 The total amount is $180.61. nt
r 76. A P 1 n 420 0.18 1000 1 4 1000(1.045)80 33,830.10 The total amount is $33,830.10.
77. 3.8% = 3.8(0.01) = 0.038 78. 124.5% = 124.5(0.01) = 1.245 79. 0.54 = 0.54(100%) = 54% 80. 95.2 = 95.2(100%) = 9520%
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Chapter 6: Ratio, Proportion, and Percent 81. 47% 47
1
47 100 100
1 82. 5.6% 5.6 100 5.6 100 5.6 10 100 10 56 1000 7 125 83.
84.
5
6 100 600 % % 120% 5 1 5
85.
43 16% x 43 0.16x 43 0.16x 0.16 0.16 268.75 x 43 is 16% of 268.75.
86.
27.5 x 25 27.5 25x 25 25 1.1 x 110% x 27.5 is 110% of 25.
87. x = 36% 1968 x = 0.36 1968 x = 708.48 708.48 is 36% of 1968. 88.
246
67 x 50 67 50 x 50 50 1.34 x 134% x 67 is 134% of 50.
p
25 100 75 100 p 25 7500 25 p 7500 25 p 25 25 300 p 75 is 300% of 25. a
90.
3 3 100 300 1 % % 37 % 8 8 1 8 2 6
75
89.
16 240 100 a 100 16 240 100a 3840 100a 3840 100 100 a 38.4 38.4 is 16% of 240.
91.
28 5 b 100 28 100 5 b 2800 5b 2800 5b 5 5 560 b 28 is 5% of 560.
92.
52 p 16 100 52 100 p 16 5200 16 p 5200 16 p 16 16 325 p 52 is 325% of 16.
93.
78 0.26 26% 300 26% of the soft drinks have been sold.
94. $96,950 7% = $96,950 0.07 = $6786.50 The house has lost $6786.50 in value. 95. Sales tax 8.75% $568 0.0875 $568 $49.70 Total price = $568 + $49.70 = $617.70 The total price is $617.70.
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ISM: Prealgebra and Introductory Algebra 96. Amount of discount 15% $23.00 0.15 $23.00 $3.45 97. commission commission rate sales $1.60 r $12.80 $1.60 r $12.80 0.125 r 12.5% r His rate of commission is 12.5%. 98. Simple interest principal rate time 6 $1400 13% 12 $1400 0.13 0.5 $91 Total amount = $1400 + $91 = $1491 The total amount is $1491. 99. Simple interest principal rate time $5500 12.5% 9 $5500 0.125 9 $6187.50 Total amount = $5500 + $6187.50 = $11,687.50 The total amount is $11,687.50. Chapter 6 Getting Ready for The Test 1. The phrase “6 is to 7 as 18 is to 21” translates to 6 18 ; A. 7 21 2.
3
5. Since 50% = 50(0.01) = 0.5, 50or 1 1 1 , choice B, 5, 50% 50 50 100 100 50 2 2 does not equal 50%. 6. Since 80% = 80(0.01) = 0.80 20 =4 0.8,4 or , choice C, 8, 80% 80 1 80 100 100 20 5 5 does not equal 80%. 7. Since 35 is half of 70, 50% of 70 is 35, B. 8. Since 8.8 is
1 10
of 88, 10% of 88 is 8.8, D.
9. Since 47 = 47, 100% of 47 is 47, A. 10. Since 7 is
1
of 28, 25% of 28 is 7, C.
4 11. Round $24.86 to $25. 10% of $25 is $2.50. 20% of $25 is 2($2.50) = $5. Of the amounts given, $5 is closest to 20% of $24.86, C. 12. amount of discount discount rate original price 10% $150 0.10 $150 $15 The amount of discount is $15, A. 13. discounted price = $150 $15 = $135 The discounted price is $135, D.
5
x 11 311 5x In the given proportion, the cross products are 3 11 and 5x. In choice A, the cross products are 3 5 and 11 x, while in choices B, C, and D, the cross products are 3 11 and 5x, so choice A is not equivalent to the given proportion. 3. “Percent” means “per hundred” so 12% means 12 12 3 or 0.12. Also, 3 4 . Thus, 100 100 4 25 25 choice D, 1.2, does not equal 12%.
Chapter 6: Ratio, Proportion, and Percent
14. 25% of a number is
1
of that number.
4 25% $40
1
$40 $10 4 The discount amount is $10, so the sale price of the shoes before tax is $40 $10 = $30, C. Chapter 6 Test 1. 85% = 85(0.01) = 0.85
4. Since 100% = 100(0.01) 1 100= 1.00 = 1, or 100% 100 , choice A, 10, does not 100 100 equal 100%.
2. 500% = 500(0.01) = 5 3. 0.6% = 0.6(0.01) = 0.006 4. 0.056 = 0.056(100%) = 5.6% 5. 6.1 = 6.1(100%) = 610% 6. 0.35 = 0.35(100%) = 35%
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Chapter 6: Ratio, Proportion, and Percent 7. 120% 120
1
1 120 1 5 100 100 1 8. 38.5% 38.5 38.5 385 77 100 1000 200 100
9. 0.2% 0.2
3
8
2 1 1 0.2 100 100 1000 500
3 100 300 % % 37.5% 8 1 8
3 7 100 700 12. 1 4 % 175% % 4 1 4 13.
3 4
3 100 300 % % 75% 4 1 4
14. 40% 40 1 40 2 100 100 5
15. Method 1: x = 42% 80 x = 0.42 80 x = 33.6 33.6 is 42% of 80. Method 2: a 42 80 100 a 100 42 80 100a 3360 100a 3360 100 100 a 33.6 33.6 is 42% of 80. 16. Method 1: 0.6% x 7.5 0.006 x 7.5 0.006x 7.5 0.006 0.006 x 1250 0.6% of 1250 is 7.5. Method 2:
248
7.5 0.6 b 100 7.5 100 b 0.6 750 0.6b
11 11 100 1100 % % 55% 10. 20 20 1 20 11.
ISM: Prealgebra and Introductory Algebra
750 0.6b 0.6 0.6 1250 b 0.6% of 1250 is 7.5. 17. Method 1: 567 x 756 567 x 756 756 756 0.75 x 75% x 567 is 75% of 756. Method 2: 567 p 756 100 567 100 756 p 56, 700 756 p 56, 700 756 p 756 756 75 p 567 is 75% of 756. 18. 12% of 320 is what number? 12% 320 x 0.12 320 x 38.4 x There are 38.4 pounds of copper. 19. 20% of what number is $11,350? 20% x $11, 350 0.20 x $11, 350 0.2x $11, 350 0.2 0.2 x $56, 750 The value of the potential crop is $56,750. 20. tax = 8.25% $354 = 0.0825 $354 $29.21 total amount = $354 + $29.21 = $383.21 The total amount of the stereo is $383.21. 21. percent of increase
amount of increase
original amount 26, 460 25, 200 25, 200 1260 25, 200 0.05 The increase in population was 5%.
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ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent
22. Amount of discount 15% $120 0.15 $120 $18 Sale price = $120 $18 = $102 The amount of the discount is $18; the sale price is $102. 23. commission 4% $9875 0.04 $9875 $395 The commission was $395. 24.
29. The ratio of $75 to $10 is
$10 30.
25. simple interest principal rate time 1 $2000 9.25% 3 2 $2000 0.0925 3.5 $647.5
75 10
15
.
2
5 4 y 1 y 2 5( y 2) 4( y 1) 5 y 10 4 y 4 y 10 4 y 6
Cumulative Review Chapters 16 1.
236 86 1 416 18 880 20, 296
2.
409 76 2 454 28 630 31, 084
nt
r 26. A P 1 n 15 0.08 13651 1 1365(1.08)5 2005.63 The total amount of $2005.63.
27. Simple interest principal rate time 6 $400 13.5% 12 $400 0.135 0.5 $27 Total amount = $400 + $27 = $427
3. 3 7 = 3 + (7) = 10 4. 8 (2) = 8 + 2 = 10 5.
x 2 1 x 2 2 1 2 x 1
6.
x43 x 44 34 x 1
amount of decrease
original amount 43,150 40,870 43,150 2280 43,150 0.053 The number of crimes decreased 5.3%.
31. Let x be the number of defective bulbs out of 510 bulbs. x defective defective 3 bulbs 85 510 bulbs 3 510 85 x 1530 85x 1530 85x 85 85 18 x There should be 18 defective bulbs in 510.
$13.77 rate $152.99 $13.77 r $152.99 0.09 r 9% r The sales tax rate is 9%.
28. percent of decrease
$75
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Chapter 6: Ratio, Proportion, and Percent
7.
8.
3(2x 6) 6 0 3 2x 3 6 6 0 6x 18 6 0 6x 12 0 6x 12 12 0 12 6x 12 6x 12 6 6 x2
3 1 6 3 7 1 2 6 4 17. 4 14 7 4 7 14 2 7 4 21 2 24 28 28 28 1 28 18.
5(x 2) 3x 5 x 5 2 3x 5x 10 3x 5x 5x 10 3x 5x 10 2x 10 2x 2 2
13 14 3 8 2 4 8 31 33 12 4 6 4 3 6 2 4 3 8 8 9 2 12 12 7 87 12
19. 2
5 x 9. 3
3 7 3 7 21 1 7 1 7 7
8 5 8 5 40 10. 8 1 5 1 5 5
Since 10 and 27 have no common factors, 10 27 is already in simplest form. 10 y 5 2 y 5y 32 16 2 16
13.
7
12
5
6
7
6
12
5
7 7 6 2 6 5 10
14. 2 7 2 10 2 2 5 4 5 10 5 7 5 7 7 8 3 8 3 5 1 15. y x 10 10 10 10 10 2 2 1 4 3 1 3 16. 2x 3y 2 5 5 5 5 5
250
7 7 8 12 7 12 8 7 7 43 427 3 2
11. 10 2 5 27 3 33
12.
2 7 1 2 5 7 3 1 15 9 15 3 9 5 15 3 3 15 10 21 15 45 45 45 16 45
21 2 2 1 20. 33 63 33 25 3 6 4 45 4 5 5 4 41 6 15 12 20 20 5 6 3 20
6
5 3 6 20 5 20 6 3 5 2 10 233 50 9
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ISM: Prealgebra and Introductory Algebra
Chapter 6: Ratio, Proportion, and Percent
21.
x x 1 2 3 2 x x 1 6 2 6 3 2 x 1 3x 6 6 3 2 3x 2x 3 3x 2x 2x 3 2x x 3 x
22.
1
3
29.
85.00 17.31 67.69
30.
38.00 10.06 27.94
31. 7.68 10 = 76.8 32. 12.483 100 = 1248.3
x
2 5 5 x 1 10 3 x 10 2 5 5 x 1 x 10 10 10 3 10 2 5 5 5x 2 30 2x 5x 2 2x 30 2x 2x 7x 2 30 7x 2 2 30 2 7x 28 7x 28 7 7 x4
33. (76.3)(1000) = 76,300 34. 853.75 10 = 8537.5 35. x y = 2.5 0.05 50 0.05 2.5 becomes 5 250 25 00 36.
2 9 4 2 36 2 38 9 9 9 9
x 4.75 100 470 0 4.75 100 4.7 4.75 False No, 470 is not a solution.
23. a.
4
b.
1
111 8 11 8 19 11 11 11 11
24. a.
3
2 5 3 2 15 2 17 5 5 5 5
38. mean
b.
6
2 7 6 2 42 2 44 7 7 7 7
39.
8
25. 0.125 125 1 1000 8 26. 0.85
85 100
2.5
5.8
58
45 41. 83 1000
15 3 17 1000 200
2500 3150
7.6
20
36 40 86 30 192 48 4 4
3.15 40.
17
27. 105.083 105
28. 17.015 17
37. 55, 67, 75, 86, 91, 91 75 86 161 median 80.5 2 2
76
50 63
29 38
5
x 7 45 7 5 x 315 5x 315 5x 5 5 63 x
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Chapter 6: Ratio, Proportion, and Percent
42.
x 1
46. 2.3% 2.3
3
3x 1 8 8(x 1) 3(3x 1) 8x 8 9x 3
8 x 3 11 x 43.
1 47. 33 % 3
7 problems x problems 6 minutes 30 minutes 7 30 6 x 210 6x 210 6x 6 6 35 x The student can complete 35 problems in 30 minutes. 1 19 1.9 45. 1.9% 1.9 100 100 1000
23 2.3 100 1000 100 100 1 100 1
3 1100 300 3 2 108 48. 108% 108 1 25 100 100
5 miles x miles 2 inches 7 inches 5 7 2 x 35 2 x 35 2 x 2 2 17.5 x 17.5 miles corresponds to 7 inches.
1
49. Wh–tanu– m– ber is 35% of 60? x = 35% 60 x = 0.35 60 x = 21 21 is 35% of 60.
44.
252
50. Wh–at nu– m– ber is 42% of 85? x = 42% 85 x = 0.42 85 x = 35.7 35.7 is 42% of 85.
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Chapter 7 Section 7.1 Practice Exercises 1. a.
b.
2. a.
6.
Class Interval (credit card balances)
Tally
Class Frequency (number of months)
$0$49
|||
3
$50$99
||||
4
$100$149
||
2
$150$199
|
1
$200$249
|
1
$250$299
|
1
Hindi has 7 symbols and each symbol represents 50 million speakers, so Hindi is spoken by 7(50) = 350 million people. Spanish has 9.5 symbols, or 9.5(50) = 475 million speakers. So, 475 million 350 million = 125 million more people speak Spanish than Hindi. The height of the bar for crustaceans is 30, so approximately 30 endangered species are crustaceans.
b. The shortest bar corresponds to arachnids, so arachnids have the fewest endangered species. 3.
7.
4. The height of the bar for 8089 is 12, so 12 students scored 8089 on the test. 5. The height of the bar for 4049 is 1, for 5059 is 3, for 6069 is 2, for 7079 is 10. So, 1 + 3 + 2 + 10 = 16 students scored less than 80 on the test.
8. a.
The lowest point on the graph corresponds to January, so the average daily temperature is the lowest during January.
b.
The point on the graph that corresponds to 25 is December, so the average daily temperature is 25F in December.
c.
The points on the graph that are greater than 70 are June, July, and August. So, the average daily temperature is greater than 70F in June, July, and August.
9. range largest data value smallest data value $102 $32 $70 The range of the salaries is $70 thousand. 10. a.
range largest data value smallest data value 67 50 17 The range of this data set is 17.
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Chapter 7: Graphs, Triangle Applications, ...
b.
mean sum of items number of items 50 1 59 3 60 3 62 5 63 2 67 1 1 3 3 5 2 1 910 15 60.7 The mean of the data set is approximately 60.7.
c.
Since there are 15 data items and the items are arranged in numerical order, the middle 15 1 16 item can be located using 8. 2 2 The median is the eighth item, which is 62.
d.
The mode has the greatest frequency, so the mode is 62.
Vocabulary, Readiness & Video Check 7.1
ISM: Prealgebra and Introductory Algebra
Exercise Set 7.1 2. South Dakota has the least number of wheat symbols, so the least amount of wheat acreage was planted in South Dakota. 4. Kansas is represented by 7.5 wheat symbols, and each symbol represents 1 million acres, so there were approximately 7.5(1 million) = 7.5 million or 7,500,000 acres of wheat planted in Kansas. 6. Each wheat symbol represents 1 million acres, so look for any states that have more than 6 million 6 wheat symbols. Texas and Kansas 1 million have more than 6 wheat symbols, so these states plant more than 6,000,000 acres of wheat. 8. North Dakota and Oklahoma each have 6 wheat symbols. The same acreage of wheat is planted in North Dakota and Oklahoma.
1. A bar graph presents data using vertical or horizontal bars.
10. 2022 has 5.5 flames and each flame represents 12,000 wildfires, so there were approximately 5.5(12,000) = 66,000 wildfires in 2022.
2. A pictograph is a graph in which pictures or symbols are used to visually present data.
12. The year with the fewest flames is 2019, so the fewest wildfires occurred in 2019.
3. A line graph displays information with a line that connects data points.
14. 2021 has 5 flames and 2022 has 5
4. A histogram is a special bar graph in which the width of each bar represents a class interval and the height of each bar represents the class frequency. 5. range = greatest data value smallest data value n 1 6. The formula
can be used to find the 2 position of the median.
7. Count the number of symbols and multiply this number by how much each symbol stands for (from the key). 8. A bar graph lets you see and visually compare the data. 9. A histogram is a special kind of bar graph.
1
flames,
2 which is
1
flame more. The increase in 2 wildfires from 2021 to 2022 was 1 (12, 000) 6000 wildfires. 2 16. Since each flame represents 12,000 wildfires, look for the years that have fewer than 60, 000 5 flames. There were fewer than 12, 000 60,000 wildfires in 2018 and 2019. 18. The shortest bar corresponds to November, so the month in which the fewest hurricanes made landfall is November. 20. The length of the bar for October is 61, so approximately 61 hurricanes made landfall in October.
10. 2021; 9.4 goals per match 11. answers may vary; for example: 6, 6, 6, 6
254
22. One of 111 hurricanes made landfall in 1 September 2007. The percent is 0.009 or 111 about 0.9%.
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24. The shortest bar corresponds to Sao Paulo, Brazil and the length of the bar is about 21.7. The metropolitan area with the smallest population is Sao Paulo, Brazil and its population is about 21.7 million or 21,700,000.
40. 29 of the adults drive 049 miles per week and 17 of the adults drive 5099 miles per week, so 29 17 = 12 more adults drive 049 miles per week than 5099 miles per week.
26. The bar between 24 and 25 corresponds to Seoul, South Korea.
42. 17 of the 100 adults surveyed drive 5099 miles 17 per week, so the ratio is . 100
28. The bar corresponding to Jakarta, Indonesia has length about 30 and the bar corresponding to Seoul, South Korea has length about 24. Thus, Jakarta, Indonesia is about 30 24 = 6 million larger than Seoul, South Korea.
44. The shortest bar corresponds to under 25, so the least number of U.S. householders were in the under 25 age range. 46. According to the graph, approximately 23 million U.S. householders were 3544 years old.
30.
48. The sum of the last three bars is 24 + 21 + 15 = 60, so approximately 60 million U.S. householders were 55 years old or older. 50. The height of the bar for 4554 is 22 and the height of the bar for 75 and over is 15. So, approximately 22 15 = 7 million more U.S. householders were 4554 years old than 75 and over.
32.
Class Interval (scores)
Tally
Class Frequency (number of games)
52.
8089
||||
4
54.
100109
||
2
Class Interval (account balances) 34. The height of the bar for 200249 miles per week is 9, so 9 of the adults drive 200249 miles per week. 36. 9 of the adults drive 200249 miles per week and 21 of the adults drive 250299 miles per week, so 9 + 21 = 30 of the adults drive 200 miles or more per week.
Tally
Class Frequency (number of people)
56.
$100$199
|||| |
6
58.
$300$399
||||
4
60.
$500$599
|
1
38. 9 of the adults drive 150199 miles per week and 9 of the adults drive 200249 miles per week, so 9 + 9 = 18 of the adults drive 150249 miles per week.
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Chapter 7: Graphs, Triangle Applications, ...
84. There are 2 + 1 + 4 + 7 + 2 + 1 = 17 data items.
62.
a.
b.
64. The point on the graph corresponding to 2017 is 8.3, so the average number of goals per match in 2017 was 8.3. 66. The lowest point on the graph corresponds to 2013, so the average number of goals per match was the lowest in 2013.
c.
3 2 4 1 5 4 6 7 7 2 8 1 17 94 17 5.5
mean
n 1 17 1 9 2 2 The data item in the ninth position is 6, so the median is 6. The data item with the greatest frequency is the mode, so the mode is 6.
86. There are 3 + 8 + 5 + 8 + 2 = 26 data items. a.
4 3 5 8 6 5 7 8 8 2 26 154 26 5.9
mean
68. The graph decreases between 2011 and 2013, so the average number of goals per match decreased from 2011 to 2013. 70. The dots for 2009, 2011, 2017, 2019, and 2021 are above the 8-level, so the average number of goals per match was greater than 8 in 2009, 2011, 2017, 2019, and 2021.
76. range largest data value smallest data value 40 11 29
c.
The data items with the greatest frequency are the mode, so the mode is 5 and 7.
88. There are 1 + 4 + 5 + 3 + 2 = 15 data items. a.
78. range largest data value smallest data value 276 129 147
b.
80. range largest data value smallest data value 10 7 3
c.
82. range largest data value smallest data value 90 40 50
256
26 1
13.5 2 2 The mean of the data items in the 13th and 66 14th positions is 6, so the median is 2 6.
72. The point on the graph corresponding to 2029 is at about 450, so the amount spent on organic food in 2029 is predicted to be $450 billion. 74. The dots for 2029 and 2030 are above the 400-level, so the amount spent on organic food is predicted to be more than $400 billion in 2029 and 2030.
n 1
b.
10 1 20 4 30 5 40 3 50 2 15 460 15 30.7
mean
n 1 15 1 8 2 2 The data item in the eighth position is 30, so the median is 30. The data item with the greatest frequency is the mode, so the mode is 30.
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
90. There are 6 + 6 + 4 + 3 + 3 = 22 data items. 5 6 10 6 15 4 20 3 25 3 22 285 22 13.0
mean
a.
n 1
c.
22 1
11.5 2 2 The mean of the data items in the eleventh 10 10 and twelfth positions is 10, so the 2 median is 10.
b.
The data items with the greatest frequency are the mode, so the mode is 5 and 10.
92. 45% of 120 is 0.45 120 = 54.
Section 7.2 Practice Exercises 1. Eight of the 100 adults prefer golf. The ratio is adults preferring golf 8 2 total adults 100 25 2. Add the percents corresponding to South America & Caribbean and India. 13% + 2% = 15% 15% of visitors to the U.S. come from South America & Caribbean and India. 3. amount percent base 0.47 91, 000, 000 42, 770, 000 Thus, 42,770,000 tourists might come from Mexico in 2027. 4.
Percent
Freshmen
30%
30% of 360 = 0.30(360) = 108
Sophomores
27%
27% of 360 = 0.27(360) = 97.2
Juniors
25%
25% of 360 = 0.25(360) = 90
Seniors
18%
18% of 360 = 0.18(360) = 64.8
94. 95% of 50 is 0.95 50 = 47.5. 2 96. 98.
5 9 10
2
100%
2 5 20
5
9 10
5 100%
% 40%
9 10 10
% 90%
10
100. The point on the low temperature graph corresponding to Thursday is 74, so the low temperature reading on Thursday was 74F. 102. The highest point on the graph of high temperature corresponds to Tuesday. The high temperature on Tuesday was 86F.
Degrees in Sector
Year
104. The difference between the graphs is the least for Friday. The high temperature was 76F and the low temperature was 70F, so the difference is 76 70 = 6F. 106. Kansas is represented by 7.5 wheat symbols, and each symbol represents 1 million acres, so there were approximately 7.5(1 million) = 7.5 million or 7,500,000 acres of wheat planted. This represents 20% of the wheat acreage in the U.S., 7.5 million so 37.5 38 million acres of 0.20 wheat were planted in the U.S.
Vocabulary, Readiness & Video Check 7.2 1. In a circle graph, each section (shaped like a piece of pie) shows a category and the relative size of the category. 2. A circle graph contains pie-shaped sections, each called a sector. 3. The number of degrees in a whole circle is 360.
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4. If a circle graph has percent labels, the percents should add up to 100. 5. 100% 6. 360 Exercise Set 7.2 2. The smallest sector corresponds to the category “own off-campus housing,” thus the least of these college students own off-campus housing. 4. 124 of the 700 total students live in off-campus rentals. 124 31 700 175 31 The ratio is . 175 6. 124 of the students live in off-campus rentals while 320 live in a parent or guardian’s home. 124 31 320 80 31 The ratio is . 80 8. The smallest sector corresponds to Australia. Thus, the smallest continent is Australia. 10. 16% + 12% = 28% 28% of the continental land area on Earth is accounted for by North and South America. 12. South America accounts for 12% of the continental land area on Earth. 12% of 57,000,000 0.12 57, 000, 000 6,840, 000 South America is 6,840,000 square miles. 14. Europe accounts for 7% of the continental land area on Earth. 7% of 57,000,000 0.07 57, 000, 000 3, 990, 000 Europe is 3,990,000 square miles. 16. The second-largest sector corresponds to dairy and eggs, so the second-greatest amount of money spent on organic food is spent on dairy and eggs. 18. Breads and grains account for 9.4% of the money spent on organic food. 0.094 400 = 37.6 It is predicted that $37.6 billion will be spent on organic breads and grains in 2028. 20. Add the percent for nonfiction (25%) to the percent for reference (17%). 25% + 17% = 42% Thus, 42% of books are nonfiction or reference. 22. The third-largest sector corresponds to children’s fiction, so the third-largest category of books is children’s fiction. 24. Reference accounts for 17% of the books. 17% of 125,600 0.17 125, 600 21, 352 The library has 21,352 reference books. 258
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ISM: Prealgebra and Introductory Algebra
Chapter 7: Graphs, Triangle Applications, …
26. Adult’s fiction accounts for 33% of the books. 33% of 125,600 0.33125, 600 41, 448 The library has 41,448 adult fiction books. 28. Nonfiction or other accounts for 25% + 3% = 28% of the books. 28% of 125,600 = 0.28 125,600 = 35,168 The library has 35,168 nonfiction or other books. 30.
Color Distribution of M&M’s Milk Chocolate Color
Percent
Degrees in Sector
Blue
24%
24% of 360 = 0.24(360) 86
Orange
20%
20% of 360 = 0.20(360) = 72
Green
16%
16% of 360 = 0.16(360) 58
Yellow
14%
14% of 360 = 0.14(360) 50
Red
13%
13% of 360 = 0.13(360) 47
Brown
13%
13% of 360 = 0.13(360) 47
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32.
ISM: Prealgebra and Introductory Algebra
Distribution of Department of the Interior Public Lands by Management Bureau of Management
Percent
Degrees in Sector
Bureau of Indian Affairs
9%
9% of 360 = 0.09(360) 32
National Park Service
11%
11% of 360 = 0.11(360) 40
Fish and Wildlife
20%
20% of 360 = 0.20(360) = 72
U.S. Forest Service
31%
31% of 360 = 0.31(360) 112
Bureau of Land Management
29%
29% of 360 = 0.29(360) 104
34. 25 5 5 52 36. 16 4 4 2 2 2 2 24 38. 105 = 3 35 = 3 5 7 40. answers may vary 42. Atlantic Ocean: 26% 264, 489, 800 0.26 264, 489, 800 68, 767, 348 square kilometers 44. Arctic Ocean: 4% 264, 489, 800 0.04 264, 489, 800 10, 579, 592 square kilometers 46. 9% of 2800 = 0.09 2800 = 252 252 members of the community would be expected to buy online daily. 48. 56% of 2800 = 0.56 2800 = 1568 1568 members of the community would be expected to buy online monthly or less than monthly.
260
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50.
percent who never buy online percent who buy monthly or less than monthly 2% 56% 2 56 1 28
52. true; answers may vary Mid-Chapter Review 1. Software developers has 5 figures and each figure represents 75,000 workers. Thus, the increase in the number of software developers is approximately 5 75,000 = 375,000. 2. Registered nurses has 2.5 figures and each figure represents 75,000 workers. Thus, the increase in the number of registered nurses is approximately 2.5 75,000 = 187,500. 3. Home health and personal care aide has the greatest number of figures, so the greatest increase is expected for home health and personal care aides.
10. The lowest point on the graph corresponds to Monday. Thus, the lowest temperature occurs on Monday, and is 82F. 11. The points on the graph that are lower than the 90 level correspond to Sunday, Monday, and Tuesday. Thus, the temperature was less than 90F on Sunday, Monday, and Tuesday. 12. The points on the graph that are higher than the 90 level correspond to Wednesday, Thursday, Friday, and Saturday. Thus, the temperature was greater than 90F on Wednesday, Thursday, Friday, and Saturday. 13. The highest points on the graph correspond to 2026 and 2027, so those years have the highest predicted percent of food purchased being organic. 14. The lowest point on the graph corresponds to 2020, so 2020 has the lowest percent of food purchased being organic. 15. The point for 2022 is below the point for 2021, so there was a decrease in the percent of organic food purchased from 2021 to 2022.
4. Food service has the second greatest number of figures, so the second greatest increase is expected for food service.
16. The steepest upward segment on the graph is between 2020 and 2021, so the greatest increase in the percent of organic food purchased is from 2020 to 2021.
5. The tallest bar corresponds to Oroville, CA. Thus, the U.S. dam with the greatest height is the Oroville Dam, which is approximately 770 feet.
17. The sector corresponding to whole milk is 35%. 35% of 200 = 0.35 200 = 70 Thus, 70 quart containers of whole milk are sold.
6. The bar whose height is between 625 and 650 feet corresponds to New Bullards Bar, CA. Thus, the U.S. dam with a height between 625 and 650 feet is the New Bullards Bar Dam, which is approximately 645 feet.
18. The sector corresponding to skim milk is 26%. 26% of 200 = 0.26 200 = 52 Thus, 52 quart containers of skim milk are sold.
7. From the graph, the Hoover Dam is approximately 725 feet and the Glen Canyon Dam is approximately 710 feet. The Hoover Dam is 725 710 = 15 feet taller than the Glen Canyon Dam. 8. There are 4 bars that are taller than the 700-feet level, so there are 4 dams with heights over 700 feet.
19. The sector corresponding to buttermilk is 1%. 1% of 200 = 0.01 200 = 2 Thus, 2 quart containers of buttermilk are sold. 20. The sector corresponding to Flavored reduced fat and skim milk is 3%. 3% of 200 = 0.03 200 = 6 Thus, 6 quart containers of flavored reduced fat and skim milk are sold.
9. The highest points on the graph correspond to Thursday and Saturday. Thus, the highest temperature occurs on Thursday and Saturday and is 100F. Copyright © 2025 Pearson Education, Inc.
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Class Interval (scores)
Class Frequency (number of quizzes)
Tally
21.
5059
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2
22.
6069
|
1
23.
7079
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3
24.
8089
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6
25.
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9. Let a = 12 and b = 16. a2 b2 c2 122 162 c2 144 256 c2 400 c2 400 c 20 c The hypotenuse is 20 feet. 10. Let a = 9 and b = 7. a2 b2 c2 92 72 c2 81 49 c2
26.
130 c2 130 c 11 c The hypotenuse is approximately 11 kilometers. 11. Let a = 7 and c = 13. a2 b2 c2 72 b2 132 49 b2 169
Section 7.3 Practice Exercises 1.
100 10 because 10 100.
2.
64 8 because 82 64.
3.
169 13 because 132 169.
4.
0 0 because 02 0.
5.
1 1 1 because . 4 2 2 4
b2 120 b 120 b 10.95 The length of the hypotenuse is exactly 120 feet or approximately 10.95 feet.
2
100 yd
12. 53 yd
2
1
Let a = 53 and b = 100. a2 b2 c2
2
3 9 3 9 because . 16 4 4 16 7. a. 10 3.162
532 1002 c2
6.
b.
2809 10, 000 c2 12,809 c2 12,809 c 113 c The diagonal is approximately 113 yards.
62 7.874
8. Recall that
49 7 and
64 8. Since 62 is
between 49 and 64, then 62 is between 49 and 64. Thus, 62 is between 7 and 8. Since 62 is closer to 64, then 62 is closer to 64, or 8. 262
Calculator Explorations 1.
1024 32
2.
676 26
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
3.
15 3.873
4.
19 4.359
5.
97 9.849
6.
4.
144 12 because 122 144.
6.
1 1 1 1 1 1 because . 64 8 8 8 8 64
8.
6 2 6 6 6 36 36 because . 81 9 3 9 9 9 81
10.
5 2.236
12.
17 4.123
14.
85 9.220
16.
35 5.916
2
2
56 7.483
Vocabulary, Readiness & Video Check 7.3 1. The square roots of 100 are 10 and 10 because 10 10 = 100 and (10)(10) = 100. 2.
100 10 only because 10 10 = 100 and 10 is positive.
3. The radical sign is used to denote the positive square root of a nonnegative number.
18. Since 27 is between 25 = 5 5 and 36 = 6 6, 27 is between 5 and 6; 27 5.20.
4. The reverse process of squaring a number is finding a square root of a number.
20. Since 85 is between 81 = 9 9 and 100 = 10 10, 85 is between 9 and 10; 85 9.22.
5. The numbers 9, 1, and
1 25
are called perfect
squares.
22.
625 25 because 252 625.
24.
18 4.243
26.
11 11 11 11 121 121 . because 169 13 13 13 13 169
28.
62 7.874
2
6. Label the parts of the right triangle. hypotenuse leg
30. Let a = 36 and b = 15.
leg
a 2 b 2 c2
7. In the given triangle, a2 c2 b2 .
362 152 c2
8. The Pythagorean Theorem can be used for right triangles. 9. The square roots of 49 are 7 and 7 since 72 49 and (7)2 49. The radical sign
means the positive square root only, so
49 7.
10. Since 38 is between 36 6 and 49 7 and 38 is much closer to 36 than to 49, we know that 38 is closer to 6 than 7. 11. The hypotenuse is the side across from the right angle. Exercise Set 7.3 2.
1296 225 c2 1521 c2 1521 c 39 c The missing length is 39 feet. 32. Let a = 3 and c = 9. a2 b2 c2 32 b2 92 9 b2 81 b2 72 b 72 b 8.485 The missing length is approximately 8.485 yards.
9 3 because 32 9. Copyright © 2025 Pearson Education, Inc.
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34. Let a = 34 and b = 70. 2
2
42.
2
a b c
342 702 c2 16
1156 4900 c2 6056 c2 6056 c 77.820 c The missing length is approximately 77.820 miles.
2
hypotenuse (leg)2 (other leg)2 22 162 4 256 260
36. Let a = 36 and b = 27. a2 b2 c2 362 272 c2 1296 729 c
16.125 The hypotenuse has length of about 16.125 units.
2
2025 c2 2025 c 45 c The missing length is 45 kilometers.
30
44. 15
hypotenuse (leg)2 (other leg)2
38.
2 2 30 15 900 225 1125 33.541 The hypotenuse has length of about 33.541 units.
12
9
hypotenuse (leg)2 (other leg)2
46.
21
92 122 81 144 225 15 The hypotenuse has length 15 units.
hypotenuse (leg)2 (other leg)2
40. 6
10
leg (hypotenuse)2 (other leg)2 102 62 100 36 64 8 The leg has length 8 units.
264
21
212 212 441 441 882 29.698 The hypotenuse has length of about 29.698 units.
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8
48.
58. leg (hypotenuse)2 (other leg)2 (60) (29.4) 2
9 2
leg (hypotenuse) (other leg)
3600 864.36 2735.64 52.3 The width of the television is about 52.3 inches.
2
92 82 81 64 17 4.123 The leg has length of about 4.123 units.
60. leg (hypotenuse)2 (other leg)2
12
hypotenuse (leg)2 (other leg)2
62. hypotenuse (leg)2 (other leg)2 (43.6) (24.5) 2
122 (22.5)2
25.5 The hypotenuse has length 25.5 units. 52. hypotenuse (leg)2 (other leg)2 12 12 11 2 1.414 The length of the diagonal is about 1.414 miles. (hypotenuse)2 (leg)2
1682 602 28, 224 3600 24, 624 156.9 The height of the antenna is about 156.9 feet. 2 2 56. hypotenuse (leg) (other leg)
2
1.774 1.77 64. 27 47.900 27 20 9 18 9 2 00 1 89 110 108 2 Yes, the aspect ratio of the television is approximately 1.78 : 1. 66.
68.
10 2 5 2 15 3 5 3 35 75 y
2
70 110 4900 12,100 17, 000 130.4 The length of the diagonal is about 130.4 yards.
2
1900.96 600.25 2501.21 50 The diagonal of the television is about 50 inches.
144 506.25 650.25
54. leg
(55)2 (47.9)2
3025 2294.41 730.59 27.0 The height of the television is about 27.0 inches.
22.5
50.
2
57 7 5 15 y 15 y
70.
3x 5 3x 5 9 9 9
72.
7 x 8x 7 x 11 7 x 11 7 11 11 11 8x 118 x 8
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74. Recall that
25 5 and
36 6. Since 27 is
5. Check: 76. Recall that
27 is closer to
25, or
27 5.20 81 9 and 100 10. Since 85 is
between 81 and 100, then 85 is between 81 and 100. Thus, 85 is between 9 and 10. Since 85 is closer to 81, then 81, or 9. Check:
1. Two triangles that have the same shape, but not necessarily the same size are congruent. false
85 9.22
2. Two triangles are congruent if they have the same shape and size. true
a2 b2 c2
3. Congruent triangles are also similar. true
202 452 0 502 400 2025 0 2500 2425 2500 No; the set does not form the lengths of the sides of a right triangle.
4. Similar triangles are also congruent. false 5. For the two similar triangles, the ratio of 5 corresponding sides is . false 6
Section 7.4 Practice Exercises 1. a.
The triangles are not congruent.
9 meters
2.
9 13 meters 13
x
6
5 9 x9 56 9x 30 9x 30 9 9 10 or 3 1 units x 3 3
7.
The ratios of corresponding sides are the same. 8.
The ratio of corresponding sides is
3.
6. Since the sides of both triangles are given, and no angle measures are given, we used SSS.
The triangles are congruent by Side-AngleSide.
b.
8
Vocabulary, Readiness & Video Check 7.4
85 is closer to
78. answers may vary 80.
n 60 5 60 n 8 300 8n 300 8n 8 8 37.5 n The height of the building is approximately 37.5 feet.
between 25 and 36, then 27 is between 25 and 36. Thus, 27 is between 5 and 6. Since 27 is closer to 25, then
5
4.
9
. 13
12 4 18 n
Exercise Set 7.4 2. The triangles are congruent by Side-Side-Side. 4. The triangles are not congruent. 6. The triangles are congruent by Angle-SideAngle. 8. The triangles are congruent by Side-Angle-Side. 8 10. 4 7 1 32 16 28 4 1 The ratio of corresponding sides is . 4 12.
6 8 4 1 6 3 4 2
4 The ratio of corresponding sides is . 3 266
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
14.
16.
18.
20.
22.
24.
x 60 3 5 x 5 3 60 5x 180 5x 180 5 5 x 36 y 14 4 7 y 7 4 14 7 y 56 7 y 56 7 7 y 8 z 22.5 9 15 z 15 9 22.5 15z 202.5 15z 202.5 15 15 z 13.5 x 14 9 8 x 8 9 14 8x 126 8x 126 8 8 x 15.75 y 33.2 9.6 8.3 y 8.3 9.6 33.2 8.3y 318.72 8.3y 318.72 8.3 8.3 y 38.4 z
26.
28.
30.
32.
34.
13 1
2 6 9 1 z 9 6 13 2 27 9z 6 2 9z 81 9z 81 9 9 z9
36.
x 13 13 26 x 26 13 13 26x 169 26x 169 26 26 x 6.5 n 8 37.5 20 n 20 37.5 8 20n 300 20n 300 20 20 n 15 28
5 x 100 x 5 28 100 5x 2800 5x 2800 5 5 x 560 The fountain is approximately 560 feet high. x 76 10 2 x 2 10 76 2x 760 2x 760 2 2 x 380 The tree is 380 feet tall. x 44 32 24 x 24 32 44 24x 1408 24x 1408 24 24 x 58.7 The shadow is approximately 58.7 feet. 3
x
5 8 38 5 x 24 5x 24 5x 5 5 4.8 x 45 4.8 = 40.2 The speed of the truck is 40.2 miles per hour. Copyright © 2025 Pearson Education, Inc.
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2 2 38. hypotenuse (leg) (other leg)
y 1 1 1 42 4 1 1 y 4 1 4 2 y 18
1002 502 10, 000 2500 12, 500 112 The length of the diagonal is approximately 112 yards. 40. Subtract the absolute values. 3.60 0.41 3.19 Attach the sign of the larger absolute value. 0.41 3.6 = 3.19
1
6 14 1 z 6 1 4 z 24 The lengths of the sides of the deck are 12 feet, 18 feet, and 24 feet.
H T H T
H H T
1.
H H
0.48 3 = 0.16 44. Let x be the width of the banner. 1 11 3 2 x 9 1 1 9 x 1 3 2 3 3 x 2 2 2 3 (3) x 3 32 2 x The width of the banner is 2 feet. n 11.6 58.7 20.8 n 20.8 58.7 11.6 20.8n 680.92 20.8n 680.92 20.8 20.8 n 32.7
Section 7.5 Practice Exercises
0.16 42. 3 0.48 3 18 18 0
46.
z
T H T
T T 8 outcomes 21 H
2. T
3 4 5 6
12 outcomes
1 2 3 4 5 6
3. The possibilities are: H, H, H, H, H, T, H, T, H, H, T, T, T, H, H, T, H, T, T, T, H, T, T, T T, H, T is one of the 8 possible outcomes, so the 1 probability is . 8
48. answers may vary
4. A 2 or a 5 are two of the six possible outcomes. 2 1 The probability is . 6 3
x 1 3 1 4 1 x 31 4 x 12
5. A blue is 2 out of the 4 possible marbles. The 2 1 probability is . 4 2
50.
268
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ISM: Prealgebra and Introductory Algebra
Vocabulary, Readiness & Video Check 7.5 1. A possible result of an experiment is called an outcome. 2. A tree diagram shows each outcome of an experiment as a separate branch.
2
8. 3
3. The probability of an event is a measure of the likelihood of it occurring. 4. Probability is calculated by number of ways that an event can occur divided by number of possible outcomes. 5. A probability of 0 means that an event won’t occur.
4
Red
7. The number of outcomes equals the ending number of branches drawn. 8. 0; Having the die land on a 7 is impossible, and the probability of something impossible is 0. Exercise Set 7.5
1
a e i uo
2
a e i o u
2.
4.
6 outcomes
Blue T H T
12. A 9 is zero of the six possible outcomes. The 0 probability is 0. 6 14. A 2 or a 3 are two of the six possible outcomes. 2 1 The probability is . 6 3 16. Three of the six possible outcomes are odd. The 3 1 probability is . 6 2
10 outcomes
1 2 3 4
18. Five of the six possible outcomes are less than 6. 5 The probability is . 6 20. A 3 is one of three possible outcomes. The 1 probability is . 3
4 outcomes
Red
6.
H T H
10.
Yellow
6. A probability of 1 means that an event is certain to occur.
Red Blue Yellow Red Blue Yellow 12 outcomes Red Blue Yellow Red Blue Yellow
1
Blue
Yellow
Red Blue Yellow Red Blue Yellow Red Blue Yellow
9 outcomes
22. Not 3 is a 1 or a 2, which are two of three 2 possible outcomes. The probability is . 3 24. An even number is a 2, which is one of three 1 possible outcomes. The probability is . 3 26. One of the seven marbles is blue. The probability 1 is . 7
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28. Three of the seven marbles are green. The 3 probability is . 7
ISM: Prealgebra and Introductory Algebra
Tree diagram for 50.52. Die 1 1
30. Three of the seven marbles are either blue or 3 yellow. The probability is . 7 32. The blood pressure was lower for 152 of the 152 19 200 people. The probability is . 200 25
2
38 152 10 200 1; answers may vary 34. 200 200 200 200 36.
7
10 38.
7
2
7 2 2 7 4 7 4 3 5 10 5 2 10 10 10 10
2
7 5 7 5 7 5 7 or 1 3 5 10 2 10 2 5 2 2 4 4
10
3
Die 2 Sum 1 2 2
3
3
4
4
5
5
6
6
7
1
3
2
4
3
5
4
6
5
7
6
8
1
4
2
5
3
6
4
7
5
8
6
9
1
5
2
6
3
7
4
8
5
9
6
10
1
6
2
7
3
8
4
9
5
10
6
11
1
7
2
8
3
9
4
10
5
11
6
12
3 3 10 310 3 5 2 6 10 6 40. 5 5 1 5 1 5 1 1 42. One of the 52 cards is the 10 of spades. The 1 probability is . 52
4
44. Four of the 52 cards are 10s. The probability is 4 1 . 52 13 46. Thirteen of the 52 cards are clubs. The 13 1 probability is . 52 4
5
48. Eight of the cards are queens or aces. The 8 2 probability is . 52 13
6
36 outcomes
50. Three of the 36 sums are 10. The probability is 3 1 . 36 12 52. One of the 36 sums is 2. The probability is
270
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1
. 36
ISM: Prealgebra and Introductory Algebra
54. answers may vary
Chapter 7 Review
Chapter 7 Vocabulary Check 1. Congruent triangles have the same shape and the same size. 2. Similar triangles have exactly the same shape but not necessarily the same size. 35.
Chapter 7: Graphs, Triangle Applications, …
5. hypotenuse 3. leg
1. Midwest has 2.5 houses, and each house represents 60,000 homes, so there were 2.5(60,000) = 150,000 new homes constructed in the Midwest. 2. South has 10 houses, and each house represents 60,000 homes, so there were 10(60,000) = 600,000 new homes constructed in the South. 3. South has the greatest number of houses, so the most new homes constructed were in the South.
4. leg
6. A triangle with one right angle is called a right triangle. 7. In the right triangle,
5. Each house represents 60,000 homes, so look for 300, 000 the regions with 5 or more houses. 60, 000 The South had 300,000 or more new homes constructed.
c
a
b 2
4. Northeast has the least number of houses, so the fewest new homes constructed were in the Northeast.
2
a b c2 is called the Pythagorean theorem.
8. A bar graph presents data using vertical or horizontal bars. 9. The possible results of an experiment are the outcomes. 10. A pictograph is a graph in which pictures or symbols are used to visually present data. 11. A tree diagram is one way to picture and count outcomes. 12. An experiment is an activity being considered, such as tossing a coin or rolling a die. 13. In a circle graph, each section (shaped like a piece of pie) shows a category and the relative size of the category. 14. The probability of an event is number of ways that event can occur . number of possible outcomes 15. A histogram is a special bar graph in which the width of each bar represents a class interval and the height of each bar represents the class frequency.
6. Each house represents 60,000 homes, so look for 300, 000 the regions with fewer than 5 houses. 60, 000 The Northeast, Midwest, and West had fewer than 300,000 new homes constructed. 7. The height of the bar representing 2010 is 30. Thus, approximately 30% of persons completed four or more years of college in 2010. 8. The tallest bar corresponds to 2020. Thus, the greatest percent of persons completing four or more years of college was in 2020. 9. The bars whose height is at a level of 20 or more are 1990, 2000, 2010, and 2020. Thus, 20% or more persons completed four or more years of college in 1990, 2000, 2010, and 2020. 10. answers may vary 11. The point on the graph corresponding to 2012 is about 960. There were approximately 960 medals awarded at the Summer Olympics in 2012. 12. The point on the graph corresponding to 2000 is about 930. There were approximately 930 medals awarded at the Summer Olympics in 2000.
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13. The point on the graph corresponding to 2016 is about 970. There were approximately 970 medals awarded at the Summer Olympics in 2016.
24.
14. The point on the graph corresponding to 2020 is about 1080. There were approximately 1080 medals awarded at the Summer Olympics in 2020. 15. The points on the graph corresponding to 2008 and 2004 are 960 and 930, respectively. There were 960 930 = 30 more medals awarded in 2008 than in 2004.
25. range largest data value smallest data value 63 3
16. The points on the graph corresponding to 2020 and 2016 are 1080 and 970, respectively. There were 1080 970 = 110 more medals awarded in 2020 than in 2016.
26. range largest data value smallest data value 8 1 7
17. The height of the bar corresponding to 4145 is 1. Thus, 1 employee works 4145 hours per week.
27. range largest data value smallest data value 95 60 35
18. The height of the bar corresponding to 2125 is 4. Thus, 4 employees work 2125 hours per week.
28. range largest data value smallest data value 99 80 19
19. Add the heights of the bars corresponding to 1620, 2125, and 2630. Thus, 6 + 4 + 8 = 18 employees work 30 hours or less per week.
29. range largest data value smallest data value 94 5
20. Add the heights of the bars corresponding to 3640 and 4145. Thus, 8 + 1 = 9 employees work 36 or more hours per week.
30. range largest data value smallest data value 2.86 0.02 2.84
Class Interval (Temperatures)
Tally
Class Frequency (Number of Months)
21.
8089
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5
22.
9099
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23.
100109
||||
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31. There are 2 + 10 + 5 + 11 + 3 = 31 data items. a.
b.
c.
272
60 2 6110 62 5 6311 64 3 31 1925 31 62.1
mean
n 1 311 16 2 2 The data item in the 16th position is 62, so the median is 62. The data item with the greatest frequency is the mode, so the mode is 63.
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
32. There are 5 + 7 + 8 + 10 + 15 = 45 data items. a.
b.
c.
60 5 61 45 2813 45 62.5
mean
b.
c.
b.
c.
b.
The data item with the greatest frequency is the mode, so the mode is 64.
c.
a.
n 1 18 1 9.5 2 2 The mean of the data items in the ninth and 12 12 tenth positions is 12, so the 2 median is 12. The data items with the greatest frequency are the mode, so the mode is 11 and 12.
mean 15 3 16 2 17 118 22 398 22 18.1 n 1
22 1
11.5 2 2 The mean of the data items in the eleventh 18 18 and twelfth positions is 18, so the 2 median is 18. The data items with the greatest frequency are the mode, so the mode is 18 and 20.
50 5 55 3 60 8 65 6 70 3 25 1495 25 59.8
mean
n 1 25 1 13 2 2 The data item in the 13th position is 60, so the median is 60. The data item with the greatest frequency is the mode, so the mode is 60.
36. There are 7 + 3 + 5 + 2 + 7 = 24 data items.
mean 11 5 12 5 13 3 14 116 3 17 1 18 233 18 12.9
34. There are 3 + 2 + 1 + 6 + 4 + 6 = 22 data items. a.
a.
n 1 45 1 23 2 2 The data item in the 23rd place is 63, so the median is 63.
33. There are 5 + 5 + 3 + 1 + 3 + 1 = 18 data items. a.
35. There are 5 + 3 + 8 + 6 + 3 = 25 data items.
10 7 12 3 14 5 16 2 18 7 24 334 24 13.9
mean
n 1
24 1
b.
12.5 2 2 The mean of the data items in the 12th and 14 14 13th positions is 14, so the 2 median is 14.
c.
The data items with the greatest frequency are the mode, so the mode is 10 and 18.
37. The largest sector corresponds to the category “Mortgage payment,” thus the largest budget item is mortgage payment. 38. The smallest sector corresponds to the category “Utilities,” thus the smallest budget item is utilities. 39. Add the amounts for mortgage payment and utilities. Thus, $975 + $250 = $1225 is budgeted for the mortgage payment and utilities. 40. Add the amounts for savings and contributions. Thus, $400 + $300 = $700 is budgeted for savings and contributions. 41.
mortgage payment $975 39 25 39 total $4000 160 25 160 39 The ratio is . 160
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Chapter 7: Graphs, Triangle Applications, ...
42.
food $700 7 100 7 total $4000 40 100 40 7 The ratio is . 40
ISM: Prealgebra and Introductory Algebra
2 2 57. hypotenuse (leg) (other leg)
43. The sector corresponding to China is 55%. 55% of 102 = 0.55 102 56 56 of the completed tall buildings are in China. 44. The sector corresponding to United Arab Emirates is 12%. 12% of 102 = 0.12 102 12 12 of the completed tall buildings are in United Arab Emirates. 45. The sector corresponding to South Korea is 4%. 4% of 102 = 0.04 102 4 4 of the completed tall buildings are in South Korea. 46. The sector corresponding to the United States is 10%. 10% of 102 = 0.10 102 10 10 of the completed tall buildings are in the United States. 47.
64 8 because 82 64.
48.
144 12 because 122 144.
49.
12 3.464
50.
15 3.873
51.
0 0 because 02 0.
52.
1 1 because 1 1.
53.
50 7.071
54.
65 8.062
55.
2 2 2 2 4 4 because . 25 5 5 5 5 25
56.
1 1 1 1 1 1 . because 100 10 10 10 10 100
122 52 144 25 169 13 The leg has length 13 units. 58. hypotenuse (leg)2 (other leg)2 202 212 400 441 841 29 The leg has length 29 units. 59. leg (hypotenuse)2 (other leg)2 2 2 14 9 196 81 115 10.7 The leg has length of about 10.7 units.
60. leg (hypotenuse)2 (other leg)2 862 662 7396 4356 3040 55.1 The leg has length of about 55.1 units. 61. hypotenuse (leg)2 (other leg)2 202 202
2
400 400 800 28.28 The diagonal is about 28.28 centimeters. 2
2
62. leg (hypotenuse)2 (other leg)2 2
2
126 90 15,876 8100 7776 88.2 The height is about 88.2 feet. 63. The triangles are congruent by Angle-SideAngle.
274
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64. The triangles are not congruent. x 65.
Red
70.
20
Blue
20 30 x 30 20 20 30x 400 30x 400 30 30 40 1 x or 13 3 3
H T
T 1 2 3 4 5
1
1 2 3 4 5 1 2 3 25 outcomes 4 5 1 2 3 4 5
2
66.
x 24 5.8 8 x 8 5.8 24 8x 139.2 8x 139.2 8 8 x 17.4
71.
3
4
x 42 5.5 7 x 7 5.5 42 7x 231 7x 231 7 7 x 33 The height of the building is approximately 33 feet. x 2 y 2 68. 10 24 26 24 x 24 10 2 y 24 26 2 24x 20 24 y 52 24x 20 24 y 52 24 24 24 24 5 13 1 x y or 2 6 6 6 5 The unknown lengths are x inch and 6 1 y 2 inches. 6 67.
H
1 2 3 45
T
1 2 3 4 5
69.
4 outcomes
H
10 outcomes
1 2 3 4 5
5
72.
1
Red Blue
2
Red Blue
3 4
5
Red Blue
10 outcomes
Red Blue Red Blue
73. One of the six possible outcomes is 4. The 1 probability is . 6 74. One of the six possible outcomes is 3. The 1 probability is . 6 75. One of the five possible outcomes is 4. The 1 probability is . 5 76. One of the five possible outcomes is 3. The 1 probability is . 5
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77. Three of the five possible outcomes are a 1, 3, or 3 5. The probability is . 5
89.
78. Two of the five possible outcomes are a 2 or a 4. 2 The probability is . 5 79. Two of the eight marbles are blue. The probability is 2 1 . 8 4 80. Three of the eight marbles are yellow. The 3 probability is . 8 81. Two of the eight marbles are red. The probability 2 1 is . 8 4 82. One of the eight marbles is green. The 1 probability is . 8 83.
36 6 because 62 36.
84.
4 16 4 4 4 16 because . 81 9 9 9 81 9
n 10 6 5 n 5 6 10 5n 60 5n 60 5 5 n 12 3
9 n 18 2 8 12
90.
3
1
91. There are 2 + 5 + 4 + 1 + 6 = 18 data items. a.
10 2 12 5 14 4 16 118 6 18 260 18 14.44
mean
2
85.
105 10.247
86.
32 5.657
b.
87. hypotenuse (leg)2 (other leg)2 662 562 4356 3136 7492 86.6
n 1 18 1 9.5 2 2 The mean of the data items in the ninth and 14 14 tenth places is 14, so the median 2 is 14.
c.
The data item with the greatest frequency is the mode, so the mode is 18.
d.
range greatest data value smallest data value 18 10 8
88. leg (hypotenuse)2 (other leg)2 242 122 576 144 432 20.8
2
2 3 8 9 2 3 8 25 26 75 n 2 3 8 25 325 n 2 4 2 25 2 325 n 25 2 25 4 13 1 n or 6 2 2 n 12
92. There are 5 + 2 + 3 + 4 + 1 = 15 data items. a.
mean
5 5 10 2 15 3 20 4 25 1 15
195 15 13 276
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ISM: Prealgebra and Introductory Algebra
n 1 15 1 8 2 2 The data item in the eighth position is 15, so the median is 15.
11. A whole circle contains 360. 90 + 10 + 20 + 100 + 100 = 320 The unknown sector must contain 360 320 = 40; C.
c.
The data item with the greatest frequency is the mode, so the mode is 5.
d.
range greatest data value smallest data value 25 5 20
12. Three of the five possible outcomes is a red 3 marble. The probability is ; C. 5
b.
Chapter 7 Getting Ready for the Test
13. None (0) of the five possible outcomes is a green 0 marble. The probability is 0; A. 5 Data sets for Exercises 14 through 16:
1. Orchard 1 has 7 symbols, and each symbol represents 4 thousand bushels, so Orchard 1 produced 7 4 = 28 thousand bushels. 28 thousand = 28,000; C.
A. range = 10 10 = 0 10 10 10 10 10 50 mean 10 5 5 median = 10
2. Orchard 5 has the most symbols, so Orchard 5 produced the most bushels; C.
B. range = 14 6 = 8 6 8 10 12 14 50 mean 10 5 5 median = 10
3. Orchard 4 has 2.5 symbols and each symbol corresponds to 28 thousand bushels, so Orchard 4 produced 2.5(4) = 10 thousand bushels; D. 4. Orchard 2 has 8.5 symbols and Orchard 4 has 2.5 symbols, which is 6 fewer symbols. Each symbol represents 4 thousand bushels, so Orchard 2 produced 6(4) = 24 thousand bushels more than Orchard 4. 24 thousand = 24,000; C. 82
64, 102 100, and 64 < 81 < 100, 5. Since then 81 is between 8 and 10; D. Also, 36 6,
25 5,
49 7, and
C. range = 12 8 = 4 8 9 10 1112 50 mean 10 5 5 median = 10 14. Data set B has the greatest range. 15. The median of all three data sets is 10; D. 16. The mean of all three data sets is 10; D.
81 9.
Data sets for Exercises 17 through 20: 6. Since 62 36, 72 49, and 36 < 40 < 49, is between 6 and 7; D.
40
7. True; the Pythagorean theorem applies only to right triangles; A. 8. False; since a triangle has three angles which sum to 180 and 90 + 90 = 180, a right triangle can have only one 90 angle; B. 9. False; a right triangle does have 3 sides, but two of the sides are called legs and one of the sides is called the hypotenuse; B.
A. range = 50 10 = 40 10 20 30 30 40 50 180 mean 30 6 6 30 30 median 30 2 mode = 30 B. range = 48 9 = 39 9 1111 30 48 109 mean 21.8 5 5 median = 11 mode = 11
10. True; the hypotenuse of a right triangle is the longest side; A.
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ISM: Prealgebra and Introductory Algebra
Chapter 7: Graphs, Triangle Applications, ... C. range = 40 20 = 20 20 25 30 35 40 150 mean 30 5 5 median = 30 mode: none
7.
D. range = 70 50 = 20 50 55 60 65 70 300 mean 60 5 5 median = 60 mode: none 17. Data sets A and C both have mean 30. 18. Data sets A and B each have one mode. 19. Data sets C and D both have range 20. 20. Data sets A and C both have median 30.
9. The line graph is above the 3 level for 2021, 2022, and 2023. Thus the inflation rate was greater than 3% in 2021, 2022, and 2023.
Chapter 7 Test 1. There are 4
8. The point on the graph corresponding to 2020 is at about 1.4. Thus, the annual inflation rate in 2020 was about 1.4%.
1
dollar symbols for the second 2 week. Each dollar symbol corresponds to $50. 1 9 $450 4 $50 $50 $225 2 2 2 $225 was collected during the second week.
10. Look for the years where the line graph is decreasing. During 20142015, 20172018, 20192020, 20212022, and 20222023 the inflation rate was decreasing. number who prefer rock music 11.
2. Week 3 has the greatest number of dollar symbols. So the most money was collected during the 3rd week. The 3rd week has 7 dollar symbols and each dollar symbol corresponds to $50, so 7 $50 = $350 was collected during week 3. 3. There are a total of 22 dollar symbols and each dollar symbol corresponds to $50, so a total amount of 22 $50 = $1100 was collected. 4. Look for the bars whose height is greater than 9. June, August, and September normally have more than 9 centimeters of precipitation. 5. The shortest bar corresponds to February. The normal monthly precipitation in February in Chicago is 3 centimeters. 6. The bars corresponding to March and November have a height of 7. Thus, during March and November, 7 centimeters of precipitation normally occurs.
278
total number 17 The ratio is . 40 number who prefer country
12.
number who prefer jazz 31 The ratio is . 22
85
200
62 44
17 40
31 22
13. The sector corresponding to under 18 is 20%. 20% of 374 million = 0.20 374 million = 74.8 million About 75 million people are expected to be in the under 18 age group in 2040. 14. The sector corresponding 85 and older is 4%. 4% of 374 million = 0.04 374 million = 14.96 million. About 15 million people are expected to be in the 85 and older age group in 2040. 15. The height of the bar for 5'8" 5'11" is 9. Thus, there are 9 students who are 5'8" 5'11" tall.
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16. Add the heights of the bars for 5'0" 5'3" and 5' 4" 5'7". There are 5 + 6 = 11 students who are 5'7" tall or shorter. 17.
Class Interval (scores)
Tally
Class Frequency (number of students)
4049
1
5059
3
6069
4
7079
||||
5
8089
|||| |||
8
4
9099
c.
The data items with the greatest frequency are the mode, so the mode is 93 and 94.
d.
range greatest data value smallest data value 94 90 4
22. There are 3 + 6 + 7 + 3 + 8 = 27 data items. a.
b.
15 3 20 6 25 7 30 3 35 8 27 710 27 26.3
mean
n 1 27 1 14 2 2 The data item in the 14th position is 25, so the median is 25.
18.
19. range greatest data value smallest data value 18 10 8 20. range greatest data value smallest data value 99 43 56
b.
157 12.530
25.
8 64 8 8 64 8 4 . because 100 10 5 10 10 10 100
2
2 2 26. hypotenuse (leg) (other leg)
42 42 16 16 32 5.66 The hypotenuse is 5.66 centimeters. 27.
11.5 2 2 The mean of the data items in the eleventh 93 93 and twelfth positions is 93, so the 2 median is 93.
range greatest data value smallest data value 35 15 20
24.
90 3 911 92 2 938 94 8 22 2041 22 92.8 22 1
d.
49 7 because 72 49.
mean
n 1
The data item with the greatest frequency is the mode, so the mode is 35.
23.
21. There are 3 + 1 + 2 + 8 + 8 = 22 data items. a.
c.
n 5 12 8 n 8 12 5 8n 60 8n 60 8 8 n 7.5
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ISM: Prealgebra and Introductory Algebra
Chapter 7: Graphs, Triangle Applications, ...
28. Let x be the height of the tower. x 48 53 4 4 3 x 4 5 48 4 23 4x 48 4 4x 276 4x 276 4 4 x 69 The tower is approximately 69 feet tall. Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow Red Blue Green Yellow
Red
Blue
29. Green
Yellow
1. 43 [3 2 (10 2)] 7 3 43 [32 5] 7 3 43 (9 5) 7 3 43 4 7 3 64 4 7 3 64 4 21 68 21 47 2. 72 [53 (6 3)] 4 2 72 [53 2] 4 2 72 (125 2) 4 2 72 123 4 2 49 123 4 2 49 123 8 172 8 180 3. x y = 3 9 = 3 + (9) = 12 4. x y = 7 (2) = 7 + 2 = 9
16 outcomes H
Cumulative Review Chapters 17
5.
3y 7 y 12 (3 7) y 12 4 y 12 4 y 12 4 4 y 3
6.
2x 6x 24 (2 6)x 24 4x 24 4x 24 4 4 x 6
H T H
30. T
T
4 outcomes 31. One of the ten possible outcomes is a 6. The 1 probability is . 10 32. Two of the ten possible outcomes are a 3 or a 4. 2 1 The probability is . 10 5 7.
280
x
1
4
6 3 x 4 6 1 6 6 3 x 4 6 6 1 6 6 3 x6 8 x66 86 x2
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
7
8.
a
16.
1
2 4 7 a 4 4(1) 2 4 7 a 4 4 4 1 2 4 14 a 4 14 14 a 4 14 a 10 9.
2 5
1 3 3
5
17. 115
8
2
24 17 7 24 1 3 17 2 5 7 3 8 24 3
2
3
8
5 3
20 15 4 4 4 20 3 3 23 7 1 8 7 20 20 20 2 3 3 3 4 8 5 4 20
11. 5.9 5
9
12. 2.8 2
10
2
4
27.88 205 = 0.136 19. (1.3)2 2.4 1.69 2.4 4.09 20. (2.7)2 (2.7)(2.7) 7.29
21.
1 1 25 25 0.25 4 4 25 100
22.
3 3 125 375 0.375 8 8 125 1000
5 23.
13.
3.500 0.068 3.432
14.
7.400 0.073 7.327
15. 0.0531 16 3186 5310 0.8496
4 decimal places
0.052 5.980 5 75 230 230 0
0.136 18. 205 27.880 20 5 7 38 6 15 1 230 1 230 0
10 8
3 decimal places 1 decimal place 3 1 4 decimal places
5.98 115 = 0.052
24 9
8
10.
0.147 0.2 0.0294
5(x 0.36) x 2.4 5 x 5 0.36 x 2.4 5x 1.8 x 2.4 5x x 1.8 x x 2.4 6x 1.8 2.4 6x 1.8 1.8 2.4 1.8 6x 4.2 6x 4.2 6 6 x 0.7
4 decimal places
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ISM: Prealgebra and Introductory Algebra
Chapter 7: Graphs, Triangle Applications, ...
24.
4(0.35 x) x 7 4 0.35 4 x x 7 1.4 4x x 7 1.4 4x x x x 7 1.4 5x 7 1.4 1.4 5x 7 1.4 5x 8.4 5x 8.4 5 5 x 1.68
c2 a2 b2 c2 22 162 c2 4 256 c2 260 c 260 16.12 The length of the hypotenuse is approximately 16.12 units. 31.
25.
80 8.944
26.
60 7.746
27.
1 1 1 1 1 1 because . 36 6 6 6 6 36
28.
4 16 4 4 4 16 because . 49 7 7 7 49 7
2
2
29. 7 in.
5 in.
Let a = 5 and c = 7. a2 b2 c2 52 b2 72 25 b2 49 b2 24 b 24 4.90
32.
8
x
5 10 8 10 5 x 80 5x 80 5x 5 5 16 x
The length of the leg is approximately 4.90 inches. 30.
Check:
16
8
x
5 10 8 16 0 5 10 8 10 0 5 16 80 80 True The solution is 16.
2
Let a = 2 and b = 16.
33.
282
x5 x 2 3 5 5(x 5) 3(x 2) 5x 25 3x 6 2x 25 6 2x 31 2x 31 2 2 31 x 2 x5 x 2 Check: 3 5 31 5 31 2 2 0 2 3 5 1 31 10 1 31 4 0 3 2 2 5 2 2 1 21 1 35 3 2 0 5 2 7 7 True 2 2 31 The solution is . 2
12 feet 12 19 feet 19
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Chapter 7: Graphs, Triangle Applications, …
ISM: Prealgebra and Introductory Algebra
34.
4 9
35. 4.6% = 4.6(0.01) = 0.046 36. 32% = 32(0.01) = 0.32 37. 0.74% = 0.74(0.01) = 0.0074 38. 2.7% = 2.7(0.01) = 0.027 39. 35% of 60 = 0.35 60 = 21 21 is 35% of 60. 40. 40% of 36 = 0.40 36 = 14.4 14.4 is 40% of 36. 41. 20.8 40% x 20.8 0.4x 20.8 0.4x 0.4 0.4 52 x 20.8 is 40% of 52. 42.
9.5 25% x 9.5 0.25x 9.5 0.25x 0.25 0.25 38 x 9.5 is 25% of 38.
43. $310 r $26.35 310r 26.35 310r 26.35 310 310 r 0.085 or 8.5% The sales tax rate is 8.5%.
nt
45. A P 1 r n 0.053 410 $4000 1 4 $4000(1.01325)40 $6772.12 46.
total $1600 $128.60 12 months 12 $1728.60 12 $144.05 The monthly payment is $144.05.
47. The list is in numerical order and there is an odd number of values, so the median is the middle value, 57. 48. The numbers are listed in order. Since there is an even number of values, the median is 47 50 97 48.5. 2 2 49. Two of the six possible outcomes are a 3 or a 4. 2 1 The probability is . 6 3 50. Three of the six possible outcomes are even. The 3 1 probability is . 6 2
44. $2 r $0.13 2r 0.13 2r 0.13 2 2 r 0.065 or 6.5% The sales tax rate is 6.5%.
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283
Chapter 8 Section 8.1 Practice Exercises 1. Figure (a) is part of a line with one endpoint, so it is a ray. It is ray AB or AB . Figure (b) has two endpoints, so it is a line segment. It is line segment RS or RS . Figure (c) extends indefinitely in two directions, so it is a line. It is line EF or EF . Figure (d) has two rays with a common endpoint, so it is an angle. It is HVT or TVH or V . 2. Two other ways to name z are RTS and STR. 3. a.
R is an obtuse angle. It measures between 90 and 180.
angles, so mz mc 135. Vocabulary, Readiness & Video Check 8.1 1. A plane is a flat surface that extends indefinitely. 2. A point has no length, no width, and no height. 3. Space extends in all directions indefinitely.
b.
N is a straight angle. It measures 180.
4. A line is a set of points extending indefinitely in two directions.
c.
M is an acute angle. It measures between 0 and 90.
5. A ray is part of a line with one endpoint.
d.
Q is a right angle. It measures 90.
4. The complement of a 29 angle is an angle that measures 90 29 = 61. 5. The supplement of a 67 angle is an angle that measures 180 67 = 113. 6. a.
my mADC mBDC 141 97 44
6. An angle is made up of two rays that share a common endpoint. The common endpoint is called the vertex. 7. A straight angle measures 180. 8. A right angle measures 90. 9. An acute angle measures between 0 and 90.
10. An obtuse angle measures between 90 and 180. 11. Parallel lines never meet and intersecting lines meet at a point.
b.
mx 79 51 28
c.
Since the measures of both x and y are between 0 and 90, they are acute angles.
7. Since a and the angle marked 109 are vertical angles, they have the same measure; so ma 109. Since a and b are adjacent angles, their measures have a sum of 180. So mb 180 109 71. Since b and c are vertical angles, they have the same measure; so mc 71.
284
8. w and x are vertical angles. w and y are corresponding angles, as are x and d. So all of these angles have the same measure: mx my md mw 45. w and a are adjacent angles, as are w and b, so ma mb 180 45 135. a and c are corresponding angles so mc ma 135. c and z are vertical
12. Two intersecting lines are perpendicular if they form right angles when they intersect. 13. An angle can be measured in degrees. 14. A line that intersects two or more lines at different points is called a transversal. 15. When two lines intersect, four angles are formed. Two of these angles that are opposite each other are called vertical angles. 16. Two angles that share a common side are called adjacent angles.
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ISM: Prealgebra and Introductory Algebra 17. WUV, VUW, U, x
Chapter 8: Geometry and Measurement 30. 20 + 70 = 90, so CAD and DAE are complementary. 63 + 27 = 90, so BAC and EAF are complementary.
18. straight angle; 180 19. 180 17 = 163
32. 38 + 142 = 180, so there are 4 pairs of supplementary angles: NMX and NMY , NMX and XMZ , YMZ and NMY , YMZ and XMZ.
20. intersect Exercise Set 8.1 2. The figure has two rays with a common endpoint. It is an angle, which can be named GHI, IHG, or H.
34. mx 150 48 102
4. The figure has one endpoint and extends indefinitely in one direction, so it is a ray. It is ray ST or ST.
38. x and the angle marked 165 are supplementary, so mx 180 165 15. y and the angle marked 165 are vertical angles, so my 165. x and z are vertical angles, so
6. The figure extends indefinitely in two directions, so it is a line. It is line AB, line t, or AB . 8. The figure has two endpoints, so it is a line segment. It is line segment PQ or PQ 10. Two other ways to name w are APC and CPA. 12. Two other ways to name y are MPQ and QPM . 14. H is an acute angle. It measures between 0 and 90. 16. T is an obtuse angle. It measures between 90 and 180. 18. M is a straight angle. It measures 180. 20. N is a right angle. It measures 90.
36. mx 36 12 48
mz mx 15. 40. x and the angle marked 44 are supplementary, so mx 180 44 136. y and the angle marked 44 are vertical angles, so my 44. x and z are vertical angles, so mz mx 136. 42. x and the angle marked 110 are vertical angles, so mx 110. x and y are corresponding angles, so my mx 110. y and z are vertical angles, so mz my 110. 44. x and the angle marked 39 are supplementary, so mx 180 39 141. y and the angle marked 39 are corresponding angles, so my 39.
22. The complement of an angle that measures 77 is an angle that measures 90 77 = 13.
y and z are supplementary, so mz 180 my 180 39 141.
24. The supplement of an angle that measures 77 is an angle that measures 180 77 = 103.
46. y can also be named CBD or DBC.
26. The complement of an angle that measures 22 is an angle that measures 90 22 = 68.
48. ABE can also be named EBA.
28. The supplement of an angle that measures 130 is an angle that measures 180 130 = 50.
52. mCBA 15
50. mEBD 45
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
54. mEBC mEBD mDBC 45 50 95
Section 8.2 Practice Exercises
56. mABE mABC mCBD mDBE 15 50 45 110 58.
Perimeter 10 m 10 m 18 m 18 m 56 meters The perimeter is 56 meters.
b.
Perimeter 125 ft 125 ft 50 ft 50 ft 350 ft
The perimeter is 350 feet.
7 1 7 1 2 7 2 5 8 4 8 4 2 8 8 8
2. P 2 l 2 w 2 32 cm 2 15 cm 64 cm 30 cm
7 4 74 7 1 60. or 3 8 4 8 1 24 2 2 7
1. a.
1
94 centimeters The perimeter is 94 centimeters.
1 1 10 5 62. 3 2 3 2 3 2 10 2 5 3 3 2 2 3 20 15 6 6 35 5 or 5 6 6
3. P = 4 s = 4 4 ft = 16 feet The perimeter is 16 feet. 4. P a b c 6 cm 10 cm 8 cm 24 centimeters The perimeter is 24 centimeters.
1 1 10 5 2 5 5 25 or 8 1 64. 3 2 3 2 3 2 3 2 3 3
5. Perimeter 6 km 4 km 9 km 4 km 23 kilometers The perimeter is 23 kilometers.
66. The three red dots divide the circle into three parts. If the three dots are about the same number of degrees apart, they divide the 360 of the circle into three parts. 360 120 3 The red dots are 120 degrees apart.
6. The unmarked horizontal side has length 20 m 15 m = 5 m. The unmarked vertical side has length 31 m 6 m = 25 m. P 15 m 31 m 20 m 6 m 5 m 25 m 102 meters The perimeter is 102 meters.
68. The complement of an angle that measures 35 is an angle that measures 90 35 = 55. 70. The complement of an angle that measures 53.13 is an angle that measures 90 53.13 = 36.87.
7. P 2 l 2 w 2 120 feet 2 60 feet 240 feet 120 feet 360 feet cost = $1.90 per foot 360 feet = $684 The cost of the fencing is $684. 8. C = d = 20 yd = 20 yd 62.8 yd The exact circumference of the watered region is 20 yards, which is approximately 62.8 yards.
72. False; answers may vary 74. True, since 5 + 175 = 180. 76. The sides of a rectangle are parallel. Thus, AB and CD, as well as AC and BD are parallel.
Vocabulary, Readiness & Video Check 8.2 1. The perimeter of a polygon is the sum of the lengths of its sides.
78. answers may vary 80. no; answers may vary
286
2. The distance around a circle is called the circumference.
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ISM: Prealgebra and Introductory Algebra 3. The exact ratio of circumference to diameter is . 4. The diameter of a circle is double its radius. 22 (or 3.14) and 3.14 or are 7 7
22 5. Both
approximations for . 6. The radius of a circle is half its diameter. 7. Opposite sides of a rectangle have the same measure, so we can just find the sum of the measures of all four sides. 8. the perimeter of a circle
16. P 2 l 2 w 2 6 ft 2 4 ft 12 ft 8 ft 20 ft The perimeter of each batter’s box is 20 feet. 18. P = 8 12 in. = 96 in. The perimeter is 96 inches. 20. P 2 l 2 w 2 70 ft 2 21 ft 140 ft 42 ft 182 ft 182 feet of fencing is needed. 22. The amount of fencing needed is 182 feet. 182 feet $2 per foot = $364 The total cost of the fencing is $364.
Exercise Set 8.2 2. P 2 l 2 w 2 14 m 2 5 m 28 m 10 m 38 m The perimeter is 38 meters.
24. a.
A pentagon has five sides.
b. P = 5 14 m = 70 m The perimeter is 70 meters.
4. P = 2 yd + 3 yd + 2 yd + 3 yd = 10 yd The perimeter is 10 yards. 6. P a b c 5 units 10 units 11 units 26 units The perimeter is 26 units. 8. Sum the lengths of the sides. P 10 m 4 m 10 m 9 m 20 m 13 m 66 m The perimeter is 66 meters. 10. All sides of a regular polygon have the same length, so the perimeter is the number of sides multiplied by the length of a side. P = 4 50 m = 200 m The perimeter is 200 meters. 12. All sides of a regular polygon have the same length, so the perimeter is the number of sides multiplied by the length of a side. P = 6 15 yd = 90 yd The perimeter is 90 yards. 14. P a b c 8 in. 12 in. 10 in. 30 in. The perimeter is 30 inches.
Chapter 8: Geometry and Measurement
26. P = 4 s = 4 3 in. = 12 in. The perimeter is 12 inches. 28. P 2 l 2 w 2 85 ft 2 70 ft 170 ft 140 ft 310 ft 310 ft $2.36 per foot = $731.60 The cost is $731.60. 30. The unmarked vertical side has length 13 in. 6 in. = 7 in. The unmarked horizontal side has length 30 in. 13 in. = 17 in. P 13 in. 13 in. 30 in. 7 in. 17 in. 6 in. 86 in. The perimeter is 86 inches. 32. The unmarked vertical side has length (2 + 4 + 3) cm = 9 cm. The width of the figure in the vertical section of height 4 cm is 16 11 = 5 cm. So the unmarked horizontal side has length (9 5) cm = 4 cm. P (16 2 11 4 4 3 9 9) cm 58 cm The perimeter is 58 centimeters. 34. The unmarked vertical side has length (22 + 5) km = 27 km. The unmarked horizontal side has length (12 + 6) km = 18 km P = (18 + 27 + 6 + 5 + 12 + 22) km = 90 km The perimeter is 90 kilometers.
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement 36. C = 2 r = 2 2.5 in. = 5 in. 15.7 in. The circumference is exactly 5 inches or approximately 15.7 inches. 38. C = d = 50 ft = 50 ft 157 ft The circumference is exactly 50 feet or approximately 157 feet. 40. C = 2 r = 2 10 yd = 20 yd 62.8 yd The circumference is exactly 20 yards or approximately 62.8 yards.
62. (72 2) 6 = 36 6 = 216 64. 41 (23 8) 41 (8 8) 41 0 4 0 0 66. a.
The first age category that 12-year-old children fit into is “Under 13,” so the maximum width is 60 yards and the maximum length is 110 yards.
b.
P 2l 2 w 2 110 yd 2 60 yd 220 yd 120 yd 340 yd The perimeter of the field is 340 yards.
42. C = 2 r = 2 18 ft = 36 ft 113.04 ft The circumference is exactly 36 feet or approximately 113.04 feet. 44. C = d = 1640 ft = 1640 ft 5149.6 ft The circumference is exactly 1640 feet or approximately 5149.6 feet.
68. The circle’s circumference is 7 in. 21.98 in. The square’s perimeter is 4 7 in. = 28 in. So the square has the greater distance around; b.
46. C = d = 150 ft = 150 ft 471 feet The circumference of the barn is 150 feet, or about 471 feet.
70. a.
48. C d 1 5 in. 2 11 in. 2 11 22 2 in. 17 in. 2 7 7 The distance around is about 17
Smaller circle: C = d = 16 m = 16 m 50.24 m Larger circle: C = d = 32 m = 32 m 100.48 m
b. Yes, when the diameter of a circle is doubled, the circumference is also doubled. 72. answers may vary
2
inches or
7 17.29 inches. 50. P = (4.5 + 7 + 9) yd = 20.5 yd The perimeter is 20.5 yards. 52. C = d = 11 m = 11 m 34.54 m The circumference is 11 meters 34.54 meters. 54. P = 4 19 km = 76 km The perimeter is 76 kilometers. 56. The unmarked vertical side has length (40 9) mi = 31 mi. The unmarked horizontal side has length (44 9) mi = 35 mi. P = (44 + 31 + 9 + 9 + 35 + 40) mi = 168 mi The perimeter is 168 miles.
74. The length of the curved section at the top is half of the circumference of a circle of diameter 6 meters. 1 1 1 C d 6 m 3 m 9.4 meters 2 2 2 The total length of the straight sides is 3 6 m = 18 m. The perimeter is the sum of these. 9.4 m + 18 m = 27.4 m The perimeter of the figure is 27.4 meters. 76. The three linear sides have lengths of 7 feet, 5 feet, and 7 feet. The length of the curved side is half of the circumference of a circle with diameter 5 feet, or 1 1 d 5 2.5 7.9 feet 2 2 7 ft + 5 ft + 7 ft + 7.9 ft = 26.9 ft The perimeter of the window is 26.9 feet.
58. 25 3 7 = 25 21 = 4 60. 6 (8 + 2) = 6 10 = 60 288
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ISM: Prealgebra and Introductory Algebra
1 1. A bh 2 1 1 12 in.8 in. 2 4 1 33 12 in. in. 2 4 4 3 33 = sq in. 21 4 49 sq in. 2 1 The area is 49 square inches. 2 A
1
(b B)h 2 1 (5 yd 11 yd)(6.1 yd) 2
48.8 sq yd The area is 48.8 square yards. 3. Split the rectangle into two pieces, a top rectangle with dimensions 12 m by 24 m, and a bottom rectangle with dimensions 6 m by 18 m. The area of the figure is the sum of the areas of these. A 12 m 24 m 6 m 18 m 288 sq m 108 sq m 396 sq m The area is 396 square meters. 4. A r 2 (7 cm)2 49 sq cm 153.86 sq cm The area is 49 square centimeters, which is approximately 153.86 square centimeters. 5. V = lwh = 7 ft 3 ft 4 ft = 84 cu ft The volume of the box is 84 cubic feet. SA 2lh 2wh 2lw 2(7 ft)(4 ft) 2(3 ft)(4 ft) 2(7 ft)(3 ft) 56 sq ft 24 sq ft 42 sq ft 122 sq ft The surface area of the box is 122 square feet.
3
1 6. V r cm 3 3 2 4 1 cu cm 3 8 1 cu cm 6 1 22 cu cm 6 7 11 cu cm 21 1 The volume is cubic centimeter, which is 6 11 approximately cubic centimeter. 21 2 1 2 SA 4r 4 cm 2 1 4 sq cm 4 sq cm 22 1 sq cm or 3 sq cm 7 7 The surface area is square centimeters, which 1 is approximately 3 square centimeters. 7 4
Section 8.3 Practice Exercises
2.
Chapter 8: Geometry and Measurement
3
4
7. V r 2h (5 in.)2 9 in. 25 sq in. 9 in. 225 cu in. 706.5 cu in. The volume is 225 cubic inches, which is approximately 706.5 cubic inches. 1 2 1 s h (3 m)2 5.1 m 3 3 1 9 sq m 5.1 m 3 15.3 cu m The volume is 15.3 cubic meters.
8. V
Vocabulary, Readiness & Vocabulary Check 8.3 1. The surface area of a polyhedron is the sum of the areas of its faces. 2. The measure of the amount of space inside a solid is its volume. 3. Area measures the amount of surface enclosed by a region.
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Chapter 8: Geometry and Measurement
4. Volume is measured in cubic units. 5. Area is measured in square units. 6. Surface area is measured in square units. 7. We don’t have a formula for an L-shaped figure, so we divide it into two rectangles, use the formula to find the area of each, and then add these two areas. 8. For each example, the exact volume is found and an approximate volume is found. Exact answers are in terms of , and approximate answers use an approximation for . Exercise Set 8.3 2. A = l w = 3.5 ft 1.2 ft = 4.2 sq ft The area is 4.2 square feet. 1 4. A b h 2 1 1 4 ft 5 ft 2 2 1 9 ft 5 ft 2 2 45 sq ft 4 1 11 sq ft 4 1 The area is 11 square feet. 4 6. A
1 2 1
b h
7 ft 5 ft 2 35 sq ft 2 1 17 sq ft 2 1 The area is 17 square feet. 2
ISM: Prealgebra and Introductory Algebra
8. A r2 (5 cm)2 25 sq cm 25 22 = sq cm 1 7 550 4 sq cm 78 sq cm 7 7 The area is 25 square centimeters 4 78 square centimeters. 7 10. A = b h = 3 cm 4.25 cm = 12.75 sq cm The area is 12.75 square centimeters. 12. A
1
14. A
1
(b B) h 12 1 5 in. 8 in. 6 in. 2 2 1 27 in. 6 in. 2 2 81 sq in. 2 1 40 sq in. 2 1 The area is 40 square inches. 2
(b B) h
2 1
(10 ft 5 ft) 3 ft 2 1 15 ft 3 ft 2 22.5 sq ft The area is 22.5 square feet. 16. A b h 5 cm 7 5 cm
3
4 31
cm cm
4 155 sq cm 4 3 38 sq cm 4 3 The area is 38 square centimeters. 4
290
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ISM: Prealgebra and Introductory Algebra 18. A = b h = 6 m 4 m = 24 sq m The area is 24 square meters.
Chapter 8: Geometry and Measurement
30. V s3 (11 mi)3 1331 cu mi The volume is 1331 cubic meters.
20. The area of the trapezoid that forms the top of the figure is 1 1 (b B) h (6 km 10 km) 4 km 2 2 32 sq km. The area of the rectangle that forms the bottom of the figure is l w = 10 km 5 km = 50 sq km. The area of the figure is the sum of these. A = 32 sq km + 50 sq km = 82 sq km The total area is 82 square kilometers. 22. The figure can be divided into two rectangles, one measuring 25 cm by 12 cm and one measuring 20 cm by 3 cm. The area of the figure is the sum of the areas of the two rectangles. A 25 cm 12 cm 20 cm 3 cm 300 sq cm 60 sq cm 360 sq cm The total area is 360 square centimeters. 24. The top of the figure is a square with sides of length 5 in., so its area is s2 (5 in.)2 25 sq in. The bottom of the figure is a parallelogram with area b h = 5 in. 4 in. = 20 sq in. The area of the figure is the sum of these. A = 25 sq in. + 20 sq in. = 45 sq in. The total area is 45 square inches. 26. r = d 2 = 5 m 2 = 2.5 m A r2
SA 6s2 6(11 mi)2 6 121 sq mi 726 sq mi The surface area is 726 square miles. 1 2 32. V r h 3 2 1 3 1 in. 9 in. 3 4 147 cu in. 16 147 22 7 cu in. 28 cu in. 16 7 8 The volume is 147 7 cubic inches 28 cubic inches. 16 8 SA r r 2 h2 r2 2 2 7 7 7 in. in. (9 in.) 2 in. 4 4 4 49 7 sq in. 1345 sq in. 49 16 167 sq in. 16 1345 16 60.00 sq in. The surface area is 49 7 1345 square inches 16 16 60.00 square inches.
34. V
(2.5 m)2 6.25 sq m 19.625 sq m
The area is 6.25 square meters 19.625 square meters. 28. V = l w h = 8 cm 4 cm 4 cm = 128 cu cm The volume is 128 cubic centimeters. SA 2lh 2wh 2lw 2 8 cm 4 cm 2 4 cm 4 cm 2 8 cm 4 cm 64 sq cm 32 sq cm 64 sq cm 160 sq cm The surface area is 160 square centimeters.
4 3 4
r3 (3 mi)3
3 36 cu mi 22 1 36 cu mi 113 cu mi 7 7 The volume is 1 36 cubic miles 113 cubic miles. 7 SA 4r 2 4(3 mi)2 36 sq mi 22 1 36 sq mi 113 sq mi 7 7 The surface area is 1 36 square miles 113 square miles. 7
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
36. r
1 1 d 10 ft 5 ft 2 2
V r 2 h (5 ft)2 6 ft 150 cu ft 22 3 150 cu ft 471 cu ft 7 7 The volume is 3 150 cubic feet 471 cubic feet. 7 1 2 1 s h (7 m)2 15 m 245 cu m 3 3 The volume is 245 cubic meters.
38. V
1 2 40. V r h 3 1 (14 ft)2 15 ft 3 980 cu ft 22 980 cu ft 3080 cu ft 7 The volume is 980 cubic feet 3080 cubic feet.
50. Including the house, the land is in the shape of a trapezoid, whose area is 1 1 (b B) h (96 ft 132 ft) 48 ft 2 2 1 228 ft 48 ft 2 5472 sq ft. The house is a rectangle, with area l w = 48 ft 24 ft = 1152 sq ft. The area of the yard is the difference between these. A = 5472 sq ft 1152 sq ft = 4320 sq ft The area of the yard is 4320 square feet. 52. V r 2 h (3 ft)2 8 ft 72 cu ft The volume is 72 cubic feet. SA 2r h 2 r 2 2(3 ft)(8 ft) 2(3 ft)2 66 sq ft The surface area is 66 square feet. 54. a.
The top part of the wall is a triangle, with 1 1 area b h 24 ft 4 ft 48 sq ft. 2 2 The bottom part is a rectangle, with area l w = 24 ft 8 ft = 192 sq ft. The area of the wall is the sum of these. A = 48 sq ft + 192 sq ft = 240 sq ft The total area is 240 square feet.
42. V s3 (5 ft)3 125 cu ft The volume is 125 cubic feet. SA 6s2 6(5 ft)2 6 25 sq ft 150 sq ft The surface area is 150 square feet. 44. A = l w = 52.7 m 3.6 m = 189.72 sq m The area of the sign is 189.72 square meters. 46. A = l w = 27.6 cm 21.5 cm = 593.4 sq cm The area of a page is 593.4 square centimeters. 1 4 r3 2 3 2 (10 in.)3 3 2000 cu in. 3 2000 22 5 cu in. 2095 cu in. 3 7 21 The volume is 2000 5 cubic inches. cubic inches 2095 3 12
48. V
292
b. Divide the area of the wall by the side area of each brick. 1 6 240 240 1440 6 1 The number of bricks needed to brick the end of the building is 1440. 56. r V
1
d
2 4
1
6 in. 3 in.
2 r3
3 4
(3 in.)3 3 36 cu in. 113.04 cu in. The volume is 36 cubic inches 113.04 cubic inches.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 58. r = d 2 = 2 cm 2 = 1 cm 2
2
A r (1 cm) sq cm 3.14 sq cm
Chapter 8: Geometry and Measurement
78. Gutters go on the edges of a house, so the situation involves perimeter.
The area of the watch face is square centimeters, or about 3.14 square centimeters.
80. Baseboards go around the edges of a room, so the situation involves perimeter.
1 2 60. V r h 3 1 (3 km)2 (3.5 km) 3 10.5 cu km 22 10.5 cu km 33 cu km 7 The volume is 10.5 cubic kilometers 33 cubic kilometers.
82. Fertilizer is spread throughout the surface of a yard, so the situation involves area.
62. A = l w = 35 ft 24 ft = 840 sq ft 840 square feet of insulation are needed. 64. V
4 3 4
84. This could be answered by dividing Utah into two rectangles, in two different ways. But it’s easier to regard the state as a 350 mile by 270 mile rectangle from which a 70 mile by 105 mile rectangle has been removed. A 350 mi 270 mi 70 mi 105 mi 94, 500 sq mi 7350 sq mi 87,150 sq mi The area of Utah is approximately 87,150 square miles. d 55 ft 27.5 ft 2 2 1 SA 4 r 2 2 1 4(3.14)(27.5 ft)2 2 4749.25 sq ft The surface area of the hemisphere is about 4749.25 sq ft.
r3
86. r
(2 cm)3
3 32 cu cm 3 32 3.14 cu cm 3 33.49 cu cm The volume is 32 cubic centimeters 3 33.49 cubic centimeters.
88. answers may vary 90. r
66. SA 2lh 2wh 2lw 2(2 in.)(2.2 in.) 2(2 in.)(2.2 in.) 2(2 in.)(2 in.) =8.8 sq in. 8.8 sq in. 8 sq in. 25.6 sq in. The surface area is 25.6 square inches. 68. 72 7 7 49 70. 202 20 20 400 72. 52 32 5 5 3 3 25 9 34 74. 12 62 11 6 6 1 36 37
d 80 ft 40 ft 2 2
A r 2 (40 ft)2 1600 sq ft 5024 sq ft The face of the tire has an area of 1600 square feet, or about 5024 square feet. 92. The semicircle at the top has radius 2.5 feet, so 1 1 its area is r2 (2.5 ft)2 3.125 sq ft. 2 2 The rectangle at the bottom has area l w = 7 ft 5 ft = 35 sq ft. The total area is 3.125 sq ft + 35 sq ft 44.8 sq ft.
76. Grass seed is planted throughout the surface of a yard, so the situation involves area.
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Chapter 8: Geometry and Measurement
94. First solid: V = lwh = (6 ft)(2 ft)(4 ft) = 48 cu ft SA 2lh 2wh 2lw 2(6 ft)(4 ft) 2(2 ft)(4 ft) 2(6 ft)(2 ft) 88 sq ft Second solid: V = lwh = (4 ft)(3 ft)(4 ft) = 48 cu ft SA 2lh 2wh 2lw 2(4 ft)(4 ft) 2(3 ft)(4 ft) 2(4 ft)(3 ft) 80 sq ft no; answers may vary Mid-Chapter Review 1. The supplement of a 27 angle measures 180 27 = 153. The complement of a 27 angle measures 90 27 = 63. 2. x and the angle marked 105 are supplementary angles, so mx 180 105 75. y and the angle marked 105 are vertical angles, so my 105. z and the angle marked 105 are supplementary angles, so mz 180 105 75. 3. x and the angle marked 52 are supplementary angles, so mx 180 52 128. y and the angle marked 52 are corresponding angles, so my 52. z and y are supplementary angles, so mz 180 my 180 52 128. 4. The sum of the measures of the angles of a triangle is 180. mx 180 90 38 52 5. d = 2 r = 2 2.3 in. = 4.6 in. 6. r
1
d 2 1 1 8 in. 2 2 1 17 in. 2 2 17 in. 4 1 4 in. 4
294
ISM: Prealgebra and Introductory Algebra 7. P = 4 s = 4 5 m = 20 m The perimeter is 20 meters. A s2 (5 m)2 25 sq m The area is 25 square meters. 8. P = a + b + c = 4 ft + 5 ft + 3 ft = 12 ft The perimeter is 12 feet. 1 1 A b h 3 ft 4 ft 6 sq ft 2 2 The area is 6 square feet. 9. C 2 r 2 5 cm 10 cm 31.4 cm The circumference is 10 centimeters 31.4 centimeters. A r2 (5 cm)2 25 sq cm 78.5 sq cm The area is 25 square centimeters 78.5 square centimeters. 10. P = 11 mi + 5 mi + 11 mi + 5 mi = 32 mi The perimeter is 32 miles. A = l w = 11 mi 4 mi = 44 sq mi The area is 44 square meters. 11. The unmarked horizontal side has length 17 cm 8 cm = 9 cm. The unmarked vertical side has length 7 cm + 3 cm = 10 cm. P = (8 + 3 + 9 + 7 + 17 + 10) cm = 54 cm The perimeter is 54 centimeters. The figure is made up of two rectangles, one with dimensions 10 cm by 8 cm, the other with dimensions 7 cm by 9 cm. A = 10 cm 8 cm + 7 cm 9 cm = 143 sq cm The area is 143 square centimeters. 12. P 2 l 2 w 2(17 ft) 2(14 ft) 34 ft 28 ft 62 ft The perimeter is 62 feet. A = l w = 17 ft 14 ft = 238 sq ft The area is 238 square feet. 13. P 2 l 2 w 2(3500 ft) 2(1400 ft) 7000 ft 2800 ft 9800 ft The perimeter is 9800 feet. A = l w = 3500 ft 1400 ft = 4,900,000 sq ft The area is 4,900,000 square feet.
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ISM: Prealgebra and Introductory Algebra
14. V s3 (4 in.)3 64 cu in. The volume is 64 cubic inches.
Chapter 8: Geometry and Measurement
4. 68 in.
5 yd 3 ft 15 ft 1 1 yd 5 yd 2 ft = 15 ft + 2 ft = 17 ft
5. 5 yd
6.
4 ft 8 in. + 8 ft 11 in. 12 ft 19 in. Since 19 inches is the same as 1 ft 7 in., we have 12 ft 19 in. = 12 ft + 1 ft 7 in. = 13 ft 7 in.
7.
4 ft 7 in. 4 16 ft 28 in. Since 28 in. is the same as 2 ft 4 in., we simplify as 16 ft 28 in. = 16 ft + 2 ft 4 in. = 18 ft 4 in.
8.
5 ft 8 in. 4 ft 20 in. 1 ft 9 in. 1 ft 9 in. 3 ft 11 in. The remaining board length is 3 ft 11 in.
1 2 1 2 16. V s h (10 cm) (12 cm) 400 cu cm 3 3 The volume is 400 cubic centimeters. 17. r V
1
d
1
2 4
2
3 mi
3
mi
2
r3
3 4 3 3 mi 3 2 9 cu mi 2 1 4 cu mi 2 9 22 1 cu mi 14 cu mi 2 7 7 The volume is 1 1 4 cubic miles 14 cubic miles. 2 7
18. The complement of an angle that measures 55 is an angle that measures 90 55 = 35. Section 8.4 Practice Exercises 6 ft
6 ft 12 in. 1 1. 6 ft 6 12 in. 72 in. 1 1 1 ft 2. 8 yd
8 yd 1
3. 18 in.
1
8 yd 3 ft 8 3 ft 24 ft 1 1 yd
18 in. 1 ft 18 ft 1.5 ft 1 12 in. 12
68 1 ft ft 1 12 in. 12
5 12 68 60 8 Thus, 68 in. = 5 ft 8 in.
SA 6 s2 6(4 in.)2 6 16 sq in. 96 sq in. The surface area is 96 square inches. 15. V = l w h = 3 ft 2 ft 5.1 ft = 30.6 cu ft The volume is 30.6 cubic feet. SA 2lh 2wh 2lw 2(3 ft)(5.1 ft) 2(2 ft)(5.1 ft) 2(3 ft)(2 ft) 30.6 sq ft 20.4 sq ft 12 sq ft 63 sq ft The surface area is 63 square feet.
68 in.
9. 2.5 m
2.5 m 1000 mm 2500 mm 1 1m
10. 3500 m = 3.500 km or 3.5 km 11. 640 m = 0.64 km 2.10 km 0.64 km 1.46 km 2.1 km = 2100 m 2100 m 640 m 1460 m 12. 18.3 hm 5 91.5 hm 13. 0.8 m = 80 cm 80 cm + 45 cm = 125 cm The scarf will be 125 cm or 1.25 m.
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Chapter 8: Geometry and Measurement
Vocabulary, Readiness & Video Check 8.4 1. The basic unit of length in the metric system is the meter. 2. The expression
1 foot
ISM: Prealgebra and Introductory Algebra
3.8 mi 5280 ft 1 1 mi 3.8 5280 ft 20, 064 ft
12. 3.8 mi
is an example of a
7216 yd 1 mi 1 1760 yd 7216 mi 1760 4.1 mi
14. 7216 yd
12 inches unit fraction. 3. A meter is slightly longer than a yard. 4. One foot equals 12 inches.
16. 129 in.
5. One yard equals 3 feet.
129 in. 1
6. One yard equals 36 inches.
8. feet; Feet are the original units and we want them to divide out. 2
5
5
2 and 5
5 ft 2 in. = 5 ft + 2 in. 10. The sum of 21 yd 4 ft is correct, but is not in a good format since there is a yard in 4 feet. Convert 4 ft = 1 yd 1 ft and add again: 21 yd + 1 yd + 1 ft = 22 yd 1 ft. 11. Since the metric system is based on base 10, we just need to move the decimal point to convert from one unit to another. 12. 1.29 cm and 12.9 mm; these two different-unit lengths are equal. Exercise Set 8.4 2. 84 in.
84 in. 1 ft 84 ft 7 ft 1 12 in. 12
4. 18 yd
18 yd 3 ft 18 3 ft 54 ft 1 1 yd
6. 36, 960 ft
8. 12
1
ft
2 10.
296
25 ft
36, 960 ft
1 ft 12 in.
129
ft 10.75 ft
12
100 ft 1 yd 100 yd 1 3 ft 3 33 yd 1 ft 3 100 9 10 9 1
18. 100 ft
7. One mile equals 5280 feet.
9. Both mean addition; 5
59 in. 1 ft 59 ft 1 12 in. 12 4 ft 11 in. 12 59 48 11
20. 59 in.
25, 000 ft 1 mi 25, 000 mi 1 5280 ft 5280 4 mi 3880 ft 5280 25, 000 21,120 3 880
22. 25, 000 ft
6 ft 12 in. 10 in. 1 1 ft 72 in. 10 in. 82 in.
24. 6 ft 10 in.
1 mi 5280 ft
1 36, 960 mi 5280 7 mi
26. 4 yd 1 ft
12.5 ft 12 in. 12.5 12 in. 150 in. 1 1 ft
28. 1 yd 2 ft
25 ft 1 yd 25 1 yd 8 yd 3 1 3 ft 3
4 yd 3 ft 1 ft 12 ft 1 ft 13 ft 1 1 yd
1 yd 3 ft 2 ft 3 ft 2 ft 5 ft 1 1 yd 5 ft 12 in. 5 ft 5 12 in. 60 in. 1 1 ft
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
30. 12 ft 7 in. 9 ft 11 in. 21 ft 18 in. 21 ft 1 ft 6 in. 22 ft 6 in. 32. 16 yd 2 ft 8 yd 2 ft 24 yd 4 ft 24 yd 1 yd 1 ft 25 yd 1 ft 34.
15 ft 5 in. 8 ft 2 in. 7 ft 3 in.
36.
14 ft 8 in. 13 ft 20 in. 3 ft 11 in. 3 ft 11 in. 10 ft 9 in.
38. 34 ft 6 in. 2 = 17 ft 3 in. 40.
15 yd 1 ft 8 120 yd 8 ft = 120 yd + 2 yd 2 ft = 122 yd 2 ft
42. 46 m
46 m 100 cm 4600 cm 1 1m
44. 14 mm 14 mm 1 cm 1.4 cm 1 10 mm 400 m
46. 400 m
1
1 km
0.4 km
1000 m
6.4 m 48. 6400 mm 6400 mm 1 m 1 1000 mm 50. 6400 cm 6400 cm 1 m 64 m 1 100 cm 52. 0.95 km 54. 5 km
0.95 km 100, 000 cm 95, 000 cm 1 1 km
5 km 1000 m 5000 m 1 1 km
56. 4.6 cm
4.6 cm 10 mm 46 mm 1 1 cm
58. 140.2 mm
140.2 mm 1
60. 0.2 m
1 dm
1.402 dm
100 mm
0.2 m 1000 mm 200 mm 1 1 Copyright © 2025 Pearson Education, Inc.
297
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
62.
14.10 cm 3.96 cm 18.06 cm
64.
30 cm 8.9 m
66.
45.3 m 2.16 dam
68.
14 cm 15 mm
30 cm 0.3 m 890 cm or 8.9 m 9.2 m 920 cm 4.53 dam 45.3 m 2.16 dam or 21.6 m 2.37 dam 23.7 m 14.0 cm 140 mm 1.5 cm or 15 mm 125 mm 12.5 cm
70. 14.1 m 4 = 56.4 m 72. 9.6 m 5 = 1.92 m Yards
Feet
Inches
74.
Four-story building
22
66
792
76.
Ostrich height
3
9
108 Meters
Millimeters
Kilometers
Centimeters
78.
Height of grizzly bear
3
3000
0.003
300
80.
Golf ball diameter
0.046
46
0.000046
4.6
82.
Distance from Houston to Dallas
396,000
396,000,000
396
39,600,000
84.
86.
1 mi 1400 ft 1 mi 4000 ft 2 mi 5400 ft 2 mi 1 mi 120 ft 3 mi 120 ft The probe is 3 mi 120 ft below the surface. 1
mi 2640 ft 2 1150 ft 1150 ft 1490 ft At their narrowest points, the Grand Canyon of the Yellowstone is 1490 feet wider than the Black Canyon of the Gunnison.
88. 3.4 m + 5.8 m = 9.2 m 9.2 m 9.20 m 8 cm 0.08 m 9.12 m The length of the tied ropes is 9.12 m.
298
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ISM: Prealgebra and Introductory Algebra
90.
14 cm 22 mm
14.0 cm 2.2 cm 11.8 cm The sediment is 11.8 cm thick now.
92. 1 yd 2 ft 25 25 yd 50 ft 25 yd 16 yd 2 ft 41 yd 2 ft The total length is 41 yd 2 ft.
114. answers may vary Section 8.5 Practice Exercises
1 ton 2000 lb tons
72 9 1 lb lb or 4 lb 2. 72 oz 72 oz 1 lb 1 16 oz 16 2 2
4837 ft 1 yd 4837 1 yd 1612 yd 1 3 ft 3 3 1 4837 feet is 1612 yards. 3
100
3. 47 oz 47 oz
1 lb
47
lb
16 oz 16 2 lb 15 oz 16 47 32 15 Thus, 47 oz = 2 lb 15 oz
43 50 4.
47 0.47 100. 100 3
3 5 15 0.15 20 20 5 100
104. Yes, a window measuring 1 meter by 0.5 meter is reasonable. 106. No, a paper clip being 4 kilometers long is not reasonable. 108. Yes, a model’s hair being 30 centimeters long is reasonable. 110.
1 6500
2000 13 1 tons or 3 tons 4 4
96. 4837 ft
102.
6500 lb
12 whole sections are removed.
86
112. answers may vary; for example, 7 m 100 cm 7m 700 cm 1 1m 7 m 1000 mm 7m 7000 mm 1 1m
1. 6500 lb
12.5 94. 9 112.5 9 22 18 45 4 5 0
98. 0.86
Chapter 8: Geometry and Measurement
45 ft 1 in. 10 ft 11 in.
8 tons 100 lb 7 tons 2100 lb 5 tons 1200 lb 5 tons 1200 lb 2 tons 900 lb
1 lb 6 oz 5. 4 5 lb 8 oz 4 lb 1 lb 16 oz 24 oz 6.
batch weight 5 lb 14 oz container weight 6 oz 5 lb 20 oz total weight 5 lb 20 oz = 5 lb + 1 lb 4 oz = 6 lb 4 oz The total weight is 6 lb 4 oz.
7. 3.41 g
Estimate: 45 ft 11 ft = 34 ft
3.41 g 1000 mg 3410 mg 1 1g
8. 56.2 cg 56.2 cg 0.562 g
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299
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
9. 3.1 dg = 0.31 g 2.50 g 0.31 g 2.19 g
or
2.5 g = 25 dg 25.0 dg 3.1 dg 21.9 dg
8. 90 oz
11, 000 lb 1 ton 1 2000 lb 11, 000 tons 2000 11 tons 2 1 5 tons 2
10. 11, 000 lb
22.9 23 10. 24 550.0 kg 48 70 48 22 0 21 6 4 12. 9.5 lb
Each bag weighs about 23 kg.
90 oz 1 lb 90 45 5 lb lb 5 lb 1 16 oz 16 8 8
9.5 lb 16 oz 9.5 16 oz 152 oz 1 1 lb
Vocabulary, Readiness & Video Check 8.5 1. Mass is a measure of the amount of substance in an object. This measure does not change.
8.3 tons 2000 lb 1 1 ton 8.3 2000 lb 16, 600 lb
14. 8.3 tons
2. Weight is a measure of the pull of gravity. 3. The basic unit of mass in the metric system is the gram.
16.
4. One pound equals 16 ounces. 5. One ton equals 2000 pounds. 6. pounds; Pounds are the units we’re converting to. 7. We can’t subtract 9 oz from 4 oz, so we borrow 1 lb (= 16 oz) from 12 lb to add to the 4 oz: 12 lb 4 oz becomes 11 lb 20 oz.
1 73 9 lb lb 8 8 73 lb 16 oz 8 1 1 lb 73 16 oz 8 146 oz 51 oz 1 lb 1 16 oz 51 lb 16 3.1875 lb 3.2 lb
18. 51 oz
8. 3 places to the right; 4 g = 4000 mg 9. 18.50 dg Exercise Set 8.5
20.
1
1 oz
oz 4
4 5 lb 16 oz 2. 5 lb 5 16 oz 80 oz 1 1 lb
7 tons 2000 lb 4. 7 tons 1 1 ton 7 2000 lb 14, 000 lb 6. 28, 000 lb
28, 000 lb
22. 2
1 1
lb
4
9 4
1 lb 16 oz
1 1 1 lb lb 4 16 64
9 lb 16 oz 9 lb 4 16 oz 36 oz
1
1 lb
24. 7 lb 6 oz 7 16 oz 6 oz 112 oz 6 oz 118 oz
1 ton 2000 lb
1 28, 000 tons 2000 14 tons 300
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4
ISM: Prealgebra and Introductory Algebra
26. 100 oz 100 oz 1 lb 100 lb 1 16 oz 16 6 lb 4 oz 16 100 96 4 100 oz = 6 lb 4 oz 28. 6 lb 10 oz 10 lb 8 oz 16 lb 18 oz 16 lb 1 lb 2 oz 17 lb 2 oz
Chapter 8: Geometry and Measurement
50. 16.23 g 16.23 g 1000 mg 1 1g 16.231000 mg 16, 230 mg 3.16 kg 1000 g 1 1 kg 3.16 1000 g 3160 g
52. 3.16 kg
4.26 cg 1 dag 1 1000 cg 4.26 dag 1000 0.00426 dag
54. 4.26 cg
30. 1 ton 1140 lb + 5 tons 1200 lb = 6 tons 2340 lb = 6 tons + 1 ton 340 lb = 7 tons 340 lb 32.
4 tons 850 lb 1 ton 260 lb 3 tons 590 lb
34.
45 lb 6 oz 26 lb 10 oz
56.
44 lb 22 oz 26 lb 10 oz 18 lb 12 oz
36. 2 lb 5 oz 5 10 lb 25 oz 10 lb 1 lb 9 oz 11 lb 9 oz 38. 5 tons 400 lb 4 tons 1 ton 400 lb 4 tons 2000 lb 400 lb 4 tons 2400 lb 4 2400 4 tons 2400 lb 4 ton lb 4 4 1 ton 600 lb 40. 820 g
820 g 1
42. 9 g
9 g 1000 mg 9 1000 mg 9000 mg 1 1g
18 kg 1000 g 18 1000 g 18, 000 g 44. 18 kg 1 1 kg 46. 112 mg 48. 4.9 g
112 mg 1
4.9 g 1
58. 2.1 g + 153 mg = 2.1 g + 0.153 g = 2.253 g or 2.1 g + 153 mg = 2100 mg + 153 mg = 2253 mg 60. 6.13 g 418 mg 6130 mg 418 mg 5712 mg or 6.13 g 418 mg = 6.13 g 0.418 g = 5.712 g 62. 4 kg 2410 g 4000 g 4.00 kg 2.41 kg or 2140 g 1590 g 1.59 kg 64.
1 kg 820 kg 0.82 kg 1000 g 1000
1g
112 g 0.112 g 1000 mg 1000
1 kg 4.9 kg 0.0049 kg 1000 g 1000
41.6 g 9.8 g 51.4 g
4.8 kg 9.3 44.64 kg
66. 8.25 g 6
8.25
g
6
1.375 6 8.250 6 22 18 45 42 30 30 0 8.25 g 6 = 1.375 g
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301
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
Object
Tons
Pounds
Ounces
68.
Statue of Libertyweight of steel
125
250,000
4,000,000
70.
A 12-inch cube of lithium (lightest metal)
or 0.016
32
512
2 125
Object
Grams
Kilograms
Milligrams
Centigrams
72.
Tablet of Topamax (epilepsy and migraine uses)
0.025
0.000025
25
2.5
74.
A golf ball
45
0.045
45,000
4500
76.
73 kg 73.0 kg 2.8 kg 2800 g 70.2 kg His new weight is 70.2 kg.
78. 177 g 6
177
g
6
29.5 6 177.0 12 57 54 30 3 0 0 Each serving weighs 29.5 g. 80.
3 tons 1500 lb 2 tons 1200 lb 5 tons 2700 lb 5 tons 1 ton 700 lb 6 tons 700 lb The total amount delivered was 6 tons 700 lb.
82.
22 lb 8 oz 7 lb 12 oz
21 lb 24 oz 7 lb 12 oz 14 lb 12 oz This is 14 lb 12 oz heavier.
84. 1900 g 6 5 = 57,000 g 57, 000 g 1 kg 57 kg 1 1000 g The weight of 5 cartons is 57 kg. 86. 0.385 kg 39 g 0.385 kg 0.039 kg 0.346 kg 0.346 kg 8 = 2.768 kg 8 boxes contain 2.768 kg of cocoa.
302
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ISM: Prealgebra and Introductory Algebra
88. 85 g 2
85
g 42.5 g
Chapter 8: Geometry and Measurement
3.
5 qt 1 gal 1 qt 2 qt 2 qt 3 qt
4.
15 gal 3 qt + 4 gal 3 qt 19 gal 6 qt = 19 gal + 1 gal 2 qt = 20 gal 2 qt The total amount of oil will be 20 gal 2 qt.
2 Each package will weigh 42.5 g.
90.
1 lb 2 oz 12 12 lb 24 oz = 12 lb + 1 lb 8 oz = 13 lb 8 oz Each carton weighs 3 lb 8 oz. 13 lb 8 oz 3 39 lb 24 oz = 39 lb + 1 lb 8 oz = 40 lb 8 oz 3 cartons weigh 40 lb 8 oz.
92.
43 lb 2 oz 3 lb 4 oz
42 lb 18 oz 3 lb 4 oz 39 lb 14 oz
39 lb 14 oz 3 117 lb 42 oz 117 lb 2 lb 10 oz 119 lb 10 oz 3 cartons contain 119 lb 10 oz of ham. 3
3 20 60 0.60 or 0.6 94. 5 5 20 100 3 96.
16
3 625 1875 0.1875 16 625 10, 000
5. 2100 ml 6. 2.13 dal
2100 ml 1
1L 1000 ml
2100 1000
L 2.1 L
2.13 dal 10 L 2.1310 L 21.3 L 1 1 dal
7. 1250 ml = 1.250 L 2.9 L = 2900 ml 1.25 L 1250 ml 2.9 L 2900 ml 4150 ml 4.15 L The total is 4.15 L or 4150 ml. 8.
28.6 L 85 143 0 2288 0 2431.0 L Thus, 2431 L can be pumped in 85 minutes.
Vocabulary, Readiness & Video Check 8.6 98. No, a full grown cat weighing 15 g is not reasonable.
1. Units of capacity are generally used to measure liquids.
100. Yes, a staple weighing 15 mg is reasonable. 102. No, a car weighing 2000 mg is not reasonable. 104. answers may vary; for example, 2 tons or 64,000 oz
2. The basic unit of capacity in the metric system is the liter. 3. One cup equals 8 fluid ounces. 4. One quart equals 2 pints.
106. true 5. One pint equals 2 cups. 108. answers may vary 6. One quart equals 4 cups. Section 8.6 Practice Exercises 7. One gallon equals 4 quarts. 43 pt 1 qt 43 1 1. 43 pt qt 21 qt 1 2 pt 2 2 2. 26 qt
26 qt 4 c 26 4 c 104 c 1 1 qt
8. When using a unit fraction, we are not changing the amount, we are changing the unit. 9. We can’t subtract 3 qt from 0 qt, so we borrow 1 gal (= 4 qt) from 3 gal to get 2 gal 4 qt. 10. 3 places to the left; 5600 ml = 5.6 L 11. 0.45 dal
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303
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
Exercise Set 8.6 2. 16 qt
4. 9 pt
16 qt 1 gal 16 gal 4 gal 1 4 qt 4
9 pt 1 qt 9 1 qt 4 qt 1 2 pt 2 2
11 c 1 pt 11 1 6. 11 c pt 5 pt 1 2c 2 2 18 pt 1 qt 18 qt 9 qt 1 2 pt 2 9 qt 1 gal 9 1 9 qt gal 2 gal 1 4 qt 4 4
8. 18 pt
10. 3 pt
3 pt 2 c 8 fl oz 3 2 8 fl oz 48 fl oz 1 1 pt 1 c
20 c 1 pt 1 qt 1 gal 1 2 c 2 pt 4 qt 20 gal 22 4 5 gal 4 1 1 gal 4
22. 70 qt 68 qt 2 qt 68 qt 1 gal 2 qt 1 4 qt 68 gal 2 qt 4 17 gal 2 qt 24. 29 pt 28 pt 1 pt 28 pt 1 qt 1 pt 1 2 pt 14 qt 1 pt 12 qt 2 qt 1 pt 12 qt 1 gal 2 qt 1 pt 1 4 qt 3 gal 2 qt 1 pt 3 gal 2 qt 1 pt 26. 3
12. 20 c
14. 7 qt
16. 6
1 2
13
28.
2 pt 1 pt 1 c
1 gal
1 pt 2 c 1 pt 1 c 1c
34. 3 qt 1 c 1 c 4 fl oz
4 gal 4 qt 1 qt 1 1 gal 4 4 qt 1 qt 16 qt 1 qt 17 qt
304
2 gal 2 qt 9 gal 3 qt 11 gal 5 qt = 11 gal + 1 gal 1 qt = 12 gal 1 qt
13
18. 4 gal 1 qt
2
qt
30. 2 c 3 fl oz 2 c 6 fl oz 4 c 9 fl oz 4 c 1 c 1 fl oz 5 c 1 fl oz 32.
13 gal 4 qt 2
4 qt 2 26 qt
1
4
13 qt 2 pt 2 c 4
gal
1
20.
13
1 1 qt 1 pt 13 22 c 4 13 c
2
4
qt
7 qt 2 pt 2 c 7 2 2 c 28 c 1 1 qt 1 pt
gal
1
1 pt
pt 2
1
1 qt 2 pt
2 qt 1 pt 3 c 1 c 4 fl oz
1 1 1 qt qt 2 2 4
2 qt 2 pt 1 c 1 c 4 fl oz 2 qt 1 pt 2 c 8 fl oz 1 c 4 fl oz 2 qt 1 pt 1 c 4 fl oz
36. 6 gal 1 pt 2 12 gal 2 pt 12 gal 1 qt 12 gal 1 qt
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
38. 5 gal 6 fl oz 2
5
6
fl oz 2 2 1 2 gal 3 fl oz 2 1 2 gal gal 3 fl oz 2 2 gal 2 qt 3 fl oz
40. 8 L
gal
Chapter 8: Geometry and Measurement
8 L 1000 ml 8000 ml 1 1L
42. 0.127 L 0.127 L 1 kl 1 1000 L 0.127 kl 1000 0.000127 kl 1500 L 1.5 L 44. 1500 ml 1500 ml 1 L 1 1000 ml 1000 46. 1.7 L
1.7 L 100 cl 1.7 100 cl 170 cl 1 1L
48. 250 L
250 L 1
1 kl 250 kl 0.25 kl 1000 L 1000
50. 39 ml 39 ml 1 L 39 L 0.039 L 1 1000 ml 1000 0.48 kl 1000 L 1 1 kl 0.48 1000 L 480 L
52. 0.48 kl
54. 1.9 L
1.9 L 1000 ml 1.9 1000 ml 1900 ml 1 1L
56. 18.5 L + 4.6 L = 23.1 L 58. 4.6 L = 4600 ml 4600 ml 1600 ml 6200 ml
1600 ml = 1.6 L 4.6 L 1.6 L 6.2 L
60.
4.8 L 283 ml
4.800 L 4800 ml 0.283 L or 283 ml 4.517 L 4517 ml
62.
6850 ml 0.3 L
6.85 L 6850 ml 300 ml or 0.30 L 6550 ml 6.55 L Copyright © 2025 Pearson Education, Inc.
305
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement 64. 290 ml 6 = 1740 ml 66. 5.4 L 3.6
5.4
L 1.5 L
3.6 Capacity
Cups
Gallons
Quarts
Pints
68.
A dairy cow’s daily milk yield
76
4 34
19
38
70.
The amount of water needed in a punch recipe
2
1 8
1 2
1
72.
54.5 L 3.8 L 50.7 L 50.7 L are needed to fill the tank.
74. 2 L 8
2
1
L 0.25 L 8 4 Each person (Avery plus 7 friends) will get 0.25 L.
76.
L
1 qt 1 pt 3 3 qt 3 pt 3 qt + 1 qt 1 pt 4 qt 1 pt 1 gal 1 pt The total amount of soup is 1 gal 1 pt. 3 c 8 fl oz 38 fl oz 24 fl oz 1 1c 24 24 fl oz 6 fl oz 4 fl oz 6 Each dish has 4 fl oz.
78. 3 c
40 cl 10 ml 40 ml 1 1 cl 400 ml 40 ml 360 ml The excess amount of water is 360 ml.
80. 40 cl 40 ml
82.
75 100
84. 86.
3 25 3 4 25 4
56 14 4 14 60 15 4 15 18
9 2 9 20 10 2 10
88. Yes, drinking 250 ml of milk with lunch is reasonable. 90. Yes, a gas tank of a car holding 60 L of gasoline is reasonable. 306
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
92. greater than; answers may vary 94. answers may vary 1.5 gal 4 qt 2 pt 2 c 8 fl oz 1 1 gal 1 qt 1 pt 1 c 1.5 4 2 2 8 fl oz 192 fl oz There are 192 fl oz in 1.5 gallons.
96. 1.5 gal
98. A indicates 0.6 cc. 100. C indicates 1.9 cc. 102. A indicates 38 u or 0.38 cc. 104. C indicates 72 u or 0.72 cc. Section 8.7 Practice Exercises 1. 1.5 cm
1.5 cm 1
2. 8 oz
1 in. 1.5 in. 0.59 in. 2.54 cm 2.54
8 oz 28.35 g 8 28.35 g 226.8 g 1 1 oz
3. 237 ml
237 ml
1 fl oz 1 29.57 ml 237 fl oz 29.57 8 fl oz
4. F
9
C 32
9
60 32 108 32 140 5 5 Thus, 60C is equivalent to 140F.
5. F 1.8 C 32 1.8 32 32 57.6 32 89.6 Therefore, 32C is the same as 89.6F. 5 (68 32) (36) 20 9 9 9 Therefore, 68F is the same temperature as 20C.
6. C
5
(F 32)
5
5 5 (F 32) (113 32) (81) 45 9 9 9 Therefore, 113F is 45C.
7. C
5
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307
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
5 8. C (F 32) 9 5 (102.8 32) 9 5 (70.8) 9 39.3 Albert’s temperature is 39.3C. Vocabulary, Readiness & Video Check 8.7 1. 1 L 0.26 gal or 3.79 L 1 gal 2. The original stated example asked for the answer in grams, so a conversion from kg to g is still needed. 3. F = 1.8C + 32; 27 4. 77; 77F = 25C Exercise Set 8.7 2. 18 L
18 L 1.06 qt 18 1.06 qt 19.08 qt 1 1L 86 mi 1.61 km 1 1 mi 86 1.61 km 138.46 km
4. 86 mi
6. 100 kg
8. 9.8 m
100 kg 2.20 lb 100 2.2 lb 220 lb 1 1 kg
9.8 m 3.28 ft 9.8 3.28 ft 32.14 ft 1 1m
10. 150 ml
150 ml
1 fl oz 1 29.57 ml 150 fl oz 29.57 5.07 fl oz
12. 15 oz
15 oz 28.35 g 15 28.35 g 425.25 g 1 1 oz Meters
Yards
Centimeters
Feet
Inches
14.
Statue of Liberty Length of Nose
1.37
1.5
137
4.5
54
16.
Blue Whale
33
36
3300
108
1296
308
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ISM: Prealgebra and Introductory Algebra
18. 250 cm
250 cm
1 in. 98.43 in. 1 2.54 cm 98.43 in. 1 ft 98.43 in. 8.20 ft 1 12 in. The rings are approximately 98.43 inches or 8.20 feet from the floor. 70 mph 1.61 km 112.7 kph 1 1 mi The speed limit is approximately 112.7 kilometers per hour.
20. 70 mph
0.5 g 0.04 oz 0.02 oz 1 1g The tablets are approximately 0.02 ounce.
22. 500 mg 0.5 g
24. 26 m
26 m 3.28 ft 85.28 ft 1 1m
26. 2.7 million kg 2, 700, 000 kg 2.20 lb 1 ton 1 1 kg 2000 lb 2970 tons 5 ft 12 in. 11 in. 1 1 ft 5 12 in. 11 in. 60 in. 11 in. 71 in. 71 in. 2.54 cm 1 1 in. 71 2.54 cm 180.34 cm Yes; the stamp is approximately correct.
28. 5 ft 11 in.
18 ft 0.30 m 5.4 m 1 1 ft The nests are up to about 5.4 meters.
30. 18 ft
32. 84 cm
84 cm 1 in.
33 in.
1 2.54 cm 33 in. 24 in. 9 in. 24 in. 1 ft 9 in. 1 12 in. 2 ft 9 in. The average two-year-old is approximately 2 ft 9 in.
Chapter 8: Geometry and Measurement 9 lb 0.45 kg 9 0.45 kg 4.05 kg 1 1 lb 4.05 kg 1000 g 4.05 kg 4050 g 1 1 kg The organ weighs approximately 4050 g.
34. 9 lb
906 kph 0.62 mi 562 mph 1 1 km The speed is approximately 562 miles per hour.
36. 906 kph
303 km 0.62 mi 187.86 mi 1 1 km The diameter is approximately 187.86 mi.
38. 303 km
2079 km 0.62 mi 1289 mi 1 1 km The distance is approximately 1289 mi.
40. 2079 km
24
4 doses 6 per day and 4 10 = 40 doses for 10 days. 12 ml 40 = 480 ml 480 ml 1 fl oz 480 ml 16.23 fl oz 1 29.57 ml 17 fluid ounces of medicine should be purchased.
42. One dose every 6 hours results in
44. A mile is longer than a kilometer; b. 46. A foot is shorter than a meter; a 48. A football field is 100 yards, which is about 90 m, since a yard is a little less than a meter; b 50. A 5-gallon gasoline can has a capacity of about 19 L (1 gal 3.79 L); a 52. The weight of a pill is about 200 mg; d 54. C
5 9 5 9 5
(F 32) (86 32)
(54) 9 30 86F is 30C. 5 5 5 56. C (F 32) (140 32) (108) 60 9 9 9 140F is 60C.
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Chapter 8: Geometry and Measurement
58. F
9
76. F 1.8C 32 1.8(127) 32 228.6 32 260.6 127C is 260.6F.
C 32
5 9
(80) 32 5 144 32 176 80C is 176F. 60. F
9
78. C
9
C 32 (225) 32 405 32 437 5 5 225C is 437F.
62. C
5
(F 32)
5
9 9 26F is 3.3C.
64. C
5
ISM: Prealgebra and Introductory Algebra
(F 32)
5
9 9 43.4F is 6.3C.
(26 32)
5
(6) 3.3
9
5
(F 32)
9 5
(58.3 32) 9 5 (26.3) 9 14.6 58.3F is 14.6C. 80. 10 2 + 9(8) = 5 + 9(8) = 5 + 72 = 77
(43.4 32)
5
(11.4) 6.3
82. 5[(18 8) 9] = 5[10 9] = 5(1) = 5
9 84. False; the boiling point of water is 212F.
66. F 1.8C 32 1.8(75) 32 135 32 167 75C is 167F.
86. Yes, an air temperature of 20F on a Vermont ski slope in winter is reasonable.
68. F 1.8C 32 1.8(48.6) 32 87.48 32 119.5 48.6C is 119.5F.
90. No, water cooled to 32C will not freeze.
70. C
5 9 5
88. No, a room temperature of 60C does not feel chilly.
63157 1.66 3600 The BSA is approximately 1.66 sq m.
92. BSA
26 in. 2.54 cm 66.04 cm 1 1 in. 13 66.04 BSA 0.49 3600 The BSA is approximately 0.49 sq m.
(F 32)
94. 26 in.
(95 32)
9 5
(63) 9 35 95F is 35C. 72. F = 1.8C + 32 = 1.8(18) + 32 = 32.4 + 32 = 64.4 18C is 64.4F. 74. C
5 9 5
(F 32) (66 32)
69 in. 2.54 cm 175.26 cm 1 1 in. 172 lb 0.45 kg 172 lb 77.4 kg 1 1 lb
96. 69 in.
77.4 175.26 1.94 3600 The BSA is approximately 1.94 sq m. BSA
9 5
(34) 9 18.9 66F is 18.9C.
310
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ISM: Prealgebra and Introductory Algebra
98. C
5 9 5
Chapter 8: Geometry and Measurement
16. The measure of the space of a solid is called its volume.
(F 32) (9010 32)
17. When two lines intersect, four angles are formed. Two of these angles that are opposite each other are called vertical angles.
9 5
(8978) 9 4988 9010F is 4988C.
18. Two of the angles from Exercise 17 that share a common side are called adjacent angles. 19. An angle whose measure is between 90 and 180 is called an obtuse angle.
100. answers may vary Chapter 8 Vocabulary Check 1. Weight is a measure of the pull of gravity.
20. An angle that measures 90 is called a right angle.
2. Mass is a measure of the amount of substance in an object. This measure does not change.
21. An angle whose measure is between 0 and 90 is called an acute angle.
3. The basic unit of length in the metric system is the meter. 4. To convert from one unit of length to another, unit fractions may be used. 5. The gram is the basic unit of mass in the metric system. 6. The liter is the basic unit of capacity in the metric system.
22. Two angles that have a sum of 180 are called supplementary angles. 23. The surface area of a polyhedron is the sum of the areas of the faces of the polyhedron. Chapter 8 Review 1. A is a right angle. It measures 90. 2. B is a straight angle. It measures 180.
7. A line segment is a piece of a line with two endpoints.
3. C is an acute angle. It measures between 0 and 90.
8. Two angles that have a sum of 90 are called complementary angles.
4. D is an obtuse angle. It measures between 90 and 180.
9. A line is a set of points extending indefinitely in two directions.
5. The complement of a 25 angle has measure 90 25 = 65.
10. The perimeter of a polygon is the distance around the polygon.
6. The supplement of a 105 angle has measure 180 105 = 75.
11. An angle is made up of two rays that share the same endpoint. The common endpoint is called the vertex.
7. mx 90 32 58
12. Area measures the amount of surface of a region.
9. mx 105 15 90
13. A ray is a part of a line with one endpoint. A ray extends indefinitely in one direction. 14. A line that intersects two or more lines at different points is called a transversal.
8. mx 180 82 98
10. mx 45 20 25 11. 47 + 133 = 180, so a and b are supplementary. So are b and c, c and d , and d and a.
15. An angle that measures 180 is called a straight angle. Copyright © 2025 Pearson Education, Inc.
311
Chapter 8: Geometry and Measurement 12. 47 + 43 = 90, so x and w are complementary. Also, 58 + 32 = 90, so y and z are complementary. 13. x and the angle marked 100 are vertical angles, so mx 100. x and y are adjacent angles, so my 180 100 80. y and z are vertical angles, so mz my 80. 14. x and the angle marked 25 are adjacent angles, so mx 180 25 155. x and y are vertical angles, so my mx 155. z and the angle marked 25 are vertical angles, so mz 25. 15. x and the angle marked 53 are vertical angles, so mx 53. x and y are alternate interior angles, so my mx 53. y and z are adjacent angles, so mz 180 my 180 53 127. 16. x and the angle marked 42 are vertical angles, so mx 42. x and y are alternate interior angles, so my mx 42. y and z are adjacent angles, so mz 180 my 180 42 138. 17. P 23 m 11
1
m 23 m 11
2 The perimeter is 69 meters.
1
m 69 m
2
21. P = 2 l + 2 w = 2 10 ft + 2 6 ft = 32 ft The perimeter is 32 feet. 22. P = 4 s = 4 110 ft = 440 ft The perimeter is 440 feet. 23. C = d = 1.7 in. 3.14 1.7 in. = 5.338 in. The circumference is 5.338 inches. 24. C 2 r 2 5 yd 10 yd 3.14 10 yd 31.4 yd The circumference is 31.4 yards. 25. A
1
(b B) h 2 1 (12 ft 36 ft) 10 ft 2 1 48 ft 10 ft 2 240 sq ft The area is 240 square feet.
26. A = b h = 21 yd 9 yd = 189 sq yd The area is 189 square yards. 27. A = l w = 40 cm 15 cm = 600 sq cm The area is 600 square centimeters. 28. A s2 (9.1 m)2 82.81 sq m The area is 82.81 square meters. 29. A r 2 (7 ft)2 49 sq ft 153.86 sq ft The area is 49 square feet 153.86 square feet. 2 2 30. A r (3 in.) 9 sq in. 28.26 sq in.
18. P = 11 cm + 7.6 cm + 12 cm = 30.6 cm The perimeter is 30.6 centimeters. 19. The unmarked vertical side has length 8 m 5 m = 3 m. The unmarked horizontal side has length 10 m 7 m = 3 m. P = (7 + 3 + 3 + 5 + 10 + 8) m = 36 m The perimeter is 36 meters. 20. The unmarked vertical side has length 5 ft + 4 ft + 11 ft = 20 ft. P (22 20 22 11 3 4 3 5) ft 90 ft The perimeter is 90 feet.
312
ISM: Prealgebra and Introductory Algebra
The area is 9 square inches 28.26 square inches. 31. A
1
bh
1
34 in. 7 in. 119 sq in. 2 2 The area is 119 square inches. 1 1 b h 20 m 14 m 140 sq m 2 2 The area is 140 square meters.
32. A
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ISM: Prealgebra and Introductory Algebra
33. The unmarked horizontal side has length 13 m 3 m = 10 m. The unmarked vertical side has length 12 m 4 m = 8 m. The area is the sum of the areas of the two rectangles. A 12 m 10 m 8 m 3 m 120 sq m 24 sq m 144 sq m The area is 144 square meters. 34. The unmarked vertical side has length 30 cm 5 cm = 25 cm. The unmarked horizontal side has length 60 cm 35 cm = 25 cm. The area is the sum of the areas of the two rectangles. A 30 cm 25 cm 35 cm 25 cm 1625 sq cm The area is 1625 square centimeters. 35. A = l w = 36 ft 12 ft = 432 sq ft The area of the driveway is 432 square feet. 36. A = 10 ft 13 ft = 130 sq ft 130 square feet of carpet are needed. 3
37. V s 1 3 2 in. 2 3 5 in. 2 125 cu in. 8 5 15 cu in. 8 The volume is 15
5
cubic inches.
8 SA 6s2 2 1 6 2 in. 2 2 5 6 in. 2 25 6 4 sq in. 75 sq in. 2 1 37 sq in. 2 The surface area is 37
Chapter 8: Geometry and Measurement 38. V = l w h = 2 ft 7 ft 6 ft = 84 cu ft The volume is 84 cubic feet. SA 2lh 2wh 2lw 2 7 ft 6 ft 2 2 ft 6 ft 2 7 ft 2 ft 84 sq ft 24 sq ft 28 sq ft 136 sq ft The surface area is 136 square feet. 39. V r 2 h (20 cm)2 50 cm 20, 000 cu cm 62,800 cu cm The volume is 20,000 cubic centimeters 62,800 cubic centimeters. 40. V
4
r3
3 4
3
1 km 3 2 1 cu km 6 11 cu km 21 The volume is 1 11 cubic kilometers cubic kilometers. 6 21
1 2 41. V s h 3 1 (2 ft)2 2 ft 3 8 cu ft 3 2 2 cu ft 3 2 The volume of the pyramid is 2 cubic feet. 3 42. V r 2 h (3.5 in.)2 8 in. 98 cu in. 307.72 cu in. The volume of the can is about 307.72 cubic inches.
1
square inches.
2
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
43. Find the volume of each drawer. V l w1 h 1 2 ft 2 ft 1 ft 2 2 3 5 3 2 cu ft 2 2 3 5 cu ft 2 The three drawers have volume 5 15 1 3 7 cubic feet. 2 2 2
53. 12.18 mm 12.18 mm 1 m 1 1000 mm 12.18 m 1000 0.01218 m
44. r = d 2 = 1 ft 2 = 0.5 ft
55.
54. 2.31 m 2.31 m 1 km 1 1000 m 2.31 km 1000 0.00231 km
V r2 h (0.5 ft)2 2 ft 0.5 cu ft The volume of the canister is 0.5 cubic feet or 1 cubic feet. 2 108 in. 108 1 ft ft 9 ft 45. 108 in.
1 46. 72 ft
48.
2
12
72 ft 1 yd 72 yd 24 yd 1 3 ft 3
47. 1.5 mi
1
12 in.
1.5 mi 5280 ft 1.5 5280 ft 7920 ft 1 1 mi
1 yd 3 ft 12 in. 1 yd 2 312 in. 18 in.
1
1 yd
1 ft
2
49. 52 ft 51 ft 1 ft 51 ft 1 yd 1 ft 1 3 ft 51 yd 1 ft 3 17 yd 1 ft 50. 46 in. 36 in. 10 in. 36 in. 1 ft 10 in. 1 12 in. 36 ft 10 in. 12 3 ft 10 in. 51. 42 m
42 m 100 cm 42 100 cm 4200 cm 1 1m
52. 82 cm
314
82 cm 10 mm 82 10 mm 820 mm 1 1 cm
4 yd 2 ft 16 yd 2 ft 20 yd 4 ft = 20 yd + 1 yd 1 ft = 21 yd 1 ft
56. 7 ft 4 in. 2 (6 ft 1 ft 4 in.) 2 (6 ft 16 in.) 2 6 16 ft in. 2 2 3 ft 8 in. 57. 8 cm = 80 mm 80 mm 15 mm or 95 mm
15 mm = 1.5 cm 8.0 cm 1.5 cm 9.5 cm
58. 4 m = 400 cm 400 cm 126 cm or 274 cm
126 cm = 1.26 m 4.00 m 1.26 m 2.74 m
59.
333 yd 1 ft 163 yd 2 ft
332 yd 4 ft 163 yd 2 ft 169 yd 2 ft The amount of material that remains is 169 yd 2 ft.
60.
5 ft 2 in. 50 250 ft 100 in. = 250 ft + 96 in. + 4 in. = 250 ft + 8 ft 4 in. = 258 ft 4 in. The sashes require 258 ft 4 in. of material.
61. 1235 km 2 2470 km 2470 km 4
2470
km 617.5 km 4 Each must drive 617.5 km.
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ISM: Prealgebra and Introductory Algebra
62.
0.8 m 0.3 m 0.24 sq m
0.8 m 30 cm
Chapter 8: Geometry and Measurement
73.
4.3 mg 5 21.5 mg
The area is 0.24 sq m. 66 oz 1 lb 66 33 1 lb lb 4 lb 63. 66 oz 1 16 oz 16 8 8 2.3 tons 2000 lb 1 1 ton 2.3 2000 lb 4600 lb
64. 2.3 tons
74. 4.8 kg = 4800 g 4800 g 4200 g or 600 g 75.
65. 52 oz 48 oz 4 oz 48 oz 1 lb 4 oz 1 16 oz 48 lb 4 oz 16 3 lb 4 oz
1 lb 12 oz 2 lb 8 oz 3 lb 20 oz = 3 lb + 1 lb 4 oz = 4 lb 4 oz The total weight was 4 lb 4 oz.
76. 38 tons 300 lb 4
66. 10,300 lb 10, 000 lb 300 lb 10, 000 lb 1 ton 300 lb 1 2000 lb 10, 000 tons 300 lb 2000 5 tons 300 lb 27 g 0.027 g 67. 27 mg 27 mg 1 g 1 1000 mg 1000 68. 40 kg
40 kg 1000 g 40 1000 g 40, 000 g 1 1 kg
69. 2.1 hg
2.1 hg 10 dag 2.110 dag 21 dag 1 1 hg
71.
6 lb 5 oz 2 lb 12 oz
72.
8 lb 6 oz 4 32 lb 24 oz = 32 lb + 1 lb 8 oz = 33 lb 8 oz
5 lb 21 oz 2 lb 12 oz 3 lb 9 oz
38
77. 28 pt
tons
300
lb 4 4 1 9 tons 75 lb 2 1 9 tons ton 75 lb 2 9 tons 1000 lb 75 lb 9 tons 1075 lb They each receive 9 tons 1075 lb. 28 pt 1 qt 28 qt 14 qt 1 2 pt 2
40 c5c 78. 40 fl oz 40 fl oz 1 c 1 8 fl oz 8 3 qt 2 pt 1 pt 1 1 qt 3 2 pt 1 pt 6 pt 1 pt 7 pt
79. 3 qt 1 pt
80. 18 qt
0.03 mg 1 dg 1 100 mg 0.03 dg 100 0.0003 dg
70. 0.03 mg
4200 g = 4.2 kg 4.8 kg 4.2 kg 0.6 kg
18 qt 2 pt 2 c 18 2 2 c 72 c 1 1 qt 1 pt
81. 9 pt 8 pt 1 pt 8 pt 1 qt 1 pt 1 2 pt 8 qt 1 pt 2 4 qt 1 pt
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315
Chapter 8: Geometry and Measurement 82. 15 qt 12 qt 3 qt 12 qt 1 gal 3 qt 1 4 qt 12 gal 3 qt 4 3 gal 3 qt
93. 85 ml 8 16 = 10,880 ml 10,880 ml 1 L 10,880 L 10.88 L 1 1000 ml 1000 There are 10.88 L of polish in 8 boxes. 94. 6 L 1300 ml 2.6 L 6 L 1.3 L 2.6 L 9.9 L Since 9.9 L is less than 10 L, yes it will fit.
3.8 L 1000 ml 1 1L 3.8 1000 ml 3800 ml
83. 3.8 L
95. 7 m
14 kl 1.4 kl 84. 14 hl 14 hl 1 kl 1 10 hl 10 85. 30.6 L
ISM: Prealgebra and Introductory Algebra
30.6 L 100 cl 30.6 100 cl 3060 cl 1 1L
7 m 3.28 ft 22.96 ft 1 1m
96. 11.5 yd 11.5 yd 1 m 10.55 m 1 1.09 yd 97. 17.5 L
17.5 L 0.26 gal 4.55 gal 1 1L
86. 2.45 ml 2.45 ml 1 L 0.00245 L 1 1000 ml
98. 7.8 L
7.8 L 1.06 qt 8.268 qt 1 1L
1 qt 1 pt 3 qt 1 pt 4 qt 2 pt = 4 qt + 1 qt = 1 gal 1 qt
99. 15 oz
15 oz 28.35 g 425.25 g 1 1 oz
100. 23 lb
23 lb 0.45 kg 10.35 kg 1 1 lb
87.
88. 3 gal 2 qt 2 6 gal 4 qt = 6 gal + 1 gal = 7 gal 89. 0.946 L = 946 ml 946 ml 210 ml or 736 ml
210 ml = 0.21 L 0.946 L 0.210 L 0.736 L
90. 6.1 L = 6100 ml 6100 ml 9400 ml or 15, 500 ml
9400 ml = 9.4 L 6.1 L 9.4 L 15.5 L
101. 1.2 mm 50 = 60 mm 60 mm 1 cm 60 mm 6 cm 1 10 mm 6 cm 1 in. 6 cm 2.36 in. 1 2.54 cm The height of the stack is approximately 2.36 in. 102. 25 5.5 mm = 137.5 mm 137.5 mm 1 cm 13.75 cm 137.5 mm 1 10 mm 13.75 cm 13.75 cm 1 in. 5.4 in. 1 2.54 cm The total thickness is approximately 5.4 in.
91.
3 gal 6 qt 1 gal 3 qt 2 gal 3 qt There are 2 gal 3 qt of tea remaining. 4 gal 2 qt 1 gal 3 qt
92. 1 c 4 fl oz 2 (8 fl oz 4 fl oz) 2 12 fl oz 2 6 fl oz Use 6 fl oz of stock for half of a recipe.
103. F 1.8C 32 1.8(42) 32 75.6 32 107.6 42C is 107.6F. 104. F = 1.8C + 32 = 1.8(160) + 32 = 288 + 32 = 320 160C is 320F.
316
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ISM: Prealgebra and Introductory Algebra
105. C
5
(F 32)
5
(41.3 32)
9 9 41.3F is 5.2C.
5
Chapter 8: Geometry and Measurement
1 2 117. V r h 13 1 2 5 in. 12 in. 3 4 2 1 21 in. 12 cu in. 3441 4 cu in. 4 441 22 1 cu in. 346 cu in. 4 7 2 1 The volume is 346 cubic inches. 2
(9.3) 5.2
9
5 5 5 106. C 9 (F 32) 9 (80 32) 9 (48) 26.7 80F is 26.7C. 107. C
5
(F 32)
9 35F is 1.7C.
5
(35 32)
9
5
(3) 1.7
9
108. F 1.8C 32 1.8(165) 32 297 32 329 165C is 329F.
118. V = l w h = 5 in. 4 in. 7 in. = 140 cu in. The volume is 140 cubic inches. SA 2lh 2wh 2lw 2 7 in. 5 in. 2 4 in. 5 in. 2 7 in. 4 in. 70 sq in. 40 sq in. 56 sq in. 166 sq in. The surface area is 166 square inches.
109. The supplement of a 72 angle is an angle that measures 180 72 = 108. 110. The complement of a 1 angle is an angle that measures 90 1 = 89. 111. x and the angle marked 85 are adjacent angles, so mx 180 85 95. 112. Let y be the angle corresponding to x at the bottom intersection. Then y and the angle marked 123 are adjacent angles, so mx my 180 123 57. 113. P = 7 in. + 11.2 in. + 9.1 in. = 27.3 in. The perimeter is 27.3 inches. 114. The unmarked horizontal side has length 40 ft 22 ft 11 ft = 7 ft. P (22 15 7 15 11 42 40 42) ft 194 ft The perimeter is 194 feet. 115. The unmarked horizontal side has length 43 m 13 m = 30 m. The unmarked vertical side has length 42 m 14 m = 28 m. The area is the sum of the areas of the two rectangles. A 28 m 13 m 42 m 30 m 364 sq m 1260 sq m 1624 sq m The area is 1624 square meters. 2
119. 6.25 ft
120. 8200 lb 8000 lb 200 lb 8000 lb 1 ton 200 lb 1 2000 lb 4 tons 200 lb
121. 5 m
The area is 9 square meters 28.26 square meters.
5 m 100 cm 500 cm 1 1m 286 mm 1 km 1 1,000,000 mm 0.000286 km
122. 286 mm
1.4 g 123. 1400 mg 1400 mg 1 g 1 1000 mg 124. 6.75 gal
6.75 gal 4 qt 27 qt 1 1 gal
125. F 1.8C 32 1.8(86) 32 154.8 32 186.8 86C is 186.8F.
2
116. A r (3 m) 9 sq m 28.26 sq m
6.25 ft 12 in. 75 in. 1 1 ft
126. C
5
(F 32)
5
9 9 51.8F is 11C.
Copyright © 2025 Pearson Education, Inc.
(51.8 32)
5
(19.8) 11
9
317
Chapter 8: Geometry and Measurement
127. 9.3 km = 9300 m 9300 m 183 m or 9117 m
183 m = 0.183 km 9.300 km 0.183 km 9.117 km
128. 35 L = 35,000 ml 35, 000 ml 700 ml 35, 700 ml
700 ml = 0.7 L 35.0 L 0.7 L 35.7 L
129.
130.
10. When parallel lines are cut by a transversal, the measures of corresponding angles are equal and the measures of alternate interior angles are equal. Thus, a has the same measure as d (vertical angles), b (corresponding angles), and c (vertical angle with b). Thus, a and c have the same measure; B. 11. Area is measured in square units; B.
3 gal 3 qt 4 gal 2 qt 7 gal 5 qt = 7 gal + 1 gal 1 qt = 8 gal 1 qt
12. Volume is measured in cubic units; C.
3.2 kg 4 12.8 kg
14. The amount of material needed for a tablecloth is a calculation of area; B.
Chapter 8 Getting Ready for the Test 1. A line is a set of points extending indefinitely in two directions; D. 2. A line segment is a piece of a line with two endpoints; B. 3. A ray is a part of a line with one endpoint; E. 4. A right angle is an angle that measures 90; A. 5. An acute angle is an angle that measures between 0 and 90; F. 6. An obtuse angle is an angle that measures between 90 and 180; C. 7. When two lines intersect, adjacent angles are supplementary (have a sum of 180). a and d are adjacent angles, so choice B is correct. 8. When two lines intersect, vertical angles (angles that are opposite each other) have the same measure. b and d re vertical angles, so choice C is correct. 9. When parallel lines are cut by a transversal, the measures of corresponding angles are equal and the measures of alternate interior angles are equal. Thus, a has the same measure as d (vertical angles), b (corresponding angles), and c (vertical angle with b). a and f are adjacent angles, so they are supplementary. e and f are corresponding angles, so they have the same measure. Thus a and e are supplementary (have a sum of 180); A. 318
ISM: Prealgebra and Introductory Algebra
13. Perimeter and circumference are measured in units; A and D.
15. The amount of trim needed to go around the edge of a tablecloth is a calculation of perimeter or circumference; A or D. 16. The amount of soil needed to fill a hole in the ground is a calculation of volume; C. Chapter 8 Test 1. The complement of an angle that measures 78 is an angle that measures 90 78 = 12. 2. The supplement of a 124 angle is an angle that measures 180 124 = 56. 3. mx 90 40 50 4. x and the angle marked 62 are adjacent angles, so mx 180 62 118. y and the angle marked 62 are vertical angles, so my 62. x and z are vertical angles, so mz mx 118. 5. x and the angle marked 73 are vertical angles, so mx 73. x and y are alternate interior angles, so my mx 73. x and z are corresponding angles, so mz mx 73. 6. d = 2 r = 2 3.1 m = 6.2 m 7. r d 2 20 in. 2 10 in.
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ISM: Prealgebra and Introductory Algebra
8. Circumference: C 2 r 2 9 in. 18 in. 56.52 in. The circumference is 18 inches 56.52 inches. Area:
Chapter 8: Geometry and Measurement 15. P 2 l 2 w 2 18 ft 2 13 ft 36 ft 26 ft 62 ft cost P $1.87 per ft 62 ft $1.87 per ft $115.94 62 feet of baseboard are needed, at a total cost of $115.94.
A r2 (9 in.)2 81 sq in. 254.34 sq in. The area is 81 square inches 254.34 square inches. 9. P 2 l 2 w 2(7 yd) 2(5.3 yd) 14 yd 10.6 yd 24.6 yd The perimeter is 24.6 yards. A = l w = 7 yd 5.3 yd = 37.1 sq yd The area is 37.1 square yards. 10. The unmarked vertical side has length 11 in. 7 in. = 4 in. The unmarked horizontal side has length 23 in. 6 in. = 17 in. P = (6 + 4 + 17 + 7 + 23 + 11) in. = 68 in. The perimeter is 68 inches. Extending the unmarked vertical side downward divides the region into two rectangles. The region’s area is the sum of the areas of these: A 11 in. 6 in. 7 in.17 in. 66 sq in. 119 sq in. 185 sq in. The area is 185 square inches. 2
11. V r h (2 in.)2 5 in. 20 cu in. 22 6 20 cu in. 62 cu in. 7 7 6 The volume is 62 cubic inches. 7 12. V l w h 5 ft 3 ft 2 ft 30 cu ft The volume is 30 cubic feet.
23 280 24 40 36 4 280 inches = 23 ft 4 in.
16. 12
1 17. 2
2 1 gal 4 qt 1 2 2 4 qt 10 qt 1 1 gal 2
gal
2
30 15 7 lb lb 18. 30 oz 30 oz 1 lb 1 lb 1 16 oz 16 8 8 2.8 tons 2000 lb 1 1 ton 2.8 2000 lb 5600 lb
19. 2.8 tons
38 pt 1 qt 1 gal 20. 38 pt 1 2 pt 4 qt 38 gal 8 19 gal 4 3 4 gal 4 21. 40 mg
40 mg 1g 40 g 0.04 g 1 1000 mg 1000
22. 2.4 kg
2.4 kg 1000 g 2.4 1000 g 2400 g 1 1 kg
23. 13. P = 4 s = 4 4 in. = 16 in. The perimeter of the frame is 16 inches.
3.6 cm
3.6 cm 10 mm 3.6 10 mm 36 mm 1 1 cm
Copyright © 2025 Pearson Education, Inc.
319
14. V = l w h = 3 ft 3 ft 2 ft = 18 cu ft 18 cubic feet of soil are needed.
ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement 4.3 g 0.43 g 24. 4.3 dg 4.3 dg 1 g 1 10 dg 10 25. 0.83 L 26.
27.
8.4 2 m 5.6 m 3 1 3 The trees will be 5.6 m tall.
34. 8.4 m
0.83 L 1000 ml 0.831000 830 ml 1 1L
3 qt 1 pt 2 qt 1 pt 5 qt 2 pt 4 qt 1 qt 2 pt 1 gal 1 qt 1 qt 1 gal 2 qt
35.
37.
28. 2 ft 9 in. 3 6 ft 27 in. 6 ft 2 ft 3 in. 8 ft 3 in.
5 95
1.
3a 6 a 4 3a a 6 a a 4 2a 6 4 2a 6 6 4 6 2a 10 2a 10 2 2
or
a5
8.0 cm 1.4 cm 6.6 cm
2.
456 m = 0.456 km 1.800 km 0.456 km 2.256 km
2x 1 3x 5 2x 2x 1 3x 2x 5 1 x5 1 5 x 5 5 6x 4
3. a.
(F 32)
(52) 9 28.9 84F is 28.9C. 33. F 1.8C 32 1.8(12.6) 32 22.68 32 54.7 12.6C is 54.7F
2 2 2 2 24 16 2 4 625 5 5 5 5 5 5 2
(84 32) b.
9 5
320
2 ft 9 in. 6 12 ft 54 in. 12 ft 4 ft 6 in. 16 ft 6 in. Thus, 16 ft 6 in. of material is needed.
14 mm = 1.4 cm
31. 1.8 km = 1800 m 1800 m 456 m or 2256 m 32. C
19 gal 4 qt 15 gal 1 qt 4 gal 3 qt Thus, 4 gal 3 qt remains in the container. 20 gal 15 gal 1 qt
Cumulative Review Chapters 18
29. 5 gal 2 qt 2 4 gal 6 qt 2 4 6 gal qt 2 2 2 gal 3 qt
80 mm 14 mm 66 mm
36. 88 m + 340 cm = 88 m + 3.40 m = 91.4 m The span is 91.4 meters
8 lb 6 oz 7 lb 22 oz 4 lb 9 oz 4 lb 9 oz 3 lb 13 oz
30. 8 cm = 80 mm
2
1 1 1 1 4 4 4 16
3
4. a.
1 1 1 1 1 3 3 3 3 27 2
b.
3 3 3 9 7 7 49 7
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
5.
4
2
2
5 1
1
2 8
2
1
3 2 4 5 3
10 13
8 1
3 10
9
14.
0.4x 9.3 2.7 0.4x 9.3 9.3 2.7 9.3 0.4x 12 0.4x 12 0.4 0.4 x 30
3 10
5 15 6 4 15 3 11 9 15 2
15. Use a2 b2 c2 where a = b = 300. 3002 3002 c2
7. 11.1x 6.3 8.9x 4.6 11.1x 8.9x 6.3 4.6 20x 10.9 8. 2.5 y 3.7 1.3y 1.9 2.5 y 1.3y 3.7 1.9 1.2 y 1.8 2 9. 5.68 (0.9) 100 5.68 0.81100 0.2 0.2 5.68 0.0081 0.2 5.6881 0.2 28.4405
10.
0.5 y 2.3 1.65 0.5 y 2.3 2.3 1.65 2.3 0.5 y 0.65 0.5 y 0.65 0.5 0.5 y 1.3
5
10 6.
13.
5
5 1
8 10
Chapter 8: Geometry and Measurement
0.12 0.96 1.08 2.16 0.5 0.5
0.77... 11. 9 7.00 63 70 63 7
90, 000 90, 000 c2 180, 000 c2 180, 000 c 424 c The length of the sidewalk is approximately 424 feet. 16. Use a2 b2 c2 where a = 200 and b = 125. 2002 1252 c2 40, 000 15, 625 c2 55, 625 c2 55, 625 c 236 c The length of the diagonal is approximately 236 feet. width 17. a. b.
7 Thus 0.7 . 9
12. 5
0.4 2.0 2 0 0
18. a.
length
5 feet 7 feet
5 7
P 2l 2 w 2(7 feet) 2(5 feet) 14 feet 10 feet 24 feet 7 feet 7 length perimeter 24 feet 24 P = 4s = 4(9 in.) = 36 in. side 9 inches 9 1 perimeter 36 inches 36 4
2 Thus 0.43 . 5
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ISM: Prealgebra and Introductory Algebra
Chapter 8: Geometry and Measurement
b.
A s2 (9 in.)2 81 sq in. perimeter area
19.
36 inches
81 sq inches
36
25. 81
4
26.
3y 1 3 3(1) 1 0 3 3 1 0 3 2 3 False No, 1 is not a solution.
20.
21.
27.
4(x 7) 8 4(5 7) 0 8 4(2) 0 8 8 8 True Yes, 5 is a solution. x 2 3 x 3 3 2 3
28.
9x 8x 1.02 9x 8x 8x 8x 1.02 x 1.02
23. Let x be the dose for a 140-lb person. 4 cc x cc 25 lb 140 lb 4 x 25 140 4 140 25 x 560 25x 560 25x 25 25 22.4 x The dose is 22.4 cc. 24. Let x be the amount for 5 pie crusts. 3c xc 2 crusts 5 crusts 3 x 2 5 3 5 2 x 15 2x 15 2x 2 2 7.5 x 5 pie crusts require 7.5 cups of flour.
322
17% 100 17% of the people surveyed drive blue cars. 38
38% 100 38% of the shoppers used only cash. 1 13 6 % x 2 13 0.065x 13 0.065x 0.065 0.065 200 x 1 13 is 6 % of 200. 2 1 54 4 2 % x 54 0.045x 54 0.045x 0.045 0.045 1200 x 1 54 is 4 % of 1200. 2
x 6 22.
9
17
29. x = 30% 9 x = 0.3 9 x = 2.7 2.7 is 30% of 9. 30. x = 42% 30 x = 0.42 30 x = 12.6 12.6 is 42% of 30. 31. percent of increase
amount of increase
original amount 45 34 34 11 34 0.32 The scholarship applications increased by 32%.
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ISM: Prealgebra and Introductory Algebra
32. percent of increase
amount of increase
original amount 19 15 15 4 15 0.27 The price of the paint increased by 27%.
33. sales tax = 85.50 0.075 = 6.4125 The sales tax is $6.41. 85.50 + 6.41 = 91.91 The total price is $91.91. 34. sales tax = 375 0.08 = 30 The sales tax is $30. 375 + 30 = 405 The total price is $405. 35. The seven numbers are listed in order. The median is the middle number, 57.
Chapter 8: Geometry and Measurement
39. The complement of a 48 angle is an angle that has measure 90 48 = 42. 40. The supplement of a 137 angle is an angle that has measure 180 137 = 43. 41. 8 ft
8 ft 12 in. 8 12 in. 96 in. 1 1 ft
42. 7 yd
7 yd 3 ft 7 3 ft 21 ft 1 1 yd
43.
8 tons 1000 lb 3 tons 1350 lb
44.
8 lb 15 oz 9 lb 3 oz
7 tons 3000 lb 3 tons 1350 lb 4 tons 1650 lb
17 lb 18 oz 17 lb 1 lb 2 oz 18 lb 2 oz
36. The five numbers in order are: 60, 72, 83, 89, 95. The median is the middle number, 83.
45. C
5
9 59F is 15C.
9
37. There is 1 red marble and 1 + 1 + 2 = 4 total 1 marbles. The probability is . 4
46. C
5
5
(F 32)
(F 32)
9 86F is 30C.
5
9
(59 32)
5
(27) 15
9
(86 32)
5
(54) 30
9
38. There are 2 nickels and 2 + 2 + 3 = 7 total coins. 2 The probability is . 7
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323
Chapter 9 Section 9.1 Practice Exercises
d.
1. Since 8 is to the right of 6 on the number line, the statement 8 < 6 is false. 2. Since 100 is to the right of 10 on the number line, the statement 100 > 10 is true. 3.
Since 21 = 21, the statement 21 21 is true.
4.
Since 21 = 21, the statement 21 21 is true. 5. Since neither 0 > 5 nor 0 = 5 is true, the statement 0 5 is false. 6. Since 25 > 22, the statement 25 22 is true. 7. a.
Fourteen is greater than or equal to fourteen is written as 14 14.
b.
Zero is less than five is written as 0 < 5.
c.
Nine is not equal to 10 is written as 9 10.
8. The integer 10 represents 10 meters below sea level. 9.
5 4
1
4
2
2 1 5 3 5 4 0
1
2.25 2
3
4
5
11 < 9 since 11 is to the left of 9 on the number line.
b.
By comparing digits in the same places, we find that 4.511 > 4.151, since 0.5 > 0.1.
c.
By dividing, we find that
7
0.875 and
8
2
0.66.... Since 0.875 > 0.66..., then 3 7 2 . 8 3
324
The irrational number is .
f.
All numbers in the given set are real numbers.
Vocabulary, Readiness & Video Check 9.1 1. The whole numbers are {0, 1, 2, 3, 4, ...}. 2. The natural numbers are {1, 2, 3, 4, 5, ...}. 3. The symbols , , and > are called inequality symbols. 4. The integers are {..., 3, 2, 1, 0, 1, 2, 3, ...}. 5. The real numbers are {all numbers that correspond to points on the number line}. 6. The rational numbers are a a and b are integers, b 0. b 7. The integer 0 is neither positive nor negative.
1
–5 –4 –3 –2 –1
11. a.
e.
1 2
10. a.
2 The rational numbers are 100, , 0, 6, 5 and 913.
8. The point on a number line halfway between 0 1 1 and can be represented by . 2 4 9. to form a true statement: 0 < 7 10. Five is greater than or equal to 4: 5 4 11. 0 belongs to the whole numbers, the integers, the rational numbers, and the real numbers; because 0 is a rational number, it cannot also be an irrational number. Exercise Set 9.1 2. Since 8 is to the right of 5 on the number line, 8 > 5.
The natural numbers are 6 and 913.
b.
The whole numbers are 0, 6, and 913.
4. Since 9 is to the left of 15 on the number line, 9 < 15.
c.
The integers are 100, 0, 6, and 913.
6. 1.13 = 1.13
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
8. Since 20 is to the right of 0 on the number line, 20 > 0.
42. –5 –4 –3 –2 –1
10. Since 0 is to the left of 100 on the number line, 0 < 100.
1 3
7 8 0
1
7 9 2 4
2
3
4
1.75
5
4.5
12. Since 360 is to the right of 180 on the number line, 360 180.
44.
14. Since 8 is to the left of 9 on the number line, 8 9 is false.
46.
16. Since 16 is to the right of 17 on the number line, 16 > 17 is true.
48.
18. 1.02 can be written as 1.020. Then comparing digits with the same place value, we have 0.000 < 0.001. Thus the statement 1.02 > 1.021 is false.
50. 7941 is a natural number, a whole number, an integer, a rational number, and a real number.
20. Rewrite the fractions with a common denominator and compare numerators. 4 44 9 45 ; 5 55 11 55 4 9 Since 44 < 45, then is true. 5 11
–5 –4 –3 –2 –1
1
0
1
2
3
4
5
is a rational number and a real number.
4
52.
1
is a rational number and a real number.
7
3 is an irrational number and a real number.
54. True; every natural number is positive. 56. False;
1
is not an integer.
2 58. True; every rational number is also a real number.
22. 13 13 has the same meaning as 13 13. 24. 5 > 3 has the same meaning as 3 < 5.
60. True; every whole number is an integer.
28. Twenty is greater than two is written as 20 > 2.
62. |12| = 12 |0| = 0 Since 12 is to the right of 0 on the number line, |12| > |0|.
30. Negative ten is less than or equal to thirty-seven is written as 10 37.
64.
26. 4 < 2 has the same meaning as 2 > 4.
32. Negative seven is not equal to seven is written as 7 7. 34. The integer 535 represents 535 feet above sea level. The integer 8 represents 8 feet below sea level. 36. The integer 136,971 represents 136,971 fewer students. 38. The integer 17 represents an ascent of 17 feet. The integer 15 represents a descent of 15 feet. 40. –5
–4
–3
–2
–1
0
1
2
3
4
5
2 2 5 5 2 2 5 5 2 2 2 Since , then 2 . 5 5 5 5
66. |5.01| = 5.01 |5| = 5 Since 5.01 is to the right of 5 on the number line, |5.01| > |5|. 68. |12| = 12 24 12 2 Since 12 is to the right of 12 on the number 24 . line, 12 2
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325
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
70. The corn production in Minnesota was 1450 million bushels, while the corn production in South Dakota was 650 million bushels. 1450 million > 650 million or 1,450,000,000 > 650,000,000 72. The corn production in Indiana was 950 million bushels, while the corn production in Minnesota was 1450 million bushels. 1450 950 = 500 The corn production in Indiana was 500 million bushels less, or 500,000,000. 74. Since 0.96 is to the left of 0.98 on the number line, 0.96 < 0.98. 76. Antares: 0.96 Spica: 0.98 Since 0.96 < 0.98, Spica is dimmer than Antares. 78. Since the dimmest star corresponds to the largest apparent magnitude, which is 1.35, the dimmest star is Regulus. 80. answers may vary
(2 + 7) + 3 = 2 + (7 + 3)
c.
(q + r) + 17 = q + (r + 17)
d.
(ab) 21 = a (b 21)
1 2
1
13. 9 3 + 9 y = 9(3 + y) 14. 6a + 6b = 6(a + b) 15. 7(a + b) = 7 a + 7 b illustrates the distributive property.
21. (7 y) 10 = y (7 10) illustrates the commutative and associative properties of multiplication. Vocabulary, Readiness & Video Check 9.2 1. x + 5 = 5 + x is a true statement by the commutative property of addition. 2. x 5 = 5 x is a true statement by the commutative property of multiplication. 3. 3(y + 6) = 3 y + 3 6 is true by the distributive property.
7. 5(x + y) = 5(x) + 5(y) = 5x + 5y 326
1 (2x) (4) 9 2 2 1x 2 9 x 11
(2x 4) 9
20. (x + 0) + 23 = x + 23 illustrates the identity element for addition.
4. Since the grouping of the numbers was changed and their order was not, the statement is true by the associative property of addition.
6. 4(5x) = (4 5) x = 20x
12.
1 19. 3 1 illustrates the multiplicative inverse 3 property.
3. Since the order of two numbers was changed but their grouping was not, the statement is true by the commutative property of multiplication.
5. (3 x) 17 3 (x 17) 3 (17 x) (3 17) x 14 x
11. (8 a b) 1(8 a b) (1)(8) (1)(a) (1)(b) 8 a b
18. 6 + (z + 2) = 6 + (2 + z) illustrates the commutative property of addition.
5 (3 6) = (5 3) 6
b.
10. 1(3 a) = (1)(3) (1)(a) = 3 + a
17. 4 (6 x) = (4 6) x illustrates the associative property of multiplication.
7y=y7
b. 4 + x = x + 4 2. a.
9. 4(x 6 y 2z) 4(x) 4(6 y) 4(2z) 4x 24 y 8z
16. 12 + y = y + 12 illustrates the commutative property of addition.
Section 9.2 Practice Exercises 1. a.
8. 3(2 + 7x) = 3(2) + (3)(7x) = 6 21x
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ISM: Prealgebra and Introductory Algebra 4. 2 (x y) = (2 x) y is a true statement by the associative property of multiplication. 5. x + (7 + y) = (x + 7) + y is a true statement by the associative property of addition. 6. The numbers
2 3 and are called reciprocals 3 2
or multiplicative inverses. 2 2 7. The numbers and are called opposites or 3 3 additive inverses. 8. order; grouping 9. 2 is outside the parentheses, so the point is made that you should distribute the 9 only to the terms within the parentheses and not also to the 2.
Chapter 9: Equations, Inequalities, and Problem Solving 20. 2(42x) = (2 42)x = 84x 1 1 22. 8 (8z) 8 8 z 1z z 24. 7 (x 4) 7 (4 x) (7 4) x 11 x 26. 3(12y) = (3 12)y = 36 27 2 7 r r 1r r 7 2 7 2 1 7 1 30. (7x) x 3 3 7 x 3
28.
32. 7(a + b) = 7(a) + 7(b) = 7a + 7b 34. 11(y 4) = 11(y) 11(4) = 11y 44 36. 5(7 + 8y) = 5(7) + 5(8y) = 35 + 40y
10. The identity element for addition is 0 because if we add 0 to any real number, the result is that real number. The identity element for multiplication is 1 because any real number times 1 gives a result of that original real number. Exercise Set 9.2 2. 8 + y = y + 8 by the commutative property of addition. 4. 2 x = x (2) by the commutative property of multiplication. 6. ab = ba by the commutative property of multiplication. 8. 19 + 3y = 3y + 19 by the commutative property of addition. 10. 3 (x y) = (3x) y by the associative property of multiplication. 12. (y + 4) + z = y + (4 + z) by the associative property of addition. 14. (3y) z = 3 (yz) by the associative property of multiplication.
38. 3(8x 1) = 3(8x) 3(1) = 24x 3 40. 2(x + 5) = 2(x) + 2(5) = 2x + 10 42. 3(z y) = 3(z) 3(y) = 3z + 3y 1 1 1 11 44. (2r 11) (2r) (11) r 2 2 2 2 46. 8(3y z 6) 8(3y) 8(z) 8(6) 24 y 8z 48 48. 4(4 2 p 5n) 16 4(4) 4(2 p) 4(5n) 16 16 8 p 20n 16 (16 16) 8 p 20n 0 8 p 20n 8 p 20n 50. (9r 5) 1(9r 5) 1(9r) (1)(5) 9r 5 52. (q 2 6r) 1(q 2 6r) 1(q) 1(2) 1(6r) q 2 6r
16. 6 + (r + s) = (6 + r) + s by the associative property of addition. 18. (r + 3) + 11 = r + (3 + 11) = r + 14
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
54.
1 7 1 1 7 (4x 2) (4x) (2) 4 2 4 4 2 1 7 1 4x 4 2 2 8 1x 2 x4
1 1 1 56. (10a 25b) (10a) (25b) 5 5 5 2a 5b
3 4 1 illustrates the multiplicative 4 3 inverse property.
86.
88. a b2 2 (5)2 2 25 23 90. x3 x2 4 (3)3 (3)2 4 27 9 4 36 4 32
58. 10(4s 6) 40 10(4s) 10(6) 40 40s 60 40 40s 20
92. gh h2 0(4) (4)2 0(4) 16 0 16 16
60. 11(5x 3) 10 11(5x) 11(3) 10 55x 33 10 55x 23
94. The opposite of
62. 0.6(2x 1) 0.1 0.6(2x) 0.6(1) 0.1 1.2x 0.6 0.1 1.2x 0.7 64. 14 z + 14 5 = 14 (z + 5) 66. 9a + 9b = 9 a + 9 b = 9(a + b) 68. (3)a + (3)y = (3) a + (3) y = 3(a + y) 70. 25x + 25y = 25 x + 25 y = 25(x + y) 72. 4(3 + 8) = 4 3 + 4 8 illustrates the distributive property. 74. 9 (x 7) = (9 x) 7 illustrates the associative property of multiplication. 76. 1 9 = 9 illustrates the identity element of multiplication. 78. 4 (8 3) = (8 3) (4) illustrates the commutative property of multiplication. 80. (a + 9) + 6 = a + (9 + 6) illustrates the associative property of addition. 82. (11 + r) + 8 = (r + 11) + 8 illustrates the commutative property of addition. 84. r + 0 = r illustrates the identity property of addition.
328
2
is
3 The reciprocal of
2
.
3 2
3 is . 3 2
96. The opposite of 4y is 4y. 1 The reciprocal of 4y is . 4y 98. The expression is the opposite of 7x or 7x. 1 The reciprocal of 7x is . 7x 100. False; the reciprocal of The opposite of
a
2 is . 2 a
a
a is . 2 2
102. “Putting on your shoes” and “putting on your socks” are not commutative, since the order in which they are performed affects the outcome. 104. “Reading the sports section” and “reading the comics section” are commutative, since the order in which they are performed does not affect the outcome. 106. “Baking a cake” and “eating the cake” are not commutative, since the order in which they are performed affects the outcome. 108. “Dialing a number” and “turning on the cell phone” are not commutative, since the order in which they are performed affects the outcome.
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Chapter 9: Equations, Inequalities, and Problem Solving 9(5 x) 3x 15 15 9 5 0 3 2 2 45 10 15 9 0 2 2 2 45 5 9 0 2 2 45 45 True 2 2 15 The solution is . 2
110. a. The property illustrated is the associative property of addition since the grouping of addition changed. b.
The property illustrated is the commutative property of addition since the order in which they are added changed.
c.
The property illustrated is the commutative property of addition since the order in which they are added changed.
Check:
112. answers may vary 114. answers may vary 3. Section 9.3 Practice Exercises 1.
2.
5(3x 1) 2 12x 6 15x 5 2 12x 6 15x 3 12x 6 15x 3 12x 12x 6 12x 3x 3 6 3x 3 3 6 3 3x 9 3x 9 3 3 x3 Check: 5(3x 1) 2 12x 6 5[3(3) 1] 2 012(3) 6 5(9 1) 2 0 36 6 5(8) 2 0 42 40 2 0 42 42 42 True The solution is 3. 9(5 x) 3x 45 9x 3x 45 9x 9x 3x 9x 45 6x 45 6x 6 6 15 x 2
3
x 1
2
x2 4 3 3 2 12 x 1 12 x 2 4 3 9x 12 8x 24 9x 12 8x 8x 24 8x x 12 24 x 12 12 24 12 x 12 3 2 x 1 x 2 Check: 4 3 3 2 (12) 1 0 (12) 2 4 3 9 1 0 8 2 10 10 True The solution is 12. 3(x 2)
4. 5
3x 6
5 3(x 2)
5(3x 6) 5 3(x 2) 5(3x 6) 3x 6 15x 30 3x 6 3x 15x 30 3x 6 12x 30 6 30 12x 30 30 36 12x 36 12x 12 12 3 x
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Check:
2. 3x 7 = 3x 1 3 1 7 = Display: 4 3 1 1 = Display: 4 Since the left side equals the right side, x = 1 is a solution.
3(x 2) 3x 6 5 3(3 2) 0 3(3) 6 5 3(5) 0 96 5 15 0 3 5 3 3
3. 5x 2.6 = 2(x + 0.8) 5 4.4 2.6 = Display: 19.4 2 ( 4.4 + 0.8 ) = Display: 10.4 Since the left side does not equal the right side, x = 4.4 is not a solution.
The solution is 3. 5.
0.06x 0.10(x 2) 0.02(8) 100[0.06x 0.10(x 2)] 100[0.02(8)] 6x 10(x 2) 2(8) 6x 10x 20 16 4x 20 16 4x 20 20 16 20 4x 36 4x 36 4 4 x9 To check, replace x with 9 in the original equation. The solution is 9.
6. 5(2 x) 8x 3(x 6) 10 5x 8x 3x 18 10 3x 3x 18 10 3x 3x 3x 18 3x 10 18 Since the statement 10 = 18 is false, the equation has no solution. 7. 6(2x 1) 14 10(x 2) 2x 12x 6 14 10x 20 2x 12x 20 12x 20 12x 12x 20 12x 12x 20 20 20 Since 20 = 20 is a true statement, all real numbers are solutions.
4. 1.6x 3.9 = 6.9x 25.6 1.6 5 3.9 = Display: 11.9 6.9 5 25.6 = Display: 60.1 Since the left side does not equal the right side, x = 5 is not a solution. 564x
5.
6. 20(x 39) = 5x 432 20 ( 23.2 39 ) = Display: 316 5 23.2 432 = Display: 316 Since the left side equals the right side, x = 23.2 is a solution. Vocabulary, Readiness & Video Check 9.3 1. x = 7 is an equation. 2. x 7 is an expression. 3. 4y 6 + 9y + 1 is an expression. 4. 4y 6 = 9y + 1 is an equation. 5.
Calculator Explorations
200x 11(649) 4 Display: 17061 ( 564 121 ) 4 = 200 121 11 649 = Display: 17061 Since the left side equals the right side, x = 121 is a solution.
1
x 1. 2x = 48 + 6x 2 12 = Display: 24 48 + 6 12 = Display: 24 Since the left side equals the right side, x = 12 is a solution.
6.
1 x
x 1
is an expression.
8
x 1
6 is an equation.
8
7. 0.1x + 9 = 0.2x is an equation. 8. 0.1x2 9 y 0.2x2 is an expression. 9. 3; distributive property, addition property of equality, multiplication property of equality
330
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10. Because both sides have more than one term, we need to apply the distributive property to make sure we multiply every single term in the equation by the LCD. 11. The number of decimal places in each number helps us determine the smallest power of 10 we can multiply through by so we are no longer dealing with decimals. 12. a. b.
If we have a true statement, then the equation has all real numbers as solutions. If we have a false statement, then the equation has no solution.
Exercise Set 9.3 2.
3x 1 2(4x 2) 3x 1 8x 4 3x 1 8x 8x 4 8x 5x 1 4 5x 11 4 1 5x 5 5x 5 5 5 x 1
4.
6.
15x 5 7 12x 15x 5 12x 7 12x 12x 3x 5 7 3x 5 5 7 5 3x 12 3x 12 3 3 x4 (5x 10) 5x 5x 10 5x 5x 5x 10 5x 5x 10 10x 10 10x 10 10 1 x
Chapter 9: Equations, Inequalities, and Problem Solving 8. 3(2 5x) 4(6x) 12 6 15x 24x 12 6 9x 12 6 9x 6 12 6 9x 6 9x 6 9 9 2 x 3 10. 4(n 4) 23 7 4n 16 23 7 4n 7 7 4n 7 7 7 7 4n 0 4n 0 4 4 n0 12.
5 6(2 b) b 14 5 12 6b b 14 7 6b b 14 7 6b 6b b 14 6b 7 7b 14 7 14 7b 14 14 7 7b 7 7b 7 7 1 b
14.
6 y 8 6 3y 13 6 y 8 7 3y 6 y 8 3y 7 3y 3y 3y 8 7 3y 8 8 7 8 3y 15 3y 15 3 3 y 5
16.
7n 5 8n 10 7n 5 7n 8n 10 7n 5 15n 10 5 10 15n 10 10 15 15n 15 15n 15 15 1 n
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Chapter 9: Equations, Inequalities, and Problem Solving
4
18.
x
8
16
5 5 5 8 4 16 5 x 5 5 5 5 4x 8 16 4x 8 8 16 8 4x 8 4x 8 4 4 x 2 20.
2
x
22. 0.40x 0.06(30) 9.8 40x 6(30) 980 40x 180 980 40x 180 180 980 180 40x 800 40x 800 40 40 x 20 3( y 3) 3( y 53)
2y 6
5(2 y 6) 5 3( y 3) 5(2 y 6) 3y 9 10 y 30 3y 9 30 10 y 30 30 3y 21 10 y 3y 21 3y 10 y 3y 21 7 y 21 7 y 7 7 3 y 5
1 x 1 x 52 4 1 4 x 1 2 4 x 4 10x 4 4x 1 10x 4 4x 4x 1 4x 6x 4 1 6x 4 4 1 4 6x 5 6x 5 6 6 5 x 6
28. 0.60(z 300) 0.05z 0.70z 205 60(z 300) 5z 70z 20, 500 60z 18, 000 5z 70z 20, 500 65z 18, 000 70z 20, 500 65z 18, 000 65z 70z 20, 500 65z 18, 000 5z 20, 500 18, 000 20, 500 5z 20, 500 20, 500 2500 5z 2500 5z 5 5 500 z 30. 14x 7 7(2x 1) 14x 7 14x 7 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. x
32.
332
5
1
1 9 3 1 2 9 x 9(1) 9 3 2x 3 9 2x 3 3 9 3 2x 12 2x 12 2 2 x6
24.
26.
x 2 x 3 x 3x x 2 3 3 3 3 2 0 Since the statement 2 = 0 is false, the equation has no solution.
34.
2(x 5) 2x 10 2x 10 2x 10 2x 10 2x 2x 10 2x 10 10 Since the statement 10 = 10 is false, the equation has no solution.
36.
5(4 y 3) 2 20 y 17 20 y 15 2 20 y 17 20 y 17 20 y 17 Since both sides of the equation are identical, the equation is an identity and every real number is a solution.
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38.
4(5 w) w 3 4(5 w) 3 3(w) 3 4(5 w) 3w 20 4w 3w 20 4w 4w 3w 4w 20 w
40. (4a 7) 5a 10 a 4a 7 5a 10 a 9a 7 10 a 9a 7 9a 10 a 9a 7 10 10a 7 10 10 10a 10 3 10a 3 10a 10 10 3 a 10 42.
9x 3(x 4) 10(x 5) 7 9x 3x 12 10x 50 7 12x 12 10x 43 12x 12 10x 10x 43 10x 2x 12 43 2x 12 12 43 12 2x 31 2x 31 2 231 x 2 5(x 1)
44.
46.
0.9x 4.1 0.4 9x 41 4 9x 41 41 4 41 9x 45 9x 45 9 9 x5
48. 3(2x 1) 5 6x 2 6x 3 5 6x 2 6x 2 6x 2 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. 50.
4(4 y 2) 2(1 6 y) 8 16 y 8 2 12 y 8 16 y 8 10 12 y 16 y 8 16 y 10 12 y 16 y 8 10 4 y 8 10 10 4 y 10 2 4 y 2 4 y 4 4 1 y 2 7
52.
3(x 1)
2 3(x 1) 4 4 2 5(x 1) 6(x 1) 5x 5 6x 6 5x 5 5x 6x 6 5x 5 x 6 5 6 x 6 6 11 x
4
4 5(x 1)
Chapter 9: Equations, Inequalities, and Problem Solving
54.
x
1
3
x 8 4 4 1 7 3 8 8 x 4 8 4 x 7x 2 6x 7x 2 7x 6x 7x 2 x 2 x 1 1 2 x x x 7 5 5 3 x x 15 7 15 5 5 3 3x 105 5x 75 3x 105 3x 5x 75 3x 105 75 2x 75 105 75 75 2x 30 2x 30 2x 2 2 15 x
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Chapter 9: Equations, Inequalities, and Problem Solving 56. 4(2 x) 1 7x 3(x 2) 8 4x 1 7x 3x 6 9 4x 4x 6 9 4x 4x 4x 6 4x 96 Since the statement 9 = 6 is false, the equation has no solution. 58. 0.01(5x 4) 0.04 0.01(x 4) 1(5x 4) 4 1(x 4) 5x 4 4 x 4 5x 4 x 5x 4 5x x 5x 4 4x 4 4x 4 4 1 x 60.
3
c. 72.
76.
78. answers may vary 80. a.
b.
64. Three times a number is 3x. 66. The difference of 8 and twice a number is 8 2x. 68. The quotient of 12 and the difference of a 12 number and 3 is . x3 x3 x5 x 3 x x 5 x 35 Since the statement 3 = 5 is false, the equation has no solution.
b. answers may vary
334
x 15 x 15 x 15 15 x 15 15 x x x x x x 2x 0 2x 0 2 2 x0 The choice is c.
62. The total length is the sum of the two lengths. x (7x 9) x 7x 9 8x 9 The total length is (8x 9) feet.
70. a.
3x 1 3x 2 3x 1 3x 3x 2 3x 1 2 Since the statement 1 = 2 is false, the equation has no solution. The choice is b.
74. x 11x 3 10x 1 2 10x 3 10x 3 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. The choice is a.
1
x 5x 8 2 1 2 3 x 2(5x 8) 2 6 x 10x 16 6 x x 10x 16 x 6 11x 16 6 16 11x 16 16 22 11x 22 11x 11 11 2 x
answers may vary
The perimeter is the sum of the lengths of the sides. x + (2x + 1) + (3x 2) = 35 x 2x 1 3x 2 35 6x 1 35 6x 11 35 1 6x 36 6x 36 6 6 x6 The lengths of the sides are: x = 6 meters 2x + 1 = 2(6) + 1 = 12 + 1 = 13 meters 3x 2 = 3(6) 2 = 18 2 = 16 meters
82. answers may vary 84.
1000(x 40) 100(16 7x) 1000x 40, 000 1600 700x 1000x 40, 000 700x 1600 700x 700x 40, 000 300x 1600 40, 000 300x 40, 000 1600 40, 000 300x 38, 400 300x 38, 400 300 300 x 128
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86.
1. 0 is a whole number, an integer, a rational number, and a real number. 2. 143 is a natural number, a whole number, an integer, a rational number, and a real number. 3.
is a rational number and a real number.
4. 1 is a natural number, a whole number, an integer, a rational number, and a real number. 5. 13 is an integer, a rational number, and a real number. 6.
9
is a rational number and a real number.
10
3 8
0.127x 2.685 0.027x 2.38 127x 2685 27x 2380 127x 2685 27x 27x 2380 27x 100x 2685 2380 100x 2685 2685 2380 2685 100x 305 100x 305 100 100 x 3.05
13.
2x 7 6x 27 2x 7 7 6x 27 7 2x 6x 20 2x 6x 6x 20 6x 4x 20 4x 20 4 4 x5
14.
3 8 y 3y 2 3 8 y 3y 3y 2 3y 3 5 y 2 3 3 5 y 3 2 5 y 5 5 y 5 5 5 y 1
Mid-Chapter Review
Chapter 9: Equations, Inequalities, and Problem Solving
1
7. is a rational number and a real number. 9 8.
5 is an irrational number and a real number.
15. 3a 6 5a 7a 8a 6 2a a 6 2a 2a a 2a 6 3a 6 3a 3 3 2 a 4b 8 b 10b 3b 3b 8 7b 3b 3b 8 3b 7b 8 4b 8 4b 4 4 2 b
16.
9. 7(d 3) 10 7 d 7 3 10 7d 2110 7d 11
17.
10. 9(z 7) 15 9 z 9 7 15 9z 63 15 9z 48 11. 4(3y 4) 12 y 4 3y 4(4) 12 y 12 y 16 12 y 12 y 12 y 16 16 12. 3(2x 5) 6x 3 2x 3 5 6x 6x 15 6x 6x 6x 15 12x 15
18.
2 5 x 3 9 3 2 3 5 x 2 3 2 9 15 x 18 5 x 6 3 1 y 8 16 8 3 8 1 y 3 8 3 16 1 y 6
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
19.
20.
10 6n 16 10 16 6n 16 16 6 6n 6 6n 6 6 1 n
22. 4(3t 4) 20 3 5t 12t 16 20 3 5t 12t 4 3 5t 12t 4 5t 3 5t 5t 7t 4 3 7t 4 4 3 4 7t 7 7t 7 7 7 t 1 23.
336
4
2(z 3)
5 z 3 2(z 3) 3 3(5 z) 3 2(z 3) 3(5 z) 2z 6 15 3z 2z 6 3z 15 3z 3z 6 5z 15 6 5z 6 15 6 5z 9 5z 9 5 5 9 z 5
2w 3
3(w 4 2)
4(2w 3)
4 3(w 2) 4(2w 3) 3w 6 8w 12 3w 6 6 8w 12 6 3w 8w 6 3w 8w 8w 6 8w 5w 6 5w 6 5 5 6 w 5
5 2m 7 5 7 2m 7 7 12 2m 12 2m 2 2 6m
21. 3(5c 1) 2 13c 3 15c 3 2 13c 3 15c 5 13c 3 15c 5 5 13c 3 5 15c 13c 8 15c 13c 13c 8 13c 2c 8 2c 8 2 2 c4
3(w 2)
24.
25. 2(2x 5) 3x 7 x 3 4x 10 4x 10 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. 26. 4(5x 2) 12x 4 8x 4 20x 8 20x 8 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. 27. 0.02(6t 3) 0.04(t 2) 0.02 2(6t 3) 4(t 2) 2 12t 6 4t 8 2 12t 6 4t 6 12t 6 4t 4t 6 4t 8t 6 6 8t 6 6 6 6 8t 0 8t 0 8 8 t0 28.
0.03(m 7) 0.02(5 m) 0.11 3(m 7) 2(5 m) 11 3m 21 10 2m 11 3m 21 21 2m 3m 21 2m 21 2m 2m 5m 21 21 5m 21 21 21 21 5m 0 5m 0 5 5 m0
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29.
30.
4( y 1) 5 4( y 1) 5(3y) 5 5 15 y 4( y 1) 15 y 4 y 4 15 y 4 y 4 y 4 4 y 19 y 4 19 y 4 19 19 4 y 19
31.
32.
Section 9.4 Practice Exercises
3y
1.
2.
5(1 x) 6 5(1 x) 6(4x) 6 6 24x 5(1 x) 24x 5 5x 24x 5x 5 5x 5x 19x 5 4x
19x 5 19 19 5 x 19 1
3
(3x 7) x 5 10 1 10 3 1010 (3x 7) 10 x 5 10 3x 7 3x 50 3x 3x 7 3x 3x 50 7 50 Since the statement 7 = 50 is false, the equation has no solution. 1
(2x 5)
2
Chapter 9: Equations, Inequalities, and Problem Solving
x 1 7 7 1 2 7 (2x 5) 7 x 1 7 7 2x 5 2x 7 2x 2x 5 2x 2x 7 5 7 Since the statement 5 = 7 is false, the equation has no solution.
Let x represent the number. 3x 6 2x 3 3x 6 2x 2x 3 2x x6 3 x 66 36 x9 The number is 9. Let x represent the number. 3(x 5) 2x 3 3x 15 2x 3 3x 15 2x 2x 3 2x x 15 3 x 15 15 3 15 x 12 The number is 12. 3. Let x represent the length of the shorter piece. Then 5x represents the length of the longer piece. Their sum is 18 feet. x 5x 18 6x 18 6x 18 6 6 x3 The shorter piece is 3 feet and the longer piece is 5(3) = 15 feet. 4. Then Let x represent the number of votesof forvotes Texas. x + 14 represents the number for California. Their sum is 94. x x 14 94 2x 14 94 2x 14 14 94 14 2x 80 2x 80 2 2 x 40 Texas will have 40 electoral votes and California will have 40 + 14 = 54 electoral votes. 5. Let x represent the number of miles driven. The cost for x miles is 0.15x. The daily cost is $28. 0.15x 28 52 0.15x 28 28 52 28 0.15x 24 0.15x 24 0.15 0.15 x 160 You drove 160 miles.
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
6. Let x represent the measure of the smallest angle. Then 2x represents the measure of the second angle and 3x represents the measure of the third angle. The sum of the measures of the angles of a triangle equals 180. x 2x 3x 180 6x 180 6x 180 6 6 x 30 If x = 30, then 2x = 2(30) = 60 and 3x = 3(30) = 90. The smallest is 30, second is 60, and third is 90. 7. If x is the first even integer, then x + 2 and x + 4 are the next two even integers. x x 2 x 4 144 3x 6 144 3x 6 6 144 6 3x 138 3x 138 3 3 x 46 If x = 46, then x + 2 = 48 and x + 4 = 50. The integers are 46, 48, 50. Vocabulary, Readiness & Video Check 9.4 1. If x is the number, then “double the number” is 2x, and “double the number, decreased by 31” is 2x 31. 2. If x is the number, then “three times the number” is 3x, and “three times the number, increased by 17” is 3x + 17. 3. If x is the number, then “the sum of the number and 5” is x + 5, and “twice the sum of the number and 5” is 2(x + 5).
6. If y is the number, then “the sum of 10 and the number” is 10 + y, and “the sum of 10 and the (10 y) number, divided by 9” is or 9 (10 + y) 9. 7. in the statement of the application 8. The original application asks for the measures of two supplementary angles. The solution of x = 43 gives us the measure of only one of the angles. 9. that the three angle measures are consecutive even integers and that they sum to 180 Exercise Set 9.4 2.
3x 1 2x 3x 1 3x 2x 3x 1 x 1 x 1 1 1 x The number is 1.
4.
4x 2 5x 2 4x 2 4x 5x 2 4x 2 x 2 2 2 x 2 2 0x The number is 0.
6. 5[x (1)] 6x 5(x 1) 6x 5x 5 6x 5x 5 5x 6x 5x 5 x The number is 5.
4. If x is the number, then “the difference of the number and 11” is x 11, and “seven times the difference of the number and 11” is 7(x 11). 5. If y is the number, then “the difference of 20 and the number” is 20 y, and “the difference of 20 20 y and the number, divided by 3” is or 3 (20 y) 3.
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8.
2(x 4) x 2x 8 x 2x 8 x x x 8 x 88
1 4 1
1
14. Let x represent the number of medals won by the Swedish team. Then x + 19 represents the number of medals won by the Norwegian team. x x 19 55 2x 19 55 2x 19 19 55 19 2x 36 2x 36 2 2 x 18 x + 19 = 18 + 19 = 37 The Swedish team won 18 medals and the Norwegian team won 37 medals.
4 1 4 1
Chapter 9: Equations, Inequalities, and Problem Solving
x
4
8
4 1
32 x 4 4 31 x 4 31 The number is . 4 10. The sum of the three lengths is 46 feet. x 3x 2 7x 46 11x 2 46 11x 2 2 46 2 11x 44 11x 44 11 11 x4 3x = 3(4) = 12 2 + 7x = 2 + 7(4) = 2 + 28 = 30 The lengths are 4 feet, 12 feet, and 30 feet. 12. Let x be the length of the shorter piece. Then 3x + 1 is the length of the longer piece. The sum of the lengths is 21 feet. x 3x 1 21 4x 1 21 4x 11 211 4x 20 4x 20 4 4 x5 3x + 1 = 3(5) + 1 = 15 + 1 = 16 The shorter piece is 5 feet and the longer piece is 16 feet.
16. Let x be the number of miles. Then the cost for x miles is 0.29x. Each day costs $24.95. 0.29x 2(24.95) 100 0.29x 49.9 100 0.29x 49.9 49.9 100 49.9 0.29x 50.1 0.29x 50.1 0.29 0.29 x 172.8 You can drive 172 whole miles on a $100 budget. 18. Let x be the number of hours. Then the total cost is 25.50x + 30. 25.5x 30 119.25 25.5x 30 30 119.25 30 25.5x 89.25 25.5x 89.25 25.5 25.5 x 3.5 You were charged for 3.5 hours. 20. Let x be the measure of the smaller angle. Then 2x 15 is the measure of the larger angle. The sum of the four angles is 360. 2x 2(2x 15) 360 2x 4x 30 360 6x 30 360 6x 30 30 360 30 6x 390 6x 390 6 6 x 65 2x 15 = 2(65) 15 = 130 15 = 115 Two angles measure 65 and two angles measure 115.
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ISM: Prealgebra and Introductory Algebra
22. Let angles B and C have measure x. Then angle A has measure x 42. The sum of the measures is 180. x x x 42 180 3x 42 180 3x 42 42 180 42 3x 222 3x 222 3 3 x 74 x 42 = 74 42 = 32 Angles B and C measure 74 and angle A measures 32. First Integer
Next Integers
Indicated Sum
24.
x
x+1
x+2
(x + 1) + (x + 2) = 2x + 3
26.
x
x+2
x+4
x + (x + 2) + (x + 4) = 3x + 6
28.
x
x+1
x+2
30.
x
x+2
x+4
x+3
x + (x + 3) = 2x + 3 x + (x + 2) + (x + 4) = 3x + 6
32. If x is the first even integer, the next consecutive even integer is x + 2. x x 2 654 2x 2 654 2x 2 2 654 2 2x 652 2x 652 2 2 x 326 The room numbers are 326 and 326 + 2 = 328. 34. Let x represent the first odd integer. Then x + 2 and x + 4 represent the next two consecutive odd integers. x x 2 x 4 675 3x 6 675 3x 6 6 675 6 3x 669 3x 669 3 3 x 223 x + 2 = 223 + 2 = 225 x + 4 = 223 + 4 = 227 Mali Republic’s code is 223, Côte d’Ivoire’s code is 225, and Niger’s code is 227. 36. Let x be the measure of the shorter piece. Then 5x + 1 is the measure of the longer piece. The measures sum to 25 feet. x 5x 1 25 6x 1 25 6x 11 25 1 6x 24 6x 24 6 6 x4 5x + 1 = 5(4) + 1 = 20 + 1 = 21 The pieces measure 4 feet and 21 feet. 340
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Chapter 9: Equations, Inequalities, and Problem Solving
38. Let x = the number. 7 1 x 8 2 8 7 8 1 x 7 8 7 2 4 x 7 4 The number is . 7 40. Let x represent the length, in miles, of I-90. Then I-80 is (x 120) miles long. The sum of the lengths is 5920 miles. x x 120 5920 2x 120 5920 2x 120 120 5920 120 2x 6040 2x 6040 2 2 x 3020 x 120 = 3020 120 = 2900 I-90 is 3020 miles long and I-80 is 2900 miles long. 42. Let x represent the floor space of the Empire State Building in thousands of square feet. Then the floor space of the Pentagon is 3x. x 3x 8700 4x 8700 4x 8700 4 4
x 2175 3x = 3(2175) = 6525 The Empire State Building has 2175 thousand square feet of floor space and the Pentagon has 6525 thousand square feet of floor space. 44. Let x = length of first piece, then 5x = length of second piece, and 6x = length of third piece. x 5x 6x 48 12x 48 12x 48 12 12 x4 5x = 5(4) = 20 6x = 6(4) =24 The first piece is 4 feet, the second piece is 20 feet, and the third piece is 24 feet.
46. The sum of the measures is 90. x (2x 3) 90 x 2x 3 90 3x 3 90 3x 3 3 90 3 3x 93 3x 93 3 3 x 31 2x 3 = 2(31) 3 = 62 3 = 59 The angles measure 31 and 59. 48. Let x be the first odd integer. Then the next three consecutive odd integers are x + 2, x + 4, and x + 6. The sum of the measures is 360. x x 2 x 4 x 6 360 4x 12 360 4x 12 12 360 12 4x 348 4x 348 4 4 x 87 x + 2 = 87 + 2 = 89 x + 4 = 87 + 4 = 91 x + 6 = 87 + 6 = 93 The angles measure 87, 89, 91, and 93. 2
50.
5 4x 5x 5 2 6 3 4x 6 5x6 3 6 4 24x 30x 5 4 24x 24x 30x 5 24x 4 6x 5 4 5 6x 5 5 9 6x 9 6x 6 6 3 x 2 3 The number is . 2
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Chapter 9: Equations, Inequalities, and Problem Solving
52. Let x represent the number of Democrat governors. Then the number of Republican governors is x + 6. The total number of governors is 50. x x 6 50 2x 6 50 2x 6 6 50 6 2x 44 2x 44 2 2 x 22 x + 6 = 22 + 6 = 28 There were 22 Democrat governors and 28 Republican governors. 54. Let x be the first even integer. The next two consecutive even integers are x + 2 and x + 4. The three integers total 48. x x 2 x 4 48 3x 6 48 3x 6 6 48 6 3x 42 3x 42 3 3 x 14 x + 2 = 14 + 2 = 16 x + 4 = 14 + 4 = 18 The boards have lengths 14 inches, 16 inches, and 18 inches. 56. Let x represent the width in inches. Then 2x + 0.23 represents the height. The sum is 9.38. x 2x 0.23 9.38 3x 0.23 9.38 3x 0.23 0.23 9.38 0.23 3x 9.15 3x 9.15 3 3 x 3.05 2x + 0.23 = 2(3.05) + 0.23 = 6.1 + 0.23 = 6.33 The width is 3.05 inches and the height is 6.33 inches. 58.
342
2(x 6) 3(x 4) 2x 12 3x 12 2x 12 2x 3x 12 2x 12 x 12 12 12 x 12 12 0 x The number is 0.
ISM: Prealgebra and Introductory Algebra
60. Let x represent the weight of the Cape York (Agpalilik) meteorite. Then 3x represents the weight of the Hoba meteorite. x 3x 88 4x 88 4x 88 4 4 x 22 3x = 3(22) = 66 The Cape York (Agpalilik) meteorite weighs 22 tons and the Hoba West meteorite weighs 66 tons. 62. If x is the first even integer, the next two even integers are x + 2 and x + 4. x x 2 x 4 48 3x 6 48 3x 6 6 48 6 3x 42 3x 42 3 3 x 14 The code is 14, 14 + 2 = 16, 14 + 4 = 18. 64. Let x represent the number of light vehicles sold in the second quarter of 2022, in thousands. Then x 100 represents the number sold in the third quarter of 2022. The total is 26,700. x x 100 26, 700 2x 100 26, 700 2x 100 100 26, 700 100 2x 26,800 2x 26,800 2 2 x 13, 400 x 100 = 13,400 100 = 13,300 In the United States, 13,400 light vehicles were sold in the second quarter of 2022 and 13,300 were sold in the third quarter. 66. Let x be the measure of the smallest angle. Then the two larger angles both measure 4x. x 4x 4x 180 9x 180 9x 180 9 9 x 20 4x = 4(20) = 80 The angles measure 20, 80, and 80.
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
68. Let x represent the number of yearly Internet complaints for 2019, in thousands. Then x + 325 represents the number for 2020. The total is 1259. x x 325 1259 2x 325 1259 2x 325 325 1259 325 2x 934 2x 934 2 2 x 467 x + 325 = 467 + 325 = 792 The number of yearly Internet complaints was 467 thousand for 2019 and 792 thousand for 2020.
2x = 2(22.5) = 45 5x = 5(22.5) = 112.5 Yes, the triangle exists and has angles that measure 22.5, 45, and 112.5. 84. One blink every 5 seconds is
1 blink
. 5 sec There are 60 60 = 3600 seconds in one hour. 1 blink 3600 sec 720 blinks 5 sec The average eye blinks 720 times each hour. 16 720 = 11,520 The average eye blinks 11,520 times while awake for a 16-hour day. 11,520 365 = 4,204,800 The average eye blinks 4,204,800 times in one year.
70. The bars ending between 30 and 45 represent the games Super Smash Bros. Ultimate and Animal Crossing: New Horizons.
86. answers may vary 72. Let x represent the sales of Pokémon Sword/ Pokémon Shield, in millions. Then x + 4.7 represents the sales of Super Smash Bros. Ultimate. The total is 56.1. x (x 4.7) 56.1 2x 4.7 56.1 2x 4.7 4.7 56.1 4.7 2x 51.4 2x 51.4 2 2 x 25.7 x + 4.7 = 25.7 + 4.7 = 30.4 Pokémon Sword/Pokémon Shield sold 25.7 million copies and Super Smash Bros. Ultimate sold 30.4 million copies. 74. answers may vary
88. answers may vary 90. Measurements may vary. Rectangle (b) best approximates the shape of the golden rectangle. Section 9.5 Practice Exercises 1. Use d = rt when d = 580 and r = 5. d rt 580 5t 580 5t 5 5 116 t It takes 116 seconds or 1 minute 56 seconds. 2.
76. answers may vary 78. Replace B by 14 and h by 22. 1 1 Bh (14)(22) 7(22) 154 2 2 80. Replace r by 15 and t by 2. r t = 15 2 = 30 82. Let x be the measure of the first angle. Then 2x is the measure of the second angle and 5x is the measure of the third angle. The measures sum to 180. x 2x 5x 180 8x 180 8x 180 8 8 x 22.5
Use A = lw when w = 18. A lw 450 l 18 450 18l 18 18 25 l The length of the deck is 25 feet. 9 3. Use F C 32 with C = 5. 5 9 F C 32 5 9 F 5 32 5 F 9 32 F 41 Thus, 5C is equivalent to 41F.
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Chapter 9: Equations, Inequalities, and Problem Solving 4. Let x be the width. Then 2x 12 is the length. The perimeter is 120 inches. P 2l 2w 120 2(2x 12) 2x 120 4x 24 2x 120 6x 24 120 24 6x 24 24 144 6x 24 x 2x 12 = 2(24) 12 = 48 12 = 36 2. The width is 24 inches and the length is 36 inches. 5.
C 2r C 2r 2 2 C C r or r 2 2
6.
7.
3. That the process of solving this equation for xdividing both sides by 5, the coefficient of xis the same process used to solve a formula for a specific variable. Treat whatever is multiplied by that specific variable as the coefficientthe coefficient is all the factors except that specific variable. Exercise Set 9.5 Use d = rt when d = 195 and t = 3. d rt 195 r 3 195 3r 3 3 65 r 4. Use V = lwh when l = 14, w = 8, and h = 3. V = lwh V = 14 8 3 V = 336
P 2l 2w P 2w 2l 2w 2w P 2w 2l P 2w 2l 2 2 P 2w P 2w l or l 2 2
6. Use A
h(B b) when A = 60, B = 7, and
b = 3. A 60
P 2a b c P c 2a b c c P c 2a b P c b 2a b b P c b 2a P c b P b c a or a 2 2
1 2 1
h(B b) h(7 3)
2 1
60 h(10) 2 60 5h 60 5h 5 5 12 h 8. Use V
A
2 a b 2A 2 2 2A a b 2A a a b a 2 A a b or b 2 A a
Vocabulary, Readiness & Video Check 9.5 1. A formula is an equation that describes known relationships among quantities.
1
Ah when V = 45 and h = 5.
3 1 V Ah 3 1 45 A 5 3 5 45 A 3 3 3 5 45 A 5 53 27 A
2. This is a distance, rate, and time problem. The speed (rate) is given in miles per hour and the time is given in hours, so the distance that we are finding must be in miles. 344
1 2
a b 8.
ISM: Prealgebra and Introductory Algebra
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ISM: Prealgebra and Introductory Algebra
10. Use A r 2 when r = 4 and 3.14 is used as an approximation for . A r2 A 3.14 42 A 3.14 16 A 50.24 12.
C 2r C 2r 2 2 C r 2
14.
T mnr T mnr mr mr T n mr
16.
18.
20.
22.
S 4lw 2wh S 4lw 4lw 2wh 4lw S 4lw 2wh S 4lw 2wh 2w 2w S 4lw h 2w
28. a.
Area = bh = 9.3(7) = 65.1 Perimeter 2(11.7) 2(9.3) 23.4 18.6 42 The area is 65.1 square feet and the perimeter is 42 feet.
b. The border goes around the edges, so it involves perimeter. The paint covers the wall, so it involves area.
A P PRT A P P PRT P A P PRT A P PRT PR PR A P T PR 1
24.
26. Use A = lw when A = 52,400 and l = 400. A lw 52, 400 400 w 52, 400 400w 400 400 131 w The width of the sign is 131 feet.
x y 13 x x y x 13 y x 13
D
Chapter 9: Equations, Inequalities, and Problem Solving
30. a.
Area
1
bh
1
(36)(27) 486 2 2 Perimeter = 27 + 36 + 45 = 108 The area is 486 square feet and the perimeter is 108 feet.
b. The fence goes around the edges of the yard, so it involves perimeter. The grass covers the yard, so it involves area.
fk
4 1 4D 4 fk 4 4D fk 4D fk f f 4D k f PR x y z w PR x y w x y z x y w PR x y w z
9 32. Use F C 32 when C = 5. 5 9 F C 32 5 9 F (5) 32 5 9 F (5) 32 5 F 9 32 F 23 Thus, 5C is equivalent to 23F.
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Chapter 9: Equations, Inequalities, and Problem Solving
1 34. Use d = rt when d = 303 and t 8 . 2 d rt 1 303 r 8 2 17 303 r 2 2 303 2 17 r 17 17 2 606 r 17 11 35 r 17 The average rate during the flight was 11 35 mph. 17
40. Use d = rt when d = 700 and r = 56. d rt 700 56t 700 56t 56 56 1 12 t 2 1 The trip will take 12 or (12.5) hours. 2
36. Let x be the width. Then 2x 10 is the length. Use P = 2 length + 2 width when P = 400. P 2 length 2 width 400 2(2x 10) 2x 400 4x 20 2x 400 6x 20 400 20 6x 20 20 420 6x 420 6x 6 6 70 x The width is 70 meters and the length is 2(70) 10 = 140 10 = 130 meters. 38. Let x represent the length of each of the equal sides. Then the shortest side is x 2. The perimeter is the sum of the lengths of the sides. x x x 2 22 3x 2 22 3x 2 2 22 2 3x 24 3x 24 3 3 x8 The sides of the triangle are 8 feet, 8 feet, and 8 2 = 6 feet.
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42. Use N = 94. N 40 T 50 4 94 40 T 50 4 54 T 50 4 T 50 13.5 T 63.5 The temperature is 63.5 Fahrenheit. 44. Use T = 65. T 50
N 40
4 N 40 65 50 4 N 40 65 50 50 50 4 N 40 15 4 N 40 4 15 4 4 60 N 40 60 40 N 40 40 100 N There are 100 chirps per minute. 46. As the air temperature of their environment decreases, the number of cricket chirps per minute decreases. 48. To find the amount of water in the tank, use 8 V r 2 h with r 4, h = 3, and 3.14. 2 V r2h 3.14(4)2 3 3.14(16) 3 150.72 The tank holds 150.72 cubic meters of water. Let x represent the number of goldfish the tank could hold. Then 2x = 150.72.
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ISM: Prealgebra and Introductory Algebra 2x 150.72 2x 150.72 2 2 x 75.36 The tank could hold 75 goldfish. 1 50. Use A bh when A = 20 and b = 5. 12 A bh 2 1 20 5 h 2 2 2 5 20 h 5 5 2 8h The height of the sail is 8 feet. 52. Use C = 2r when r = 4000 and 3.14. C = 2r C = 2 3.14 4000 C = 25,120 Thus, 25,120 miles of rope is needed to wrap around the Earth. 54. The perimeter is the sum of the lengths of the sides. x (2x 8) (3x 12) 82 x 2x 8 3x 12 82 6x 20 82 6x 20 20 82 20 6x 102 6x 102 6 6 x 17 x feet = 17 feet (2x 8) feet (2 17 8) feet (34 8) feet 26 feet (3x 12) feet (317 12) feet (5112) feet 39 feet
Chapter 9: Equations, Inequalities, and Problem Solving
58. Let x be the length of the sides of the square pen. Then 2x 15 is the length of the sides of the triangular pen. The perimeters are equal. 4x 3(2x 15) 4x 6x 45 4x 6x 6x 45 6x 2x 45 2x 45 2 2 x 22.5 2x 15 = 2(22.5) 15 = 45 15 = 30 The square’s side length is 22.5 units and the triangle’s side length is 30 units. 60. Use d = rt when d = 150 and r = 45. d rt 150 45t 150 45t 45 45 1 3 t or t 3 hr 20 min 3 If he left at 4 A.M., then he will arrive at 4 A.M. + 3 hr 20 min = 7:20 A.M. 9 62. Use F C 32 when F = 78. 5 9 F C 32 5 9 78 C 32 5 9 78 32 C 32 32 5 9 46 C 5 55 9 46 C 95 230 C 9 5 25 C 9
9
56. Use A = lw when A = 6875 and w = 6875. A lw 6875 l 6875 25l 25 25 275 l The length of the ribbon was 275 meters.
5 Normal room temperature is about 25 C. 9
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
9 64. Use F C 32 when F = 12. 5 9 F C 32 5 9 12 C 32 5 9 12 32 C 32 32 95 20 C 5 5 5 9 (20) C 9 95 100 C 9 11.1 C The low temperature is approximately 11.1C. 66. Use V
4
r3 when r
4
r3
15 and = 3.14.
2
3 V
30
4
(3.14)(15)3 14,130
3 3 The volume of the sphere is 14,130 cubic inches. 9 68. Use F C 32 when F = 227. 5 9 F C 32 5 9 227 C 32 5 9 227 32 C 32 32 59 259 C 5 5 5 9 259 C 9 9 5 144 C The average temperature on Jupiter is 144C. 70. 8% = 0.08 72. 0.5% = 0.005 74. 0.03 = 0.03(100%) = 3%
78.
F P V F B(P V ) (P V ) P V B(P V ) F BP BV F BP BV BP F BP BV F BP BV F BP B B BP F V B
80. Use A = bh. If the base is doubled, the new base is 2b. If the height is doubled, the new height is 2h. A = (2b)(2h) = 2 2 b h = 4bh The area is multiplied by 4. 9 82. Let x be the temperature. Use F C 32 5 when F = C = x. 9 F C 32 5 9 x x 32 5 9 9 9 x x x 32 x 5 5 5 5 9 x x 32 5 5 4 x 32 5 45 5 x 32 4 5 4 x 40 They are the same when the temperature is 40. 84. ☐ O ☐ Δ O ☐ Δ ☐ O☐ Δ ☐ ☐Δ ☐ O
76. 5 = 5(100%) = 500%
348
B
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
miles hour miles 5280 feet 1 hour 20 hour 1 mile 3600 seconds 88 feet/second 3 88 Use d = rt when d = 1300 and r . 3
86. 20
d rt 88 1300 t 3 3 (1300) 3 88 88
Section 9.6 Practice Exercises 1. x 2
–5 –4 –3 –2 –1 0 1 2 3 4 5
2. 5 > x or x < 5 –5 –4 –3 –2 –1 0 1 2 3 4 5
3. 3 x < 1 –5 –4 –3 –2 –1 0 1 2 3 4 5
4.
t 88 3
x 6 11 x 6 6 11 6 x 5 –5 –4 –3 –2 –1 0 1 2 3 4 5
5. 3x 12
44.3 t It will take about 44.3 seconds. 88. Use d = rt when d = 238,860 and r = 186,000. d rt 238,860 186, 000t 238,860 186, 000t 186, 000 186, 000 1.3 t It will take 1.3 seconds. 90. Let d = 2 then r = 1. 15 feet 12 inches 15 feet 180 inches, so 1 1 foot h = 180. Use V r 2h when r = 1 and h = 180. V ()(1)2 (180) 180 565.5
3x 12 3 3 x 4 –5 –4 –3 –2 –1 0 1 2 3 4 5
6. 5x 20 5x 20 5 5 x 4 –5 –4 –3 –2 –1 0 1 2 3 4 5
7.
The volume of the column is 565.5 cubic inches. 1 92. Use V r 2h when V = 565.2 and r = 6. 3 1 V r2h 3 1 565.2 62 h 3 3 3 1 62 h 2 565.2 6 62 3 15 h 94. Use I = PRT when I = 3750, P = 25,000 and R = 0.05. I PRT 3750 25, 000(0.05)T 3750 1250T 3750 1250T 1250 1250 3T
3x 11 13 3x 1111 13 11 3x 24 3x 24 3 3 x8 {x|x 8} –10–8 –6 –4 –2 0 2 4 6 8 10
8.
2x 3 4(x 1) 2x 3 4x 4 2x 3 4x 4x 4 4x 2x 3 4 2x 3 3 4 3 2x 1 2x 1 2 6 1 x 2
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
1 x x
8. x < 6
2
|6| = 6 is not a solution. 1 2
9. x < 4.01 4.1 is not a solution.
–5 –4 –3 –2 –1 0 1 2 3 4 5
9.
10. x 3 4 is not a solution.
3(x 5) 1 5(x 1) 7 3x 15 1 5x 5 7 3x 14 5x 2 3x 14 5x 5x 2 5x 2x 14 2 2x 14 14 2 14 2x 12 2x 12 2 2 x6 {x|x 6}
11. An open circle indicates > or <; a closed circle indicates or . 12. addition property of equality 13. {x|x 2}
10. Let x be the unknown number. 35 2x 15 35 2x 35 15 35 2x 20 2x 20 2 2 x 10 All numbers less than 10 make the statement true. 11. Let x represent the minimum sales. 600 0.04x 3000 0.04x 2400 x 60, 000 Alex must have minimum sales of $60,000. Vocabulary, Readiness & Video Check 9.6
14. The multiplication property of inequality is applied at this step when we divide by the coefficient of x. The coefficient is positive and so the inequality symbol remains the same, but if the coefficient had been negative, the direction of the inequality symbol would have been reversed. 15. is greater than; > Exercise Set 9.6 2.
–5 –4 –3 –2 –1
3. 6x < 7(x + 9) is an inequality.
4. 6. 8. 10. 12.
4. 5y 2 38 is an inequality. 9 x2 5. is an equation. 7 14 6.
9 7
x2
7. x 3 5 is not a solution. 350
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
2
3
4
5
0
1
2
3
4
5
0
2
4
6
8
10
2 3
x4 1 x 4 4 1 4 x 3
14.
– 5 –4 –3 –2 –1
is an expression.
14
1
1. 6x 7(x + 9) is an expression. 2. 6x = 7(x + 9) is an equation.
0
16.
3 m 5 3 3 m 3 5 m 8 –10 –8 –6 –4 –2
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ISM: Prealgebra and Introductory Algebra
18.
Chapter 9: Equations, Inequalities, and Problem Solving 30. 0.3x 2.4 0.3x 2.4 0.3 0.3 x8 {x|x < 8}
3 7x 10 8x 3 7x 8x 10 8x 8x 3 x 10 3 x 3 10 3 x7 7 –10 –8 –6 –4 –2
20.
0
2
4
6
–10 –8
8
32.
7x 3 9x 3x 7x 3 6x 7x 3 7x 6x 7x 3 x 3 x 1 1 3 x or x 3 –5 –4 –3 –2 –1
0
1
2
3
4
5
0
1
2
3
4
5x 20 5 5 x 4 {x|x > 4} 0
1
2
3
4
5
y0 (1)( y) (1)(0) y0 {y|y 0}
40.
–5 –4 –3 –2 –1
0
1
2
3
4
5
0
2
4
6
8
10
5
x 8 6 65 6 x (8) x
2
4
6
8
10
11 x 4 11 4 x 4 4 15 x {x|x < 15}
38.
–5 –4 –3 –2 –1
5
0
6x 5 6 6 5 x 6 5 x x 6
5
24. 5x 20
5 6
–2
36. 6x 5
–5 –4 –3 –2 –1
28.
–4
34. 10(x 2) 9x 1 10x 20 9x 1 x 20 1 x 20 20 1 20 x 21 {x|x 21}
22. 3x 9 3x 9 3 3 x 3 {x|x > 3}
26.
–6
10
48
3 y9 4 34 4 y 9 3 4 3 y 12 {y|y 12} 6(2 z) 12 12 6z 12 12 6z 12 12 12 6z 0 6z 0 6 6 z0 {z|z 0}
5 48 x x 5 – 48 5 –10 –8 –6 –4 –2
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
42.
2x 1 4x 5 2x 1 4x 4x 5 4x 2x 1 5 2x 11 5 1 2x 4 2x 4 2 2 x2 {x|x 2}
44.
44 8x 2x xx 10x
50.
4 x x 10x x 4 11x 52.
4 11x 11 11 4 x 11 x x 4 11
46.
7x 4 3(4 x) 7x 4 12 3x
7x 4 3x 12 3x 3x 4x 4 12 4x 4 4 12 4 4x 8 4x 8 4 4 x 2 {x|x < 2} 48.
352
5(x 2) 3(2x 1) 5x 10 6x 3 5x 10 5x 6x 3 5x 10 x 3 10 3 x 3 3 7 x {x|x 7}
3(5x 4) 4(3x 2) 15x 12 12x 8 15x 12 12x 12x 8 12x 3x 12 8 3x 12 12 8 12 3x 4 3x 4 3 3 4 x 3 4 x x 3 7(x 2) x 4(5 x) 12 7x 14 x 20 4x 12 8x 14 4x 32 8x 14 4x 4x 32 4x 4x 14 32 4x 14 14 32 14 4x 18 4x 18 4 4 9 x 2 9 x x 2
54. 2(x 4) 3x (4x 1) 2x 2x 8 3x 4x 1 2x 5x 8 2x 1 5x 8 2x 2x 1 2x 3x 8 1 3x 8 8 1 8 3x 9 3x 9 3 3 x3 {x|x > 3}
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ISM: Prealgebra and Introductory Algebra
56.
58.
1
1 (x 5) (2x 1) 2 3 1 1 6 (x 5) 6 (2x 1) 2 3 3(x 5) 2(2x 1) 3x 15 4x 2 3x 15 3x 4x 2 3x 15 x 2 15 2 x 2 2 13 x {x|x > 13}
Chapter 9: Equations, Inequalities, and Problem Solving
64. Convert heights to inches. 6'8" 6 12 8 80 6'6" 6 12 6 78 6'0" 6 12 0 72 5'9" 5 12 9 69 6'5" 6 12 5 77 Let x be the height of the center. x 80 78 72 69 77 5 x 299 77 x 5299 5 5 77 5 x 299 385 x 299 299 385 299 x 86 86" 7 ' 2" The center should be at least 7' 2".
6x 2 3(x 4) 6x 2 3x 12 6x 2 3x 3x 12 3x 3x 2 12 3x 2 2 12 2 3x 14 3x 14 3 3 14 x 3 14 x x 3
66. Let x represent the number of people. Then the cost is 40 + 15x. 40 15x 860 40 15x 40 860 40 15x 820
60. Let x be the number. 5x 1 9 5x 1 1 9 1 5x 10 5x 10 5 5 x 2 All numbers less than 2 make this statement true. 62. Use P = a + b + c when a = x, b = 4x, c = 12, and P 87. x 4x 12 87 5x 12 87 5x 12 12 87 12 5x 75 5x 75 5 5 x 15 4x 4(15) = 60 The maximum lengths of the other two sides are 15 inches and 60 inches.
15x 820 15 15 820 x 54 15 They can invite at most 54 people. 68. Let x represent the number of minutes. 5.3x 200 5.3x 200 5.3 5.3 200 x 38 5.3 The person must bicycle at least 38 minutes. 70. Let x be the number. 1 x4 2 2 1 x44 24 2 1 x6 2 1 2 x 26 2 x 12 All numbers greater than or equal to 12 make the statement true. 72. 43 4 4 4 64
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Chapter 9: Equations, Inequalities, and Problem Solving
ISM: Prealgebra and Introductory Algebra
4. A linear inequality in one variable can be written in the form ax + b < c, (or >, , ). 5. The solutions to the equation x + 5 = x + 5 are all real numbers.
74. 07 0 0 0 0 0 0 0 0 3
2 2 2 2 8 76. 3 3 3 3 27 78. The brand name Google is worth about $280 billion or $280,000,000,000.
6. The solution to the equation x + 5 = x + 4 is no solution.
80. The bars that end between 200 and 300 correspond to Google and Apple. The brands worth greater than $200 billion and less than $300 billion were Google and Apple.
7. If both sides of an inequality are multiplied or divided by the same positive number, the direction of the inequality symbol is the same.
82. The longest bar corresponds to Amazon, so the brand worth the most in 2023 was Amazon.
8. If both sides of an inequality are multiplied by the same negative number, the direction of the inequality symbol is reversed. 9. Two numbers whose sum is 0 are called opposites.
84. Since m n, then 2m 2n. 86. If x < y, then x > y.
10. Two numbers whose product is 1 are called reciprocals.
88. No; answers may vary 90. Let x be the score on the final exam. Since the final counts as three tests, the final course 85 95 92 3x average is . 6 85 95 92 3x 90 6 272 3x 90 272 63x 6 6(90) 6 272 3x 540 272 2x 272 540 272 3x 268 3x 268 3 3 x 89.3 The final exam score must be at least 89.3 to get an A. 92. answers may vary
Chapter 9 Review 1. Since 8 is to the left of 10 on the number line, 8 < 10. 2. Since 7 is to the right of 2 on the number line, 7 > 2. 3. Since 4 is to the right of 5 on the number line, 4 > 5. 4. Since
12 6 is to the right of 8 on the number 2
12 line,
8.
2
5. Since |7| = 7 is to the left of |8| = 8 on the number line, |7| < |8|. 6. Since |9| = 9 is to the right of 9 on the number line, |9| > 9. 7. |1| = 1
Chapter 9 Vocabulary Check 1. A linear equation in one variable can be written in the form ax + b = c.
8. Since |14| = 14 and (14) = 14, |14| = (14).
2. Inequalities that have the same solution are called equivalent inequalities.
9. Since 1.2 is to the right of 1.02 on the number line, 1.2 > 1.02.
3. An equation that describes a known relationship among quantities is called a formula.
354
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ISM: Prealgebra and Introductory Algebra
10. Since
3 2
6
and
4
the number line,
6
is to the left of
4 3 2
Chapter 9: Equations, Inequalities, and Problem Solving
3
on
4 3
. 4
20. 6 1 = 6 illustrates the identity element for multiplication. 21. 3(8 5) = 3 8 + 3 (5) illustrates the distributive property.
11. Four is greater than or equal to negative three is written as 4 3.
22. 4 + (4) = 0 illustrates the additive inverse property.
12. Six is not equal to five is written as 6 5.
23. 2 + (3 + 9) = (2 + 3) + 9 illustrates the associative property of addition.
13. 0.03 is less than 0.3 is written as 0.03 < 0.3. 14. Since 83 is to the left of 1500 on a number line, 83 < 1500.
24. 2 8 = 8 2 illustrates the commutative property of multiplication. 25. 6(8 + 5) = 6 8 + 6 5 illustrates the distributive property.
15. a.
The natural numbers are 1, 3.
b.
The whole numbers are 0, 1, 3.
c.
The integers are 6, 0, 1, 3.
26. (3 8) 4 = 3 (8 4) illustrates the associative property of multiplication. 1
1 The rational numbers are 6, 0, 1, 1 , 3, 2 9.62.
27. 4
e.
The irrational number is .
28. 8 + 0 = 8 illustrates the element identity for addition.
f.
The real numbers are all numbers in the set.
d.
1 illustrates the multiplicative inverse 4 property.
5
29.
16. a.
The natural numbers are 2, 5.
b.
The whole numbers are 2, 5.
c.
The integers are 3, 2, 5.
d.
11 The rational numbers are 3, 1.6, 2, 5, 2 , 15.1.
e. f.
The irrational numbers are
5, 2.
The real numbers are all numbers in the set.
17. Since 4 < 2, the most negative number is 4 which corresponds to Friday. Thus, Friday showed the greatest loss. 18. The greatest positive number is +5 which corresponds to Wednesday. Thus, Wednesday showed the greatest gain. 19. 6 + 5 = 5 + (6) illustrates the commutative property of addition.
2 x4 x 53 3 2 3 x 4 3 x 3 3 5x 12 2x 5x 12 5x 2x 5x 12 3x 12 3x 3 3 4 x 7
30.
5 x 1 x 7 8 8 5 8 x 1 8 x 8 8 7x 8 5x 7x 8 7x 5x 7x 8 2x 8 2x 2 2 4 x
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
31.
(5x 1) 7x 3 5x 1 7x 3 5x 1 7x 7x 3 7x 2x 1 3 2x 11 3 1 2x 4 2x 4 2 2 x2
32.
4(2x 1) 5x 5
36.
8x 4 5x 5 8x 4 8x 5x 5 8x 4 3x 5 4 5 3x 5 5 9 3x 9 3x 3 3 3 x 33.
34.
6(2x 5) 3(9 4x) 12x 30 27 12x 12x 12x 30 12x 27 12x 30 27 Since the statement 30 = 27 is false, the equation has no solution. 3(8 y 1) 6(5 4 y) 24 y 3 30 24 y 24 y 3 24 y 30 24 y 24 y 3 30 Since the statement 3 = 30 is false, the equation has no solution.
3(2 z) 35. z 5 3(2 z) 5 5 z 5 3(2 z) 5z 6 3z 5z 6 3z 3z 5z 3z 6 8z 6 8z 8 8 3 z 4
356
4(n 2) n 5 4(n 2) 5 5(n) 5 4(n 2) 5n 4n 8 5n 4n 8 4n 5n 4n 8 9n 8 9n 9 9 8 n 9
37. 0.5(2n 3) 0.1 0.4(6 2n) 5(2n 3) 1 4(6 2n) 10n 15 1 24 8n 10n 16 24 8n 10n 16 8n 24 8n 8n 2n 16 24 2n 16 16 24 16 2n 40 2n 40 2 2 n 20 38. 1.72 y 0.04 y 0.42 172 y 4 y 42 168 y 42 168 y 42 168 168 y 0.25 5(c 1)
39.
6 5(c 1)
2c 3
6(2c 3) 6 5(c 1) 6(2c 3) 5c 5 12c 18 5c 5 5c 12c 18 5c 5 7c 18 5 18 7c 18 18 23 7c 23 7c 7 7 23 c 7
6
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
40.
9 5a 3(6a 1) 9 5a 18a 3 9 5a 5a 18a 3 5a 9 23a 3 9 3 23a 3 3 6 23a 6 23a 23 23 6 a 23
41. If the sum of two numbers is 10 and one number is x, the other number is 10 x, which is choice b. 42. If Mandy is 5 inches taller than Melissa and x represents Mandy’s height, then Melissa’s height can be represented by x 5, which is choice a. 43. The measures of complementary angles sum to 90. If an angle measures x, then the measure of its complement is (90 x), which is choice b. 44. The measures of supplementary angles sum to 180. If an angle measures (x + 5), then the measure of its supplement is [180 (x + 5)] = (180 x 5) = (175 x), which is choice c. 45. Let x represent the length of the side of the square base. Then 10x + 50.5 represents the height. The sum is 7327. x 10x 50.5 7327 11x 50.5 7327 11x 50.5 50.5 7327 50.5 11x 7276.5 11x 7276.5 11 11 x 661.5 10x 50.5 10(661.5) 50.5 6615 50.5 6665.5 The height is 6665.5 inches. 46. Let x represent the length of the short piece. Then 2x represents the length of the long piece. The lengths sum to 12. x 2x 12 3x 12 3x 12 3 3 x4 2x = 2(4) = 8 The short piece is 4 feet and the long piece is 8 feet.
Chapter 9: Equations, Inequalities, and Problem Solving
47. Let x represent the number of national battlefields. Then 3x 2 represents the number of national memorials. The total is 42. x 3x 2 42 4x 2 42 4x 2 2 42 2 4x 44 4x 44 4 4 x 11 3x 2 = 3(11) 2 = 33 2 = 31 In 2022, there were 11 national battlefields and 31 national memorials. 48. Let x be the first integer. Then x + 1 and x + 2 are the next two consecutive integers. Their sum is 114. x x 1 x 2 114 3x 3 114 3x 3 3 114 3 3x 117 3x 117 3 3 x 39 x + 1 = 39 + 1 = 38 x + 2 = 39 + 2 = 37 The integers are 39, 38, and 37. 49.
50.
x x2 3 x 3 3(x 2) 3 x 3x 6 x 3x 3x 6 3x 2x 6 2x 6 2 2 x3 The number is 3. 2(x 6) x 2x 12 x 2x 2x 12 2x x 12 3x 12 3x 3 3 4 x The number is 4.
Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
57. C d C d C d
51. Use P = 2l + 2w when P = 46 and l = 14. P 2l 2w 46 2(14) 2w 46 28 2w 46 28 28 2w 28 18 2w 18 2w 2 2 9w
58.
52. Use V = lwh when V = 192, l = 8, and w = 6. V lwh 192 8 6 h 192 48h 192 48h 48 48 4h 53.
r vst 5 r 5 vst 5 5 r 5 vst r 5 vst vt vt r5 s vt
55.
2 y 5x 7 2 y 2 y 5x 2 y 7 5x 2 y 7 5x 2 y 7 5 5 2y 7 x 5
56.
358
59. Use V = lwh when V = 900, l = 20 and h = 3. V lwh 900 20 w 3 900 60w 900 60w 60 60 15 w The width is 15 meters.
y mx b y b mx b b y b mx y b mx x x yb m x
54.
3x 6 y 2 3x 3x 6 y 3x 2 6 y 3x 2 6 y 3x 2 6 6 3x 2 y 6
C 2r C 2r 2 2 C r 2
60. Let x be the width. Then the length is x + 6. Use P = 2 length + 2 width when P = 60. P 2 length 2 width 60 2(x 6) 2x 60 2x 12 2x 60 4x 12 60 12 4x 12 12 48 4x 48 4x 4 4 12 x x + 6 = 12 + 6 = 18 The dimensions of the billboard are 12 feet by 18 feet. 61.
Use d = rt when d = 10K or 10,000 m and r = 125. d rt 10, 000 125t 10, 000 125t 125 125 80 t 80 1 The time is 80 minutes or 1 hours or 60 3 1 hour and 20 minutes.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
9 62. Use F C 32 when F = 134. 5 9 F C 32 5 9 134 C 32 95 134 32 C 32 32 5 9
Chapter 9: Equations, Inequalities, and Problem Solving
70.
102 C 5 5 5 9 102 C 9 C 9 5 56.7 Thus, 134F is equivalent to 56.7C.
2
71.
63. 64. 65.
–5 –4 –3 –2 –1
0
1
2
3
4
5
–5 –4
0
1
2
3
4
5
–3 –2 –1
x 5 4 x 5 5 4 5 x 1 {x|x 1}
{y|y 15} 73.
2(x 5) 2(3x 2) 2x 10 6x 4 2x 10 6x 6x 4 6x 8x 10 4 8x 10 10 4 10 8x 14 8x 14 8 8 7 x 4 7 x x 4
74.
4(2x 5) 5x 1 8x 20 5x 1 8x 20 5x 5x 1 5x 3x 20 1 3x 20 20 1 20 3x 19 3x 19 3 3 19 x 3 19 x x 3
x7 2 x77 27 x 5 {x|x > 5} 67. 2x 20 2x 20 2 2 x 10 {x|x 10}
69.
5x 7 8x 5 5x 7 8x 8x 5 8x 3x 7 5 3x 7 7 5 7 3x 12 3x 12 3 3 x 4 {x|x < 4}
y6 3 3 2 3 y 6 2 3 2 y9 {y|y > 9}
72. 0.5 y 7.5 0.5 y 7.5 0.5 0.5 y 15
66.
68. 3x 12 3x 12 3 3 x 4 {x|x < 4}
x 4 6x 16 x 4 6x 6x 16 6x 5x 4 16 5x 4 4 16 4 5x 20 5x 20 5 5 x4 {x|x 4}
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359
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
75. Let x be the sales. Her weekly earnings are 175 + 0.05x. 175 0.05x 300 175 0.05x 175 300 175 0.05x 125 0.05x 125 0.05 0.05 x 2500 She must have weekly sales of at least $2500. 76. Let x be his score on the fourth round. 76 82 79 x 80 4 237 x 80 4 237 x 4 4 80 4 237 x 320 237 x 237 320 237 x 83 His score must be less than 83. 77. 6x 2x 1 5x 11 8x 1 5x 11 8x 1 5x 5x 11 5x 3x 1 11 3x 1 1 111 3x 12 3x 12 3 3 x4 78.
2(3y 4) 6 7 y 6y 8 6 7 y 6y 8 6y 6 7 y 6y 8 6 y 8 6 6 y 6 14 y
79. 4(3 a) (6a 9) 12a 12 4a 6a 9 12a 3 10a 12a 3 10a 10a 12a 10a 3 2a 3 2a 2 2 3 a 2
360
x
80.
2 5 3 x 2 2 52 3 x 7 3 x 3 3 7 3 x 21
81. 2( y 5) 2 y 10 2 y 10 2 y 10 Since both sides of the equation are identical, the equation is an identity and every real number is a solution. 82. 7x 3x 2 2(2x 1) 4x 2 4x 2 4x 2 4x 4x 2 4x 2 2 Since the statement 2 = 2 is false, there is no solution. 83. Let x be the number. 6 2x x 7 6 2x x x 7 x 6 x 7 6 x 6 7 6 x 13 The number is 13. 84. Let x be the length of the shorter piece. Then 4x + 3 is the length of the longer piece. The lengths sum to 23. x 4x 3 23 5x 3 23 5x 3 3 23 3 5x 20 5x 20 5 5 x4 4x + 3 = 4(4) + 3 = 16 + 3 = 19 The shorter piece is 4 inches and the longer piece is 19 inches.
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ISM: Prealgebra and Introductory Algebra
85.
86.
V
1
Chapter 9: Equations, Inequalities, and Problem Solving
4. 9 + x = x + 9 by the commutative property of addition; C.
Ah
3 1 3V 3 Ah 3 3V Ah 3V Ah A A 3V h A
5. 5(4x) = (5 4)x by the associative property of multiplication; B. 6.
5
8. 23 2 2 2 8; C. 9. 7 is the opposite of 7; A. 10. 6x + 2 + 4x 10 does not contain an equal sign, so it is an expression; B.
9 0
2
4
6
8 10
87. 5x 20 5x 20 5 5 x 4 {x|x > 4} –5 –4 –3 –2 –1
is the reciprocal of 5; B.
7. 3 + 2(8) = 3 + (16) = 13; C.
4x 7 3x 2 4x 7 3x 3x 2 3x x7 2 x77 27 x9 {x|x > 9} –10 –8 –6 –4 –2
1
11. 6x + 2 = 4x 10 contains an equal sign, so it is an equation; A. 12. 2(x 1) = 12 contains an equal sign, so it is an equation; A.
0
1
2
3
4
5
88. 3(1 2x) x (3 x) 3 6x x 3 x 3 5x 3 x 3 5x x 3 x x 3 6x 3 3 6x 3 3 3 6x 0 6x 0 6 6 x0 {x|x 0}
1 13. 7 x 22 does not contain an equal sign, 2 so it is an expression; B. 14. Subtracting 100z from 8m translates to 8m 100z; B. 15. Subtracting 7x 1 from 9y translates to 9y (7x 1); C. 16.
7x 6 7x 9 7x 7x 6 7x 7x 9 69 Since the statement 6 = 9 is false, the equation has no solution; B.
5
17.
1. The number halfway between 2 and 3 on a 1 number line is 2 2.5; C and D. 2
2y 5 2y 5 2y 2y 5 2y 2y 5 5 5 Since the statement 5 = 5 is true, all real numbers are solutions of the equation; A.
18.
11x 13 10x 13 11x 10x 13 10x 10x 13 x 13 13 x 13 13 13 13 x0 The solution of the equation is 0; C.
–5 –4 –3 –2 –1
0
1
2
3
4
Chapter 9 Getting Ready for the Test
2. The number halfway between 2 and 3 on a 1 number line is 2 2.5; C and D. 2 3. 5(2x 1) = 10x 5 by the distributive property; A.
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361
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
19.
20.
x 15 x 15 x x 15 x x 15 2x 15 15 2x 15 15 15 15 2x 0 2x 0 2 2 x0 The solution of the equation is 0; C.
5. 6 8 = 8 6 illustrates the commutative property of multiplication.
5(3x 2) (x 20) 5(3x 2) 1(x 20) 5 3x 5 2 1 x (1)(20) 15x 10 x 20 The correct choice is B.
8. The opposite of 9 is 9.
6. 6(2 + 4) = 6 2 + (6) 4 illustrates the distributive property. 7.
1
of 3.
3 10. 7 2(5 y 3) 7 10 y 6 7 6 10 y 1 10 y 10 y 1 11. 4(x 2) 3(2x 6) 4x 8 6x 18 4x 6x 8 18 2x 10 12.
4(n 5) (4 2n) 4n 20 4 2n 4n 20 2n 4 2n 2n 2n 20 4 2n 20 20 4 20 2n 16 2n 16 2 2 n 8
13.
2(x 3) x 5 3x 2x 6 2x 5 2x 2x 6 2x 2x 5 65 Since the statement 6 = 5 is false, there is no solution.
Chapter 9 Test 1. The absolute value of negative seven is greater than five is written as |7| > 5. 2. The sum of nine and five is greater than or equal to four is written as (9 + 5) 4. 3. a.
The natural numbers are 1, 7.
b.
The whole numbers are 0, 1, 7.
c.
The integers are 5, 1, 0, 1, 7.
d.
The rational numbers are 5, 1,
1 , 0, 1, 7, 4
11.6. e.
The irrational numbers are
7, 3.
f.
The real numbers are 5, 1,
1
, 0, 1, 7,
4 7, 3.
4. 8 + (9 + 3) = (8 + 9) + 3 illustrates the associative property of addition.
362
(6) 1 illustrates the multiplicative inverse 6 property.
9. The reciprocal of
8x x2 21. 1 3 10 8x x 2 30 1 30 3 10 8x 30 30 1 3(x 2) 3 10 8x 30 3 x 3 2 80x 30 3x 6 The correct choice is D.
11.6,
1
14. 4z 1 z 1 z 3z 1 1 z 3z 1 z 1 z z 2z 1 1 2z 1 1 1 1 2z 0 2z 0 2 2 z0
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ISM: Prealgebra and Introductory Algebra 2(x 6)
15.
19. Let x be the number. 2 x x 35 3 3 2 x x 35 3 3 5 x 35 3 3 5 3 x 35 5 3 5 x 21 The number is 21.
x5
2(x 36) 3(x 5) 3 2(x 6) 3(x 5) 2x 12 3x 15 2x 12 2x 3x 15 2x 12 x 15 12 15 x 15 15 27 x
3
16.
1
x
3
x4 2 4 x x 4 2 x 2 x 4 x 2 x x 4 x 2 2x 4 2 4 2x 4 4 6 2x 6 2x 2 2 3 x
20. A = lw = (35)(20) = 700 The area of the deck is 700 square feet. To paint two coats of water seal means covering 2 700 = 1400 square feet. 1 gal 1400 sq ft 7 gal 200 sq ft 7 gallons of water seal are needed.
2
17.
Chapter 9: Equations, Inequalities, and Problem Solving
21. Use y = mx + b when y = 14, m = 2, and b = 2. y mx b 14 2x (2) 14 2 2x (2) 2 12 2x 12 2x 2 2 6x
0.3(x 4) x 0.5(3 x) 0.3(x 4) 1.0x 0.5(3 x) 3(x 4) 10x 5(3 x) 3x 12 10x 15 5x 7x 12 15 5x 7x 12 5x 15 5x 5x 12x 12 15 12x 12 12 15 12 12x 3 12x 3 12 12 1 x 0.25 4
22.
V r 2h V r 2h r2 r2 V h r2
23.
18. 4(a 1) 3a 7(2a 3) 4a 4 3a 14a 21 4 7a 14a 21 4 7a 14a 14a 2114a 4 7a 21 4 7a 4 21 4 7a 25 7a 25 7 7 25 a 7
3x 4 y 10 3x 4 y 3x 10 3x 4 y 10 3x 4 y 10 3x 4 4 3x 10 y 4
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363
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
24.
3x 5 7x 3 3x 5 3x 7x 3 3x 5 4x 3 5 3 4x 3 3 8 4x 8 4x 4 4 2 x {x|x < 2} –5 –4 –3 –2 –1
0
28. Let x = one area code, then 2x = other area code. x 2x 1203 3x 1203 3x 1203 3 3 x 401 2x = 2(401) = 802 The area codes are 401 and 802. 1
25. 0.3x 2.4 0.3x 2.4 0.3 0.3 x 8 {x|x 8} 26.
27.
5(x 1) 6 3(x 4) 1 5x 5 6 3x 12 1 5x 11 3x 11 5x 11 3x 3x 11 3x 2x 11 11 2x 1111 1111 2x 22 2x 22 2 2 x 11 {x|x 11} 2(5x 1) 2 3 2(5x 1) 3 3(2) 3 2(5x 1) 6 10x 2 6 10x 2 2 6 2 10x 4 10x 4 10 10 2 x 5 2 x x 5
2
3
4
5
Cumulative Review Chapters 19 1. a.
(4) = 4
b. |5| = 5 c. 2. a.
|6| = 6 |0| < 2 since |0| = 0 and 0 < 2.
b. |5| = 5 c.
|3| > |2| since 3 > 2.
d. |5| < |6| since 5 < 6. e.
|7| > |6| since 7 > 6.
3. 2[3 2(1 6)] 5 2[3 2(5)] 5 2(3 10) 5 2(7) 5 14 5 19 4.
3 4 3 22 64
3 1 22 64
3 1 22 2 3 1 4 2 8 2 4
5. 18 + 10 = 8 6. 8 + (11) = 19 7. 12 + (8) = 4 8. 2 + 10 = 8 9. (3) + 4 + (11) = 1 + (11) = 10
364
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
10. 0.2 + (0.5) = 0.3
2
1 1 16. 2 y x 2 3 2 1 1 2 3 4 2 1 3 4 8 3 12 12 11 12 2
11. 2 10 2 11 10 3 3 11 3 11 11 3 22 30 33 33 22 30 33 8 33 8 33
2 3 22 14 2 1114 28 3 5 17. 4 1 5 11 5 11 5 11 5 5 1 1 5 5 25 1 6 18. 2 2 2 2 4 4 2 2 1 1 7 5 7 2 14 19. 2 2 3 5 15 3 2 3 2
2 39 2 8 39 5 40 5 8 40 16 39 40 40 16 39 40 23 40 23 40
12.
13. 2
x
x
2 3 x 6 x 6 x
3 14. 5
2
x 2
1
5 1
3
1 3
20. 3 3 3 1 18 10 5 3 5 3 18 3 5 10 293 5 25 27 25 2 1 25
3
3
3
3
5 2 x 10 x 10 x 2 1 2 2 2 2 2 x
2
1 1 2 15. 2x y 2 21 13 2 2 9 1 1 9 9 1 9 9 8 9
21. (1.3)2 2.4 1.69 2.4 4.09 22. 1.2(7.3 9.3) = 1.2(2) = 2.4 23.
49 7 since 72 7 7 49.
24.
64 8 since 82 8 8 64.
25.
2 2 4 2 2 4 . since 25 5 5 5 25 5
26.
3 3 3 3 9 9 . since 100 10 10 10 10 100
2
2
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365
ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
27.
28.
1.2x 5.8 8.2 10(1.2x 5.8) 10(8.2) 12x 58 82 12x 58 58 82 58 12x 24 12x 24 12 12 x2
3
1.3x 2.6 9.1 10(1.3x 2.6) 10(9.1) 13x 26 91 13x 26 26 91 26 13x 65 13x 65 13 13 x 5
42.
30. Since 4 = 4, the statement 4 4 is true. 31. Since 8 = 8, the statement 8 8 is true. 32. Since 4 = 4, the statement 4 4 is true. 33. Since neither 23 < 0 nor 23 = 0 is true, the statement 23 0 is false. 34. Since 8 < 0 is true, the statement 8 0 is true. 35. Since 0 < 23 is true, the statement 0 23 is true. 36. Since neither 0 < 8 nor 0 = 8 is true, the statement 0 8 is false. 37. 5(3 2z) 5(3) (5)(2z) 15 10z 38. 4(2x 1) = 4(2x) (4)(1) = 8x + 4 1 2
1 (6x) (14) 10 2 2 3x 7 10 3x 17
(6x 14) 10
1
40. 9 2(5x 6) 9 2(5x) 2(6) 9 10x 12 10x 9 12 10x 21
366
6a 2
3 2(a 3)
3(6a 2) 3 2(a 3) 3(6a 2) 2a 6 18a 6 2a 6 18a 18a 6 18a 16a 6 6 16a 6 6 6 6 16a 0 16a 0 16 16 a0
29. Since 8 = 8, the statement 8 8 is true.
39.
2(a 3)
41.
x x 3 2 5 x x 10 10(3) 2 5 5x 2x 30 7x 30 7x 30 7 7 30 x 7
43. Let x represent the number of beetle species. Then x 4100 represents the number of fly species. The total number of these two species is 43,300. x x 4100 43, 300 2x 4100 43, 300 2x 4100 4100 43, 300 4100 2x 47, 400 2x 47, 400 2 2 x 23, 700 x 4100 = 23,700 4100 = 19,600 There are 23,700 beetle species and 19,600 fly species. 44. Let x be the number. 3(x 2) 9x 3x 6 9x 3x 3x 6 9x 3x 6 6x 6 6x 6 6 1 x The number is 1.
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ISM: Prealgebra and Introductory Algebra
Chapter 9: Equations, Inequalities, and Problem Solving
45. Use d = rt when d = 31,680 and r = 400. d rt 31, 680 400t 31, 680 400t 400 400 79.2 t It will take the ice 79.2 years to reach the lake. 46. V lwh V lwh lh lh V w lh
50.
47. 1 > x or x < 1 –5 –4 –3 –2 –1
48. 3x 30 3x 30 3 3 x 10 {x|x > 10}
49. 2(x 3) 5 3(x 2) 18 2x 6 5 3x 6 18 2x 11 3x 12 x 11 12 x 1 x 1 1 1 x 1 {x|x 1}
0
1
2
3
4
5
10 x 6x 10 10 x x 6x x 10 10 5x 10 10 10 5x 10 10 20 5x 20 5x 5 5 4 x {x|x > 4}
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367
Chapter 10 Section 10.1 Practice Exercises
15. (m5n10 )(mn8 ) (m5 m) (n10 n8 ) m6 n18
1. 34 3 3 3 3 81
m6n18
2. 71 7 3. (2)3 (2)(2)(2) 8
9 2 11 9 2 11 16. (x y)(4x y ) (1 4) (x x ) ( y y ) 4x11 y12
4. 23 (2 2 2) 8
4 10 410 940 17. (9 ) 9
2
2 2 2 4 5. 3 3 3 9
18. (z6 )3 z63 z18 7 7 7 7 7 19. (xy) x y x y
6. (0.3)4 (0.3)(0.3)(0.3)(0.3) 0.0081 20. (3y)4 34 y4 81y4 7. 5 6 5 36 180 2
8. a.
4 2 3 3 4 3 2 3 1 3 21. (2 p q r) (2) ( p ) (q ) (r )
When x is 4, 3x2 3 42 3(4 4) 316 48.
b. When x is 2,
8 p12q6r3 22. (a4b)7 (1a4b)7 (1)7 (a4 )7 (b1)7
x4 (2)4 8 8 (2)(2)(2)(2) 8 16 8 2.
1a28b7 a28b7 6
2
2 6 2 12 24. 5x6 5 (x ) 25x , y
9. 73 72 732 75 10. x4 x9 x49 x13
11. r5 r r51 r6
9 91 (3)10 13. (3) (3) (3)
14. (6x3 )(2x9 ) (6 x3 ) (2 x9 ) (6 2) (x3 x9 )
y3
92 ( y3 )2
96
5 56 14 (2)
27. 10
3
5 125 1410
(2)
(2)
12x12
368
81y6
y73 y4
59 26.
9 y3 y7
25. 12. s6 s2 s3 s623 s11
r r6 23. , s 0 s s6
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4
(2)
16
0
ISM: Prealgebra and Introductory Algebra
28.
7a4b11 a4 b11 7 1 1 ab a b 7 (a41) (b111)
Chapter 10: Exponents and Polynomials
10. In 5 32, the base for the exponent 2 is 3. 11. In 5x2, the base for the exponent 2 is x.
7a3b10 12. In (5x)2, the base for the exponent 2 is 5x. 0
29. 8 1
13. Video Notebook Number 4 can be written as
30. (2r s) 1
42 1 42 , which is similar to Video
31. (7) 1
Notebook Number 7, 4 32, and shows why the negative sign should not be considered part of the base when there are no parentheses.
2 0
0
32. 70 1 70 11 1
14. The properties allow us to reorder and regroup factors and put factors with common bases together, making it easier to apply the product rule.
33. 7 y0 7 y0 7 1 7 34.
x7 x4
x74 x3
15. Be careful not to confuse the power rule with the product rule. The power rule involves a power raised to a power (exponents are multiplied), and the product rule involves a product (exponents are added).
4 4 4 4 4 16 35. (3y ) 3 ( y ) 81y 3
x3 x3 x 36. 4 43 64
16. Remember to raise the 2 (or any number) to the power along with the variables.
Vocabulary, Readiness & Video Check 10.1 1. Repeated multiplication of the same factor can be written using an exponent. 2. In 52, the 2 is called the exponent and the 5 is called the base. 3. To simplify x2 x7 , keep the base and add the exponents. 4. To simplify (x 3 ) 6 , keep the base and multiply the exponents. 5. The understood exponent on the term y is 1. 6.
17. the quotient rule 18. No, Video Notebook Number 30 is a fraction and the quotient rule is not needed to simplify. Exercise Set 10.1 2. 32 3 3 9 4. (3)2 (3)(3) 9 6. 43 (4 4 4) 64 8. (4)3 (4)(4)(4) 64 2
1, the exponent is 0.
1 1 1 1 9 9 81 10. 9
7. In 32 , the base for the exponent 2 is 3.
12. 9 22 9(2 2) 9 4 36
8. In (3)2, the base for the exponent 2 is 3.
14. When x is 2, x3 (2)3 (2)(2)(2) 8.
9. In 42, the base for the exponent 2 is 4.
16. When x is 5,
If x
4x2 4(5)2 4(5)(5) 4(25) 100. Copyright © 2025 Pearson Education, Inc.
369
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials 18. When x = 2 and y = 1,
11
4x2 y3 4(2)2 (1)3 4(2)(2)(1)(1)(1) 4(4)(1) 16.
20. When y is 3, 10 10 10 10 10 . 3 3 3(3)(3)(3) 3(27) 81 3y 3(3) 22. y2 y y21 y3
2
xy x2 y2 x2 y2 48. 2 7 49 7 3
xy4 3 43 3 12 x y x y 50. 3 3 3 3 9 27z 3z (3) (z ) 52. Area (radius)2 (5 y centimeters)2 (5 y centimeters)(5y centimeters) =(5)2 ( y)2 square centimeters
24. (5)7 (5)6 (5)76 (5)13 3
q q11 46. 11 t t
2
3
2
26. (2z )(2z ) (2)(2)(z z ) 4z
32
4z
5
25 y12 square centimeters 25 y2 square centimeters
28. (a2b)(a13b17 ) (a2 a13 )(b b17 ) (a213 )(b117 )
54. Area (radius)2 (height)
a15b18
(4x meters)2 (5x3 meters) (4x meters)(4 x meters)(5x3 meters)
3 3 19 3 19 3 30. (7a b )(7a b) (7 7)(a a )(b b) 49a319b31
(4 4 5)(x x x3) cubic meters 80 x113 cubic meters
49a22b4
80 x5 cubic meters 5
6
4
5
6
4
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or 80x5 cubic meters y10 56.
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10
(9 2)( y y ) square meters 18 y710 square meters 18 y17 square meters 36. ( y7 )5 y75 y35 6
6
6
6 6
38. (ab) a b a b
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x8 y6 60. xy5
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Copyright © 2025 Pearson Education, Inc.
x
36
65
7 1
7
x y x y
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials 4
68. 2x0 2 x0 2 1 2
5x12 y13
70. 30 40 1 1 0
104.
72. (0.4)2 (0.4)(0.4) 0.16
x5 y 7
x12
y13
125 137
5 5 7 5x x y
y
7 6
5x y
106. 9 12 = 9 + (12) = 3
3
2 2 2 8 2 74. 3 3 3 3 27
108. 5 (10) = 5 + 10 = 15 110. 15 (21) = 15 + 21 = 6
76. y4 y y41 y5
112. The expression x14 x23 can be simplified by adding the exponents; a.
78. x2 x15 x9 x2159 x26 4
4
2ab ab 4a4b44 4 a4b44 4 3yz 3 y z 81y z 102. 6 yz
66. (4 y)0 1
4
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41
15 y
82. ( y2 z2 )( y15 z13 ) ( y2 y15 )(z2 z13 ) y215 z213 y17 z15
5
114. The expression
can be simplified by x17 subtracting the exponents; b.
116. answers may vary 118. answers may vary
84. (3s5t)(7st10 ) (3)(7)(s5 s)(t t10 ) 21s51t110
120. s 6x2 6(5 meters)2 6(5)2 square meters 6(25) square meters 150 square meters
21s6t11 5 11 511 t55 86. (t ) t
88. (2ab)4 (2)4 a4b4 16a4b4 90. (3xy2a3 )3 (3)3 x3 ( y2 )3 (a3 )3 27x3 y23a33
x35
122. The surface area of a cube measures the amount of material needed to cover a cube, so to find the amount of material needed to cover an ottoman, the surface area of the cube should be used. 124. answers may vary
3 6 9
27x y a
92.
b6 63 b3 b b3 x9
5x9 94.
126. answers may vary.
x
5
x
128. b9ab4a b9a4a b13a 91
5x
8
4b 4 4 4b 4 4b4 16a16b 130. (2a ) (2 )(a ) 16a
5x
0
96. (5ab) 1
132.
y15b y6b
y15b6b y9b
Section 10.2 Practice Exercises 5 5 5 5 5 5 98. (2ab) 2 a b 32a b
100. 7 7 49 1 48 2
0
1. 53
1 53
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1 125
371
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials 4
71
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b.
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Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
d. 20. a.
1
6.002 106 6, 002, 000
7.
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b.
Chapter 10: Exponents and Polynomials
8 8104 4(3) 10 2 103 2
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8.
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2
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11. A negative exponent has nothing to do with the sign of the simplified result. 12. power of a product rule, power rule for exponents, negative exponent definition, quotient rule for exponents 13. When you move the decimal point to the left, the sign of the exponent will be positive; when you move the decimal point to the right, the sign of the exponent will be negative. 14. the exponent on 10
6. 230, 000 1000 2.310
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7. (3.26 106 )(2.51013 ) 8.151019 8. (8.76 104 )(1.237 109 ) 1.083612 106
Exercise Set 10.2 2. 62
1
62
1 36
Vocabulary, Readiness & Video Check 10.2 1. The expression x3 equals
, choice b. x3 1
2. The expression 54 equals 625 , choice d. 3. The number 3.021103 is written in scientific notation, choice a. 4. The number 0.0261 is written in standard form, choice a. 1 5 5. 5x2 5 2 2 x x 6. 3x3 3
1 x3
2
4.
1
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2
2
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3
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373
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
10.
1
q5
q5
32.
p p3q5
r5 12.
14.
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1
x1
p
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2 32 22
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25x6
6 4
25x y
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ISM: Prealgebra and Introductory Algebra
1 5 2
50. (4 x )
52.
a5b
4
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x
3
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42
2 10
4 x
Chapter 10: Exponents and Polynomials
x10
15r6 s
16
60. x10
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64. (34 )(70 )(2) (81)(1)(2) 162 (rs)3
66.
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5
1
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r s
r34 s36
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r3s3 (r 2 s3 )2 (r 2 )2 (s3 )2 3 3
232 x41 25 x5 1 5 5 2 x 1 5 32x 5 1 2
s
5 r4 s3 3r6(4) s1(3) 3r64 s13
a15b3
68.
(3x2 y2 )2 (3)2 (x2 )2 ( y2 )2 (xyz)2 x24 y24 z2 2 (3) x y x2 y2 z2 (3)2 x4(2) y4(2) z2 2 42 42 2
3 2
(3) x y (3)2 x2 y2 z2
6 y x3
216 y2
Copyright © 2025 Pearson Education, Inc.
z
z2
2 2 2
(3) x y z2 9x2 y2
375
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials (a6 )4 (b2 )4
(a6b2 )4 70.
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90. 6.02214199 1023
602, 214,199, 000, 000, 000, 000, 000
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98. (2.5106 )(2 106 ) (2.5 2)(106 106 ) 5 1066
1 72. A 2 bh 3 4 1 5x 2 7 meters meters x 3 5x 4 1 = square meters 2 7 x 20x3 square meters 14xx 3 20 square meters 14 x 10 x31 square meters 7 10 4 x square meters 7 10 4 square meters 7x
5 100 5 1 5
102.
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9.460 1016 108. 7w + w 2w = 6w 110. 6z + 20 3z = 6z 3z + 20 = 9z + 20
86. 4.8 106 0.0000048
112. 10y 14 y 14 = 10y y 14 14 = 9y 28
88. 9.07 1010 90, 700, 000, 000
114. 8.5 billion 8, 500, 000, 000 8.5109 116. 155.557 million 155, 557, 000 1.55557 108
376
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ISM: Prealgebra and Introductory Algebra
118. 7 micrometers 7 106 meter
Chapter 10: Exponents and Polynomials
Section 10.3 Practice Exercises
120. 28 micrometers 28106 meter
1.
6x6 4x5 7 x3 9x2 1
2.8105 meter 0.000028 meter 122.
Term
Coefficient
7 x3
7
8a9a3 a11a5
9x2
9
8a6 a6
6x6
6
4x5
4
1
1
(2a3 )3 a3 a11a5 23 (a3 )3 a3 a11a5
6
9a 124. If 7
1
1
, the exponent is 2. 49 72
2. 15x3 2x2 5
126. answers may vary 128. a.
The term 15x3 has degree 3.
1 5 1 52 25 1 1 Since , the statement “ 51 52 ” is 5 25 false. 51
The term 2x2 has degree 2. The term 5 has degree 0 since 5 is 5x0. 3. a. b.
The degree of the polynomial 9x 3x6 5x4 2 is 6. The polynomial is
neither a monomial, binomial, or trinomial.
1
1 1 b. 1 5 5 5 2 1 1 2 25 5 5 52 Since 5 < 25, the statement 1 2 1 1 “ ” is true. 5 5 c.
The degree of the binomial 6x + 14 is 1.
c.
4. a. b.
From part a, the statement “ a1 a2 for all nonzero numbers” is false.
The degree of the trinomial 10x2 6x 6 is 2. When x = 1, 2x + 10 = 2(1) + 10 = 2 + 10 = 12. When x = 1, 6x2 11x 20 6(1)2 11(1) 20 6 11 20 25.
5. When t = 2 seconds, 130. answers may vary 132. a4m a5m a4m5m am 2z 3 3 2z 3 6z 134. (3y ) 3 ( y ) 27 y
16t2 592.1 16(2)2 592.1 16(4) 592.1 64 592.1 528.1 feet. When t = 4 seconds, 16t2 592.1 16(4)2 592.1 16(16) 592.1 256 592.1 336.1 feet. 6. 6y + 8y = (6 + 8)y = 2y Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
7. 14 y2 3 10 y2 9 14 y2 10 y2 3 9
c.
3a3 a4 a4 3a3 0a2 0a1 0a0 a4 3a3 0a2 0a 0a0
4 y2 6
Vocabulary, Readiness & Video Check 10.3
8. 7 x3 x3 7 x3 1x3 8x3 9. 23x2 6x x 15 23x2 7 x 15 10.
2 3 1 1 3 x x 2 x3 x 7 4 2 8 2 1 1 3 x3 x3 x x 2 7 2 4 8 4 7 2 3 x3 x3 x x 2 14 14 8 8 3 3 1 x x2 14 8
2. A monomial is a polynomial with exactly one term. 3. A trinomial is a polynomial with exactly three terms. 4. The numerical factor of a term is called the coefficient. 5. A number term is also called a constant.
11. Area 5 x x x 4 5 x x 8 x 5x x2 20 x2 8x x2 x2 5x 8x 20
6. The degree of a polynomial is the greatest degree of any term of the polynomial. 7. 3; x2, 3x, 5
2x2 13x 20 12.
1. A binomial is a polynomial with exactly two terms.
Terms of Polynomial
Degree of Term
8. The degree of the polynomial is the greatest degree of any of its terms, so we need to find the degree of each term first.
2x3 y2
3 + 2 or 5
9. the replacement value for the variable
4
0
8xy
1 + 1 or 2
3
3x y
3 + 1 or 4
5xy 2
1 + 2 or 3
10. simplifying it 11. 2; 9ab 12. 2; 0 y2
The degree of the polynomial is 5. 13. 11ab 6a2 ba 8b2 (11 1)ab 6a2 8b2 10ab 6a2 8b2 14. 7x2 y2 2 y2 4 y2 x2 x2 y2 5x2 7x2 y2 4x2 y2 2 y2 y2 x2 5x2
Exercise Set 10.3 2.
Term
Coefficient
2x3
2
x
1
4
4
3x2 y2 y2 6x2 15. a.
x2 9 x2 0x1 9 or x2 0x 9
b.
9m3 m2 5 9m3 m2 0m1 5 9m3 m2 0m 5
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ISM: Prealgebra and Introductory Algebra
4.
Term
Coefficient
9.7 x7
9.7
3x5
3
x3
1
1 x2
14
4
8. a 5a2 3a3 4a4 This is a polynomial of degree 4. None of these. 10. 7r 2 2r 3r5 3r5 7r 2 2r This is a trinomial of degree 5. 12. 5 y6 2 This is a binomial of degree 6.
b.
18. a.
b.
2x 10 = 2(0) 10 = 0 10 = 10
x2 3x 4 (0)2 3(0) 4 0 0 4 4 x2 3x 4 (1)2 3(1) 4 1 3 4 6 2x3 3x2 6 2(0)3 3(0)2 6 006 6 2x3 3x2 6 2(1)3 3(1)2 6 2(1) 3(1) 6 2 36 1 2
2
2
26. 16t 2650 16(12) 2650 16(144) 2650 2304 2650 346 After 12 seconds, the height of the object is 346 feet. 28. 2025 corresponds to x = 25.
b. 2x 10 = 2(1) 10 = 2 10 = 12 16. a.
2 2 22. 16t 200t 16(10.3) 200(10.3) 16(106.09) 2060 1697.44 2060 362.56 After 10.3 seconds, the height of the rocket is 362.56 feet.
24. 16t 2 2650 16(7)2 2650 16(49) 2650 704 2650 1866 After 7 seconds, the height of the object is 1866 feet.
6. 6 y 4 6 y1 4 This is a binomial of degree 1.
14. a.
Chapter 10: Exponents and Polynomials
4.3x2 138x 291 4.3(25)2 138(25) 291 4.3(625) 138(25) 291 2687.5 3450 291 6428.5 The polynomial predicts 6428.5 million Internet users in the world in 2025. 30. 2025 corresponds to x = 10. 5x2 191x 2839 5(10)2 191(10) 2839 5(100) 191(10) 2839 500 1910 2839 4249 The polynomial predicts 4249 thousand visitors to the park in 2025. 32. 14y 30y = (14 30)y = 16y 34. 18x3 4x3 (18 4)x3 22x3 36. 8x2 4 11x2 20 8x2 11x2 4 20
2
20. 16t 200t 16(5) 200(5) 16(25) 1000 400 1000 600 After 5 seconds, the height of the rocket is 600 feet.
(8 11)x2 4 20 19x2 16 38. 12k3 9k3 11 (12 9)k3 11 3k3 11 40. 5y + 7y 6y = (5 + 7 6)y = 6y
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Chapter 10: Exponents and Polynomials
54. 2a2b 2a2b1 has degree 2 + 1 = 3.
42. 7.6 y 3.2 y2 8 y 2.5 y2 3.2 y2 2.5 y2 7.6 y 8 y
10a4b 10a4b1 has degree 4 + 1 = 5.
0.7 y2 0.4 y
9ab 9a1b1 has degree 1 + 1 = 2.
2 4 1 x 23x2 x4 5x2 5 44. 5 15 2 4 1 4 x x 23x2 5x2 5 5 15 2 1 4 x (23 5)x2 5 5 15 6 1 4 x (23 5)x2 5 15 15 7 x4 18x2 5 15 46.
ISM: Prealgebra and Introductory Algebra
5 3 1 3 1 9 x x x 14x2 16 5 8 8 4 16 1 1 9 3 2 3 x 14x x x 16 5 3 9 81 42 8 6 16 x 14x2 x 16 16 16 8 8 5 3 3 1 x 14x2 x 16 16 8
6 6a0b0 has degree 0. 2a2b 10a4b 9ab 6 is a polynomial of
degree 5. 56. 9xy 7 y xy 6 y 9xy xy 7 y 6 y (9 1)xy (7 6) y 10xy y 58. 3a2 9ab 4b2 7ab 3a2 9ab 7ab 4b2 3a2 (9 7)ab 4b2 3a2 16ab 4b2 2 2 3 3 2 60. 17a b 16ab 3a 4ba b a 2 2 2 3 17a b 16ab ab 3a 4ba3
17a22b (1621)ab32 3a33 4ba3 17a b 17ab 3a 4ba 62. 18x4 2x3 y3 1 2 y3 x3 17x4 18x4 17x4 2x3 y3 2x3 y3 1 (18 17)x4 (2 2)x3 y3 1
48. 7x 4x2 4x2 18 16
x4 0x3 y3 1
4x2 4x2 7x 18 16
x4 1
(4 4)x2 7x 18 16 8x2 7x 34
64. 5x2 2 5x2 0x 2
50. 9x + 10 + 3x + 12 + 4x + 15 + 2x + 7 = 9x + 3x + 4x + 2x + 10 + 12 + 15 + 7 = (9 + 3 + 4 + 2)x + 10 + 12 + 15 + 7 = 18x + 44 52. y 4 has degree 4.
66. x3 8 x3 0x2 0x 8 68. 6m3 3m 4 6m3 0m2 3m 4 70. 11z 4z 4 4z4 0z3 0z 2 11z 0
6 y3 x 6 y3 x1 has degree 3 + 1 = 4. 2x2 y2 has degree 2 + 2 = 4.
72. 9 y5 y2 2 y 11 9 y5 0 y4 0 y3 y2 2 y 11
5 y2 has degree 2. 3 3x0 y0 has degree 0.
74. 9 6(5x + 1) = 9 30x 6 = 30x + 3
y4 6 y3 x 2x2 y2 5 y2 3 is a polynomial of degree 4.
76. 3(w 7) 5(w 1) 3w 21 5w 5 3w 5w 21 5 (3 5)w 21 5 2w 16 78. answers may vary 80. answers may vary
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Chapter 10: Exponents and Polynomials
82. x4 x9 x4 x9 84. a b3 a2 b7 a b3 a2 b7 86. x5 y2 y2 x5 x5 y2 x5 y2 2x5 y2 or 2 y2 x5 88. a. b.
The height of the bar for 2030 is about 395. The graph predicts that the size of the market for wearable technology in 2030 will be $395 billion or $395,000,000,000. 2030 corresponds to x = 6. 2.2x2 21.8x 180 2.2(6)2 21.8(6) 180 2.2(36) 130.8 180 79.2 130.8 180 390 The polynomial model predicts that the size of the market for wearable technology in 2030 will be $390 billion or $390,000,000,000.
c.
Yes, the answers are about the same, as they should be; answers may vary.
90. 2025 corresponds to x = 8. 7.2x2 78.6x 522 7.2(8)2 78.6(8) 522 7.2(64) 628.8 522 460.8 628.8 522 1611.6 The number of connected wearable devices worldwide in 2025 is predicted to be 1611.6 million or 1,611,600,000. 92. answers may vary 94. 16t 2 200t 0 when t 12.5 seconds. 96. 7.75x 9.16x2 1.27 14.58x2 18.34 9.16x2 14.58x2 7.75x 1.27 18.34 (9.16 14.58)x2 7.75x 1.27 18.34 5.42x2 7.75x 19.61 Section 10.4 Practice Exercises 1. (3x5 7x3 2x 1) (3x3 2x) 3x5 7x3 2x 1 3x3 2x 3x5 (7x3 3x3 ) (2x 2x) 1 3x5 4x3 1 2. (5x2 2x 1) (6x2 x 1) 5x2 2x 1 6x2 x 1 (5x2 6x2 ) (2x x) (11) x2 x 3.
9 y2 6 y 5 4y 3 9 y2 2 y 8
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Chapter 10: Exponents and Polynomials 4. (9x 5) (4x 3) (9x 5) [(4x 3)] (9x 5) (4x 3) 9x 5 4x 3 5x 8 5. (4x3 10x2 1) (4x3 x2 11) (4x3 10x2 1) (4x3 x2 11) 4x3 10x2 1 4x3 x2 11 4x3 4x3 10x2 x2 111 8x3 11x2 12 6.
2 y2 2 y 7 (6 y2 3 y 2)
2 y2 2 y 7 6 y2 3y 2 4 y2 y 5
7. [(4x 3) (12x 5)] (3x 1) 4x 3 12x 5 3x 1 4x 12x 3x 3 5 1 13x 9 8. (2a2 ab 6b2 ) (3a2 ab 7b2 ) 2a2 ab 6b2 3a2 ab 7b2 a2 b2 9. (5x2 y2 3 9x2 y y2 ) (x2 y2 7 8xy2 2 y2 ) 5x2 y2 3 9x2 y y2 x2 y2 7 8xy2 2 y2 6x2 y2 4 9x2 y 8xy2 y2 Vocabulary, Readiness & Video Check 10.4 1. 9y 5y = 14y 2. 6m5 7m5 13m5 3. x + 6x = 7x 4. 7z z = 6z 5. 5m2 2m 5m2 2m 6. 8 p3 3 p2 8 p3 3 p2 7. 3y2 and 2 y2 ; 4y and y 8. Video Notebook Number 2 is a subtraction example, so the signs of the polynomial being subtracted must change when parentheses are removed. 9. We’re translating to a subtraction problem. Order matters when subtracting, so we need to be careful that the order of the expressions is correct. 10. because the operation is addition
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Exercise Set 10.4
Chapter 10: Exponents and Polynomials
20. (7 y2 5) (8 y2 12)
2. ( y 2) (3y 5) y 2 3y 5 y 3y 2 5 2y 3 4. (3x 8) (4x2 3x 3) 3x 8 4x2 3x 3 4x2 3x 3x 8 3 2
4x 5 6. (3x2 7) (3x2 9) 3x2 7 3x2 9 3x2 3x2 7 9 6x2 16
(7 y2 5) (8 y2 12) 7 y2 5 8 y2 12 7 y2 8 y2 5 12 y2 7 22. 4 ( y 4) 4 ( y 4) 4 y4 y44 y 8 24. (7x2 4x 7) (8x 2) (7x2 4x 7) (8x 2)
8. (7x2 2x 9) (3x2 5)
7x2 4x 7 8x 2
7x2 2x 9 3x2 5
7x2 4x 8x 7 2
7x2 3x2 2x 9 5
7x2 12x 5
4x2 2x 4
2 2 26. (6 y 3y 4) (9 y 3y)
10. (9.6 y3 2.7 y2 8.6) (1.1y3 8.8 y 11.6)
(6 y2 3y 4) (9 y2 3y)
9.6 y3 2.7 y2 8.6 1.1y3 8.8 y 11.6
6 y2 3y 4 9 y2 3y
9.6 y3 1.1y3 2.7 y2 8.8 y 8.6 11.6
6 y2 9 y2 3y 3y 4
10.7 y3 2.7 y2 8.8 y 3
15 y2 6 y 4
5 1 3 5 3 4 12. 7 n2 6 m 20 7 n2 12 m 10 4 2 5 1 3 2 5 3 n m n m 7 6 20 7 12 10 4 2 3 2 5 5 1 3 n n m m 7 7 6 12 20 10 4 2 3 2 10 5 1 6 n n m m 7 7 12 12 20 20 1 2 5 7 n m 7 12 20 14. 7x3 3 2x 3 7
28. (0.3y2 0.6 y 0.3) (0.5 y2 0.3) 2
(0.3y2 0.6 y 0.3) (0.5 y 0.3) 0.3y2 0.6 y 0.3 0.5 y2 0.3 0.3y2 0.5 y2 0.6 y 0.3 0.3 0.8 y2 0.6 y 0.6 1 2 2 4 2 1 x x x x 2 30. 3 3 7 21 21 2 2 4 1 1 x2 x x2 x 3 7 21 21 3 1 2 2 4 2 1 2 x x x x 3 7 21 21 3 1 4 2 1 2 x2 x2 x x 3 21 7 21 32 7 2 4 2 6 1 x x x x 3 21 21 21 21 3 2 7 2 x x 21 21 3 1 2 1 2 x x 7 3 3
9x3 4 16. 2x3 3x2 x 4 5x3 2x2 3x 2 7x3 x2 2x 2 18. (4 5a) (a 5) (4 5a) (a 5) 4 5a a 5 5a a 4 5 6a 9
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7a2 9a 6 (11a2 4a 2)
32.
34.
7a2 9a 6 11a2 4a 2 4a2 5a 4
5x3 4x2 6x 2 5x3 4x2 6x 2 (3x3 2x2 x 4) 3x3 2x2 x 4 2x3 2x2 7x 2
36. (2 y 20) (5 y 30) 2 y 20 5 y 30 2 y 5 y 20 30 7 y 10 38. (7 y 7) ( y 6) (7 y 7) ( y 6) 7y 7 y 6 6 y 13 40. (26 y 17) (20 y 10) 26 y 17 20 y 10 6y 7 42. (5 y2 3 y 1) (2 y2 y 1) (5 y2 3y 1) (2 y2 y 1) 5 y2 3 y 1 2 y2 y 1 3y2 4 y 2 44. (2x2 7x 9) (x2 x 10) (3x2 30) (2x2 7x 9) (x2 x 10) (3x2 30) 2x2 7x 9 x2 x 10 3x2 30 6x 31 46. (m2 3) (m2 13) (6m2 m 1) (m2 3) (m2 13) (6m2 m 1) m2 3 m2 13 6m2 m 1 4m2 m 17 48. ( y2 4 y 1) y y2 4 y 1 y y2 5 y 1 50. (3x2 5x 2) (x2 6x 9) 3x2 5x 2 x2 6x 9 2x2 11x 11 52. (7 y2 9 y 8) (5 y2 8 y 2) (7 y2 9 y 8) (5 y2 8 y 2) 7 y2 9 y 8 5 y2 8 y 2 2 y2 y 10 54. (x2 7x 1) (7x 5) (4x2 2x 2) (x2 7x 1) (7x 5) (4x2 2x 2) x2 7x 1 7x 5 4x2 2x 2 3x2 16x 4
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Chapter 10: Exponents and Polynomials
56. (3y2 y 4) (2 y2 6 y 10) ( y2 9) (3y2 y 4) (2 y2 6 y 10) ( y2 9) 3 y2 y 4 2 y2 6 y 10 y2 9 4 y2 5 y 3 58. (3x 2 6 y) (7x 2 y) 3x 2 6 y 7x 2 y 10x 4 5 y 60.
(7a2 3b2 10) (2a2 b2 12) 7a2 3b2 10 2a2 b2 12 9a2 4b2 22
62. (a2 ab 4b2 ) (6a2 8ab b2 ) a2 ab 4b2 6a2 8ab b2 7a2 7ab 3b2 64. (3x2 y 6xy x2 y2 5) (11x2 y2 1 5 yx2 ) 3x2 y 6xy x2 y2 5 11x2 y2 1 5x2 y 2x2 y 6xy 10x2 y2 4 66. (x 4) (5x) (x2 6x 2) (x2 ) x 4 5x x2 6x 2 x2 2x2 2x 2 The perimeter is (2x2 2x 2) centimeters. 3 7 3 3 7 3 68. (3y 4) (3y 1) y y y (2 y 3) 3y 4 3y 1 2y 4y 2y 2 y 3 2 4 2 51 y6 4 51 The perimeter is 4 y 6 units. 70. (13x 7) (2x 2) (13x 7) (2x 2) 13x 7 2x 2 11x 9 The remaining piece is (11x 9) inches long. 72.
[(7.9 y4 6.8 y3 3.3y) (6.1y3 5)] (4.2 y4 1.1y 1) (7.9 y4 6.8 y3 3.3y) (6.1y3 5) (4.2 y4 1.1y 1) 7.9 y4 6.8 y3 3.3y 6.1y3 5 4.2 y4 1.1y 1 3.7 y4 0.7 y3 2.2 y 4
74. 7x(x) (7 1)(x x) 7x2 76. 6r3 (7r10 ) (6 7)(r3 r10 ) 42r13 78. z2 y(11zy) (111)( y y)(z2 z) 11y2 z3 80. Since 9 + 3 = 12, 9 y7 3y7 12 y7 is a true statement.
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ISM: Prealgebra and Introductory Algebra
82. Since 3 + 7 = 10 and 2 1 = 3, 3y5 7 y5 2 y2 y2 10 y5 3y2 is a true statement. 84. 5x + 5x = 10x; c 86. (15x 3) (5x 3) = 15x 3 5x + 3 = 10x; c 88. a.
2y + y = 2y + 1y = 3y
b.
2 y y 2 y1 y1 2 y11 2 y2
c.
2y y = 2y 1y = 3y
d.
(2 y)( y) (2 y1)(1 y1) 2 y11 2 y2 ; answers may vary
90. a.
x + x = 1x + 1x = 2x
b.
x x x1 x1 x11 x2
c.
x x = 1x 1x = 2x
d.
(x)(x) (x1)(x1) x11 x2 ; answers may vary
92. (4x2 17x 149) (6x2 13x 342) 4x2 17x 149 6x2 13x 342 2x2 4x 491 Section 10.5 Practice Exercises 1. 10x 9x (10 9)(x x) 90x2 2. 8x3 (11x7 ) (8 11)(x3 x7 ) 88x10 3. (5x4 )(x) (5 1)(x4 x) 5x5 4.
4x(x2 4x 3) 4x(x2 ) 4x(4x) 4x(3) 4x3 16x2 12x
5. 8x(7x4 1) 8x(7x4 ) 8x(1) 56x5 8x 6.
2x3 (3x2 x 2) 2x3 (3x2 ) 2x3 (x) 2x3 (2) 6x5 2x4 4x3
7. a.
(x 5)(x 10) x(x 10) 5(x 10) x x x 10 5 x 5 10 x2 10x 5x 50 x2 15x 50
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b.
Chapter 10: Exponents and Polynomials
(4x 5)(3x 4) 4x(3x 4) 5(3x 4) 4x(3x) 4x(4) 5(3x) 5(4) 12x2 16x 15x 20 12x2 x 20
8. (3x 2 y)2 (3x 2 y)(3x 2 y) 3x(3x) 3x(2 y) (2 y)(3x) (2 y)(2 y) 9x2 6xy 6xy 4 y2 9x2 12xy 4 y2 9. (x 3)(2x2 5x 4) x(2x2 ) x(5x) x(4) 3(2x2 ) 3(5x) 3(4) 2x3 5x2 4x 6x2 15x 12 2x3 x2 11x 12 10.
y2 4 y 5 3y2 1 y2 4 y 5 4 3 3y 12 y 15 y2 3y4 12 y3 16 y2 4 y 5
11.
4x2 x 1 3x2 6x 2
8x2 2x 2 24x3 6x2 6x 4 12x 3x3 3x2 12x4 21x3 17x2 4x 2 Vocabulary, Readiness & Video Check 10.5 1. The expression 5x(3x + 2) equals 5x 3x + 5x 2 by the distributive property, choice c. 2. The expression (x + 4)(7x 1) equals x(7x 1) + 4(7x 1) by the distributive property, choice c. 3. The expression (5 y 1)2 equals (5y 1)(5y 1), choice c. 4. The expression 9x 3x equals 27x2 , choice b. 5. x3 x5 x35 x8 6. x2 x6 x26 x8 7. x3 x5 cannot be simplified. 8. x2 x6 cannot be simplified. 9. x7 x7 x77 x14 Copyright © 2025 Pearson Education, Inc.
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Chapter 10: Exponents and Polynomials
10. x11 x11 x1111 x22
22. 9x3 (5x2 12) 9x3 (5x2 ) 9x3 (12) 45x5 108x3
7 7 7 7 7 7 11. x x 1x 1x (11)x 2x 11
12. x
11
x
11
1x
11
1x
(11)x
11
24. 4b2 (3b3 12b2 6) 4b2 (3b3 ) 4b2 (12b2 ) 4b2 (6) 12b5 48b4 24b2
11
2x
13. No; the monomials are unlike terms. 2 3 2 2 3 26. 4xy (7x 3x y 9 y ) 4xy2 (7x3 ) 4xy2 (3x2 y2 ) 4xy2 (9 y3 ) 28x4 y2 12x3 y4 36xy5
14. distributive property, product rule 15. Three times: first (a 2) is distributed to a and 7, and then a is distributed to (a 2) and 7 is distributed to (a 2).
3 2 2 3 28. x(6 y 5xy x y 5x ) x(6 y3 ) x(5xy2 ) x(x2 y) x(5x3 ) 6xy3 5x2 y2 x3 y 5x4
16. Yes; the parentheses have been removed for the vertical format, but every term in the first polynomial is still distributed to every term in the second polynomial.
30.
Exercise Set 10.5 2. 6x 3x2 (6 3)(x x2 ) 18x3 4. (x6 )(x) (1 1)(x6 x) x7
32. (x 2)(x 9) x(x 9) 2(x 9) x(x) x(9) 2(x) 2(9) x2 9x 2x 18 x2 11x 18
6. 9t6 (3t5 ) (9 3)(t6 t5 ) 27t11 8. (5.2x4 )(3x4 ) (5.2 3)(x4 x4 ) 15.6x8
3 7 1 4 3 1 7 4 10. y y ( y y ) 4 7 4 7 3 y11 28
34. ( y 10)( y 11) y( y 11) (10)( y 11) y( y) y(11) (10)( y) (10)(11)
12. (x)(5x4 )(6x7 ) (1 5 6)(x x4 x7 ) 30x12
3 2 36. x x 5 5 2 3 2 x x x 5 25 3 5 3 2 x(x) x (x) 5 5 5 5 3 6 2 2 x x x 5 5 25 6 2 1 x x 5 25
y2 11y 10 y 110 y2 y 110
14. 2x(6x 3) 2x(6x) 2x(3) 12x2 6x
1 2 y (9 y2 6 y 1) 3 1 1 1 y2 (9 y2 ) y2 (6 y) y2 (1) 3 3 3 4 3 1 2 3y 2 y y 3
2 2 16. 5 y( y y 10) 5 y(3 y ) 25 y( y) 5 y(10) 5 y 5 y 50 y
18. 3a(2a 7) 3a(2a) (3a)(7) 6a2 21a 20. 4x(5x2 6x 10) 4x(5x2 ) 4x(6x) 4x(10) 20x3 24x2 40x
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Chapter 10: Exponents and Polynomials
38. (5x2 2)(6x2 2) 5x2 (6x2 2) 2(6x2 2) 5x2 (6x2 ) 5x2 (2) 2(6x2 ) 2(2) 30x4 10x2 12x2 4 30x4 22x2 4 40. (8x 3)(2x 4) 8x(2x 4) (3)(2x 4) 8x(2x) 8x(4) (3)(2x) (3)(4) 16x2 32x 6x 12 16x2 38x 12 42. (3 2a)(2 a) 3(2 a) (2a)(2 a) 3(2) 3(a) (2a)(2) (2a)(a) 6 3a 4a 2a2 6 7a 2a2 44. (6x 7)2 (6x 7)(6x 7) 6x(6x 7) (7)(6x 7) 6x(6x) 6x(7) (7)(6x) (7)(7) 36x2 42x 42x 49 36x2 84x 49 46. (x 3)(x2 5x 8) x(x2 5x 8) 3(x2 5x 8) x(x2 ) x(5x) x(8) 3(x2 ) 3(5x) 3(8) x3 5x2 8x 3x2 15x 24 x3 8x2 7x 24 48. (a 2)(a3 3a2 7) a(a3 3a2 7) 2(a3 3a2 7) a(a3 ) a(3a2 ) a(7) 2(a3 ) 2(3a2 ) 2(7) a4 3a3 7a 2a3 6a2 14 a4 a3 6a2 7a 14 50. (3 b)(2 5b 3b2 ) 3(2 5b 3b2 ) b(2 5b 3b2 ) 3(2) 3(5b) 3(3b2 ) b(2) b(5b) b(3b2 ) 6 15b 9b2 2b 5b2 3b3 3b3 14b2 13b 6 52. (x2 4)2 (x2 4)(x2 4) x2 (x2 4) (4)(x2 4) x2 (x2 ) x2 (4) (4)(x2 ) (4)(4) x4 4x2 4x2 16 x4 8x2 16
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Chapter 10: Exponents and Polynomials
54.
56.
4x 7 5x 1 4x 7 20x2 35x 20x2 31x 7
70. (t 3)(t 2 5t 5) t(t2 5t 5) 3(t2 5t 5) t(t2 ) t(5t) t(5) 3(t2 ) 3(5t) 3(5) t3 5t 2 5t 3t2 15t 15 t3 2t 2 10t 15
8x2 2x 4 4x 5 2 40x 10x 20 3 32x 8x2 16x
72. (x 4)(x 4) x(x 4) 4(x 4) x(x) x(4) 4(x) 4(4) x2 4x 4x 16 x2 8x 16
32x3 32x2 26x 20
2
The area is (x2 8x 16) square feet.
3x x 2 x2 2x 1
58.
3 74. ( y 1) ( y 1)[( y 1)( y 1)] ( y 1)[ y( y 1) (1)( y 1)]
3x2 x 2 6x 2x2 4x 3x4 x3 2x2
( y 1)[ y2 y y 1] y( y2 2 y 1) (1)( y2 2 y 1)
3
3x4 5x3 3x2 3x 2
y3 2 y2 y y2 2 y 1
5 5 6 60. 4.2x(2x ) (4.2 2)(x x ) 8.4x
62. 5x(x2 3x 10)
y3 3y2 3y 1 3
2
5x(x ) (5x)(3x) (5x)(10) 5x3 15x2 50x 2
64. (3y 4)( y 11) 3y( y 11) 4( y 11) 3y( y) 3y(11) 4( y) 4(11) 3y2 33y 4 y 44 3y2 37 y 44 66. m 2 m 1 9 9 1 2 1 m m m 9 19 2 9 2 1 m(m) m (m) 9 9 9 9 2 2 2 1 m m m 9 9 81 1 2 m2 m 9 81 68. (7 y 2)2 (7 y 2)(7 y 2) 7 y(7 y 2) 2(7 y 2) 7 y(7 y) 7 y(2) 2(7 y) 2(2) 49 y2 14 y 14 y 4
2 2
76. (4 p) 4 p 16 p 2 2
2
2
2 2
78. (7m ) (7) (m ) 49m 80. a. b.
4
(8x 3) (5x 2) 8x 3 5x 2 3x 1 (8x 3)(5x 2) 8x(5x 2) (3)(5x 2) 8x(5x) 8x(2) (3)(5x) (3)(2) 40x2 16x 15x 6 40x2 31x 6
answers may vary 82. (2x 1) + (10x 7) = 2x 1 + 10x 7 = 12x 8 84. (2x 1)(10x 7) 2x(10x 7) (1)(10x 7) 2x(10x) 2x(7) (1)(10x) (1)(7) 20x2 14x 10x 7 20x2 24x 7 86. (2x 1) (10x 7) (2x 1) (10x 7) 2x 110x 7 8x 6
49 y2 28 y 4
390
2
The volume is ( y 3y 3y 1) cubic meters.
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88. The areas of the smaller rectangles are: 2
xxx x 3 = 3x 2 x = 2x 23=6 The area of the figure is x2 3x 2x 6 x2 5x 6.
Chapter 10: Exponents and Polynomials
4. (2x 9)2 (2x 9)(2x 9) (2x)(2x) (2x)(9) 9(2x) 9(9) 4x2 18x 18x 81 4x2 36x 81 5. ( y 3)2 y2 2( y)(3) 32 y2 6 y 9
90. The area of the figure is (3x + 1)(3x + 1). The areas of the smaller rectangles are: 3x 3x 9x2 1 3x 3x 1 3x 3x 11 1 The area of the figure is 9x2 3x 3x 1 9x2 6x 1.
6. (r s)2 r2 2(r)(s) s2 r 2 2rs s2 7. (6x 5)2 (6x)2 2(6x)(5) 52 36x2 60x 25 8. (x2 3y)2 (x2 )2 2(x2 )(3y) (3y)2 x4 6x2 y 9 y2
92. 5a 6a (5 6)(a a) 30a2
9. (x 9)(x 9) x2 92 x2 81
94. (5x 2 y)2 (5x 2 y)(5x 2 y) 5x(5x 2 y) 2 y(5x 2 y) 5x(5x) 5x(2 y) 2 y(5x) 2 y(2 y) 25x2 10xy 10xy 4 y2
10. (5 4 y)(5 4 y) 52 (4 y)2 25 16 y2
25x2 20xy 4 y2 96. a.
(2 3)2 (5)2 25
13. (2x2 6 y)(2x2 6 y) (2x2 )2 (6 y)2
22 32 4 9 13 b.
4x4 36 y2
(8 10)2 (18)2 324 14. (7x 1)2 (7x)2 2(7x)(1) 12
82 102 64 100 164
c.
2
1 1 2 1 2 1 11. x x x x 9 3 3 3 12. (3a b)(3a b) (3a)2 b2 9a2 b2
49x2 14x 1
no; answers may vary
15. (5 y 3)(2 y 5) (5 y)(2 y) (5 y)(5) (3)(2 y) (3)(5)
Section 10.6 Practice Exercises 1. (x 7)(x 5) (x)(x) (x)(5) (7)(x) (7)(5) x2 5x 7x 35
10 y2 25 y 6 y 15 10 y2 19 y 15 16. (2a 1)(2a 1) (2a)2 (1)2 4a2 1
x2 2x 35 2. (6x 1)(x 4) 6x(x) 6x(4) (1)(x) (1)(4) 6x2 24x x 4 6x2 25x 4
2
2
1 1 1 2 17. 5 y (5 y) 2(5 y) 9 9 10 19 25 y2 y 9 81
3. (2 y2 3)( y 4) 2 y3 8 y2 3y 12
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Vocabulary, Readiness & Video Check 10.6
14.
1. (x 4)2 x2 2(x)(4) 42
x
2 5
x2 8x 16 x2 16 The statement is false.
x
1
1 2 2 x2 x x
5
5 5 1 2 x2 x 5 25
25
16. (0.3 2a)(0.6 5a) 0.18 1.5a 1.2a 10a2 0.18 2.7a 10a2
2. The statement is true. 3. (x 4)(x 4) x2 42 x2 16 x2 16 The statement is false. 4. (x 1)(x3 3x 1) x(x3 3x 1) 1(x3 3x 1) x4 3x2 x x3 3x 1 x4 x3 3x2 4x 1 The product is a polynomial of degree 4, not 5. The statement is false.
18. (x 4 y)(3x y) 3x2 xy 12xy 4 y2 3x2 11xy 4 y2 20. (x 7)2 x2 2(7)(x) 72 x2 14x 49 22. (7x 3)2 (7x)2 2(7x)(3) 32 49x2 42x 9 2 2 2 24. (5a 2) (5a) 2(5a)(2) 2
5. a binomial times a binomial
25a2 20a 4
6. FOIL order for multiplication, distributive property
26. (x2 0.3)2 (x2 )2 2(x2 )(0.3) (0.3)2 x4 0.6x2 0.09
7. Multiplying gives you four terms, and the two like terms will always subtract out. 8. multiplying the sum and difference of the same two terms, squaring a binomial, and the FOIL order for multiplication when multiplying a binomial and a binomial Exercise Set 10.6 2. (x 5)(x 1) x2 x 5x 5 x2 6x 5
28.
2 2 3 y y 2 2( y) 3 3 4 4 4 9 2 6 y y 4 16 3 9 y2 y 2 16
30. (5b 4)2 (5b)2 2(5b)(4) 42 25b2 40b 16
4. ( y 12)( y 4) y 4 y 12 y 48 2
32. (6s 2)2 (6s)2 2(6s)(2) 22
y2 8 y 48
36s2 24s 4 6. (3y 5)(2 y 7) 6 y2 21y 10 y 35 6 y2 11y 35 8. (2x 9)(x 11) 2x2 22x 9x 99 2x2 31x 99 10. (6x 2)(x 2) 6x2 12x 2x 4 6x2 10x 4
16s2 16sy 4 y2 36. (3n 5m)2 (3n)2 2(3n)(5m) (5m)2 9n2 30mn 25m2 38. (7x3 6)2 (7x3 )2 2(7x3 )(6) 62 49x6 84x3 36
12. ( y2 3)(5 y 6) 5 y3 6 y2 15 y 18
392
34. (4s 2 y)2 (4s)2 2(4s)(2 y) (2 y)2
40. (b 3)(b 3) b2 32 b2 9
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
42. (x 8)(x 8) x2 82 x2 64 44. (7x 5)(7x 5) (7x)2 52 49x2 25 46. (a2 6)(a2 6) (a2 )2 62 a4 36 48. (3x2 1)(3x2 1) (3x2 )2 12 9x4 1 50. (8 7x)(8 7x) 82 (7x)2 64 49x2 2
2 2 2 2 52. 10x 10x (10x) 7 7 7 4 100x2 49
54. (2x y)(2x y) (2x)2 y2 4x2 y2 56. (5m 4n)(5m 4n) (5m)2 (4n)2 25m2 16n2 58. (a 5)(a 7) a2 7a 5a 35 a2 12a 35 60. (b 2)2 b2 2(b)(2) 22 b2 4b 4 62. (6a 7)(6a 5) 36a2 30a 42a 35 36a2 72a 35 64. (x 10)(x 10) x2 102 x2 100 66. (4a 2)2 (4a)2 2(4a)(2) 22 16a2 16a 4 68. (3x 2)(4x 2) 12x2 6x 8x 4 12x2 2x 4 a
a
2
a a2 2 2 16 y 70. 4 y 4 y (4 y) 4 2 2 2 72. (3y 13)( y 3) 3y2 9 y 13y 39 3y 2 22 y 39 74. (x2 8)(x2 8) (x2 )2 82 x4 64 76. (3x 5)(3x 5) (3x)2 52 9x2 25
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393
Chapter 10: Exponents and Polynomials
ISM: Prealgebra and Introductory Algebra
78. (4x 9 y)2 (4x)2 2(4x)(9 y) (9 y)2 16x2 72xy 81y2 80. (x 2)(x2 4x 2) x(x2 4x 2) (2)(x2 4x 2) x3 4x2 2x 2x2 8x 4 x3 6x2 10x 4 82. (3x 2)(x 4) 3x2 12x 2x 8 3x2 14x 8 The area of the canvas is (3x2 14x 8) square inches. 60 y6
60 y6
3 62
84.
y 80 y2 80 y2 4 6a8 y 6 a8 y 86. 3 a4 y 3a4 y
3 4 3y4 y 4 4
2a84 y11 2a4 y0 2a4 6 6 88. 48ab 48 a b 32ab3 32 a b3 3 a11b63 2 3 a0b3 2 3b3 2 2 2 90. (a b)(a b) a b which is choice a. 2 2 2 2 2 2 92. (a b) (a b) (a 2ab b )(a 2ab b ) a2 (a2 2ab b2 ) 2ab(a2 2ab b2 ) b2 (a2 2ab b2 )
a4 2a3b a2b2 2a3b 4a2b2 2ab3 a2b2 2ab3 b4 a4 2a2b2 b4 which is choice e. 94. Since (5x3)2 25x6 , 2(5x3 )(2) 20x3, and 22 4, (5x3 2)2 25x6 20x3 4 is a true statement. 96. (3x 5)(3x 5) (2x)2 ((3x)2 52 ) (2x)2 (9x2 25) 4x2 9x2 25 4x2 5x2 25 The area is (5x2 25) square miles.
394
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
98. (3x 4)(3x 4) (4x2 ) (3x)2 42 4x2 9x22 16 4x2 5x 16
14. (7x 1)(7x 1) (7x)2 12 49x2 1 15. (x
The area is (5x2 16) square centimeters. 100. (2 y 11)(2 y 11) (2 y 11)2 2
(2 y) 2(2 y)(11) 11 4 y2 44 y 121
2
The area is (4 y2 44 y 121) square units. 102. answers may vary
y ) x
79 59
1 9 3
17. (3 x ) 3 (r7 s5 )6 (2r4 s4 )4
106. answers may vary Mid-Chapter Review 1. (5x2 )(7x3 ) (5 7)(x2 x3 ) 35x5
13 93
x
y
y45
x63
3 27
3 x
x27
33
x27
27
r58s14 16 r58 16s14
19. (7x2 2x 3) (5x2 9)
4. (4) (4)(4) 16 2
2
2
5. (x 5)(2x 1) 2x 1x 10x 5 2x 9x 5 6. (3x 2)(x 5) 3x2 15x 2x 10
(7x2 2x 3) (5x2 9) 7x2 2x 3 5x2 9 2x2 2x 6 20. (10x2 7x 9) (4x2 6x 2)
3x 13x 10 2
(10x2 7x 9) (4x2 6x 2)
7. (x 5) + (2x + 1) = x 5 + 2x + 1 = 3x 4 8. (3x 2) + (x + 5) = 3x 2 + x + 5 = 4x + 3
10x2 7x 9 4x2 6x 2 6x2 13x 11 2 2 2 21. 0.7 y 1.2 1.8 y 6 y 1 2.5 y 6 y 0.2
9 12
x3 y10
63 45
24 r44 s44 r 42 s 30 16r16 s16 r42(16) s30(16) 16
3. 42 (42 ) 16
7x y
x
r76 s56
2. (4 y2 )(8 y7 ) (4 8)( y2 y7 ) 32 y9
9.
y
1 1 2 16. 8 2 64 8
18.
104. answers may vary
7 5 9
7 x93 y1210 7x6 y2
2 8
0 6
6
10. 20a b 20 a22b82 10a b 10b 7 7 14a2b2 14 7 6 2
2
11. (12m n ) 12 m
72 62
n
14 12
144m n
12. (4 y9 z10 )3 43 y93 z103 64 y27 z30
2 2 22. 7.8x 6.8x 3.3 0.6x 0.9 2 8.4x 6.8x 4.2
23. (3y2 6 y 1) ( y2 2) (3y2 6 y 1) ( y2 2) 3y2 6 y 1 y2 2 2 y2 6 y 1
2 2 2 13. (4 y 3)(4 y 3) (4 y) 3 16 y 9
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395
Chapter 10: Exponents and Polynomials
ISM: Prealgebra and Introductory Algebra
1 1 24. (z 2 5) (3z2 1) 8z2 2z (z2 5) (3z2 1) 8z2 2z 2 2 1 2 2 2 z 5 3z 1 8z 2z 2 11 2 6z 2z 2 25. (x 4)2 x2 2(x)(4) 42 x2 8x 16 26. ( y 9)2 y2 2( y)(9) 92 y2 18x 81 27. (x + 4) + (x + 4) = x + 4 + x + 4 = 2x + 8 28. (y 9) + (y 9) = y 9 + y 9 = 2y 18 29. 7x2 6xy 4( y2 xy) 7x2 6xy 4 y2 4xy 7x2 10xy 4 y2 30. 5a2 3ab 6(b2 a2 ) 5a2 3ab 6b2 6a2 a2 3ab 6b2 31. (x 3)(x2 5x 1) x(x2 5x 1) (3)(x2 5x 1) x(x2 ) x(5x) x(1) (3)(x2 ) (3)(5x) (3)(1) x3 5x2 x 3x2 15x 3 x3 2x2 16x 3 32. (x 1)(x2 3x 2) x(x2 3x 2) 1(x2 3x 2) x(x2 ) x(3x) x(2) 1(x2 ) 1(3x) 1(2) x3 3x2 2x x2 3x 2 x3 2x2 5x 2 33. (2x 7)(3x 10) 6x2 20x 21x 70 6x2 x 70 34. (5x 1)(4x 5) 20x2 25x 4x 5 20x2 21x 5 35. (2x 7)(x2 6x 1) 2x(x2 6x 1) (7)(x2 6x 1) 2x(x2 ) 2x(6x) 2x(1) (7)(x2 ) (7)(6x) (7)(1) 2x3 12x2 2x 7x2 42x 7 2x3 19x2 44x 7
396
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
36. (5x 1)(x2 2x 3) 5x(x2 2x 3) (1)(x2 2x 3) 5x(x2 ) 5x(2x) 5x(3) (1)(x2 ) (1)(2x) (1)(3) 5x3 10x2 15x x2 2x 3 5x3 9x2 17x 3 2
5 5 2 5 2 25 37. 2x 2x (2x) 4x 81 9 9 9
2
3 3 2 3 38. 12 y 12 y (12 y) 7 7 7 9 144 y2 49
Section 10.7 Practice Exercises 1.
25x3 5x2 5x2
25x3 5x2
5x2 5x2
5x 1
7 2 7 2 2. 24x 12x 4x 24x 12x 4x 4x2 4x2 4x2 4x2 1 6x5 3 x
3.
12x3 y3 18xy 6 y 12x3 y3 18xy 6 y 3xy 3xy 3xy 3xy 2 4x2 y2 6 x
x7 4. x 5 x2 12x 35 x2 5x 7x 35 7x 35 0 Thus,
x2 12x 35 x 7. x5
4x 3 5. 2x 1 8x 2x 7 8x2 4x 6x 7 6x 3 4 4 8x2 2x 7 . Thus, 4x 3 4 or 4x 3 2x 1 2x 1 2x 1 2
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
6. 15 2x2 2x2 0x 15 2x 6 2 x 3 2x 0x 15 2x2 6x 6x 15 6x 18 3 15 2x2 3 2x 6 or Thus, x3 x 3 3 2x 6 x 3 . 7.
5 x 9x3 3x 2
9x3 0x2 x 5
a4
a 64 a2
p8 4.
5.
6.
p3 y2
83
p
p
y2
21 1 y y y 1
y a3
y
a3
a1
a
a31 a 2
Exercise Set 10.7
x2 x 1
2.
15x2 9x5
4.
6.
15x2
8.
5
9x 15x 9x 4 x x
8x3 4x2 6x 2 8x3 4x2 6x 2 2 23 22 2 2 4x 2x 3x 1 6x5 3x4 3x
x2 0x x2 x x 1 x 1 0
x
8. x 1 x3 0x2 0x 1 x3 x2
4
6x5
3x4
3x
3x
4
4
2x 1
14m2 27m3 14m2 27m3 27m2 2m 7m 7m 7m 7
6a2 4a 12 6a2 4a 12 2 2 2 10. 2a 2a 2a 2a2 2 6 3 2 a a
x3 1 x2 x 1. x 1
12a3 36a 15
Vocabulary, Readiness & Video Check 10.7
12.
3a
3 1. In 6 18, the 18 is the dividend, the 3 is the quotient, and the 6 is the divisor. x2 2. In x 1 x2 3x 2, the x + 1 is the divisor, the x2 3x 2 is the dividend, and the x + 2 is the
quotient. 398
5
8. Filling in missing powers helps us keep like terms lined up and our work clear and neat.
2
3 Thus, 5 x 9x 3x2 2x 1 3 . 3x 2 3x 2
Thus,
a6
7. the common denominator
3x 2 2 3x 2x 1
3x 2 9x 0x x 5 9x3 6x2 6x22 x 6x 4x 3x 5 3x 2 3 3
3.
Copyright © 2025 Pearson Education, Inc.
12a3
36a
15
3a 3a 3a 5 2 4a 12 a
ISM: Prealgebra and Introductory Algebra
x2 14. x 5 x 7x 10 x2 5x 2x 10 2x 10 0
4x2 x 5 2 24. x 3 4x 11x 8x 10 4x3 12x2 x2 8x x2 3x 5x 10 5x 15 5
2
x2 7x 10 x5
Chapter 10: Exponents and Polynomials
3
x 2
5 4x3 11x2 8x 10 2 4x x 5 x3 x3
3x 2 2 16. x 2 3x 8x 4 3x2 6x 2x 4 2x 4 0 2 3x 8x 4 3x 2 x2
a7 26. a 7 a 0a 49 a2 7a 7a 49 7a 49 0 a2 49 a 7 a7 2
3x 2
18. x 1 3x x 4 3x2 3x 2x 4 2x 2 2 2 3x x 4 2 3x 2 x 1 x 1 2
x2 4x 16 28. x 4 x3 0x2 0x 64 x3 4x2 4x2 0x 4x2 16x 16x 64 16x 64 0
2x2 3x 4 20. 2x 3 4x 12x2 x 14 4x3 6x2 3
x3 64 2 x 4x 16 x4
6x2 x 6x2 9x 8x 14 8x 12 2 4x3 12x2 x 14 2 2x 2 3x 4 2x 3 2x 3 x5 22. 3x 2 3x2 17x 7 3x2 2x 15x 7 15x 10 3
30. 7 5x2 5x2 0x 7 5x 15 2 x 3 5x 0x 7 5x2 15x 15x 7 15x 45 38 2 7 5x 5x 15 38 x3 x3
3x2 17x 7 3 x 5 3x 2 3x 2 Copyright © 2025 Pearson Education, Inc.
399
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials y4
3x2 12x 59 42. x 4 3x3 0x2 11x 12 3x3 12x2 12x2 11x 12x2 48x 59x 12 59x 236 224 3x3 11x 12 2 224 3x 12x 59 x4 x4
32. 2 y 5 2 y 3y 15 2 y2 5 y 8 y 15 8 y 20 5 2
3y 2 y2 15 y 4 5 2y 5 2y 5 m3n2 mn4 34.
mn
m3 n 2 mn
mn4
mn
2
m nn
3
7x 21 2 44. x 3 7x 0x 3 7x2 21x 21x 3 21x 63 60 3 7x2 60 7x 21 x3 x 3
6w 2 36. 3w 4 18w2 18w 8 18w2 24w 6w 8 6w 8 0 18w2 18w 8 6w 2 3w 4
x3 x2 x
3 3 2 2 3 3 2 2 38. 11x y 33xy x y 11x y 33xy x y 11xy 11xy 11xy 11xy xy 2 2 x y 3 11
3 2 46. x x
x5 x4
2x2 x 1 40. x 2 2x3 3x2 3x 4 2x3 4x22 x 3x
x4 x3 x4 x3 0
x6 x3 3 2 x x x x3 x2
x2 2x x 4 x 2 6
48.
12
6, so 12 = 2 6.
2
2x3 3x2 3x 4 2 x 1 6 2x x2 x2
400
x6 0x5 0x4 x3 0x2 0x 0 x6 x5 x5 0x4
50.
20 4
5, so 20 = 4 5.
52.
9x2 x, so 9x2 9x x. 9x
54.
36x2 18x, so 36x2 2x 18x. 2x
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ISM: Prealgebra and Introductory Algebra
56.
36x5 12x3 6x2 36x5 12x3 6x2 2 (2x)(3x) 5 6x 3 2 36x 12x 6x 2 6x2 6x2 6x 3 6x 2x 1 The length of the swimming pool is (6x3 2x 1) feet.
7x 10 58. 7x 20 49x2 70x 200 49x2 140x 70x 200 70x 200 0 The width of the ping-pong table is (7x 10) inches.
62.
5
5x 5
15
Chapter 10 Review 1. In 32, the base is 3 and the exponent is 2. 2. In (5)4 , the base is 5 and the exponent is 4. 3. In 54, the base is 5 and the exponent is 4. 4. In x6, the base is x and the exponent is 6. 5. 83 8 8 8 512 6. (6)2 (6)(6) 36 7. 62 (6 6) 36 8. 43 40 64 1 65
60. answers may vary 5x 15
Chapter 10: Exponents and Polynomials
x 3 which is choice b.
5
9. (3b)0 1 10. 8b 1 8b
Chapter 10 Vocabulary Check 1. A term is a number or the product of numbers and variables raised to powers.
11. y2 y7 y27 y9
2. The FOIL method may be used when multiplying two binomials.
12. x9 x5 x95 x14
3. A polynomial with exactly 3 terms is called a trinomial. 4. The degree of polynomial is the greatest degree of any term of the polynomial. 5. A polynomial with exactly 2 terms is called a binomial. 6. The coefficient of a term is its numerical factor. 7. The degree of a term is the sum of the exponents on the variables in the term. 8. A polynomial with exactly 1 term is called a monomial.
5 6 56 11 13. (2x )(3x ) 6x 6x 3 4 34 7 14. (5 y )(4 y ) 20 y 20 y 4 2 42 8 15. (x ) x x 3 5 35 15 16. ( y ) y y 6 4 4 64 24 17. (3y ) 3 y 81y 3 3 3 33 9 18. (2x ) 2 x 8x
x9 19.
9. Monomials, binomials, and trinomials are all examples of polynomials. 10. The distributive property is used to multiply 2x(x 4).
x4 z12 20.
z5
94
x
5
x
z125 z7
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401
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials a5b4 21.
22.
23.
a
ab
51 41
b
3
1 3 35. (5) 125 5
4 3
a b
3 5 x4 y6 41 61 x y x y xy
2
3 41 106 1 3 4 x3 y4 x y x y 12 4 4 12xy6
0
37. 2 2
1 71 82 x6 y6 x y 4 4 8xy2 4 3
7
3
25. 5a (2a ) 5a (2 a
4
x5 39.
)
x3 z4
5a7 (8a12 ) 40a712 40a19
40. z4
2 2 2 26. (2x) (9x) (2 x )(9x)
41.
4x (9x) 36x21 2
36x3
42.
0 0 0 27. (5a) 7 8 111 3
r3 r4 y2 y5
5(3)
x
x 4(4)
z r
2 62
12
5a 25a 5 a ; c 30. b3 32 b6 b 1 1 2 31. 7 72 49 1 1 2 32. 7 2 49 7
y
2(5) y3
34. (2x)
402
4
3
3(2) 4(5) 3 x3 y4 y ) 44. 2 5 (x x y (x1 y1)3 x13 y13
x3 y3
45.
46.
x4
1
1 1 4 4 2 x (2x) 16x4 4
8
4
2
4 33. 2x
z
3(4) r1 r
3
6 2
8
2 43. bc (b11c2(3) )4 (b0c1)4 c4 bc 3
28. 8x0 90 8 1 1 8 1 9
9
2 4
43
3x4 33 x43 27x12 3 13 29. 4 y ; b 3 4 y 64 y
2
1 1 17 1 4 1 16 16 2 1 1 1 1 38. 6 7 7 6 1 6 7 42 42 42
3x4 y10
7
3
36. 3
2x7 y8 24.
2
x4 y6 x2 y7 a5b5
x3 y3
x42 y 67 x6 y 13 5(5) (5)5
a
b
a5b5 4 47. 0.00027 2.7 10
Copyright © 2025 Pearson Education, Inc.
10 10
a b
1 x6 y13 a10
b10
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
48. 0.8868 8.868101 49. 80,800, 000 8.08107 50. 868, 000 8.68105 51. 159, 000, 000 1.59 108 52. 150, 000 1.5105 53. 8.67 105 867, 000 54. 3.86 103 0.00386 55. 8.6 104 0.00086 56. 8.936105 893, 600 57. 1.431281015 1, 431, 280, 000, 000, 000 58. 11010 0.0000000001 59. (8104 )(2 107 ) 16 103 0.016 60.
8104 2 10
7
4 10
11
400, 000, 000, 000
61. The degree of ( y5 7x 8x4 ) is 5. 62. The degree of (9 y2 30 y 25) is 2. 63. The degree of (14x2 y 28x2 y3 42x2 y2 ) is 2 + 3 or 5. 64. The degree of (6x2 y2 z2 5x2 y3 12xyz) is 2 + 2 + 2 or 6. 65. 16t 2 4000 16(0)2 4000 0 4000 4000 16t 2 4000 16(1)2 4000 16 4000 3984
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
16t 2 4000 16(3)2 4000 144 4000 3856 16t2 4000 16(5)2 4000 400 4000 3600 t
0 seconds
1 second
3 seconds
5 seconds
16t 2 4000
4000 feet
3984 feet
3856 feet
3600 feet
66. 2x2 20x 2(1)2 20(1) 2 20 22 2x2 20x 2(3)2 20(3) 18 60 78 2x2 20x 2(5.1)2 20(5.1) 52.02 102 154.02 2x2 20x 2(10)2 20(10) 200 200 400 x
1
3
5.1
10
2x2 20x
22
78
154.02
400
67. 7a2 4a2 a2 (7 4 1)a2 2a2 68. 9 y y 14 y (9 114) y 4 y 69. 6a2 4a 9a2 6a2 9a2 4a 15a2 4a 70. 21x2 3x x2 6 21x2 x2 3x 6 22x2 3x 6 71. 4a2b 3b2 8q2 10a2b 7q2 4a2b 10a2b 3b2 8q2 7q2 6a2b 3b2 q2 72. 2s14 3s13 12s12 s10 cannot be combined. 73. (3x2 2x 6) (5x2 x) 3x2 2x 6 5x2 x 3x2 5x2 2x x 6 8x2 3x 6 74.
(2x5 3x4 4x3 5x2 ) (4x2 7x 6) 2x5 3x4 4x3 5x2 4x2 7x 6 2x5 3x4 4x3 9x2 7x 6
404
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ISM: Prealgebra and Introductory Algebra
75. (5 y2 3) (2 y2 4) (5 y2 3) (2 y2 4)
Chapter 10: Exponents and Polynomials
87. 2(x3 9x2 x)
5 y2 3 2 y2 4
(2)(x3 ) (2)(9x2 ) (2)(x)
7 y2 1
2x3 18x2 2x
76. (2m7 3x4 7m6 ) (8m7 4m2 6x4 ) (2m7 3x4 7m6 ) (8m7 4m2 6x4 ) 2m7 3x4 7m6 8m7 4m2 6x4 6m7 3x4 7m6 4m2
88. 3a(a2b ab b2 ) (3a)(a2b) (3a)(ab) (3a)(b2 ) 3a3b 3a2b 3ab2 89. (3a3 4a 1)(2a)
77. (3x2 7xy 7 y2 ) (4x2 xy 9 y2 ) (3x2 7xy 7 y2 ) (4x2 xy 9 y2 ) 3x2 7xy 7 y2 4x2 xy 9 y2 x2 6xy 2 y2
(3a3)(2a) (4a)(2a) (1)(2a) 6a4 8a2 2a 90. (6b3 4b 2)(7b) (6b3)(7b) (4b)(7b) (2)(7b)
78. (8x6 5xy 10 y2 ) (7x6 9xy 12 y2 ) (8x6 5xy 10 y2 ) (7x6 9xy 12 y2 ) 2 6 2 6 8x 5xy 10 y 7x 9xy 12 y
42b4 28b2 14b 91. (2x 2)(x 7) 2x2 14x 2x 14 2x2 12x 14
x6 4xy 2 y2 2
92. (2x 5)(3x 2) 6x2 4x 15x 10
2
79. (9x 6x 2) (4x x 1)
6x2 11x 10
9x 6x 2 4x x 1 2
2
5x2 5x 1 2
93. (4a 1)(a 7) 4a2 28a a 7 2
4a2 27a 7
2
80. [(x 7x 9) (x 4)] (4x 8x 7) (x2 7x 9) (x2 4) (4x2 8x 7)
94. (6a 1)(7a 3) 42a2 18a 7a 3 42a2 11a 3
x2 7x 9 x2 4 4x2 8x 7 2x2 x 20
95. (x 7)(x3 4x 5)
81. 6(x + 5) = 6(x) + 6(5) = 6x + 30
x(x3 4x 5) 7(x3 4x 5)
82. 9(x 7) = 9(x) + 9(7) = 9x 63
x(x3 ) x(4x) x(5) 7(x3 ) 7(4x) 7(5)
83. 4(2a + 7) = 4(2a) + 4(7) = 8a + 28
x4 7x3 4x2 23x 35
84. 9(6a 3) = 9(6a) + 9(3) = 54a 27 85. 7x(x2 5) (7x)(x2 ) (7x)(5) 7x 35x 3
2 2 86. 8 y(4 y 6) (8 y)(4 y ) (8 y)(6)
32 y3 48 y
x4 4x2 5x 7x3 28x 35
96. (x 2)(x5 x 1) x(x5 x 1) 2(x5 x 1) x(x5 ) x(x) x(1) 2(x5 ) 2(x) 2(1) x6 x2 x 2x5 2x 2 5 2 6 x 2x x 3x 2
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Chapter 10: Exponents and Polynomials
ISM: Prealgebra and Introductory Algebra
97. (x2 2x 4)(x2 2x 4) x2 (x2 2x 4) 2x(x2 2x 4) 4(x2 2x 4) x4 2x3 4x2 2x3 4x2 8x 4x2 8x 16 x4 4x3 4x2 16 98. (x3 4x 4)(x3 4x 4) x3 (x3 4x 4) 4x(x3 4x 4) 4(x3 4x 4) x6 4x4 4x3 4x4 16x2 16x 4x3 16x 16 x6 8x4 16x2 16 99. (x 7)3 (x 7)(x 7)2 (x 7)(x2 2(x)(7) 72 ) (x 7)(x2 14x 49) x(x2 14x 49) 7(x2 14x 49) x3 14x2 49x 7x2 98x 343 x3 21x2 147x 343 100. (2x 5)3 (2x 5)(2x 5)2 (2x 5)((2x)2 (2)(2x)(5) 52 ) (2x 5)(4x2 20x 25) 2x(4x2 20x 25) (5)(4x2 20x 25) 8x3 40x2 50x 20x2 100x 125 8x3 60x2 150x 125 101. (x 7)2 x2 2(x)(7) 72 x2 14x 49 102. (x 5)2 x2 2(x)(5) 52 x2 10x 25 103. (3x 7)2 (3x)2 2(3x)(7) 72 9x2 42x 49 104. (4x 2)2 (4x)2 2(4x)(2) 22 16x2 16x 4 105. (5x 9)2 (5x)2 2(5x)(9) 92 25x2 90x 81 106. (5x 1)(5x 1) (5x)2 12 25x2 1 107. (7x 4)(7x 4) (7x)2 42 49x2 16 108. (a 2b)(a 2b) a2 (2b)2 a2 4b2 109. (2x 6)(2x 6) (2x)2 62 4x2 36
406
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ISM: Prealgebra and Introductory Algebra
110. (4a2 2b)(4a2 2b) (4a2 )2 (2b)2 16a4 4b2
Chapter 10: Exponents and Polynomials
a2 3a 8 117. a 2 a a 2a 6 a3 2a2 3
2
111. (3x 1)2 (3x)2 2(3x)(1) 12
3a2 2a 3a2 6a 8a 6 8a 16 22
9x 6x 1 The area is (9x2 6x 1) square meters. 2
112. (5x 2)(x 1) 5x2 5x 2x 2 5x2 3x 2 The area is (5x2 3x 2) square miles. x2 21x 49 x2 21x 49 2 2 2 113. 7x2 7x 7x 7x 1 3 7 2 7 x x 3 2 3 2 114. 5a b 15ab 20ab 5a b 15ab 20ab 5ab 5ab 5ab 5ab a2 3b 4
a 1 115. a 2 a a 4 a2 2a a4 a2 6 a2
a 1
3b2 4b 118. 3b 2 9b 3 18b 2 8b 1 9b 6b 3
2
2
12b 8b 12b2 8b 9b3 18b2 8b 1 3b 2
1 1 3b2 4b 3b 2
2x3 x2 0x 2 119. 2x 1 4x 4x3 x2 4x 3
2
a2 a 4
22 a3 a2 2a 6 a2 3a 8 a2 a2
4
4x4 2x3 2x3 x2 2x3 x2 4x 3 4x 2
6 a2
4x 116. x 5 4x 20x 7 4x2 20x 7 4x2 20x 7 7 4x x5 x5
4x4 4x3 x2 4x 3 2x 1
2
1
3
2
2x x 2
1 2x 1
x2 16x 117 120. x 6 x3 10x2 21x 18 x3 6x2 16x2 21x 16x2 96x 117x 18 117x 702 684 x3 10x2 21x 18 684 2 x 16x 117 x6 x6
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Chapter 10: Exponents and Polynomials 15x3 3x2 60 121.
122.
60
(8 y2 3y 1) (3y2 2) 8 y2 3y 1 3y2 2 5 y2 3y 1 134. (2x 5)(3x 2) 6x2 4x 15x 10
21a3b6 3a 3 21a3b6 3a 3 3 3 3 3 7a3b6 a 1
135. 4x(7x2 3) 4x(7x2 ) 4x(3) 28x3 12x
The length of a side is (7a3b6 a 1) units.
136. (7x 2)(4x 9) 28x2 63x 8x 18 28x2 71x 18
123. 33 (3)(3)(3) 27 1 1 1 1 1 124. 2 2 2 2 8
x3 x2 18x 18
2 3 5 13 25 4 7 125. (4xy )(x y ) 4x y 4x y
18x9 27x3
2 93 2x6 x 3 3
3a 3 43 27a12 3 a 127. 2 23 6 b b b 128. (2x
y )
a3b6 129.
91 a5b2
4
4 44 34
2
x
2
y
x16
16 y12
9a3(5)b6(2) 9a2b8
130. ( y2 4) (3y2 6) y2 4 3y2 6 2x2 10 131. (6x 2) (5x 7) 6x 2 5x 7 11x 5
2
3
4
3
2a 4a 5 140. 8a 2a 4a 5 8a 3 3 2a 2a 2a3 2a3 2a3 2 5 4a 1 2a3 2 a x3 141. x 5 x2 2x 10 x2 5x 3x 10 3x 15 25 10 x 2 2x x 3 25 x5 x5
132. (5x2 2x 6) (x 4) (5x2 2x 6) (x 4) 5x2 2x 6 x 4 5x2 3x 2
408
2
138. (5x 4) (5x) 2(5x)(4) 4 25x2 40x 16
139. (6x 3)(6x 3) (6x)2 32 36x2 9
4 3
4 3 4
6x2 11x 10
137. (x 3)(x2 4x 6) x(x2 4x 6) (3)(x2 4x 6) x3 4x2 6x 3x2 12x 18
3
126.
133. (8 y2 3y 1) (3y2 2)
3x2 3x2 3x2 20 5x 1 x2 20 The width is 5x 1 feet. 2 x 3x2
3x2
15x3
ISM: Prealgebra and Introductory Algebra
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ISM: Prealgebra and Introductory Algebra
13. ( y)( y)( y) (1) y (1) y (1) y
2x2 7x 5 142. 2x 3 4x3 8x2 11x 4 4x3 6x2 14x2 11x 14x2 21x 10x 4 10x 15 19
(1)3 y3 1 y3 y3 ; F 14. y y y (1) y (1) y (1) y [(1) (1) (1)]y 3 y; D
4x3 8x2 11x 4 2x2 7x 5 19 2x 3 2x 3
Chapter 10 Getting Ready for the Test 1. To simplify x2 x5 , the exponents are added: x2 x5 x25 x7 ; C.
25
(x ) x
10
x ; A.
3. The expression x2 x5 cannot be simplified; E. 4. To simplify x
5
x2
x5 x2
Chapter 10 Test 1. 25 2 2 2 2 2 32 2. (3)4 (3)(3)(3)(3) 81 3. 34 (3 3 3 3) 81
2. To simplify (x2)5, the exponents are multiplied; 2 5
Chapter 10: Exponents and Polynomials
4. 43
1 43
1 64
5. (3x2 )(5x9 ) 3(5)(x2 x9 ) 15x29 15x11
, the exponents are subtracted: y7
x52 x3; D.
6.
5. Add 20y and 4y: 20y + 4y = (20 + 4)y = 24y; F.
7.
6. Subtract 20y and 4y: 20y 4y = (20 4)y = 16y; C. 7. Multiply 20y and 4y: 20 y 4 y 20 4 y y 80 y2; E.
y2 r8 r3
72
y r
5
1
8(3) r5
4x2 y3 8.
y
r5 2
3 4 x y
20 y 20 y 8. Divide 20y and 4y: 4 y 5 1 5; I. 4 y 1 9. 51 ; C 5
42 x22 y32 x32 y42 16x4 y6
x6 y8 16x46 y6(8)
16x2 y14
16 y14 x2
1 1 3 10. 2 3 ; D 2 8 11. y + y + y = 1y + 1y + 1y = (1 + 1 + 1)y = 3y; C 12. y y y y1 y1 y1 y111 y3; B
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
2 3 19. [(8x 7x 5) (x 8)] (4x 2)
62 x4 y1 9.
62 x4 y1 63 x3 y7 63 x3 y7 623 x4(3) y17 61 x1 y8 1 1 1 6 x y8 1 6xy8
(8x2 7x 5) (x3 8) (4x 2) 8x2 7x 5 x3 8 4x 2 x3 8x2 3x 5 20. (3x 7)(x2 5x 2) 3x(x2 5x 2) 7(x2 5x 2) 3x(x2 ) 3x(5x) 3x(2) 7(x2 ) 7(5x) 7(2) 3x3 15x2 6x 7x2 35x 14 3x3 22x2 41x 14
5
10. 563, 000 5.6310 11. 0.0000863 8.6310 12. 1.5 10
3
2 2 21. 3x (2x 3x 7) 2 3x (2x2 ) 3x2 (3x) 3x2 (7) 6x4 9x3 21x2
5
0.0015
22. (x 7)(3x 5) 3x2 5x 21x 35 3x2 16x 35
13. 6.2310 62, 300 4
14. (1.2 105 )(3107 ) 3.6 102 0.036 4xy2 7xyz x3 y 2
15. a.
2
1 1 2 1 2 23. 3x 1 3x (3x) 9x 5 5 25 5 24. (4x 2)2 (4x)2 2(4x)(2) 22 16x2 16x 4
Term
Numerical Coefficient
Degree of Terms
4xy 2
4
3
7xyz
7
3
x3 y
1
4
26. (x2 9b)(x2 9b) (x2 )2 (9b)2 x4 81b2
2
2
0
27. 16t 2 1001 16(0)2 1001 1001 ft
2
25. (8x 3)2 (8x)2 2(8x)(3) (3)2 64x2 48x 9
16(1)2 1001 16 1001 985 ft
3
b. The degree of 4xy 7xyz x y is 3 + 1 or 4.
16(3)2 1001 144 1001 857 ft
16. 5x2 4x 7x2 11 8x
16(5)2 1001 400 1001 601 ft
(5 7)x2 (4 8)x 11 2x2 12x 11
28. (2x 3)(2x 3) (2x)2 (3)2 3
2
3
17. (8x 7x 4x 7) (8x 7x 6) 8x3 7x2 4x 7 8x3 7x 6 8x3 8x3 7x2 4x 7x 7 6 16x3 7x2 3x 13 3
18.
2
5x x 5x 2 (8x3 4x2 x 7)
3
4x2 9 The area is (2x 3)(2x + 3) or (4x2 9) square inches. 2
5x x 5x 2 8x3 4x2 x 7 3x3 5x2 4x 5
410
4x2 2xy 7x 4x2 2xy 7x 29. 8xy 8xy 8xy 8xy x 1 7 2y 4 8y
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials 7. 2(3 7x) 15 2(3) 2(7x) 15 6 14x 15 14x 9
x2 30. x 5 x 7x 10 x2 5x 2x 10 2x 10 0 2
8. 5(x + 2) = 5 x + 5 2 = 5x + 10 9. 2(x 5) 4(2x 2) 2x 10 8x 8 2x 8x 10 8 6x 18
x2 7x 10 x2 x5
10. 2( y 3z 1) (2)( y) (2)(3z) (2)(1) 2 y 6z 2
9x2 6x 4 31. 3x 2 27x 0x2 0x 8 27x3 18x2 3
18x2 0x 18x2 12x 12x 8 12x 8 16 3 27x 8 2 6x 4 16 9x 3x 2 3x 2
11.
y 12. 7
Cumulative Review Chapters 110 1. 92 81
13.
2. 5 125 3
3. 34 81 4. 33 27 5. a.
7+x
b.
15 x
c.
2x
y 5 2 6 y 5 8 y 5 5 8 5 y 3 20
7 y
20 7 7 y 140
7(x 2) 9x 6 7x 14 9x 6 7x 7x 14 9x 7x 6 14 2x 6 14 6 2x 6 6 8 2x 8 2x 2 2 4 x
5
14. 6(2a 1) (11a 6) 7 12a 6 11a 6 7 a 12 7 a 12 12 7 12 a 19
e.
x2
15.
6. a.
x+3
d.
x
or x 5
b.
3x
c.
2x
d.
10 x
e.
5x + 7
3
a9 5 5 3 5 a 9 3 5 3 a 15
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411
ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
2
16.
y 16
d. 3, 2, 0,
3 3 2 3 y 16 2 3 2 y 24 3y 1 1 11 2 3y 3 3 11 2 y 33
5
30. 2 2
2
is a rational number and a real number.
5 1 5 5 1 y 25
32.
0.6x 10 1.4x 14 10(0.6x 10) 10(1.4x 14) 6x 100 14x 140 100 8x 140 40 8x 40 8x 8 8 5 x
5y
20. 0.250
1
125
250
1 8
1
1000 5
21. 43.5 43
4 43
1
10
2
75
10
22. 10.75 10
43
1 2
3
100
10
3
4 4 83 83 23. 105.083 105 105 1000 1000 7 7 31 24. 31.07 31 100 100 25. x y = 2.8 0.92 = 1.88 26. x y = 1.2 7.6 = 8.8 27. xy = 2.3 0.44 = 1.012 28. xy = 6.1 0.5 = 3.05 11, 112
b. 0, 11, 112
412
1
0.25x 0.10(x 3) 1.1 0.25x 0.10x 0.30 1.1 0.35x 0.30 1.1 0.35x 1.4 x4
1000
c.
1 5 2 2
31.
19. 0.125
29. a.
All numbers in the given set are real numbers.
1
5y 1
2
e. f.
18.
, 11, 112
4
2
17.
1
33. 2(x 4) 4x 12 2x 8 4x 12 8 2x 12 20 2x 10 x The number is 10. 34. 3(x 2) 5x 10 3x 6 5x 10 6 2x 10 4 2x 4 2x 2 2 2 x The number is 2. 35. 30 2 2x 140 60 2x 140 2x 80 x 40 The length is 40 feet.
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ISM: Prealgebra and Introductory Algebra
Chapter 10: Exponents and Polynomials
36. Use A = l w with A = 2016 and l = 63. 2016 63w 2016 63w 63 63 32 w The width is 32 feet.
2x3 42. y
38.
–4
–3
b.
c. 40. a.
b.
c.
y
–1
0
1
2
3
4
y
4x 1 25 y6 1 27x21
4
5
46. 9x3 x3 (9 1)x3 8x3 47. 5x2 6x 9x 3 5x2 3x 3 48. 2x x2 5x 4x2 x2 4x2 2x 5x 5x2 7 x 49. 7 x(x2 2x 5) 7x(x2 ) 7 x(2x) 7x(5) 7x3 14x2 35x
4
50. 2x(x2 x 1) 2x(x2 ) (2x)(x) (2x)(1) 2x3 2x2 2x
(9 y5 )2 92 y52 81y10
5 2 5 2 51. 9x 12x 3x 9x 12x 3x
1 5 15 6 y 5 y y y y y
3x2
3
2 8 2 3 27 3 3 3 2
6
3 3 3 45. 9x x (9 1)x 10x
–2
t t t 4 16 2 2
3
2
3
x7 x4 x74 x11 4
2
44. (3x7 )3 33 (x7 )3 33 x21
3x 4 2x 6 3x 2x 4 2x 2x 6 x 4 6 x 4 4 6 4 x 10 {x|x 10}
39. a.
2 3 2 2 6 2 2 (x ) 2 x y
43. (5 y3 )2 52 y32 52 y6
37. 4x 7 9 4x 16 x4 {x|x 4} –5
2
2
3 2
32
(8x ) 8 (x ) 64x
64x
6
3x2 3x2 1 3x3 4 x
3x2
7 2 7 2 52. 4x 12x 2x 4x 12x 2x
2x
3
2x 2x 2x 2x6 6x 1
3a2 33 a23 33 a6 b3 41. b 27a6 b3 b3
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Chapter 11 Section 11.1 Practice Exercises 1. a.
6.
45 = 3 3 5 75 = 3 5 5 GCF = 3 5 = 15
2 7. 5
b. 32 = 2 2 2 2 2 33 = 3 11 There are no common prime factors; thus, the GCF is 1. c.
2. a.
b.
3.
a.
b.
The GCF is y4 , since 4 is the smallest exponent to which y is raised.
6x2 2 3 x2 9x4 3 3 x4
a2
5
1
a2 (2a3 4a 1)
5
8. 6a3b 3a3b2 9a2b4 3a2b(2a ab 3b3 ) 9. 7(p + 2) + q(p + 2) = (p + 2)(7 + q)
11. ab 7a 2b 14 (ab 7a) (2b 14) a(b 7) 2(b 7) (b 7)(a 2) 12. 28x3 7x2 12x 3 (28x3 7x2 ) (12x 3) 7x2 (4x 1) 3(4x 1) (4x 1)(7x2 3)
12x5 1 2 2 3 x5
(2xy 5 y2 ) (4x 10 y) y(2x 5 y) 2(2x 5 y)
GCF 3 x2 3x2
(2x 5 y)( y 2)
16y = 1 2 2 2 2 y
14. 3x2 4xy 3x 4 y (3x2 4xy) (3x 4 y) x(3x 4 y) 1(3x 4 y) (3x 4 y)(x 1)
The GCF of a5 , a, and a3 is a.
2
ab .
The GCF of terms 10y and 25 is 5. 10y + 25 = 5 2y + 5 5 = 5(2y + 5) 4
9
4
b. The GCF of x and x is x . x4 x9 x4 (1) x4 (x5 ) x4 (1 x5 ) 10x3 8x2 2x 2x(5x2 ) 2x(4x) 2x(1) 2x(5x2 4x 1)
414
1
13. 2xy 5 y2 4x 10 y
Thus, the GCF of a5b4, ab3 , and a3b2 is
5.
5
a3
The GCF is x1 or x, since 1 is the smallest exponent on x.
The GCF of b4 , b3 , and b2 is b2.
a.
4
( p q)(7xy3 1)
40 y4 2 2 2 5 y4 GCF = 2 2 y = 4y
4.
a5
10. 7xy3 ( p q) ( p q) 7xy3 ( p q) 1( p q)
14 = 2 7 24 = 2 2 2 3 60 = 2 2 3 5 GCF = 2
20 y6 1 2 2 5 y6
c.
4x3 12x 4x(x2 3)
15. 3xy 4 x 12 y (3xy x) (12 y 4) x(3y 1) 4(3 y 1) (3 y 1)(x 4) 16. 2x 2 x3 3x2 2(x 1) x2 (x 3) There is no common binomial factor that can now be factored out. This polynomial is not factorable by grouping. 17. 15xz 15 yz 5xy 5 y2 5(3xz 3yz xy y2 ) 5[3z(x y) y(x y)] 5(x y)(3z y) Vocabulary, Readiness & Video Check 11.1 1. Since 5 4 = 20, the numbers 5 and 4 are called factors of 20.
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ISM: Prealgebra and Introductory Algebra
2. The greatest common factor of a list of integers is the largest integer that is a factor of all the integers in the list. 3. The greatest common factor of a list of common variables raised to powers is the variable raised to the least exponent in the list. 4. The process of writing a polynomial as a product is called factoring. 5. A factored form of 7x + 21 + xy + 3y is 7(x + 3) + y(x + 3). false 6.
A factored form of 3x3 6x x2 2 is 3x(x2 2). false
Chapter 11: Factoring Polynomials
8. The GCF of x3 , x2 , and x5 is x2 . 10. The GCF of y8 , y10 , and y12 is y8. 12. The GCF of p7 , p8 , and p9 is p7 . The GCF of q, q2 , and q3 is q. Thus, the GCF of p7 q, p8q2 , and p9q3 is p7q. 14. 20y = 2 2 5 y 15 = 3 5 GCF = 5 16. 32x5 2 2 2 2 2 x5
7. 14 = 2 7
18x2 2 3 3 x2
8. 15 = 3 5
GCF 2 x2 2x2
9. The GCF of 18 and 3 is 3. 18. 21x3 1 3 7 x3 14x 2 7 x GCF 7 x 7x
10. The GCF of 7 and 35 is 7. 11. The GCF of 20 and 15 is 5. 12. The GCF of 6 and 15 is 3.
20. 15 y2 3 5 y2
13. The GCF of a list of numbers is the largest number that is a factor of all numbers in the list.
5 y7 5 y7
14. There is no need to factor the variable parts. The GCF of common variable factors is the variable raised to the smallest exponent.
GCF 5 y2 5 y2
15. When factoring out a GCF, the number of terms in the other factor should be the same as the number of terms in your original polynomial. 16. Look for a GCF other than 1 or 1; four terms Exercise Set 11.1
20 y3 1 2 2 5 y3
22. 7x3 y3 7 x3 y3 21x2 y2 1 3 7 x2 y2 14xy4 2 7 x y4 GCF 7 x y2 7xy2 24. 40x7 y2 z 2 2 2 5 x7 y2 z
2. 36 = 2 2 3 3 90 = 2 3 3 5 GCF = 2 3 3 = 18
64x9 y 2 2 2 2 2 2 x9 y GCF 2 2 2 x7 y 8x7 y 26. 18a + 12 = 6(3a + 2)
4. 30 = 2 3 5 75 = 3 5 5 135 = 3 3 3 5 GCF = 3 5 = 15
28. 42x 7 = 7(6x 1) 30. y5 6 y4 y4 ( y 6)
6. 15 = 3 5 25 = 5 5 27 = 3 3 3 GCF = 1 since there are no common prime factors.
32. 5x2 10x6 5x2 (1 2x4 ) 34. 10xy 15x2 5x(2 y 3x)
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Chapter 11: Factoring Polynomials
ISM: Prealgebra and Introductory Algebra
3 2 2 38. 12x 16x 8x 4x(3x 4x 2)
68. 6x 7 y 42 xy 6x 42 xy 7 y 6(x 7) y(x 7) (x 7)(6 y)
40. x9 y6 x3 y5 x4 y3 x3 y3
70. 2x3 x2 10x 5 x2 (2x 1) 5(2x 1)
36. 7x + 21y 7 = 7(x + 3y 1)
(2x 1)(x2 5)
x3 y3 (x6 y3 y2 x 1) 42. 14x3 y 7x2 y 7xy 7xy(2x2 x 1) 44. 9 y6 27 y4 18 y2 6 3(3y6 9 y4 6 y2 2) 5 3 2 2 46. 2 y7 4 y y y 5 7 5 5 1 6 4 y(2 y 4 y 3y 2) 5
72. 7x 21 x3 2x2 7(x 3) x2 (x 2) The polynomial is not factorable by grouping. 74. 5xy 15x 6 y 18 5x( y 3) 6( y 3) ( y 3)(5x 6) 76. 6m2 5mn 6m 5n m(6m 5n) 1(6m 5n) (6m 5n)(m 1)
48. x( y2 1) 3( y2 1) ( y2 1)(x 3)
78. 90 15 y2 18x 3xy2 3(30 5 y2 6x xy2 ) 3[5(6 y2 ) x(6 y2 )]
50. 8(x + 2) y(x + 2) = (x + 2)(8 y) 52. q(b3 5) (b3 5) q(b3 5) 1(b3 5) (b3 5)(q 1) 54. 7y 21 = 7(y) + (7)(3) = 7(y + 3)
3(6 y2 )(5 x) 80. 16x2 4xy2 8xy 2 y3 2(8x2 2xy2 4xy y3) 2[2x(4x y2 ) y(4x y2 )] 2(4x y2)(2x y)
56. 5 y3 y6 y3 (5) ( y3 )( y3 ) 82. ( y 3)( y 6) y2 6 y 3y 18 y2 9 y 18
y3[5 ( y3)] y3(5 y3)
84. (x 5)(x 10) x2 10x 5x 50 x2 5x 50
58. 5m6 10m5 5m3 5m3 (m3 ) (5m3 )(2m2 ) (5m3 )(1) 5m3[m3 (2m2 ) 1] 5m3(m3 2m2 1) 60. x3 4x2 3x 12 x2 (x 4) 3(x 4) (x 4)(x2 3) 62. xy y 2x 2 y(x 1) 2(x 1) (x 1)( y 2) 64. 16x3 28x2 12x 21 4x2 (4x 7) 3(4x 7) (4x 7)(4x2 3) 66. 8w2 7wv 8w 7v w(8w 7v) 1(8w 7v) (8w 7v)(w 1)
416
86. 4 5 = 20 4+5=9 4 and 5 have a product of 20 and a sum of 9. 88. 2 (8) = 16 2 + (8) = 10 2 and 8 have a product of 16 and a sum of 10. 90. 3 3 = 9 3 + 3 = 0 3 and 3 have a product of 9 and a sum of 0. 92. 9 4 = 36 9 + 4 = 5 9 and 4 have a product of 36 and a sum of 5. 94. 8a 24 = 8(a 3), which is choice b.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
96. (x + 5)(x + y) is factored.
45x2 900x 11, 520
c.
15(3x2 ) 15(60x) 15(768)
98. Since 3x(a + 2b) + 2(a + 2b) = (a + 2b)(3x + 2), the given expression is not factored. 100. a.
11x2 99x 3828 11(1)2 99(1) 3828 11 99 3828 3740 It is expected that 3740 thousand students will graduate from U.S. high schools in 2026.
b. 2030 corresponds to x = 5. 11x2 99x 3828 11(5)2 99(5) 3828 11(25) 99(5) 3828 275 495 3828 3608 It is expected that 3608 thousand students will graduate from U.S. high schools in 2030.
15(3x2 60x 768) 45x2 900x 11, 520
d.
15(3x2 ) 15(60x) 15(768) 15(3x2 60x 768) 104. The area of a rectangle is lw. The area of the shaded region is the area of the small rectangle subtracted from the area of the large rectangle. 12x(x2 ) 2x 12x3 2x 2x(6x2 1) Area width length
106.
4n 24n 4n(n3 6) 4
The length is (n3 6) units. 108. answers may vary
c.
d.
102. a.
11x2 99x 3828 11 x2 11(9x) 11 348
110. answers may vary
11(x2 9x 348)
Section 11.2 Practice Exercises
69% of 3608 thousand = 0.69(3608 thousand) = 2489.52 thousand In 2030, it is expected that 2489.52 thousand U.S. high school graduates will immediately pursue higher education.
Factors of 20
Sum of Factors
1, 20
21
2, 10
12
4, 5
9
Thus, x2 12x 20 (x 10)(x 2).
45x2 900x 11, 520 45(6)2 900(6) 11, 520 45(36) 900(6) 11, 520 1620 5400 11, 520 15, 300 The annual U.S. bottled water consumption in 2021 was approximately 15,300 million gallons.
b.
1.
2025 corresponds to x = 10. 45x2 900x 11, 520 45(10)2 900(10) 11, 520 45(100) 900(10) 11, 520 4500 9000 11, 520 16, 020 Annual U.S. bottled water consumption is predicted to be 16,020 million gallons in 2025.
2. a.
Factors of 22
Sum of Factors
1, 22
23
2, 11
13
x2 23x 22 (x 1)(x 22) b.
Factors of 50
Sum of Factors
1, 50
51
2, 25
27
5, 10
15
x2 27x 50 (x 2)(x 25)
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3.
Sum of Factors
1, 36
35
1, 36
35
2, 18
16
2, 18
16
3, 12
9
Two factors of 4 whose sum is 3 are 4 and 1.
3, 12
9
x(x2 3x 4) x(x 4)(x 1)
4, 9
5
4, 9
5
6, 6
0
4. a.
4(x2 6x 9) 4(x 3)(x 3) or 4(x 3)2 b.
5x 3 (x 1)(x 6) Vocabulary, Readiness & Video Check 11.2
Two factors of 40 whose sum is 3 are 8 and 5. q 3q 40 (q 8)(q 5)
b. Two factors of 48 whose sum is 2 are 8 and 6. y2 2 y 48 ( y 8)( y 6) 5. Since there are no two numbers whose product is 15 and whose sum is 6, the polynomial is prime.
b.
x3 3x2 4x x(x2 3x 4)
10. 5x5 25x4 30x3 5x3 (x2 5x 6)
2
6. a.
9. a.
Two factors of 9 whose sum is 6 are 3 and 3.
x2 5x 36 (x 9)(x 4)
1. To factor x2 7x 6, we look for two numbers whose product is 6 and whose sum is 7. true 2. We can write the factorization (y + 2)(y + 4) also as (y + 4)(y + 2). true 3. The factorization (4x 12)(x 5) is completely factored. false 4. The factorization (x + 2y)(x + y) may also be written as (x 2 y)2. false 2 5. x 9x 20 (x 4)(x 5)
Two factors of 14 y2 whose sum is 9y are 2y and 7y.
2 6. x 12x 35 (x 5)(x 7)
x2 9xy 14 y2 (x 2 y)(x 7 y)
2 7. x 7x 12 (x 4)(x 3)
Two factors of 30b2 whose sum is 13b are 3b and 10b.
2 8. x 13x 22 (x 2)(x 11)
a2 13ab 30b2 (a 3b)(a 10b) 7. Two factors of 12 whose sum is 8 are 6 and 2. x4 8x2 12 (x2 6)(x2 2) 8. 48 14x x2 x2 14x 48 Two factors of 48 whose sum is 14 are 6 and 8. x2 14x 48 (x 6)(x 8)
418
4x2 24x 36 4(x2 6x 9)
Factors of 36
2 9. x 4x 4 (x 2)(x 2) 2 10. x 10x 24 (x 6)(x 4)
11. 15 is positive, so its factors would have to be either both positive or both negative. Since the factors need to sum to 8, both factors must be negative. 12. Since the sum of the factors is 3, the factors are 2 and 5 (2 + 5 = 3). (In other words, the factor with the smaller absolute value is negative so that the sum is positive.) If we accidentally choose factors whose sum is 3, simply “switch” the signs of the factors.
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Exercise Set 11.2 2. Two factors of 8 whose sum is 6 are 4 and 2. x2 6x 8 (x 4)(x 2)
Chapter 11: Factoring Polynomials
26. 3x3 12x2 36x 3x(x2 4x 12) 3x(x 6)(x 2) 28. x2 4xy 77 y2 (x 11y)(x 7 y)
4. Two factors of 11 whose sum is 12 are 11 and 1. y2 12 y 11 ( y 11)( y 1)
30. x2 19x 60 (x 4)(x 15) 4 2 2 2 32. x 5x 14 (x 7)(x 2)
6. Two factors of 25 whose sum is 10 are 5 and 5. x2 10x 25 (x 5)(x 5) or (x 5)2 8. Two factors of 30 whose sum is 1 are 6 and 5. x2 x 30 (x 6)(x 5) 10. Two factors of 32 whose sum is 4 are 8 and 4.
34. r 2 10r 21 (r 7)(r 3) 2 2 36. x xy 6 y (x 3y)(x 2 y)
38. 4x2 4x 48 4(x2 x 12) 4(x 4)(x 3) 40. 2x4 24x2 70 2(x4 12x2 35) 2(x2 7)(x2 5)
x2 4x 32 (x 8)(x 4) 12. Since there are no two numbers whose product is 5 and whose sum is 7, the polynomial is prime.
42. x2 x 42 (x 7)(x 6) 44. x2 4x 10 is prime.
2
14. Two factors of 8 y whose sum is 6y are 4y and 2y.
46. x2 9x 14 (x 7)(x 2)
x2 6xy 8 y2 (x 4 y)(x 2 y) 16. Two factors of 70 whose sum is 3 are 10 and 7. y4 3y2 70 ( y2 10)( y2 7) 18. 17 18n n2 n2 18n 17 Two factors of 17 whose sum is 18 are 17 and 1.
48. 3x3 3x2 126x 3x(x2 x 42) 3x(x 7)(x 6) 50. 3x2 y 9xy 45 y 3y(x 2 3x 15) 52. x2 4x 32 (x 8)(x 4)
17 18n n2 n2 18n 17 (n 17)(n 1)
54. x2 3xy 4 y2 (x 4 y)(x y)
20. 6q 27 q2 q2 6q 27 Two factors of 27 whose sum is 6 are 3 and 9.
56. 50 20t 2t 2 2t 2 20t 50
2
2(t2 10t 25)
2
6q 27 q q 6q 27 (q 3)(q 9) 22. Two factors of 18b2 whose sum is 9b are 3b and 6b. a2 9ab 18b2 (a 3b)(a 6b)
2(t 5)(t 5) or 2(t 5)2 58. x3 3x2 28x x(x2 3x 28) x(x 7)(x 4) 6 5 4 4 2 60. 3x 30x 72x 3x (x 10x 24)
24. 3x2 30x 63 3(x2 10x 21) 3(x 7)(x 3)
3x4 (x 6)(x 4) 62. 7a3b 35a2b2 42ab3 7ab(a2 5a 6b2 ) 7ab(a 3b)(a 2b)
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64. 48 20n 2n2 2n2 20n 48 2(n2 10n 24) 2(n 6)(n 4) 66. x2 8x 7 1(x2 8x 7) 1(x 7)(x 1) 68.
70.
1 2 5 1 y y 8 ( y2 5 y 24) 3 3 3 1 ( y 8)( y 3) 3
90. The factors of c must sum to 4. Since c is positive, both factors must have the same sign. 1 + (3) = 4; (1)(3) = 3 2 + (2) = 4; (2)(2) = 4 The possible values of c are 3 and 4. 92. The factors of 15 must sum to b. Since 15 is positive and b is positive, both factors of 15 must be positive. 1 + 15 = 16; (1)(15) = 15 3 + 5 = 8; (3)(5) = 15 The possible values of b are 8 and 16. 94. x2n 5xn 6 (xn 2)(xn 3)
a2b3 ab2 30b b(a2b2 ab 30) b(ab 5)(ab 6)
Section 11.3 Practice Exercises
72. (3x 2)(x 4) 3x2 12x 2x 8 2 3x 14x 86
1. a.
Factors of 10: 10 = 1 10, 10 = 2 5 5x2 27x 10 (5x 2)(x 5)
74. (4z 7)(7z 1) 28z 4z 49z 7 2
28z2 53z 7 b. 76. ( y 5x)(6 y 5x) 6 y2 5xy 30xy 25x2
78. To factor x2 13x 42, think of two numbers whose product is 42 and whose sum is 13.
2. a.
80. answers may vary
b. Factors of 6x2: 6x2 6x x, 6x2 3x 2x Factors of 1: 1 = 1 1 6x2 5x 1 (3x 1)(2x 1)
4x3 24x2 32x 4x(x2 6x 8) 4x(x 4)(x 2) 2
3. a.
3x2 14x 5 (3x 1)(x 5)
84. 16t 96t 112 16(t 6t 7) 16(t 7)(t 1) 1
Factors of 3x2: 3x2 3x x Factors of 5: 5 = 5 1, 5 = 1 5
2
Factors of 2x2: 2x2 2x x Factors of 12: 12 = 1 12, 12 = 2 6, 12 = 3 4 2x2 11x 12 (2x 3)(x 4)
82. P 2l 2w 2(12x2 ) 2(2x3 16x) 24x2 4x3 32x
1
88. y2 (x 1) 2 y(x 1) 15(x 1)
2
1 86. x x x x or x 4 2 2 2 1
Factors of 4x2: 4x2 4x x, 4x2 2x 2x Factors of 5: 5 = 1 5 4x2 12x 5 (2x 5)(2x 1)
6 y2 25xy 25x2
2
Factors of 5x2: 5x2 5x x
b.
Factors of 35x2 : 35x2 35x x, 35x2 5x 7 x Factors of 4: 4 = 1 4, 4 = 1 4, 4 = 2 2 35x2 4x 4 (5x 2)(7x 2)
(x 1)( y 2 y 15) (x 1)( y 5)( y 3) 2
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ISM: Prealgebra and Introductory Algebra
4. a.
Factors of 14x2: 14x2 14x x, 14x2 7 x 2x
Factors of 2 y2: 2 y2 2 y y, 2 y2 2 y y 14x2 3xy 2 y2 (7x 2 y)(2x y) b. Factors of 12a2: 12a2 12a a, 12a2 6a 2a, 12a2 3a 4a Factors of 3b2: 3b2 3b b, 3b2 3b b
12a2 16ab 3b2 (6a b)(2a 3b) 5. Factors of 2x4: 2x4 2x2 x2 Factors of 7: 7 = 7 1, 7 = 7 1 2x4 5x2 7 (2x2 7)(x2 1) a.
3x3 17x2 10x x(3x2 17x 10) Factors of 3x2: 3x2 3x x Factors of 10: 10 = 1 10, 10 = 2 5 x(3x2 17x 10) x(3x 2)(x 5)
b.
6xy2 33xy 18x 3x(2 y2 11y 6)
Chapter 11: Factoring Polynomials
4. 5x2 61x 12 factors as (5x + 1)(x + 12), which is choice a. 5. Consider the factors of the first and last terms and the signs of the trinomial. Continue to check possible factors by multiplying until we get the middle term of the trinomial. 6. If the GCF has been factored out, then neither binomial can contain a common factor other than 1 or 1. This helps limit our choice of factors for one or both binomials since we cannot choose factors that would give the terms in either binomial a common factor. Exercise Set 11.3 2. 2 y2 2 y y 25 = 5 5 2 y2 15 y 25 (2 y 5)( y 5) 4. 6 y2 2 y 3y 10 = 5 2 6 y2 11y 10 (2 y 5)(3y 2) 6. 8 y2 2 y 4 y 55 = 5 11 8 y2 2 y 55 (2 y 5)(4 y 11)
Factors of 2 y2: 2 y2 2 y y Factors of 6: 6 = 1 6, 6 = 1 6, 6 = 2 3, 6 = 2 3
8. Factors of 3x2: 3x2 3x x Factors of 4: 4 = 1 4, 4 = 2 2
3x(2 y2 11y 6) 3x(2 y 1)( y 6)
3x2 8x 4 (3x 2)(x 2)
7. 5x2 19x 4 1(5x2 19x 4) Factors of 5x2: 5x2 5x x Factors of 4: 4 = 4 1, 4 = 4 1, 4 = 2 2 1(5x2 19x 4) 1(x 4)(5x 1)
10. Factors of 21x2: 21x2 21x x, 21x2 3x 7 x Factors of 10: 10 = 1 10, 10 = 2 5 21x2 41x 10 (7x 2)(3x 5) 12. Factors of 36r 2: 36r2 36r r, 36r2 18r 2r, 36r2 12r 3r, 36r2 9r 4r, 36r 2 6r 6r
Vocabulary, Readiness & Video Check 11.3 1. 2x2 5x 3 factors as (2x + 3)(x + 1), which is choice d.
Factors of 24: 24 = 1 24, 24 = 1 24, 24 = 2 12, 24 = 2 12, 24 = 3 8, 24 = 3 8, 24 = 4 6, 24 = 4 6 36r 2 5r 24 (9r 8)(4r 3)
2. 7 x2 9x 2 factors as (7x + 2)(x + 1), which is choice b. 3. 3x2 31x 10 factors as (3x + 1)(x + 10), which is choice c. Copyright © 2025 Pearson Education, Inc.
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14. Factors of 3x2: 3x2 3x x Factors of 63: 63 = 1 63, 63 = 1 63, 63 = 3 21, 63 = 3 21, 63 = 7 9, 63 = 7 9 3x2 20x 63 (3x 7)(x 9) 16. Factors of 12x2: 12x2 12x x, 12x2 6x 2x, 12x2 4x 3x
12x2 17x 5 (12x 5)(x 1) 2
2
2a2 11ab 5b2 (2a b)(a 5b) 32. Factors of 3r 2: 3r 2 3r r Factors of 8: 8 = 8 1, 8 = 8 1, 8 = 2 4, 8 = 2 4
34. Factors of 42a 2: 42a2 42a a, 42a2 21a 2a, 42a2 14a 3a,
18. y 8 y 9 8 y y 9 2
Factors of 5b2 : 5b2 5b b
3r 2 10r 8 (3r 2)(r 4)
Factors of 5: 5 = 1 5 2
30. Factors of 2a2: 2a2 2a a
42a2 7a 6a 2
Factors of 8 y : 8 y 8 y y, 8 y 4 y 2 y Factors of 9: 9 = 1 9, 9 = 1 9, 9 = 3 3 8 y2 y 9 ( y 1)(8 y 9) 20. Factors of 8x2: 8x2 8x x, 8x2 4x 2x
Factors of 6: 6 = 6 1, 6 = 3 2 42a2 43a 6 (7a 6)(6a 1) 36. Factors of 3n2: 3n2 3n n Factors of 5: 5 = 1 5 3n2 20n 5 is prime.
Factors of 3y 2 : 3y2 3y y 8x2 14xy 3y2 (4x y)(2x 3y) 22. Factors of 25n2: 25n2 25n n, 25n2 5n 5n Factors of 6: 6 = 6 1, 6 = 6 1, 6 = 2 3, 6 = 2 3
38. Factors of 24x2 : 24x2 24x x, 24x2 12x 2x, 24x2 8x 3x, 24x2 6x 4x Factors of 15: 15 = 1 15, 15 = 3 5 24x2 49x 15 (8x 3)(3x 5) 40. 8a3 14a2 3a a(8a2 14a 3)
25n2 5n 6 (5n 2)(5n 3)
Factors of 8a2: 8a2 8a a, 8a2 4a 2a 24. 7 x 12 x x 7 x 12 2
2
Factors of x2: x2 x x Factors of 12: 12 = 1 12, 12 = 2 6, 12 = 3 4 x 7x 12 (x 3)(x 4)
Factors of 3: 3 = 1 3 8a3 14a2 3a a(4a 1)(2a 3) 42. 12x2 14x 10 2(6x2 7x 5)
2
Factors of 6x2: 6x2 6x x, 6x2 3x 2x Factors of 5: 5 = 1 5, 5 = 1 5
26. Factors of 2x2: 2x2 2x x Factors of 72: 72 = 1 72, 72 = 1 72, 72 = 2 36, 72 = 2 36, 72 = 3 24, 72 = 3 24, 72 = 4 18, 72 = 4 18, 72 = 6 12, 72 = 6 12, 72 = 8 9, 72 = 8 9 2x2 7x 72 (2x 9)(x 8)
Factors of 4t 2: 4t2 4t t, 4t 2 2t 2t Factors of 7: 7 = 1 7, 7 = 1 7
422
44. 16t 15t 2 15 15t 2 16t 15 Factors of 15t 2 : 15t2 15t t, 15t2 5t 3t Factors of 15: 15 = 15 1, 15 = 15 1, 15 = 5 3, 15 = 5 3 15t2 16t 15 (5t 3)(3t 5)
28. 3t 4t 2 7 4t 2 3t 7
4t2 3t 7 (4t 7)(t 1)
12x2 14x 10 2(3x 5)(2x 1)
2 2 46. 8x y 34xy 84 y 2 y(4x 17x 42)
Factors of 4x2: 4x2 4x x, 4x2 2x 2x
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ISM: Prealgebra and Introductory Algebra Factors of 42: 42 = 42 1, 42 = 42 1, 42 = 21 2, 42 = 21 2, 42 = 14 3, 42 = 14 3, 42 = 7 6, 42 = 7 6 8x2 y 34xy 84 y 2 y(4x 7)(x 6) 2
Chapter 11: Factoring Polynomials
62. 6x3 28x2 16x 2x(3x2 14x 8) Factors of 3x2: 3x2 3x x Factors of 8: 8 = 1 8, 8 = 2 4 6x3 28x2 16x 2x(3x 2)(x 4)
2
48. Factors of 6x : 6x 6x x, 6x 3x 2x 2
Factors of 10: 10 = 1 10, 10 = 1 10, 10 = 5 2, 10 = 5 2 6x2 11x 10 (3x 2)(2x 5) 50. 5x2 75x 60 5(x2 15x 12) x2 15x 12 is prime, so 5(x2 15x 12) is a factored form of 5x2 75x 60.
Factors of 6x2: 6x2 6x x, 6x2 3x 2x Factors of 12: 12 = 1 12, 12 = 2 6, 12 = 3 4 12x3 34x2 24x 2x(3x 4)(2x 3) 4
3
2 2
66. 42x 99x y 15x y
3x2 (14x2 33xy 5 y2 ) Factors of 14x2: 14x2 14x x, 14x2 7x 2x
52. Factors of 54a 2: 5a2 54a a, 54a2 27a 2a, 54a2 18a 3a,
Factors of 5 y2: 5 y2 y 5 y,
54a2 9a 6a
5 y2 y 5 y
Factors of 8b2 : 8b2 8b b, 8b = 8b b, 8b = 4b 2b, 8b = 4b 2b
42x4 99x3 y 15x2 y2 3x2 (2x 5 y)(7x y)
54a2 39ab 8b2 (9a 8b)(6a b) 54.
3 2 2 64. 12x 34x 24x 2x(6x 17x 12)
2
2
x 4x 21 1(x 4x 21) 1(x 3)(x 7)
56. 6x3 31x2 5x x(6x2 31x 5)
68. 15x2 26x 8 1(15x2 26x 8) Factors of 15x2: 15x2 15x x, 15x2 5x 3x Factors of 8: 8 = 1 8, 8 = 2 4 15x2 26x 8 1(3x 4)(5x 2) 70. 9q4 42q3 49q2 q2 (9q2 42q 49)
Factors of 6x2: 6x2 6x x, 6x2 3x 2x
Factors of 9q2: 9q2 9q q, 9q2 3q 3q
Factors of 5: 5 = 1 5
Factors of 49: 49 = 1 49, 49 = 7 7
6x3 31x2 5x x(x 5)(6x 1)
9q4 42q3 49q2 q2 (3q 7)(3q 7) or
58. Factors of 36x2: 36x2 36x x, 36x2 9x 4x, 36x2 6x 6x Factors of 14: 14 = 1 14, 14 = 1 14, 14 = 7 2, 14 = 7 2 36x2 55x 14 (4x 7)(9x 2) 60. 24 y2 x 7 yx 5x x(24 y2 7 y 5) Factors of 24 y2: 24 y2 24 y y,
q2 (3q 7)2 2 2 72. 3x 8x 16 1(3x 8x 16)
Factors of 3x2: 3x2 3x x Factors of 16: 16 = 1 16, 16 = 1 16, 16 = 8 2, 16 = 8 2, 16 = 4 4 3x2 8x 16 1(3x 4)(x 4) 74. 5 55x 50x2 50x2 55x 5 5(10x2 11x 1)
24 y2 12 y 2 y, 24 y2 8 y 3y, 24 y2 6 y 4 y Factors of 5: 5 = 1 5, 5 = 1 5
Factors of 10x2: 10x2 10x x, 10x2 5x 2x Factors of 1: 1 = 1 1
24 y2 x 7 yx 5x x(8 y 5)(3y 1)
5 55x 50x2 5(10x 1)(x 1)
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Chapter 11: Factoring Polynomials
76. Factors of 4r 4: 4r4 4r2 r2 , 4r2 2r 2 2r2 Factors of 15: 15 = 1 15, 15 = 1 15, 15 = 3 5, 15 = 3 5 4
2
2
100. Factors of 2z 2: 2z2 2z z Factors of 7: 7 = 1 7, 7 = 1 7 (2z 1)(z 7) 2z2 13z 7
2
4r 17r 15 (4r 3)(r 5) 2 2
(2z 1)(z 7) 2z2 13z 7 (2z 7)(z 1) 2z2 5z 7
2 2
78. Factors of 3a b : 3a b 3ab ab Factors of 1: 1 = 1 1 There is no combination that gives the correct middle term, so 3a2b2 12ab 1 is prime. 5
3 2
(2z 7)(z 1) 2z2 5z 7 Since b is positive, the possible values are 5 and 13. 2 102. (3x 2)(x 2) 3x 8x 4
4
80. 5m 26m h 5mh 4 2 2 4 m(5m 26m h 5h )
(3x 5)(x 1) 3x2 8x 5 Therefore, c can be 4 or 5.
Factors of 5m4 : 5m4 5m2 m2 Factors of 5h4: 5h4 5h2 h2
104. answers may vary
5m5 26m3h2 5mh4 m(5m2 h2 )(m2 5h2 )
Section 11.4 Practice Exercises 1.
82. (2x 9)(2x 9) (2x)2 92 4x2 81 84. (x 3)2 x2 2x(3) 32 x2 6x 9 2 2 2 86. (3x 5) (3x) 2(3x)(5) 5
9x2 30x 25 88. The two shortest bars correspond to Snapchat and Pinterest, so these two apps have the lowest percent of U.S. users.
5x2 61x 12
5 12 = 60 1 60 = 60 1 + 60 = 61 5x2 61x 12 5x2 x 60x 12 x(5x 1) 12(5x 1) (5x 1)(x 12) 2. a.
3x2 14x 8 3x2 12x 2x 8 3x(x 4) 2(x 4) (x 4)(3x 2)
90. answers may vary 92. a, b, and d are not factored forms of 4x2 19x 12. The middle term on a is 14x, on b is 16x, and on d is 16x.
b.
94. 2(22 y 7) 2(3y2 ) 44 y 14 6 y2 6 y2 44 y 14 2(3y2 22 y 7) 2(3y 1)( y 7) 2 96. 27x 2x
1 1 1 3x 9x 9 3 3
(a 3)3(3x2 28x 25)
12x2 19x 5 12 5 = 60 15 4 = 60 15 + 4 = 19 12x2 19x 5 12x2 15x 4x 5 3x(4x 5) 1(4x 5) (4x 5)(3x 1) 2
98. 3x2 (a 3)3 28x(a 3)3 25(a 3)3
3x2 14x 8 3 8 = 24 12 2 = 24 12 + 2 = 14
3. a.
15 2 = 30 3 10 = 30 3 + (10) = 13
(a 3)3(3x 25)(x 1)
424
2
30x 26x 4 2(15x 13x 2)
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ISM: Prealgebra and Introductory Algebra
30x2 26x 4 2(15x2 13x 2) 2(15x2 3x 10x 2) 2[3x(5x 1) 2(5x 1)] 2(5x 1)(3x 2) b.
6x2 y 7xy 5 y y(6x2 7x 5) 6 5 = 30 3 10 = 30 3 + (10) = 7
Chapter 11: Factoring Polynomials
4. a = 4, b = 8, c = 3 a c = 4 3 = 12 2 6 = 12 and 2 + 6 = 8; choice d. 5. This gives us a four-term polynomial, which may be factored by grouping. Exercise Set 11.4 2. x2 5x 3x 15 x(x 5) 3(x 5) (x 5)(x 3)
6x2 y 7xy 5 y y(6x2 7x 5) y(6x2 3x 10x 5) y[3x(2x 1) 5(2x 1)] y(2x 1)(3x 5) 4. 12 y5 10 y4 42 y3 2 y3 (6 y2 5 y 21) 6 21 = 126 14 9 = 126 14 + (9) = 5
2
4.
z 10z 7z 70 z(z 10) 7(z 10) (z 10)(z 7)
6. 4x2 9x 32x 72 x(4x 9) 8(4x 9) (4x 9)(x 8) 8. 2 y4 10 y2 7 y2 35 2 y2 ( y2 5) 7( y2 5) ( y2 5)(2 y2 7)
12 y5 10 y4 42 y3
10. a.
2 y3(6 y2 5 y 21) 2 y3(6 y2 14 y 9 y 21) 2 y3[2 y(3y 7) 3(3y 7)]
2 12 = 24 2 + 12 = 14 2 and 12 are numbers whose product is 24 and whose sum is 14.
b. 14x = 2x + 12x
2 y3(3y 7)(2 y 3)
c.
5. 16x2 24x 9 16 9 = 144 12 12 = 144 12 + 12 = 144
8x2 14x 3 8x2 2x 12x 3 2x(4x 1) 3(4x 1) (4x 1)(2x 3)
16x2 24x 9 16x2 12x 12x 9 4x(4x 3) 3(4x 3) (4x 3)(4x 3) (4x 3)2
12. a.
10 3 = 30 10 + (3) = 13 10 and 3 are numbers whose product is 30 and whose sum is 13.
b. 13x = 10x 3x
Vocabulary, Readiness & Video Check 11.4 c. 1. a = 1, b = 6, c = 8 ac=18=8 4 2 = 8 and 4 + 2 = 6; choice a. 2. a = 1, b = 11, c = 24 a c = 1 24 = 24 8 3 = 24 and 8 + 3 = 11; choice c. 3. a = 2, b = 13, c = 6 a c = 2 6 = 12 12 1 = 12 and 12 + 1 = 13; choice b.
6x2 13x 5 6x2 10x 3x 5 2x(3x 5) 1(3x 5) (3x 5)(2x 1)
14. 15x2 11x 2 15 2 = 30 5 6 = 30 5 + 6 = 11 15x2 11x 2 15x2 5x 6x 2 5x(3x 1) 2(3x 1) (3x 1)(5x 2)
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ISM: Prealgebra and Introductory Algebra
28. 3x3 8x2 4x x(3x2 8x 4) 3 4 = 12 2 6 = 12 2+6=8
16. 8x2 x 9 8 9 = 72 9 8 = 72 9 + 8 = 1 8x2 x 9 8x2 9x 8x 9 x(8x 9) 1(8x 9) (8x 9)(x 1) 18. 30x2 23x 3 30 3 = 90 18 5 = 90 18 + (5) = 23 30x2 23x 3 30x2 18x 5x 3 6x(5x 3) 1(5x 3) (5x 3)(6x 1)
3x3 8x2 4x x(3x2 8x 4) x(3x2 2x 6x 4) x[x(3x 2) 2(3x 2)] x(3x 2)(x 2) 30. 4 y2 2 y 12 2(2 y2 y 6) 2 6 = 12 3 4 = 12 3 + (4) = 1 4 y2 2 y 12 2(2 y2 y 6) 2(2 y2 3y 4 y 6) 2[ y(2 y 3) 2(2 y 3)] 2(2 y 3)( y 2)
20. 2x2 7x 3 23=6 1 6 = 6 1 + (6) = 7 2x 7x 3 2x x 6x 3 x(2x 1) 3(2x 1) (2x 1)(x 3) 2
2
2
2
22. 20x 25x 4 25x 20x 4 25 4 = 100 10 10 = 100 10 + 10 = 20 25x2 20x 4 25x2 10x 10x 4 5x(5x 2) 2(5x 2) (5x 2)(5x 2) or (5x 2)2
32. 25x 12 12x2 12x2 25x 12 12 12 = 144 16 9 = 144 16 + (9) = 25 12x2 25x 12 12x2 16x 9x 12 4x(3x 4) 3(3x 4) (3x 4)(4x 3) 34. 30a2 38a 20 2(15a2 19a 10) 15 10 = 150 25 6 = 150 25 + (6) = 19 30a2 38a 20 2(15a2 19a 10)
24. 6x 11x 10 6 10 = 60 4 15 = 60 4 + (15) = 11 2
2(15a2 25a 6a 10) 2[5a(3a 5) 2(3a 5)] 2(3a 5)(5a 2)
6x2 11x 10 6x2 4x 15x 10 2x(3x 2) 5(3x 2) (3x 2)(2x 5) 26. 21x2 13x 2 21 2 = 42 6 7 = 42 6 + (7) = 13
36. 10a3 17a2 3a a(10a2 17a 3) 10 3 = 30 15 2 = 30 15 + 2 = 17 10a3 17a2 3a a(10a2 17a 3)
21x2 13x 2 21x2 6x 7x 2 3x(7 x 2) 1(7 x 2) (7x 2)(3x 1) 426
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a(10a2 15a 2a 3) a[5a(2a 3) 1(2a 3)] a(2a 3)(5a 1)
ISM: Prealgebra and Introductory Algebra
38. 30x3 155x2 25x 5x(6x2 31x 5) 6 5 = 30 1 30 = 30 1 + (30) = 31 30x3 155x2 25x 5x(6x2 31x 5) 5x(6x2 x 30x 5) 5x[x(6x 1) 5(6x 1)] 5x(6x 1)(x 5) 40. 6r2t 7rt2 t3 t(6r2 7rt t 2 ) 6 t 2 6t 2 t 6t 6t2 t + 6t = 7t 6r2t 7rt2 t3 t(6r2 7rt t2 ) t(6r2 rt 6rt t2 ) t[r(6r t) t(6r t)] t(6r t)(r t) 42. 36 1 = 36 There are no factors of 36 which sum to 6, so 36z2 6z 1 is prime. 2
2
2
6 5b2 30b2 6b 5b 30b2
x2 14x 33 x2 3x 11x 33 x(x 3) 11(x 3) (x 3)(x 11) or (3 x)(11 x) 50. 5 12x 7 x2 7 x2 12x 5 7 5 = 35 5 7 = 35 5 + (7) = 12 7x2 12x 5 7x2 5x 7 x 5 x(7x 5) 1(7x 5) (7x 5)(x 1) or (5 7x)(1 x) 2
30a2 5ab 25b2 5(6a2 ab 5b2 ) 5(6a2 6ab 5ab 5b2 ) 5[6a(a b) 5b(a b)] 5(a b)(6a 5b) 46. 20s4 61s3t 3s2t2 s2 (20s2 61st 3t2 ) 20 3t 2 60t 2 t 60t 60t2 t + 60t = 61t
2
54. (x 7)(x 7) (x 7)2 x2 2(x)(7) 72 x2 14x 49 56. (8 y 9)(8 y 9) (8 y)2 92 64 y2 81 58. (2z 1)2 (2z)2 2(2z)(1) 12 4z2 4z 1
7 4 y2 28 y2 4 y 7 y 28 y2 4y + 7y = 11y 3(7x2 11xy 4 y2 ) 3(7x2 4xy 7 xy 4 y2 ) 3[x(7x 4 y) y(7x 4 y)] 3(7x 4 y)(x y) The perimeter is 21x2 33xy 12 y2 or 3(7x + 4y)(x + y). 62. x2n 6xn 10xn 60 xn (xn 6) 10(xn 6)
2 2
20s 61s t 3s t
(x n 6)(xn 10)
s2 (20s2 61st 3t2 ) s2 (20s2 st 60st 3t 2 ) s2[s(20s t) 3t(20s t)] s (20s t)(s 3t) 2
2
52. ( y 5)( y 5) y 5 y 25
60. 3(7x2 11xy 4 y2 ) 21x2 33xy 12 y2
6b + (5b) = b
3
48. 33 14x x2 x2 14x 33 1 33 = 33 3 11 = 33 3 + 11 = 14
2
44. 30a 5ab 25b 5(6a ab 5b )
4
Chapter 11: Factoring Polynomials
64. 12 25 = 300 30 10 = 300 30 + (10) = 40 12x2n 40xn 25 12x2n 30xn 10xn 25 6xn (2xn 5) 5(2xn 5) (2xn 5)(6xn 5) Copyright © 2025 Pearson Education, Inc.
427
ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
Section 11.5 Practice Exercises
8. a.
8n2 40n 50 2(4n2 20n 25) 2[(2n)2 2 2n 5 52 ]
Since 36 62 and 12x = 2 6 x,
1. a.
2(2n 5)2
x2 12x 36 is a perfect square trinomial.
b. Since 100 102 and 20x = 2 10 x,
2. a.
3x(4x2 28x 49)
Since 9x2 (3x)2 and 25 52 , but
3x(2x 7)2
b. Since 4x2 (2x)2 , but 11 is not a perfect square, the polynomial is not a perfect square trinomial.
3x[(2x)2 2 2x 7 72 ]
2 2 2 9. x 9 x 3 (x 3)(x 3)
10. a2 16 a2 42 (a 4)(a 4) 2
Since 25x2 (5x)2 and 1 12 , and
3 3 3 c c c 11. c 25 5 5 5
2 5x 1 = 10x which is the opposite of 10x, the trinomial is a perfect square trinomial.
12. s2 9 is prime since it is the sum of two squares.
b. Since 9x2 (3x)2 and 49 72 , and 2 3x 7 = 42x which is the opposite of 42x, the trinomial is a perfect square trinomial. 4. x2 16x 64 (x)2 2 x 8 82 (x 8)
2 2
9
2
13. 9s2 1 (3s)2 12 (3s 1)(3s 1) 14. 16x2 49 y2 (4x)2 (7 y)2 (4x 7 y)(4x 7 y)
( p2 9)( p2 9)
(3r 4s)2 2
2
15. p4 81 ( p2 )2 92
2
5. 9r 2 24rs 16s2 (3r)2 2 3r 4s (4s)2
4
12x3 84x2 147x
x2 20x 100 is a perfect square trinomial.
20x 2 3x 5, the polynomial is not a perfect square trinomial.
3. a.
b.
2
( p2 9)( p 3)( p 3) 16. 9x3 25x x(9x2 25)
6. 9n 6n 1 (3n ) 2 3n 11
x[(3x)2 52 ] x(3x 5)(3x 5)
2
(3n2 1)2 17. 48x4 3 3(16x4 1) 7. Notice that this trinomial is not a perfect square trinomial. 9x2 (3x)2 and 4 22 , but 2 3x 2 = 12x and 12x is not the middle term 15x. Factor by grouping. 9 4 = 36 3 12 = 36 3 + 12 = 15 9x2 15x 4 9x2 3x 12x 4 3x(3x 1) 4(3x 1) (3x 1)(3x 4)
428
3[(4x2 )2 12 ] 3(4x2 1)(4x2 1) 3(4x2 1)(2x 1)(2x 1) 18. 9x2 100 1(9x2 100)
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1[(3x)2 102 ] 1(3x 10)(3x 10)
ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
19. 121 m2 112 m2 (11 m)(11 m) or
12. 81b2 (9b)2 13. 36 p4 (6 p2 )2
121 m2 m2 121 1(m2 121)
14. 4q4 (2q2 )2
1(m2 112 ) 1(m 11)(m 11)
15. No, it just means it won’t factor into a binomial squared. It may or may not be factorable.
Calculator Explorations x2 2x 1
x2 2x 1
(x 1)2
x=5
16
14
16
x = 3
16
14
16
x = 2.7
2.89
0.89
2.89
x = 12.1
171.61
169.61
171.61
1
1
1
x=0
The first and last terms are squares, a2 and b2, and the middle term is 2 a b or 2 a b. 17. in order to recognize the binomial as a difference of squares and also to identify the terms to use in the special factoring formula 18. A prime polynomial is one that can’t be factored further (using rational real numbers). Exercise Set 11.5
Vocabulary, Readiness & Video Check 11.5
2. Since 121 112 and 22x 2 11 x,
1. A perfect square trinomial is a trinomial that is the square of a binomial. 2
16.
2
2. The term 25 y written as a square is (5y) .
x2 22x 121 is a perfect square trinomial.
4. Since 16 42 but 4y 2 4 y, y2 4 y 16 is not a perfect square trinomial.
3. The expression x2 10x 25 y2 is called a perfect square trinomial.
6. Since 4 22 and 4p = 2 2 p, p2 4 p 4 is a perfect square trinomial.
4. The expression x2 49 is called a difference of two squares.
8. Since 144 122 but 20n 2 12 n,
5. The factorization (x + 5y)(x + 5y) may also be written as (x 5 y)2. 6. The factorization (x 5y)(x + 5y) may also be written as (x 5 y)2. false
n2 20n 144 is not a perfect square trinomial.
10. 25x2 (5x)2 but 2 y2 is not a perfect square, so 25x2 20xy 2 y2 is not a perfect square trinomial. 12. 36a2 (6a)2 , b2 (b)2 , and 12ab = 2 6a b,
7. The trinomial x2 6x 9 is a perfect square trinomial. false
so 36a2 12ab b2 is a perfect square trinomial.
8. The binomial y2 9 factors as ( y 3)2. false
14. x2 18x 81 x2 2 x 9 92 (x 9)2
9. 64 82
16. x2 12x 36 x2 2 x 6 62 (x 6)2
10. 9 32
18. 25x2 20x 4 (5x)2 2 5x 2 22 (5x 2)2
11. 121a2 (11a)2 Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
20. m4 10m2 25 (m2 )2 2 m2 5 52
44. 1 y2 1(1 y2 ) 1(12 y2 ) 1(1 y)(1 y)
(m2 5)2 22. 3 y2 6 y 3 3( y2 2 y 1)
or
3( y 2 y 11 ) 2
2
1 y2 y2 1 y2 12 ( y 1)( y 1)
3( y 1)2 24. 9 y2 48 y 64 (3y)2 2 3y 8 82 (3y 8)2
46. n4 16 (n2 )2 42 (n2 4)(n2 4)
26. 4x2 y2 28xy 49 (2xy)2 2 2xy 7 72
(n2 4)(n2 22 ) (n2 4)(n 2)(n 2)
(2xy 7)2 28. y3 12 y2 36 y y( y2 12 y 36) y( y2 2 y 6 62 ) y( y 6)2
48. x2 225 y2 x2 (15 y)2 (x 15 y)(x 15 y) 50. 32t 2 50 2(16t 2 25) 2[(4t)2 52 ] 2(4t 5)(4t 5)
30. Since 1 12 and x4 (x2 )2 , but 16x2 2 1 x2 , the polynomial is not a perfect square trinomial. Since there are no factors of 1 1 = 1 that sum to 16, the trinomial is prime. 32. 25x2 60xy 36 y2 (5x)2 2 5x 6 y (6 y)2 (5x 6 y)2
52. 36x2 y 25 y y(36x2 25) y[(6x)2 52 ] y(6x 5)(6x 5) 54. 25 y4 100 y2 25 y2 ( y2 4) 25 y2 ( y2 22 ) 25 y2 ( y 2)( y 2)
34. x2 36 x2 62 (x 6)(x 6) 36. 100 t 2 102 t 2 (10 t)(10 t) or
56. x3 y 4xy3 xy(x2 4 y2 ) xy[x2 (2 y)2 ] xy(x 2 y)(x 2 y)
100 t 2 t2 100 1(t2 100)
58. 225a2 81b2 9(25a2 9b2 )
1(t2 102 ) 1(t 10)(t 10)
9[(5a)2 (3b)2 ] 9(5a 3b)(5a 3b)
38. 9t 2 1 1(9t 2 1)
60. 12x2 27 3(4x2 9)
1[(3t)2 12 ] 1(3t 1)(3t 1)
3[(2x)2 32 ] 3(2x 3)(2x 3)
40. 36 y2 25 (6 y)2 52 (6 y 5)(6 y 5) 42. 49 y2 1 is the sum of two squares, which is prime.
430
62. 49a2 16 (7a)2 42 (7a 4)(7a 4) 64. 169a2 49b2 (13a)2 (7b)2 (13a 7b)(13a 7b)
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ISM: Prealgebra and Introductory Algebra
66. a2b2 16 (ab)2 42 (ab 4)(ab 4) 2
68. y
2
1 1 2 1 y y y 4 16 4 4 1
Chapter 11: Factoring Polynomials
86.
4a 2 0 4a 2 2 0 2 4a 2 4a 2 4 4 1 a2
4 2 2 70. 100 81 n2 102 9 n 2 2 10 n 10 n 9 9
2 88. x
72. 49 y2 100z2 (7 y)2 (10z)2 (7 y 10z)(7 y 10z)
90. ( y 6)2 z2 [( y 6) z][( y 6) z] ( y 6 z)( y 6 z)
74. x2 10xy 25 y2 x2 2 x 5 y (5 y)2
1
x2
25
2[(9a2 )2 2 9a2 2 22 ]
n5 (n2 1)(n2 1)
94. (x2 2x 1) 36 y2 (x2 2 x 112 ) 36 y2 (x 1)2 (6 y)2 [(x 1) 6 y][(x 1) 6 y] (x 1 6 y)(x 1 6 y)
98. 9x2 (3x)2 and 25 52.
n5 (n2 1)(n2 12 )
(3x 5)2 (3x)2 2 3x 5 52
n5 (n2 1)(n 1)(n 1) 80. 100x3 y 49xy3 xy(100x2 49 y2 ) xy[(10x)2 (7 y)2 ] xy(10x 7 y)(10x 7 y)
84.
y5 0 y 55 05 y 5 3x 9 0 3x 9 9 0 9 3x 9 3x 9 3 3 x3
96. x2n 121 (xn )2 112 (xn 11)(xn 11)
n5[(n2 )2 12 ]
82.
(n 8)(m2 32 ) (n 8)(m 3)(m 3)
2(9a2 2)2 9 5 5 4 78. n n n (n 1)
92. m2 (n 8) 9(n 8) (n 8)(m2 9)
(x 5 y)2 76. 162a4 72a2 8 2(81a4 36a2 4)
1 2 1 1 x x 5 5 5
9x2 30x 25, so the number 30 makes the given expression a perfect square trinomial. 100. (x 3y)2 x2 2 x 3y (3y)2 x2 6xy 9 y2 102. The difference of two squares is the result of multiplying binomials of the form (x + a) and (x a), so multiplying (5 + y) by (5 y) results in the difference of two squares. 104. A perfect square trinomial factoring formula is represented by the square. 106. a.
Let t = 2. 841 16t2 841 16(2)2 841 16(4) 841 64 777 After 2 seconds, the height of the object is 777 feet.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
b.
Let t = 5.
Mid-Chapter Review
841 16t 841 16(5) 841 16(25) 841 400 441 After 5 seconds the height of the object is 441 feet. 2
c.
2
When the object hits the ground, its height is zero feet. Thus, to find the time, t, when the object’s height is zero feet above the ground, we set the expression 841 16t equal to 0 and solve for t.
176. Factoring methods may vary. 1. x2 x 12 (x 3)(x 4) 2.
x2 10x 16 (x 8)(x 2)
3. x2 2x 1 x2 2 x 112 (x 1)2
2
4. x2 6x 9 x2 2 x 3 32
841 16t 0
(x 3)2
2
292 (4t)2 0 (29 4t)(29 4t) 0 29 4t 0 or 29 4t 0 4t 29 29 4t t 7.25 7.25 t Discard t = 7.25 since time cannot be negative. The object hits the ground after approximately 7 seconds. d.
5. x2 x 6 (x 2)(x 3) 6. x2 x 2 (x 2)(x 1) 7. x2 x 6 (x 3)(x 2) 8. x2 7x 12 (x 3)(x 4)
841 16t2 292 (4t)2 (29 4t)(29 4t) 9. x2 7x 10 (x 5)(x 2)
108. a.
Let t = 2. 784 16t2 784 16(2)2 720 After 2 seconds the performer’s height is 720 feet.
b.
12. 3x2 75 3(x2 25) 3(x2 52 ) 3(x 5)(x 5)
16(49 t 2 ) 0 16(7 t)(7 t) 0 7 t 0 or 7 t 0 7t t 7 Discard t = 7 since time cannot be negative. The performer reaches ground level after 7 seconds.
13. x2 3x 5x 15 x(x 3) 5(x 3) (x 3)(x 5)
784 16t2 16(49 t2 )
16. x2 3x 28 (x 7)(x 4)
16(72 t2 ) 16(7 t)(7 t)
432
2(x2 72 ) 2(x 7)(x 7)
When the performer reaches ground level, the height is 0. 784 16t2 0
d.
11. 2x2 98 2(x2 49)
Let t = 5. 784 16t2 784 16(5)2 384 After 5 seconds the performer’s height is 384 feet.
c.
10. x2 x 30 (x 6)(x 5)
14. 3y 21 xy 7x 3( y 7) x( y 7) ( y 7)(3 x) 15. x2 6x 16 (x 8)(x 2)
3 2 2 17. 4x 20x 56x 4x(x 5x 14) 4x(x 7)(x 2)
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
18. 6x3 6x2 120x 6x(x2 x 20) 6x(x 5)(x 4) 2
37. 12x3 y 243xy 3xy(4x2 81) 38. 6x3 y2 8xy2 2xy2 (3x2 4)
2
19. 12x 34x 24 2(6x 17x 12) 2(3x 4)(2x 3) 2
2
2
39. 2xy 72x3 y 2xy(1 36x2 ) 2xy[1 (6x)2 ] 2xy(1 6x)(1 6x)
2
20. 24a 18ab 15b 3(8a 6ab 5b ) 3(2a b)(4a 5b) 2
2
2
40. 2x3 18x 2x(x2 9)
2
21. 4a b (2a) b (2a b)(2a b)
2x(x2 32 ) 2x(x 3)(x 3)
22. x2 25 y2 x2 (5 y)2 (x 5 y)(x 5 y)
41. x3 6x2 4x 24 x2 (x 6) 4(x 6)
2
23. 28 13x 6x (4 3x)(7 2x)
(x 6)(x2 4) (x 6)(x 2)(x 2)
24. 20 3x 2x2 (5 2x)(4 x) 25. 4 2x x
3 2 2 42. x 2x 36x 72 x (x 2) 36(x 2)
2
is prime.
(x 2)(x2 36) (x 2)(x 6)(x 6)
26. a a2 3 is prime. 27. 6 y2 y 15 (3y 5)(2 y 3)
43. 6a3 10a2 2a2 (3a 5)
28. 4x2 x 5 (4x 5)(x 1)
44. 4n2 6n 2n(2n 3)
29. 18x3 63x2 9x 9x(2x2 7x 1)
45. 3x3 x2 12x 4 x2 (3x 1) 4(3x 1) (3x 1)(x2 4)
30. 12a3 24a2 4a 4a(3a2 6a 1)
46. x3 2x2 3x 6 x2 (x 2) 3(x 2) (x 2)(x2 3)
31. 16a2 56a 49 (4a)2 2 4a 7 72 (4a 7)2
47. 6x2 18xy 12 y2 6(x2 3xy 2 y2 ) 6(x 2 y)(x y)
32. 25 p2 70 p 49 (5 p)2 2 5 p 7 72 (5 p 7)2
2
2
2
2
48. 12x 46xy 8 y 2(6x 23xy 4 y ) 2(x 4 y)(6x y)
33. 14 5x x2 (7 x)(2 x)
49. 5(x + y) + x(x + y) = (x + y)(5 + x) 34. 3 2x x (3 x)(1 x) 2
50. 7(x y) + y(x y) = (x y)(7 + y)
4 3 2 2 2 35. 3x y 6x y 72x y 3x y(x 2x 24)
2
51. 14t 9t 1 (7t 1)(2t 1)
3x y(x 6)(x 4) 2
3
2 2
3
2
2
36. 2x y 8x y 10xy 2xy(x 4xy 5 y ) 2xy(x 5 y)(x y)
52. 3t 2 5t 1 is prime. 53. 3x2 2x 5 1(3x2 2x 5) 1(3x 5)(x 1)
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
54. 7x2 19x 6 1(7x2 19x 6) 1(7x 2)(x 3)
2 2 72. 3x 10xy 8 y (3x 2 y)(x 4 y)
73. x4 14x2 32 (x2 2)(x2 16) 55. 1 8a 20a2 (110a)(1 2a)
(x2 2)(x 4)(x 4) 74. x4 22x2 75 (x2 3)(x2 25)
56. 1 7a 60a2 (1 5a)(1 12a)
(x2 3)(x 5)(x 5)
4 2 2 2 57. x 10x 9 (x 9)(x 1) (x 3)(x 3)(x 1)(x 1)
75. answers may vary 76. Yes; 9x2 81y2 9(x2 9 y2 )
58. x4 13x2 36 (x2 9)(x2 4) (x 3)(x 3)(x 2)(x 2)
Section 11.6 Practice Exercises 1. (x 7)(x + 2) = 0 x 7 0 or x 2 0 x7 x 2 The solutions are 7 and 2.
59. x2 23x 120 (x 15)(x 8) 60. y2 22 y 96 ( y 16)( y 6) 2 2 2 2 61. 25 p 70 pq 49q (5 p) 2 5 p 7q (7q)
(5 p 7q)2 2
2
2
62. 16a 56ab 49b (4a) 2 4a 7b (7b) (4a 7b)2
2
63. x2 14x 48 is prime.
2. (x 10)(3x + 1) = 0 x 10 0 or 3x 1 0 x 10 3x 1 1 x 3 1 The solutions are 10 and . 3 3. a.
64. 7x2 24xy 9 y2 (7x 3y)(x 3y) 65. x2 x 30 1(x2 x 30) 1(x 5)(x 6) 2
b. x(4x 3) = 0 x = 0 or 4x 3 0 4x 3 3 x 4
2
66. x 6x 8 1(x 6x 8) 1(x 2)(x 4) 67. 3rs s 12r 4 s(3r 1) 4(3r 1) (3r 1)(s 4) 68. x3 2x2 x 2 x2 (x 2) 1(x 2) (x 2)(x2 1) 69. 4x2 8xy 3x 6 y 4x(x 2 y) 3(x 2 y) (x 2 y)(4x 3)
y(y + 3) = 0 y = 0 or y 3 0 y 3 The solutions are 0 and 3.
3 The solutions are 0 and . 4 4.
x2 3x 18 0 (x 6)(x 3) 0 x 6 0 or x 3 0 x6 x 3 The solutions are 6 and 3.
70. 4x2 2xy 7 yz 14xz 2x(2x y) 7z( y 2x) 2x(2x y) 7z(2x y) (2x y)(2x 7 z) 71. x2 9xy 36 y2 (x 12 y)(x 3y) 434
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ISM: Prealgebra and Introductory Algebra
9x2 24x 16
5.
9. (x 3)(3x2 20x 7) 0
9x2 24x 16 0 (3x 4)(3x 4) 0 3x 4 0 or 3x 4 0 3x 4 3x 4 4 4 x x 3 3 4 The solution is . 3
x3 0 or 3x2 20x 7 0 x 3 (3x 1)(x 7) 0 3x 1 0 or x 7 0 3x 1 x7 1 x 3 1 The solutions are 3, , and 7. 3
x(x 4) 5
6. a.
10. 5x3 5x2 30x 0
x 4x 5 2
5x(x2 x 6) 0 5x(x 3)(x 2) 0 5x 0 or x 3 0 or x 2 0 x0 x 3 x2 The solutions are 0, 3, and 2.
x 4x 5 0 (x 5)(x 1) 0 x 5 0 or x 1 0 x5 x 1 The solutions are 5 and 1. 2
Vocabulary, Readiness & Video Check 11.6
x(3x 7) 6
b.
3x2 7x 6 3x2 7x 6 0 (3x 2)(x 3) 0 3x 2 0 or x 3 0 3x 2 x 3 2 x 3 2 The solutions are and 3. 3 7.
1. An equation that can be written in the form ax2 bx c 0, (with a 0), is called a quadratic equation. 2. If the product of two numbers is 0, then at least one of the numbers must be 0. 3. The solutions to (x 3)(x + 5) = 0 are 3, 5. 4. If a b = 0, then a = 0 or b = 0. 5. One side of the equation must be a factored polynomial and the other side must be zero.
3x2 6x 72 0 3(x2 2x 24) 0 3(x 6)(x 4) 0 x6 0 or x 4 0 x 6 x4 The solutions are 6 and 4.
8.
Chapter 11: Factoring Polynomials
6. Because no matter how many factors you have in a multiplication problem, it’s still true that for a zero product, at least one of the factors must be zero. Exercise Set 11.6
2x 18x 0 3
2x(x2 9) 0 2x(x 3)(x 3) 0 2x 0 or x 3 0 or x 3 0 x0 x3 x 3 The solutions are 0, 3, and 3.
2. (x + 3)(x + 2) = 0 x3 0 or x 2 0 x 3 x 2 The solutions are 3 and 2. 4. (x + 4)(x 10) = 0 x40 or x 10 0 x 4 x 10 The solutions are 4 and 10.
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Chapter 11: Factoring Polynomials 6. (x 11)(x 1) = 0 x 11 0 or x 1 0 x 11 x 1 The solutions are 11 and 1.
22.
x3 x2 The solutions are 3 and 2.
8. x(x 7) = 0 x = 0 or x 7 0 x7 The solutions are 0 and 7. 10. 2x(x + 12) = 0 2x 0 or x 12 0 x0 x 12 The solutions are 0 and 12.
24.
x2 3x 0 x(x 3) 0 x = 0 or x 3 0 x3 The solutions are 0 and 3.
26.
x2 15x 0 x(x 15) 0 x = 0 or x 15 0 x 15 The solutions are 0 and 15.
12. (3x 2)(5x + 1) = 0 3x 2 0 or 5x 1 0 3x 2 5x 1 2 1 x x 3 5 2 1 The solutions are and . 3 5
28.
The solutions are
1
and
9 16.
4 3
(x 3)(x 3) 0 x 3 0 or x 3 0 x3 x 3 The solutions are 3 and 3. 2
436
30.
x 5x 24 2 x 5x 24 0 (x 8)(x 3) 0 x 8 0 or x 3 0 x8 x 3 The solutions are 8 and 3.
32.
(x 3)(x 8) x
.
4
2 1 x x 0 9 4 2 1 x 0 or x 0 9 4 1 2 x x 4 9 2 1 The solutions are and . 4 9
x2 11x 24 x x2 10x 24 0 (x 4)(x 6) 0
18. (x + 1.7)(x + 2.3) = 0 x 1.7 0 or x 2.3 0 x 1.7 x 2.3 The solutions are 1.7 and 2.3. 20.
x2 9 x 9 0 2
14. (9x + 1)(4x 3) = 0 9x 1 0 or 4x 3 0 9x 1 4x 3 1 3 x x 9
x2 5x 6 0 (x 3)(x 2) 0 x 3 0 or x 2 0
x40 or x 6 0 x 4 x 6 The solutions are 4 and 6.
x2 2x 63 0 (x 9)(x 7) 0 x9 0 or x 7 0 x 9 x7 The solutions are 9 and 7. Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
34.
x(4x 11) 3
Chapter 11: Factoring Polynomials
44.
4x 11x 3 2
4x 11x 3 0 (4x 1)(x 3) 0 4x 1 0 or x 3 0 4x 1 x3 1 x 4 1 The solutions are and 3. 4 2
36.
46. (2x 9)(x2 5x 36) 0 (2x 9)(x 9)(x 4) 0 2x 9 0 2x 9 9 x 2
36x2 x 21 0 (4x 3)(9x 7) 0 4x 3 0 or 9x 7 0 4x 3 9x 7 3 7 x x
4 y( y2 9) 0 4 y( y 3)( y 3) 0 4 y 0 or y 3 0 or y 3 0 y 3 y0 y 3 The solutions are 0, 3, and 3.
The solution is
.
(2x 5)(4x2 20x 25) 0 2
2
x4 0 x4
, 9, and 4.
(2x 5)[(2x) 2 2x 5 5 ] 0 (2x 5)(2x 5)(2x 5) 0 2x 5 0 2x 5 5 x 2 5 The solution is . 2
x 26 11x x2 11x 26 0 (x 13)(x 2) 0 x 13 0 or x 2 0 x 13 x2 The solutions are 13 and 2.
50.
9x2 7x 2 9x2 7x 2 0 (x 1)(9x 2) 0 x 1 0 or 9x 2 0 x 1 9x 2 2 x 9 2 The solutions are 1 and . 9
52.
10 42.
9
or
2
48.
4 y3 36 y 0
3
x9 0 x 9
2
9 3 7 The solutions are and . 4 9
40. 5(3 4x) 9 15 20x 9 20x 6 6 x 20 3 x 10
or
The solutions are
4
38.
4 y2 81 0 (2 y 9)(2 y 9) 0 2 y 9 0 or 2 y 9 0 2y 9 2 y 9 9 9 y y 2 2
12x2 7x 12 0 (4x 3)(3x 4) 0 4x 3 0 or 3x 4 0 4x 3 3x 4 4 3 x x4 3 3 4 The solutions are and . 4 3
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
54.
3x2 6x 9 0
66.
3(x2 2x 3) 0 3(x 3)(x 1) 0 x 3 0 or x 1 0 x3 x 1 The solutions are 3 and 1.
2x2 9x 5 0 (2x 1)(x 5) 0 2x 1 0 or x 5 0 2x 1 x 5 1 x 2 1 and 5. The solutions are 2
56. ( y 5)( y 2) 28 y2 7 y 10 28 y2 7 y 18 0 ( y 9)( y 2) 0 y 9 0 or y 2 0 y 9 y 2 The solutions are 9 and 2. 58.
9x2 14x 5 0 (9x 5)(x 1) 0 9x 5 0 or x 1 0 9x 5 x 1 5 x 9 5 The solutions are and 1.
x3 14x2 49x 0
9 70.
x2 22x 121 0 (x 11)(x 11) 0 x 11 0 x 11 The solution is 11.
72.
74.
5
5
5 4 5 3 20 15 5 9 4 12 3 36 36 36
3 12 312 36 7 17 7 17 119
76. After factoring, such as (x 4)(x + 2) = 0, set each factor equal to zero and solve. x 4 0 or x 2 0 x4 x 2 The solutions should be 4 and 2.
0 6 y2 9 y 0 3y(2 y 3) 3y 0 or 2 y 3 0 y0 2y 3 3 y 2
78. answers may vary, for example, x(x + 2) = 0
3 The solutions are 0 and . 2
80. answers may vary, for example x(x 1)(x 2) 0 x(x2 3x 2) 0
3
3x 27x 0 3x(x2 9) 0 3x(x 3)(x 3) 0 3x 0 or x 3 0 or x 3 0 x0 x3 x 3 The solutions are 0, 3, and 3.
438
2 3 2 7 3 3 14 9 23 3 7 3 7 7 3 21 21 21
9 12
62. 9 y 6 y2
64.
4x2 20x 5x2 6x 5
68.
x(x2 14x 49) 0 x(x 7)(x 7) 0 x = 0 or x 7 0 x7 The solutions are 0 and 7. 60.
2x2 12x 1 4 3x
x3 3x2 2x 0
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ISM: Prealgebra and Introductory Algebra
82. a.
b.
c. 84.
h 16t 2 64
Time, x (in seconds)
Height, y (feet)
0
16(0)2 100(0) 0
1
16(1)2 100(1) 84
2
16(2)2 100(2) 136
3
16(3)2 100(3) 156
4
16(4)2 100(4) 144
5
16(5)2 100(5) 100
6
16(6)2 100(6) 24
7
16(7)2 100(7) 84
When the rocket strikes the ground, its height is 0 feet, so it strikes the ground between 6 and 7 seconds. The maximum height of the rocket listed in the table is 156 feet, after 3 seconds. (2x 3)(x 6) (x 9)(x 2)
2x2 12x 3x 18 x2 2x 9x 18 2x2 9x 18 x2 7x 18
0 16t 2 64 0 16(t2 4) 0 16(t 2)(t 2) t 2 0 or t 2 0 t2 t 2 It takes the diver 2 seconds to reach the pool. 2. Let x be the number. x2 2x 63 x2 2x 63 0 (x 9)(x 7) 0 x 9 0 or x 7 0 x9 x 7 The numbers are 9 and 7. 3. Let x be the width. Then x + 5 is the length. Area length width 176 (x 5) x 176 x2 5x 0 x2 5x 176 0 (x 16)(x 11) x 16 0 or x 11 0 x 16 x 11 The width is 11 feet and the length is x + 5 = 11 + 5 = 16 feet. 4. Let x be the first odd integer. Then x + 2 is the next consecutive odd integer. x(x 2) x (x 2) 23
x2 16x 0 x(x 16) 0 x = 0 or x 16 0 x 16 The solutions are 0 and 16. 86.
Chapter 11: Factoring Polynomials
x2 2x2 2x 25 x 25
(x 6)(x 6) (2x 9)(x 4) x2 6x 6x 36 2x2 8x 9x 36 x2 36 2x2 x 36 0 x2 x 0 x(x 1) x = 0 or x 1 0 x 1 The solutions are 0 and 1.
x2 25 0 (x 5)(x 5) 0 x5 0 or x 5 0 x 5 x5 x 2 3 x27 The two consecutive odd integers are 5 and 3 or 5 and 7. 5. Let x be the length of one leg. Then x 7 is the length of the other leg. x2 (x 7)2 132
Section 11.7 Practice Exercises 1. The diver will be at a height of 0 when he or she reaches the pool.
x2 x2 14x 49 169 2x2 14x 120 0 2(x2 7x 60) 0 2(x 5)(x 12) 0
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Chapter 11: Factoring Polynomials x5 0 or x 12 0 x 5 x 12 Discard x = 5 since length cannot be negative. If x = 12, then x 7 = 5. The lengths of the legs are 5 meters and 12 meters. Vocabulary, Readiness & Video Check 11.7 1. In applications, the context of the stated application needs to be considered. Each translated equation resulted in both a positive and a negative solution, and a negative solution is not appropriate for any of the stated applications. Exercise Set 11.7 2. Let x be the width of the rectangle, then the length is 2x. 4. Let x be the first even integer, then the next consecutive even integer is x + 2. 6. Let x be the height of the trapezoid, then the base is 5x 3. 8. Area length width 84 (x 3)(x 2) 0 x2 x 90 0 (x 10)(x 9) x 10 0 or x 9 0 x 10 x9 For x = 10, x + 3 and x 2 are negative. Since lengths cannot be negative, x = 9 is the only solution that works in this context. x + 3 = 9 + 3 = 12 x2=92=7 The length is 12 inches and the width is 7 inches. 10. The perimeter is the sum of the lengths of the sides.
x2 4x 21 0 (x 3)(x 7) 0 x 3 0 or x 7 0 x3 x 7
440
For x = 7, 2x and 2x + 5 are negative. Since lengths cannot be negative, x = 3 is the only solution that works in this context. 2x = 2(3) = 6 2x + 5 = 2(3) + 5 = 6 + 5 = 11 x2 3 32 3 9 3 12 The sides have lengths 6 feet, 11 feet, and 12 feet. 2 12. Area (radius)
25 x2 25 x2 2 25 x 0 x2 25 0 (x 5)(x 5) x 5 0 or x 5 0 x5 x 5 Discard x = 5 since length cannot be negative. The radius is 5 kilometers. 14. The compass will hit the ground when its height is 0. 0 16t2 400
84 x2 x 6
(2x 5) (x2 3) 2x 29 x2 4x 8 29
ISM: Prealgebra and Introductory Algebra
0 16(t 2 25) 0 16(t 5)(t 5) t 5 0 or t 5 0 t 5 t 5 Since the time t cannot be negative, we discard t = 5. The compass hits the ground after 5 seconds. 16. Let x be the width. Then x + 9 is the length. Area length width 112 (x 9) (x) 112 x2 9x 0 x2 9x 112 0 (x 7)(x 16) x 7 0 or x 16 0 x7 x 16 Discard x = 16 since length cannot be negative. The width is 7 inches and the length is x + 9 = 7 + 9 = 16 inches.
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ISM: Prealgebra and Introductory Algebra
18. D
1
n(n 3)
Chapter 11: Factoring Polynomials
28.
2 1
x2 10x 25 100 x2 10x 75 0 (x 5)(x 15) 0 x 5 0 or x 15 0 x5 x 15 Discard x = 15. The original square had sides of 5 meters.
(15)(15 3) 2 1 (15)(12) 2 90 A polygon with 15 sides has 90 diagonals. D
20.
14
1
n(n 3)
30. Let x be the length of the shorter leg. Then the length of the other leg is x + 9.
2 1
n(n 3) 2 1 2 14 2 n(n 3) 2 28 n(n 3)
452 x2 (x 9)2 2025 x2 x2 18x 81 0 2x2 18x 1944 0 2(x2 9x 972) 0 2(x 27)(x 36) x 27 0 or x 36 0 x 27 x 36 Discard x = 36 since length cannot be negative. x + 9 = 27 + 9 = 36 The lengths of the legs are 27 centimeters and 36 centimeters.
0 n2 3n 28 0 (n 7)(n 4) n 7 0 or n 4 0 n7 n 4 Discard n = 4. The polygon has 7 sides. 22. Let x be the number. x x2 182 x2 x 182 0 (x 13)(x 14) 0 x 13 0 or x 14 0 x 13 x 14 The number is 13 or 14.
32.
x2 x 420 x x 420 0 (x 20)(x 21) 0 x 20 0 or x 21 0 x 20 x 21 Discard x = 21. The page numbers are 20 and x + 1 = 20 + 1 = 21. 2
26. Use the Pythagorean theorem where x + 10 = hypotenuse, x = one leg, and 30 = other leg. (x 10)2 x2 302 x 20x 100 x2 900 20x 800 x 40 x + 10 = 50 The guy wire is 50 feet.
area 60
24. Let x be the first number. Then x + 1 is the next consecutive number. x(x 1) 420
2
(x 5)2 100
1
base height
2 1
x (x 2) 2 1 2 60 2 x(x 2) 2 120 x(x 2) 0 x2 2x 120 0 (x 12)(x 10) x 12 0 or x 10 10 x 12 x 10 Discard x = 10. The base is 12 millimeters. 34. Let x be the length of the shorter leg. Then x + 10 is the length of the longer leg and 2x 10 is the length of the hypotenuse. (2x 10)2 x2 (x 10)2 4x2 40x 100 x2 x2 20x 100 2x2 60x 0 2x(x 30) 0 2x 0 or x 30 0 x0 x 30 Discard x = 0 since the length must be positive. The shorter leg of the triangle has length 30 miles.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
Discard x = 16 since length is positive. The length is 20 inches and the width is 1 1 x 2 (20) 2 10 2 8 inches. 2 2
36. When the object reaches the ground, the height is 0. h 16t 2 961 0 16t 2 961 0 1(16t2 961) 0 1(4t 31)(4t 31) 4t 31 0 or 4t 31 0 4t 31 4t 31 31 31 t t 4 4 31 Discard t since time must be positive. 4 31 The object reaches the ground after or 4 7.75 seconds.
42. Since the product of the two numbers is 34 more than their sum, x(x + 2) = x + (x + 2) + 34. x(x 2) x (x 2) 34 x2 2x 2x 36 x2 36 0 (x 6)(x 6) 0 x6 0 or x 6 0 x 6 x6 Since the number of visitors cannot be negative, x = 6. Then x + 2 = 6 + 2 = 8. In 2020, there were 6 million visitors to Natchez Trace Parkway and 8 million visitors to Lake Mead National Recreation Area.
38. Use A P(1 r)2 when P = 2000 and A = 2420. 2420 2000(1 r)2
44. Let C = 120 in C
2420 2000(1 2r r2 )
n(n 1) 2 1) n(n 2 120 2 2 240 n(n 1) 120
0 2000r2 4000r 420 2
0 20(100r 200r 21) 0 20(10r 1)(10r 21) 10r 1 0 or 10r 21 0 10r 1 10r 21 1 21 r r 10 10 21 Discard r since the interest rate must be 10 1 positive. The interest rate is 0.10 10%. 10 1 area length width 1 160 x x 2 2 1 160 x2 2x 2 1 2 160 2 x2 2x 2 2 320 x 4x
x 2 is the width.
2
0 x2 4x 320 0 (x 16)(x 20) x 16 0 or x 20 0 x 16 x 20 442
.
2
2420 2000 4000r 2000r2
40. Let x be the length. Then
n(n 1)
240 n2 n 0 n2 n 240 0 (n 16)(n 15) n 16 0 or n 15 0 n 16 n 15 Discard n = 15 since the number of telephones cannot be negative. 16 telephones are handled by the switchboard. 46. From the graph, there were approximately 700 thousand visitors to White Sands National Park in 2022. 48. From the graph, there were approximately 420 thousand visitors to White Sands National Park in 2020. 50. Look for the years where the graphed line for White Sands is lower than that for Mammoth Cave. The number of visitors to White Sands National Park was less than the number of visitors to Mammoth Cave National Park in 2012, 2014, 2015, and 2016. 52.
24 32
38 48
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3 4
ISM: Prealgebra and Introductory Algebra
54.
Chapter 11 Vocabulary Check
15 3 5 5 27 3 9 9 45
56.
50
1. An equation that can be written in the form ax2 bx c 0 (with a not 0) is called a quadratic equation.
5 9
9 5 10 10
58. Let x be the rate of the slower boat. Then x + 7 is the rate of the faster boat. x2 (x 7)2 172
2. Factoring is the process of writing an expression as a product. 3. The greatest common factor of a list of terms is the product of all common factors.
x x 14x 49 289 2
Chapter 11: Factoring Polynomials
2
2x2 14x 240 0 2(x 7x 120) 0 2(x 8)(x 15) 0 x 8 0 or x 15 0 x8 x 15 Discard 15 since the rate cannot be negative. The slow boat’s rate is 8 mph and the fast boat’s rate is x + 7 = 8 + 7 = 15 mph. 2
4. A trinomial that is the square of some binomial is called a perfect square trinomial. 5. In a right triangle, the side opposite the right angle is called the hypotenuse. 6. In a right triangle, each side adjacent to the right angle is called a leg. 7. The Pythagorean theorem states that (leg)2 (leg)2 (hypotenuse)2 .
60. Let x be one of the numbers. Since the numbers sum to 20, the other number is 20 x. x2 (20 x)2 218
Chapter 11 Review
x 400 40x x 218 2
2
1. 5m + 30 = 5 m + 5 6 = 5(m + 6)
2x2 40x 182 0 2(x2 20x 91) 0 2(x 13)(x 7) 0 x 13 0 or x 7 0 x 13 x7 If x = 13, then 20 13 = 7. If x = 7, then 20 7 = 13. The numbers are 13 and 7. 62. Dimensions of garden: length = 24 2x width = 16 2x area of garden length width 180 (24 2x)(16 2x) 180 384 80x 4x
2
0 4x2 80x 204 0 4(x2 20x 51) 0 4(x 17)(x 3) x 17 0 or x 3 0 x 17 x3 Discard x = 17 since the walkway must be less than the width of the total space. The walkway is 3 yards wide. 64. answers may vary
2. 6x2 15x 3x 2x 3x 5 3x(2x 5) 3.
4x5 2x 10x4 2x 2x4 2x 1 2x 5x3 2x(2x4 1 5x3)
4.
20x3 12x2 24x 4x 5x2 4x 3x 4x 6 4x(5x2 3x 6)
5. 3x(2x + 3) 5(2x + 3) = (2x + 3)(3x 5) 6.
5x(x 1) (x 1) 5x(x 1) 1(x 1) (x 1)(5x 1)
7. 3x2 3x 2x 2 3x(x 1) 2(x 1) (x 1)(3x 2) 8. 3a2 9ab 3b2 ab 3a(a 3b) b(3b a) (a 3b)(3a b) 9. 10a2 5ab 7b2 14ab 5a(2a b) 7b(b 2a) (2a b)(5a 7b)
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Chapter 11: Factoring Polynomials
10. 6x2 10x 3x 5 2x(3x 5) 1(3x 5) (3x 5)(2x 1)
ISM: Prealgebra and Introductory Algebra
25. Factors of 6x2: 6x2 6x x, 6x2 3x 2x Factors of 4 y2: 4 y2 4 y y, 4 y2 4 y y, 4 y2 2 y 2 y
11. x2 6x 8 (x 4)(x 2)
6x2 5xy 4 y2 (3x 4 y)(2x y) 12. x 11x 24 (x 8)(x 3) 2
26. x2 x 2 is prime.
13. x x 2 is prime. 2
27. 2 39 = 78 3 26 = 78 3 + (26) = 23
14. x 5x 6 (x 6)(x 1) 2
15. x2 2x 8 (x 4)(x 2) 16. x2 4xy 12 y2 (x 6 y)(x 2 y) 17. x 8xy 15 y (x 5 y)(x 3y) 2
2
2 2 18. 72 18x 2x 2x 18x 72
2(x2 9x 36) 2(x 3)(x 12) 2
2
19. 32 12x 4x 4x 12x 32 4(x2 3x 8) 20. 5 y3 50 y2 120 y 5 y( y2 10 y 24) 5 y( y 6)( y 4)
2x2 23x 39 2x2 3x 26x 39 x(2x 3) 13(2x 3) (2x 3)(x 13) 28. 18 20 y2 360 y2 15 y 24 y 360 y2 15 y (24 y) 9 y 18x2 9xy 20 y2 18x2 15xy 24xy 20 y2 3x(6x 5 y) 4 y(6x 5 y) (6x 5 y)(3x 4 y) 3
2
2
29. 10 y 25 y 60 y 5 y(2 y 5 y 12) 2 12 = 24 3 8 = 24 3 + 8 = 5 10 y3 25 y2 60 y 5 y(2 y2 5 y 12)
21. To factor x 2x 48, think of two numbers whose product is 48 and whose sum is 2. 2
22. The first step to factor 3x2 15x 30 is to factor out the GCF, 3. 23. Factors of 2x2: 2x x Factors of 6: 6 = 1 6, 6 = 2 3 2x2 13x 6 (2x 1)(x 6)
5 y(2 y2 3 y 8 y 12) 5 y[ y(2 y 3) 4(2 y 3)] 5 y(2 y 3)( y 4) 30. 60 y3 39 y2 6 y 3y(20 y2 13y 2) 20 2 = 40 5 8 = 40 5 + (8) = 13 60 y3 39 y2 6 y 3y(20 y2 13y 2) 3y(20 y2 5 y 8 y 2) 3y[5 y(4 y 1) 2(4 y 1)]
24. Factors of 4x2: 4x2 4x x, 4x2 2x 2x Factors of 3: 3 = 1 3, 3 = 1 3
3y(4 y 1)(5 y 2)
4x2 4x 3 (2x 3)(2x 1)
31. (x2 2) (x2 4x) (3x2 5x) x2 2 x2 4x 3x2 5x 5x2 9x 2 5 2 = 10 1 10 = 10 444
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ISM: Prealgebra and Introductory Algebra 1 + (10) = 9 2
2
5x 9x 2 5x x 10x 2 x(5x 1) 2(5x 1) (5x 1)(x 2) The perimeter is 5x2 9x 2 or (5x + 1)(x 2).
Chapter 11: Factoring Polynomials
41. x2 81 x2 92 (x 9)(x 9) 42. x2 12x 36 x2 2 x 6 62 (x 6)2 43. 4x2 9 (2x)2 32 (2x 3)(2x 3)
32. 2(2x2 3) 2(6x2 14x) 4x2 6 12x2 28x 16x2 28x 6
2 2 2 2 44. 9t 25s (3t) (5s) (3t 5s)(3t 5s) 2 2 2 45. 16x (4x) and y are both perfect squares,
2(8x2 14x 3) 8 3 = 24 2 12 = 24 2 + (12) = 14
but they are added instead of subtracted, so the polynomial is prime. 46. n2 18n 81 n2 2 n 9 92 (n 9)2
2(8x2 14x 3) 2(8x2 2x 12x 3) 2[2x(4x 1) 3(4x 1)] 2(4x 1)(2x 3)
47. 3r 2 36r 108 3(r 2 12 36) 3(r 2 2 r 6 62 ) 3(r 6)2
The perimeter is 16x2 28x 6 or 2(4x 1)(2x 3). 33. Since 9 3 and 6x = 2 3 x, x 6x 9 is a perfect square trinomial. 2
2
34. Since 64 82 , but 8x 2 8 x, x2 8x 64 is not a perfect square trinomial.
48. 9 y2 42 y 49 (3y)2 2 3y 7 72 (3y 7)2 49. 5m8 5m6 5m6 (m2 1) 5m6 (m2 12 ) 5m6 (m 1)(m 1)
2 2 2 35. 9m (3m) and 16 4 , but 2 3m 4 12m,
so 9m2 12m 16 is not a perfect square trinomial. 36. Since 4 y2 (2 y)2 and 49 72 and
50.
4x2 28xy 49 y2 (2x)2 2 2x 7 y (7 y)2 (2x 7 y)2
2 2 3 2 2 51. 3x y 6xy 3y 3y(x 2xy y )
2 2y 7 = 28y, 4 y2 28 y 49 is a perfect square trinomial. 2
2
2
37. Yes; x 9 or x 3 is the difference of two squares. 38. No; x2 16 or x2 42 is the sum of two squares. 39. Yes; 4x2 25 y2 or (2x)2 (5 y)2 is the difference of two squares. 40. No; 9a3 1 is not the difference of two squares [note: 9a2 1 or (3a)2 12 is the difference of two squares.]
3y(x y)2 52. 16x4 1 (4x2 )2 12 (4x2 1)(4x2 1) [(2x)2 12 ](4x2 1) (2x 1)(2x 1)(4x2 1) 53. (x + 6)(x 2) = 0 x6 0 or x 2 0 x 6 x2 The solutions are 6 and 2. 54. (x 7)(x + 11) = 0 x 7 0 or x 11 0 x7 x 11 The solutions are 7 and 11.
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445
ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials 55. 3x(x + 1)(7x 2) = 0 3x 0 or x 1 0 or 7x 2 0 2 x0 x 1 x 7 2 The solutions are 0, 1, and . 7 56. 4(5x + 1)(x + 3) = 0 5x 1 0 or x 3 0 1 x 3 x 5 1 The solutions are and 3. 5 57.
58.
x2 8x 7 0 (x 7)(x 1) 0 x7 0 or x 1 0 x 7 x 1 The solutions are 7 and 1. x2 2x 24 0 (x 4)(x 6) 0 x40 or x 6 0 x 4 x6 The solutions are 4 and 6. x2 10x 25
59.
x2 10x 25 0 (x 5)2 0 x5 0 x 5 The solution is 5. x(x 10) 16
60.
x2 10x 16 x 10x 16 0 (x 2)(x 8) 0 x 2 0 or x 8 0 x2 x8 The solutions are 2 and 8. 2
61.
446
(3x 1)(9x2 6x 1) 0 (3x 1)(3x 1)(3x 1) 0 3x 1 0 1 The solution is . 3
62.
63.
56x2 5x 6 0 (7x 2)(8x 3) 0 7x 2 0 or 8x 3 0 2 3 x x 7 8 2 3 The solutions are and . 7 8 m2 6m m2 6m 0 m(m 6) 0 m = 0 or m 6 0 m6 The solutions are 0 and 6. r 2 25
64.
r 2 25 0 (r 5)(r 5) 0 r 5 0 or r 5 0 r 5 r 5 The solutions are 5 and 5. 65. answers may vary, for example (x 4)(x 5) 0 x2 9x 20 0 66. answers may vary, for example [ x (1)][ x (1)] 0 (x 1)(x 1) 0 x2 2x 1 0 67. Let x be the width. Then 2x is the length. perimeter 2 length 2 width 24 2(2x) 2x 24 4x 2x 24 6x 4 x 8 = 2x The dimensions are 4 inches by 8 inches. The choice is c. 68. Let x be the width. Then 3x + 1 is the length. area length width 80 (3x 1) x 80 3x2 x 0 3x2 x 80 0 (3x 16)(x 5)
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ISM: Prealgebra and Introductory Algebra 3x 16 0 x
16
Chapter 11: Factoring Polynomials
or x 5 0 x5
72. Let x be the height. Then 4x is the base. 1 area base height 2 1 162 4x x 2 162 2x2
3 16 since length cannot be 3 negative. The dimensions are 5 meters by 3x + 1 = 3(5) + 1 = 15 + 1 = 16 meters. The choice is d. Discard x
0 2x2 162 0 2(x2 81) 0 2(x 9)(x 9) x 9 0 or x 9 0 x9 x 9 Discard x = 9 since length cannot be negative. The height is 9 yards and the base is 4x = 4(9) = 36 yards.
69. area side2 81 x2 0 x2 81 0 (x 9)(x 9) x 9 0 or x 9 0 x9 x 9 Discard x = 9 since length cannot be negative. The length of each side is 9 units.
73. Let x be the first positive integer. Then x + 1 is the next consecutive integer. x(x 1) 380 x2 x 380
70. The perimeter is the sum of the sides. (2x 3) (3x 1) (x2 3x) (x 3) 47
x2 x 380 0 (x 20)(x 19) 0 x 20 0 or x 19 0 x 20 x 19 Discard x = 20 since it is not positive. The integers are 19 and 19 + 1 = 20.
x 3x 7 47 2
x 3x 40 0 (x 5)(x 8) 0 2
x 5 0 or x 8 0 x5 x 8 Discard x = 8 since, for example, x + 3 would be negative. 2x + 3 = 2(5) + 3 = 10 + 3 = 13 3x + 1 = 3(5) + 1 = 15 + 1 = 16 x2 3x 52 3(5) 25 15 10 x+3=5+3=8 The lengths of the sides are 13 units, 16 units, 10 units, and 8 units. 71. Let x be the width. Then 2x 15 is the length. area length width 500 (2x 15) x 500 2x2 15x
74.
Let x be the first positive even integer. Then x + 2 is the next consecutive even integer. x(x 2) 440 2
x2 2x 440
x 2x 440 0 (x 22)(x 20) 0 x 22 0 or x 20 0 x 22 x 20 Discard x = 22 since it is not positive. The integers are 20 and 20 + 2 = 22. h 16t 2 440t
75. a.
0 2x2 15x 500 0 (2x 25)(x 20) 2x 25 0 or x 20 0 25 x 20 x 2 25 Discard x since length cannot be 2 negative. The width is 20 inches and the length is 2 20 15 = 40 15 = 25 inches.
2800 16t2 440t 16t 2 440t 2800 0 8(2t 2 55t 350) 0 8(2t 35)(t 10) 0 2t 35 0 or t 10 0 35 t 10 t or 17.5 2 The rocket reaches a height of 2800 feet at 10 seconds on the way up and at 17.5 seconds on the way down.
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ISM: Prealgebra and Introductory Algebra
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b. The height is 0 when the rocket reaches the ground. h 16t 2 440t 0 16t 440t 0 8t(2t 55) 8t 0 or 2t 55 0 t0 2t 55 t 27.5 The rocket reaches the ground again after 27.5 seconds. 2
87. 16x2 24x 9 (4x)2 2 4x 3 32 (4x 3)2 88. 5x2 20x 20 5(x2 4x 4) 5(x2 2 x 2 22 ) 5(x 2)2 89.
76. Let x be the length of the longer leg. Then x + 8 is the length of the hypotenuse and x 8 is the length of the shorter leg. (x 8)2 (x 8)2 x2 x2 16x 64 x2 16x 64 x2
x2 2x 15
90.
0 x 32x 0 x(x 32) x = 0 or x 32 0 x 32 The longer leg is 32 centimeters. 2
77. 6x + 24 = 6 x + 6 4 = 6(x + 4)
2x2 x 28 0 (2x 7)(x 4) 0 2x 7 0 or x 4 0 7 x4 x 2 7 The solutions are and 4. 2
x 2x 15 0 (x 3)(x 5) 0 2
x 3 0 or x 5 0 x 3 x5 The solutions are 3 and 5.
78. 7x 63 = 7 x 7 9 = 7(x 9)
91. 2x(x + 7)(x + 4) = 0 2x 0 or x 7 0 or x0 x 7
79. 11x(4x 3) 6(4x 3) = (4x 3)(11x 6)
92.
x2 5x 6 0 (x 3)(x 2) 0 x 3 0 or x 2 0 x3 x2 The solutions are 3 and 2.
81. 3x3 4x2 6x 8 x2 (3x 4)2 2(3x 4) (3x 4)(x 2)
2
2
83. 2x 2x 24 2(x x 12) 2(x 4)(x 3) 3 2 2 84. 3x 30x 27x 3x(x 10x 9) 3x(x 9)(x 1)
85. 4x2 81 (2x)2 92 (2x 9)(2x 9) 86. 2x2 18 2(x2 9) 2(x2 32 ) 2(x 3)(x 3)
448
x(x 5) 6 x2 5x 6
80. 2x(x 5) (x 5) 2x(x 5) 1(x 5) (x 5)(2x 1)
82. xy 2x y 2 x( y 2) 1( y 2) ( y 2)(x 1)
x40 x 4
93.
x2 16x x2 16x 0 x(x 16) 0 x = 0 or x 16 0 x 16 The solutions are 0 and 16.
94. The perimeter is the sum of the sides. (x2 3) 2x (4x 5) 48 x2 6x 8 48 x 6x 40 0 (x 4)(x 10) 0 x 4 0 or x 10 0 x4 x 10 Discard x = 10 since length, such as 2
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 2x = 2(10) = 20 cannot be negative. x2 3 42 3 16 3 19 2x = 2 4 = 8 4x + 5 = 4 4 + 5 = 16 + 5 = 21 The lengths are 19 inches, 8 inches, and 21 inches. 95. Let x be the length. Then x 4 is the width. area length width 12 x(x 4)
Chapter 11: Factoring Polynomials
Chapter 11 Test 1. 9x2 3x 3x(3x 1) 2. x2 11x 28 (x 7)(x 4) 3. 49 m2 72 m2 (7 m)(7 m) 4.
y2 22 y 121 y2 2 y 11112
12 x2 4x
( y 11)2
0 x2 4x 12 0 (x 6)(x 2) x 6 0 or x 2 0 x6 x 2 Discard x = 2 since length cannot be negative. The length is 6 inches and the width is x 4 = 6 4 = 2 inches.
5. x4 16 (x2 )2 (4)2 (x2 4)(x2 4) (x2 4)[(x)2 (2)2 ] (x2 4)(x 2)(x 2) 6. 4(a + 3) y(a + 3) = (a + 3)(4 y)
Chapter 11 Getting Ready for the Test 7. x2 4 is prime. 1. 10x 70x 2x 14x 4
3
2
2x 5x3 2x 35x2 2x x 2x 7 2x(5x3 35x2 x 7) The greatest common factor is 2x; B. 2. All of the choices are equivalent to 9 y3 18 y 2 . Since choice D is a difference, not a product, it is not a factored form of 9 y3 18 y2 .
8. y2 8 y 48 ( y 12)( y 4) 9. 3a2 3ab 7a 7b 3a(a b) 7(a b) (a b)(3a 7) 10. 3x2 5x 2 (3x 2)(x 1) 11. 180 5x2 5(36 x2 )
3. (x 1)(x + 5) is a factored expression; A.
5(62 x2 ) 5(6 x)(6 x)
4. z(z + 12)(z 12) is a factored expression; A. 5. y(x 6) + 1(x 6) is a sum, not a product, so it is not a factored expression; B.
12. 3x3 21x2 30x 3x(x2 7x 10) 3x(x 5)(x 2)
6. m m 5 5 is a difference, not a product, so it is not a factored expression; B.
13. 6t 2 t 5 6t2 5t 6t 5 t(6t 5) 1(6t 5) (6t 5)(t 1)
7. 4x2 16 4 x2 4 4 4(x2 4) Choice B is the correct factored form.
14. xy2 7 y2 4x 28 y2 (x 7) 4(x 7)
8. x2 64 is the sum of two squares with no common factors, so it cannot be factored using real numbers; A.
(x 7)( y2 4) (x 7)( y 2)( y 2)
9. Choice C is an incorrect next step.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
15. x x5 x(1 x4 ) x[12 (x2 )2 ]
22.
x(1 x22 )(1 x2 ) x(1 x )(1 x)(1 x) 16. x2 14xy 24 y2 x2 12xy 2xy 24 y2 x(x 12 y) 2 y(x 12 y) (x 12 y)(x 2 y)
t 2 2t 15 0 (t 3)(t 5) 0 t 3 0 or t 5 0 t 3 t 5 The solutions are 3 and 5. 6x2 15x
23.
6x2 15x 0 3x(2x 5) 0 3x 0 or 2x 5 0 5 x0 x 2
17. (x 3)(x + 9) = 0 x 3 0 or x 9 0 x3 x 9 The solutions are 3 and 9.
The solutions are 0 and
5
.
2 18.
x2 5x 14
24. Let x be the height. Then the base is x + 9. 1 area base height 2 1 68 (x 9) x 2 1 2 68 2 (x 9) x 2 136 x(x 9)
x2 5x 14 0 (x 7)(x 2) 0 x7 0 or x 2 0 x 7 x2 The solutions are 7 and 2. 19.
x(x 6) 7 x2 6x 7
136 x2 9x
x2 6x 7 0 (x 7)(x 1) 0 x7 0 or x 1 0 x 7 x 1 The solutions are 7 and 1.
0 x2 9x 136 0 (x 17)(x 8) x 17 0 or x 8 0 x 17 x8 Discard x = 17 since length cannot be negative. The height is 8 feet and the base is x + 9 = 8 + 9 = 17 feet.
20. 3x(2x 3)(3x + 4) = 0 3x 0 or 2x 3 0 or 3x 4 0 3 4 x0 x x 2 3 3 4 The solutions are 0, , and . 2 3 21.
5t3 45t 0 5t(t 2 9) 0 5t[(t)2 (3)2 ] 0 5t(t 3)(t 3) 0 5t 0 or t 3 0 or t 3 0 t0 t 3 t3 The solutions are 0, 3, and 3.
450
25.
(x 1)(x 2) 54 x 2x x 2 54 x2 x 56 0 (x 8)(x 7) 0 x 8 0 or x 7 0 x 8 x7 Discard x = 8 since length cannot be negative. x1=71=6 x+2=7+2=9 The width of the rectangle is 6 units and the length is 9 units. 2
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ISM: Prealgebra and Introductory Algebra
26. The object is at a height of 0 when it reaches the ground. h 16t 2 784
Chapter 11: Factoring Polynomials
4. x3 y3 (2)3 43 8 64 72 5.
8 n 1 8 1 n 11 9 n
6.
8x 72 8x 72 8 8
0 16t 2 784 0 16(t2 49) 0 16(t 7)(t 7) t 7 0 or t 7 0 t7 t 7 Discard t = 7 since time cannot be negative. The object reaches the ground after 7 seconds. 27. Let x be the length of the shorter leg. Then x + 10 is the length of the hypotenuse and x + 10 5 = x + 5 is the length of the longer leg. x2 (x 5)2 (x 10)2 2
2
x 9 7. 2y 6 + 4y + 8 = 2y + 4y 6 + 8 = 6y + 2 8. 5a 3 + a + 2 = 5a + a 3 + 2 = 4a 1 9. 4x + 2 5x + 3 = 4x 5x + 2 + 3 = x + 5
2
x x 10x 25 x 20x 100 x2 10x 75 0 (x 15)(x 5) 0 x 15 0 or x 5 0 x 15 x 5 Discard x = 5 since length cannot be negative. The shorter leg is 15 cm, the longer leg is 15 + 5 = 20 cm, and the hypotenuse is 15 + 10 = 25 cm. 28. The height of the object is 0 when it reaches the ground. h 16t2 1089 0 16t2 1089 0 1(16t2 1089) 0 1[(4t)2 332 ] 0 1(4t 33)(4t 33) 4t 33 0 or 4t 33 0 33 33 t or 8.25 t 4 4 33 Discard t since time cannot be negative. 4 The object reaches the ground after 8.25 seconds. Cumulative Review Chapters 111
10. 2x + 5x 6 = 7x 6 11.
3(3x 5) 10x 9x 15 10x 9x 9x 15 10x 9x 15 x
12.
2(7x 1) 15x 14x 2 15x 14x 14x 2 15x 14x 2 x
13. 3(2x 6) 6 0 6x 18 6 0 6x 12 0 6x 12 12 0 12 6x 12 6x 12 6 6 x2 14. 4(x 3) 12 0 4x 12 12 0 4x 0 4x 0 4 4 x0
1. x 2 y z 3 2(5) (4) 3 10 4 3 2. 5x y = 5(2) 4 = 10 4 = 14 3. 7 x2 7 (4)2 7 16 9 Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 11: Factoring Polynomials
15.
16.
21.
x 1 x 7 2 3 x 1 x 7 6 6 2 3 3(x 1) 2(x 7) 3x 3 2x 14 x 3 14 x 17
23. (53 )6 536 518
17. a. b. c. 18. a.
22. 3x 2 y 5 3x 2 y 5 3x 2 y 5 3 3 2y 5 x 3
24. (79 )2 792 718 25. ( y8 )2 y82 y16 11 3 113 x33 26. (x ) x
Nine is less than or equal to eleven is translated as 9 11.
27.
(x3 )4 x x7
x7
Three is not equal to four is translated as 3 4.
x12 x1 x7 x121 x7 x13
x7 x137
Five is greater than or equal to one is translated as 5 1.
3(x 4) 3x 12 3x 12 3x 12 3x 3x 12 3x 3x 12 12 12 Since this results in a true statement, all real numbers are solutions.
x34 x1
Eight is greater than one is translated as 8 > 1.
x6
b. Two is not equal to negative four translates as 2 4. 19.
V lwh V lwh wh wh V V l or l wh wh
x5 x2 3 5 x5 x 2 15 15 5 3 5(x 5) 3(x 2) 5x 25 3x 6 2x 25 6 2x 31 2x 31 2 2 31 x 2
( y3 )9 y 28.
y6
20. 2(x 5) 2x 9 2x 10 2x 9 10 9 Since 10 = 9 is a false statement, there is no solution.
y6 y27 y1 y6 y271 y6 y28 y6
y286 y22 3 6 6
29. ( y z )
452
y39 y1
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y
36 66
z
18 36
y z
y18
z36
ISM: Prealgebra and Introductory Algebra
4 2
30. (xy
31.
)
x7 (x4 )3
x7 x43
2 5
y
(y ) 32.
x
2
4 2
(y
)
x7
x12
10
x
y
44.
x2
x127
103
y
20a2b3 5ab 10b
1
x12(7)
10(3)
y3 3 y y
y8
2 8
1
Chapter 11: Factoring Polynomials
5ab
1 x19
20a2b3
10b 5ab 5ab 5ab 2 2 4ab 1 a
5ab
45. 5(x 3) y(x 3) (x 3)(5 y) 46. 9(y 2) + x(y 2) = (y 2)(9 + x)
13
y 47. x4 5x2 6 x4 2x2 3x2 6
33. 3x + 7x = (3 + 7)x = 4x
x2 (x2 2) 3(x2 2)
34. y + y = 2y
(x2 2)(x2 3)
35. 11x2 5 2x2 7 11x2 2x2 5 7
48. x4 4x2 5 x4 x2 5x2 5
13x2 2 2
x2 (x2 1) 5(x2 1)
2
2
36. 8 y y 4 y y 8 y 4 y y y
(x2 1)(x2 5)
2
12 y 2 y2
49. 6x2 2x 20 2(3x2 x 10) 2(3x2 6x 5x 10) 2[3x(x 2) 5(x 2)] 2(x 2)(3x 5)
37. (2x y)2 (2x y)(2x y) 2x(2x) 2x( y) ( y)(2x) ( y)( y) 4x2 2xy 2xy y2 4x 4xy y 2
50. 10x2 25x 10 5(2x2 5x 2)
2
5(2x2 4x x 2) 5[2x(x 2) 1(x 2)] 5(x 2)(2x 1)
38. (3x 1)2 (3x 1)(3x 1) 3x(3x) 3x(1) 1(3x) 1(1)
51. The height is 0 when the diver reaches the ocean.
9x2 3x 3x 1
h 16t 2 144
9x 2 6x 1
0 16t 2 144 0 16(t 2 9) 0 16(t 3)(t 3) t 3 0 or t 3 0 t 3 t 3 Discard t = 3 since time cannot be negative. The diver reaches the ocean after 3 seconds.
39. (t 2)2 t 2 2(t)(2) 22 t2 4t 4 40. (x 4)2 x2 2(x)(4) 42 x2 8x 16 41. (x2 7 y)2 (x2 )2 2(x2 )(7 y) (7 y)2 x4 14x2 y 49 y2 52. 2
2
2 2
2
2
42. (x 7 y) (x ) 2(x )(7 y) (7 y) 4
2
2x
x2 2x 120 0 (x 12)(x 10) 0 x 12 0 or x 10 0 x 12 x 10
4xy
The number is 12 or 10.
2
x 14x y 49 y 8x2 y2 16xy 2x 43.
4xy
8x2 y2
4xy 2xy 4
16xy 4xy 1
x2 2x 120
2y
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453
Chapter 12 2x2 5x 12
Section 12.1 Practice Exercises 1. a. b.
2. a.
x3 43 1 1 5x 1 5(4) 1 20 1 21
3.
x 8 0 x 8
x2 5x 4 0 (x 4)(x 1) 0 x40 or x 1 0 x 4 x 1 When x = 4 or x = 1, the expression x3 is undefined. 2 x 5x 4 x2 3x 2 The denominator of is never 0, 5 so there are no values of x for which this expression is undefined.
x4 x3 5x 5
x3 (x 1) 5(x 1)
x2 11x 18 4.
x2 x 2
x3
b.
454
(x 4)(2x 3)
(4 x)(4 x) (x 4)(2x 3) (1)(x 4)(4 x) 2x 3 (1)(x 4)
(x 9)(x 2) (x 1)(x 2)
2x 3 2x 3 or x4 x4
9. 3x 7 (3x 7) 3x 7 or x6 x6 x6 3x 7 3x 7 (x 6) x 6 Vocabulary, Readiness & Video Check 12.1 1. A rational expression is an expression that can be P written in the form Q , where P and Q are polynomials and Q 0. 2. The expression
3. The expression
x3 3 x x3 3 x
simplifies to 1.
simplifies to 1.
4. A rational expression is undefined for values that make the denominator 0.
5
x9
5. The expression
x4 x4 1 1 4 x (1)(x 4) 1
is undefined for x = 2.
6. The process of writing a rational expression in lowest terms is called simplifying. a a a 7. For a rational expression, . b b b 8.
x x7
x4 x4 1 4 x x4
7x x2
x 1
2 5. x 10x 25 (x 5)(x 5) x 5 x(x 5) x x2 5x x5 x5 1 6. x2 25 (x 5)(x 5) x 5
7. a.
undefined.
c.
16 x2
x 3 (3) 3 6 6 3 5x 1 5(3) 1 15 1 14 7
x When x = 8, the expression x 8 is
b.
8.
cannot be simplified.
9. Since 3 + x = x + 3,
3 x can be simplified. x3
10. Since 5 x = 1(x 5),
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5 x can be simplified. x5
ISM: Prealgebra and Introductory Algebra
11.
x2 cannot be simplified. x 8
12. replacement values of variables; by evaluating the expression for different replacement valuesvariables are replaced with these values and the expression is simplified 13. Rational expressions are fractions and are therefore undefined if the denominator is zero; if a denominator contains variables, set it equal to zero and solve. 14. Although x is a factor in the numerator, it is not a factor in the denominatorfactor means to write as a product, and the denominator is a difference, not a product. 15. We would need to write parentheses around the numerator or denominator if it had more than one term because the negative sign needs to apply to the entire numerator or denominator.
Chapter 12: Rational Expressions 14. 5x 2 0 5x 2 2 x 5 x 1 2 is undefined for x . 5x 2 5 16.
19x3 2 x2 x
20.
x 8 2 8 10 x 1 2 1 3
7 y 1 7(2) 1 14 1 15 4. 5 y 1 2 1 3 3 6.
z z 5
2
8.
5
5 5 1 4 (5) 5 25 5 20
x5 x2 4x 8
2
25 22 4(2) 8
10. 5x 0 x0 3 is undefined for x = 0. 5x 12. x 9 0 x9 5x 1 is undefined for x = 9. x9
7 7 488 4
is undefined for x = 0 and x = 1.
18. Since 9 0, there are no values of x for which 9 y5 y3 is undefined. 9
Exercise Set 12.1 2.
x2 x 0 x(x 1) 0 x = 0 or x 1 0 x 1
x2 5x 14 0 (x 7)(x 2) 0 x 7 0 or x 2 0 x7 x 2 11x2 1 x2 5x 14
is undefined for x = 7 and x = 2.
22. 2x2 15x 27 0 (2x 9)(x 3) 0 2x 9 0 or x 3 0 2x 9 x 3 9 x 2 9 x is undefined for x 2x2 15x 27 x = 3. 24.
y9 y9 1 9 y y9
26.
y 9 1( y 9) 1(9 y) 1 9 y 9 y 9 y
28.
3 3 1 9x 6 3(3x 2) 3x 2
30.
x5 2
x 25
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2
and
x5 1 (x 5)(x 5) x 5
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
32.
3x 9 3(x 3) can’t be simplified. 4x 16 4(x 4)
34.
4x 4 y 4(x y) 4 x y x y 9x 99
36.
x2 11x
9(x 11) x(x 11)
x 3 38.
x2 6x 9
x
(x 3)(x 3)
1 x3
2 40. 12x 4x 1 (6x 1)(2x 1)
2x 1 x4 10x3 42.
44.
x2 17x 70
2x 1
x3 (x 10) (x 7)(x 10)
6x 1
4x2 24x 4x(x 6) 4x x6 x6
5x2 500 5(x2 100) 35x 350 35(x 10) 5(x 10)(x 10) 5 7(x 10) x 10 7
x2 16 x 8x 16 2
456
(x 4)(x 4) (x 4)(x 4)
2x( y 1) 3( y 1)
2x( y 2) 3( y 2) ( y 1)(2x 3) ( y 2)(2x 3) y 1 y2
62. x 11 (x 11) x4 x4 x 11 x4 x 11 (x 4) x 11 x 4 8 y 1 (8 y 1) y 15 y 15 8 y 1 y 15 8 y 1 ( y 15) 8 y 1 y 15
66.
2xy 4x 3y 6
64.
2 49 y2 ( y 49) y7 y7 ( y 7)( y 7) y7 ( y 7)
54.
60.
x7
2 48. 4x 4x 1 (2x 1)(2x 1) 2x 1 (2x 1)(x 5) x5 2x2 9x 5
52.
2xy 2x 3y 3
x3
2 46. 2x 7x 4 (2x 1)(x 4) 2x 1 (x 1)(x 4) x 1 x2 3x 4
50.
ab ac b2 bc a(b c) b(b c) bc bc (a b)(b c) bc ab
58. xy 6x 2 y 12 x( y 6) 2( y 6) y( y 6) y2 6 y (x 2)( y 6) y( y 6) x2 y
9
x3
56.
x4 x4
100 x2 100 x2 (10 x) x 10 (10 x)(10 x) (10 x) 10 x 1 10 x The given answer is correct.
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ISM: Prealgebra and Introductory Algebra 2 68. 2 15x 8x (2 x)(1 8x) (8x 1)(8x 1) 64x2 1 (x 2)(1 8x) (8x 1)(1 8x) x2 8x 1 The given answer is correct.
70.
72.
1
46 2 23 23 6 86. 54 2 27 27 5 The given answer is incorrect. 88. answers may vary 90. no; answers may vary 92. a.
5 2 2 5 2 27 5 27 5 27 7
Chapter 12: Rational Expressions
7 2 7 2 7 8 1 2 2 2 1 4
8 74.
2
8 5 8 8 8 8 64 15 8 15 5 15 5 75
76. Use S
H D 2T 3R
with H = 152, D = 24,
b.
B T = 1, R = 22, and B = 522. 152 24 2(1) 3(22) 244 S 0.467 522 522 Carlos Correa’s 2022 slugging percentage was 0.467. H D 2T 3R with H = 158, D = 39, B T = 0, R = 28, and B = 527. 158 39 2(0) 3(28) 281 S 0.533 527 527 José Altuve’s 2022 slugging percentage was 0.533.
78. Use S
80. Use S
H D 2T 3R
with H = 172, D = 37, B T = 1, R = 32, and B = 578. 172 37 2(1) 3(32) 307 S 0.531 578 578 Manny Machado’s 2022 slugging percentage was 0.531.
82. José Altuve had the second-greatest slugging percentage in 2022. 7m 9
84.
9 m 7 7 7 7 The given answer is incorrect.
7m
9
c.
250(100) 10, 000 100 25, 000 10, 000 100 35, 000 100 350 The cost per fax machine when producing 100 fax machines is $350. C
250(1000) 10, 000 1000 250, 000 10, 000 1000 260, 000 1000 260 The cost per fax machine when producing 1000 fax machines is $260. C
decrease; answers may vary
94. Use B
703w
with w = 198 and h = 68. h2 703w 703(198) 139,194 B 30 4624 h2 682 The body-mass index is approximately 30. R C R with R = 151.7 and C = 129.0. R C 151.7 129.0 22.7 P R 151.7 151.7
96. Use P
Ford’s gross profit margin in 2022 was 15%. Section 12.2 Practice Exercises 1. a.
16 y 1 16 y 1 16 y 3 x2 3 x2 3x2
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457
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
5a3 2b2 5a3 2b2 3b3 15a 3b3 15a 5 a3 2 b2 3 b32 3 5 a 2a 9b
b.
3x 6
b.
3(x 2) 7x2
7x2
3 3 14 x 2x2 2 7 x2 (x 2) 2
3.
4x 8 3x2 5x 2 7x2 14x 9x2 1
9. 288 sq in.
4(x 2) (3x 1)(x 2) 7x(x 2) (3x 1)(3x 1) 4(x 2)(3x 1)(x 2) 7x(x 2)(3x 1)(3x 1) 4(x 2) 7x(3x 1)
4.
6
10x 4 x2 4
3x 12 2
(x 4)2
2
5x3 2x2 10x 4 x2 2 3 x2 x 4 5x 2x2 2(5x 2)(x 2) (x 2)(x 2) x2 (5x 2) 2 2 x (x 2)
3x2 10x 8 9x 12 7. 7x2 14 21 3x 10x 8 21 7x 14 9x 12 (3x 4)(x 2) 3 7 7(x 2) 3(3x 4) 1
458
2 sq ft
144 sq in.
600, 003 sq ft
9 sq ft 1 sq yd
12. 102.7 feet/second 102.7 feet 3600 seconds 1 mile 1 second 1 hour 5280 feet 102.7 3600 miles/hour 5280 70.0 miles/hour or 70.0 mi/hr Vocabulary, Readiness & Video Check 12.2 1. The expressions
x 2y
reciprocals. 2.
a c ac b d bd
3.
a c ad b d bc
x x x2 4. 7 6 42 5.
x 7
8. a.
1
1 sq ft
11. 66, 667 sq yd 66, 667 sq yd
3x 12 6 (x 4)(x 4) 2 2 3 3(x 4) x4 9
288 sq in.
3.5 sq ft 144 sq in. 504 sq in. 1 1 sq ft
7x2 2 y 7 x x 2 y 7xy 2 3 x 3 6 2y 6 x
5.
6.
10. 3.5 sq ft
x
(x 4)2
2
3 x 2x 10 x2 6x 5 x2 7x 12 (3 x) 2(x 5) (x 5)(x 1) (x 4)(x 3) 1(x 3) 2 (x 1)(x 4)(x 3) 2 (x 1)(x 4)
c.
2.
7x2
x3 7 x 3 x 3 (x 3) x x3 x 7 7x
x 6
6 7
x3 7 (x 3) 7 7 x x 3 x (x 3) x
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and
2y x
are called
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
6. Yes, multiplying and simplifying rational expressions often require polynomial factoring. Video Notebook Number 2 alone involves factoring out a GCF, factoring a trinomial with a 1, and factoring a difference of squares.
16.
7. Dividing rational expressions is exactly like dividing fractions. Therefore, to divide by a rational expression, multiply by its reciprocal.
9 y4 18.
8. Multiplication and division of rational expressions are performed similarlyboth involve multiplicationbut there are important differences. Note the operation first to see whether you multiply by the reciprocal or not. 9. We’re converting to cubic feet, so we want cubic feet in the numerator. We want cubic yards to divide out, so cubic yards is in the denominator. Exercise Set 12.2
65 x x x
5x
y2
6y
9 y4
3 2 6y y 3 3 3 3 y y y y 3 2 y y y 3 3 y 2 9y 2
2
7a b
20.
2 2. 9x 4 y3 3 3 4 x x y 12 3 x x x y x y 3x
6x2
x2 9x 20 x2 11x 28 x2 15x 44 x2 12x 35 (x 4)(x 5) (x 7)(x 4) (x 4)(x 11) (x 7)(x 5) (x 4)(x 7) (x 11)(x 7)
2 2
21a b
3b2
14b
1
10x 3 12 5 2 6 2 x x x 4 9x3 y2 6. 3 9x3 y2 y3 5 1 5 y 18xy 18xy 9 x x x y y y y y 9 2 x y y y y y x x 2 x2 2
7a b 14b 3ab2 21a2b2 27 7 a a bb 3 3 7 a a b b b b 2 7 3 3 b b
4x 24
20x 10.
x6
4(x 6) 4 5x
5 x6
(x 3)2 22.
5
5x 15 25
mn
m 2
m mn
m nm n
25 5 5x 15 (x 3)(x 3) 5 5 5 5(x 3) x3 1 x3
2 9x5 27x 9x5 3b 3a 2 3b 3a a2 b2 27x2 24. a b
1
9x2 x3 3(a b) (a b)(a b) 3 9x2 x3 ab
x
(m n)(m n)
(x 3)2
2
x(x 1) 8 2 x 2 x2 x 16 2x 8 x 1 8 x 1 1 (m n)2
12.
5
14 9b2
4.
8.
2
(m n)2
m m(m n)
26.
mn
mn
a 4a 4 a 3 (a 2)(a 2) a 3 a 3 14. a 2 (a 2)(a 2) a 2 a 2 a2 4 2
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m2 mn m
(m n)2
m m n m2 mn (m n)(m n) m mn m(m n) mn mn
459
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions x2 3x 18 x 3 x2 2x 8 28. 2 x x2 2x 8 2 x x2 3x 18 x 3 (x 4)(x 2) 2 x (x 6)(x 3) x 3 (x 4)(x 2) (x 2) (x 6)(x 3) x4 x6 x3
x3
x 3 (x 3)2 40. 2 x 9 (x 3)2 x2 9 5x 15 x3 (x 3)(x 3) (x 3)(x 3) 5(x 3) x3 5(x 3) 3y 42.
30.
32.
x 1 20x 100 2 2x 5x 3 2x 3 x 1 2x 3 (x 1)(2x 3) 20x 100 x 1 2x 3 (x 1)(2x 3) 20(x 5) 1 20(x 5) 6x 6 9x 9 6x 6 10 5 10 5 9x 9 6(x 1) 5 2 5 9(x 1) 3 2 2 3 3 4 3 3x2 12x
34.
6
9 2x 8
3x(x 4)
3 2 3 3 x 22 9x 4
3 3 2(x 4)
3(x 2) 4 2 2 10 4(x 2) 3 10
x2 y2 3x2 6x 38. 3x2 3xy 3x2 2xy y2 (x y)(x y) 3x(x 2) 3x(x y) (3x y)(x y) x2 3x y
460
x2 9 x2 9 3 x 12xy 3y (x 3)(x 3) 3 x 12xy 3y (x 3)(x 3) 3 4xy (x 3) x3 4x x3 4x
12xy
b2 2b 3
44.
3y
b2 4
b2(b b 2 b2 6b 8 3)(b 1) (b 2)(b 2) (b 2)(b 1) (b 2)(b 4) (b 3)(b 2) (b 2)(b 4)
9x 18 4x2 11x 6 46. 4x2 3x x2 4 9(x 2) (4x 3)(x 2) x(4x 3) (x 2)(x 2) 9 x
48.
36. 3x 6 4x 8 3x 6 8 20 8 20 4x 8
3 x
5x 15
36n2 64 3n2 5n 12 3n2 10n 8 n2 9n 14 36n2 64 n2 9n 14 3n2 10n 8 3n2 5n 12 2 4(9n 16) (n 7)(n 2) (3n 4)(n 2) (3n 4)(n 3) 4(3n 4)(3n 4) (n 7)(n 2) (3n 4)(n 2) (3n 4)(n 3) 4(n 7) n3
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
50. 1008 square inches
1008 square inches
1 7 square feet 52.
2
Chapter 12: Rational Expressions
1 square foot 144 square inches
2 square yards 9 square feet 144 square inches 1 1 square yard 1 square feet 2592 square inches
square yards
2 cubic yards 27 cubic feet 1728 cubic inches 1 1 cubic yard 1 cubic foot 93, 312 cubic inches
54. 2 cubic yards
56.
58.
10 feet 10 feet 1 mile 3600 seconds 1 second 1 second 5280 feet 1 hour 6.8 miles per hour 3.6
3.6 square yards 9 square feet 1 1 square yard 32.4 square feet
square yards
3, 705, 793 square feet 1 square yard 1 9 square feet 411, 755 square yards
60. 3, 705, 793 square feet
200 miles 5280 feet 1 hour 1 hour 1 mile 3600 seconds 293.3 ft/sec
62. 200 miles per hour
64.
3
6
15 15 66.
3 6 9 3 3 3 15 15 3 5 5
4 8 4 8 4 4 3 3 3 3 3
68. 3 1 3 3 2 5 2 2 2 2 2 2
70.
2 2 4 1 2 3 4 12 3 7 The statement is false.
72. 7 3 21 21 a a a2 a The statement is false.
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461
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
74.
2x 2x 4x2 5x 3 5x 3 (5x 3) 2
3.
2x2 5x x2
4x 6 x2
4x2
The area is square meters. (5x 3)2 2 2 x 9 x 2x 1 2x 3 76. 2 x 1 2x2 9x 9 1 x x2 9 x2 2x 1 1 x 2 x 1 2x2 9x 9 2x 3 (x 3)(x 3) (x 1)(x 1) (x 1) (x 1)(x 1) (2x 3)(x 3) 2x 3 (x 3)(x 1) (2x 3)2
LCD 32 5 9 5 45
5 11 , 6x3 18x5
b.
6x3 2 3 x3 and 18x5 2 32 x5
2 2
x y xy 8x 8 y y x 4x 4 y 3y 3x 8 xy(xy 1) 8(x y) y x 4(x y) 3( y x) 8 xy(xy 1) 12
LCD 2 2 x5 2 9 x5 18x5
5.
3a
, 7a a5 a5 LCD = (a + 5)(a 5) 7x2
6.
80. answers may vary 82. 30 gallons per minute 30 gallons 1 minute 1 cubic foot 1 minute 60 seconds 7.48 gallons 0.067 cubic foot per second The flow rate is not in violation of the regulations.
5x ,
(x 4)2 3x 12 (x 4)2 (x 4)(x 4) and 3x 12 = 3(x 4) LCD 3(x 4)(x 4) 3(x 4)2
7.
y5 2
y 2y 3
,
y4 2
y 3y 2
y2 2 y 3 ( y 3)( y 1)
Section 12.3 Practice Exercises
2.
x2 2x2 5x 4x 6 x2 2x2 x 6 x2 (2x 3)(x 2) x2 2x 3
9 32 and 15 = 3 5
1.
2x2 5x (4x 6)
2 7 , 9 15
4. a.
x2 y2 xy 3y 3x y x 78. 4x 4 y 8x 8 y 8
y 2 3y 2 ( y 2)( y 1)
8x x 8x x 9x 3x 3y 3y y 3y 3y
LCD = (y + 3)(y 1)(y 2)
3x 7 3x 7 1 or 1 3x 7 3x 7 3x 7 1
8.
6
,
9
x4 4 x (4 x) = (x 4) LCD = (x 4) or (4 x) 2x 4x2 y 8x3 y 9. 1 5y 5y 5 y 4x2 y 20x2 y2 2x
462
2x
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ISM: Prealgebra and Introductory Algebra 3
10.
3
2
x 25
Chapter 12: Rational Expressions 8 y 16 8( y 2) 8y 16 8 y2 y2 y2 y2
6.
(x 5)(x 5) 3 x3 (x 5)(x 5) x 3 3(x 3) (x 5)(x 5)(x 3) 3x 9 (x 5)(x 5)(x 3)
9
8.
y9
x2 9x 10.
y5 9 y5 y 4 y9 y9 y9
4x 14
x7
x7
x2 9x (4x 14)
x7 x 9x 4x 14 x7 x2 5x 14 x7 (x 7)(x 2) x7 x 2 2
Vocabulary, Readiness & Video Check 12.3 7 1.
2.
9
11 11 11 7 2 5 11 11 11 a
3.
2
b
c b
ac
12.
3y
y 3y 10 2
b
4.
a c ac b b b
5.
5 6 x 5 (6 x) x x x
3x 1
14.
x 5x 6 2
6. in order to carry out the subtraction properlyparentheses make sure each term in the numerator is affected by the subtraction, not just the first term 7. We completely factor denominatorsincluding coefficientsso we can determine the greatest number of times each unique factor occurs in any one denominator for the LCD. 8. To write an equivalent rational expression, we multiply the numerator of the rational expression by the same expression as the denominator. This means we’re multiplying the original rational expression by a factor of one and therefore not changing the value of the original expression. Exercise Set 12.3 x 1 2. 7 4.
6 7
x 1 6 7
x7 7
3 p 11p 3 p 11p 14 p 7 p 2q 2q 2q 2q q
16.
18.
y 3y 10 2
2x 7
3y 6
6
y 3y 10 3( y 2) ( y 5)( y 2) 3 y5 2
3x 1 (2x 7) x2 5x 6 x 5x 6 3x 1 2x 7 2 x 5x 6 x6 (x 6)(x 1) 1 x 1 2
4 p 3 3p 8 4p 3 3p 8 7 p 5 2p 7 2p 7 2p7 2p 7 6x2 25 2x2 6x2 (25 2x2 ) 2x 5 2x 5 2x 5 6x2 25 2x2 2x 5 4x2 25 2x 5 (2x 5)(2x 5) 2x 5 2x 5
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463
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
20.
7x 1 2x 21 7x 1 (2x 21) x4 x4 x4 7x 1 2x 21 x4 5x 20 x4 5(x 4) x4 5
22. 4y = 2 2 y
42.
LCD 2 2 2 y 8 y 5
46.
4x 1 4x 1 (4x 1) y 4xy y 3x 6 3(x 2) 3(x 2) y 3y(x 2) 5 y
2
2x 10
2(x 5)
x 2x 3x 3
2
(5 y) 2
2
2(x 5) 2
10 2 y 4(x2 5)
6m 5 52.
30. 7x 14 = 7(x 2) (x 2)2 (x 2)(x 2) LCD 7(x 2)(x 2) 7(x 2)2 32. 4x 12 = 4(x 3) 2x2 12x 18 2(x2 6x 9) 2(x 3)2 LCD 4(x 3)2 34. 7 3x = (3x 7) LCD = 3x 7 or 7 3x
x(x 2x 3) 5x x(x 3)(x 1) 5x(x 5) x(x 3)(x 1)(x 5) 2
5x2 25x x(x 3)(x 1)(x 5)
(6m 5)4 (3x2 9)4
24m 20
12x2 36
5x 9x 5x 9x 45x2 7 7 77 49
56. x 3 2x 1 x 3 (2x 1) 4 4 4 x 3 2x 1 4 x 4 4 58. 2x 3x 2x 3x x3 8x x3 8x x3 8x x 2 x(x 8) 1 2 x 8
36. x2 4x 3 (x 3)(x 1) x2 10x 21 (x 3)(x 7) LCD = (x + 3)(x + 1)(x + 7) 2
38. 4x 5x 1 (4x 1)(x 1) 3x2 2x 1 (3x 1)(x 1) LCD = (4x + 1)(x + 1)(3x + 1)(x 1) 40. x2 25 (x 5)(x 5)
54.
3x2 9
5x
28. x + 5 = x + 5 4x + 20 = 4(x + 5) LCD = 4(x + 5)
60.
2x x3 8x
3x
2x
3 x 8x x3 8x
3x 15x 3x (x 5) 2
LCD 3x2 (x 5)(x 5) 464
2
5x
50.
26. x 1 = x 1 x+5=x+5 LCD = (x 1)(x + 5)
2
5 y
5
24. 6y = 6 y = 2 3 y 4y + 12 = 4(y + 3) = 2 2 (y + 3) LCD = 3 4 y (y + 3) = 12y(y + 3)
3
24 y4
9 y5 8 y4 72 y9 5 8 yx 40 yx 44. 4 y2 x 4 y2 x 8 yx 32 y3 x2 9 y5 5
48.
3y5 2 2 2 y5
38 y4
3
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x3 8x 3x
2
3
ISM: Prealgebra and Introductory Algebra
62.
64.
12x 6 4x2 13x 3 x2 3x 4x2 1 6(2x 1) (4x 1)(x 3) x(x 3) (2x 1)(2x 1) 6(4x 1) x(2x 1)
86. 8 = 2 2 2 12 = 2 2 3 LCD = 2 2 2 3 = 24 24 = 8 3 24 = 12 2 You should buy 3 packages of hot dogs and 2 packages of buns.
9
88. answers may vary
3 9 3 2 9 6 3 10 5 10 5 2 10 10 10
66. 68.
Chapter 12: Rational Expressions
11 5 11 3 5 5 33 25 58 15 9 15 3 9 5 45 45 45 7
3
7 3 3 5 21 15 36 2 30 18 30 3 18 5 90 90 90 5
90. answers may vary Section 12.4 Practice Exercises y 3y y 3 3y 3 y 3 y 0 0 5 15 5 3 15 15 15
1. a.
5
b.
70. 14x 2 7 x 2
2
8x
6x3 2 3 x3
LCD 2 3 7 x3 42x3; c
72. answers may vary
3 y 3 y 74. x x x Choice c is correct. 76.
2 x
8y 1(2 x)
5 10x 5(x 3) 2. x2 9 x 3 (x 3)(x 3) (x 3)(x 3) 10x 5(x 3) (x 3)(x 3) 10x 5x 15 (x 3)(x 3) 5x 15 (x 3)(x 3) 5(x 3) (x 3)(x 3) 5 x3 10x
3 y 3 x 3 x 3 x x x y x y y Choice a is correct. 8y
78.
11 5 5x 11 4 10x2 8x 5x 10x2 4 25x 44 40x2 40x2 25x 44 40x2
8y x2
x 3 1(x 3) x 3 3 x 80. (x 2) x 2 x 2 x2 x2 x3
82. The perimeter of a trapezoid is the sum of the lengths of its sides. x4 5 x 1 5 x3 x3 x3 x3 x 4 5 x 1 5 x3 2x 15 x3 2x 15 The perimeter is inches. x3
3.
5 7x
2
5(x 1) 2(7x) x 1 7x(x 1) (x 1)(7x) 5(x 1) 2(7x) 7x(x 1) 5x 5 14x 7x(x 1) 19x 5 7x(x 1)
84. answers may vary
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465
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
4.
10
15
x6
6 x
x
5. 2
x5
10
3. The exercise is adding two rational expressions with denominators that are opposites of each other. Recognizing this special case can save us time and effort. If we recognize that one denominator is 1 times the other denominator,
15
x 6 (x 6) 10 15 x6 x6 10 (15) x6 25 x6
we may save many steps. Exercise Set 12.4 2.
2 x 1 x5 2(x 5) x 1(x 5) x 5 2x 10 x x5 3x 10 x5
4. 6.
4
3x
4
16 3x2 4x(3x 2)
6x x 4x6x 4 2
x 2
x 4
x
(x 2)(x 2) (x 2)(x 2) 6x(x 2) x(x 2) (x 2)(x 2)(x 2) (x 2)(x 2)(x 2) 6x(x 2) x(x 2)
(x 2)2 (x 2) 6x2 12x x2 2x
(x 2) (x 2) 7x2 10x
(x 2)2 (x 2) x(7x 10)
(x 2)2 (x 2)
Vocabulary, Readiness & Video Check 12.4 1.
3
6
7x 1
2
2.
x
466
x2
1 x xx
2 x
2
x2 x2
14
6 3x
5 4x 5 4x x 4 x2 16 x 4 (x 4)(x 4) 5(x 4) 4x (x 4)(x 4) (x 4)(x 4) 5x 20 4x (x 4)(x 4) 9x 20 (x 4)(x 4)
5
3 5 x 3 5x ; b 7 7x 7 x 7x
4c 5 8d d 5 d 5 5 d 20c 8d 2 5d 5d 20c 8d 2 5d
8d
8 3 8 3 8. x 4 3x 12 x 4 3(x 4) 8 1 x4 x 4 7 x4 10.
2
14 18x 14 18x 3x2 x 3x2 x 3x 3x2 3x2 3x2 14
6. 7.
4c d
3x
3x2 2x 12x 8 x(3x 2) 4(3x 2) 4(4) 3x(x) x(3x 2)(4) 4(3x 2)(x)
15 8 15 6 8 7 7a 6a 7a 6 6a 7 90 56 42a 42a 146 42a 73 21a
; a
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ISM: Prealgebra and Introductory Algebra
5
12.
2 1) (2 y 1)( y2 ) 2 2 y 1 y (2 y y 102y 5 y3 y (2 y 1) 15
14.
16.
y( y2 )
5(2 y 1)
y
y4
5
a7
20 4 y
5
7a
15
20.
x2
5
5
a7 5
22.
5x
7
y5
1 y5 1 6 y 1( y 5) y 5 1( y 5) 6y y 5 y5 7y 5 y5
1
6y
2x 3
7 2x 3 7
3
1 3(2x 3) 2x 3 1(2x 3) 6x 9 7 2x 3 2x 3 7 (6x 9) 2x 3 7 6x 9 2x 3 16 6x 2x 3 2(8 3x) 2x 3
26. y 1 2 y 5 ( y 1)3 2 y 5 y 3y y 3 3y 3y 3 2 y 5 3y 3y 3y 3 (2 y 5) 3y 3y 3 2 y 5 3y 5 y 8 5y 8 or 3y 3y
5x x2
5x 2 2 1 x x 1 x2
6y
3
20
(a 7) 5 a7 a7 0 a 7 0
7
7
y2 (2 y 1)
7 9 7 18. 9 2 2 2 25x 1 (25x2 1) 25x 1 1 25x 9 7 2 2 25x 1 25x 1 16 25x2 1 16 25x2 1 7
24.
5 10 y 3y
y 4 ( y 4) 15 20 y4 y4 5 y4 5 y4
Chapter 12: Rational Expressions
7x 28.
x3
4x 9 x3
7 5x3 x2
30.
7x (4x 9)
x3 7x 4x 9 x3 3x 9 x 3 3(x 3) x3 3
5x 11x2 5x 11x2 3 6 2 6 23 5x 33x2 6 6 5x 33x2 6
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467
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
5x
32.
2
(x 2)
34.
5x 3(x 2) 3 x 2 (x 2)(x 2) (x 2)(x 2) 5x 3(x 2) (x 2)2 5x 3x 6 (x 2)2 5x 3x 6 (x 2)2 2x 6 (x 2)2 2(x 3) (x 2)2
1 2 1 (3y) 2( y 5) y 5 3y ( y 5)(3y) 3y( y 5) 3y 2 y 10 3y( y 5) 3y( y 5) 3y 2 y 10 3y( y 5) 5 y 10 3y( y 5)
5
42.
(x 1)(x 5) (x 5)2 5(x 5) 2(x 1) (x 1)(x 5)(x 5) (x 5)2 (x 1) 5x 25 2x 2 (x 5)2 (x 1) 3x 23 (x 5)2 (x 1) 44.
x 5 2 x 4 x 4x 4 x(x 2) 5(x 2) (x 2)(x 2)(x 2) (x 2)2 (x 2) 2
x2 2x 5x 10 (x 2)2 (x 2) x2 7x 10 (x 2)2 (x 2) 1
46.
2x 2 y
5( y 2) 3y( y 5) 6 36.
38.
x
1
6 x
x x
6 x
48.
x
y
( y)(2) x y 2(x y) (x y)(2) 1 2 y 2(x y)
9x 27x (x 10x) (x 10)(x 3) 2
2
9x2 27x x2 10x (x 10)(x 3) 8x2 17x (x 10)(x 3) x(8x 17) (x 10)(x 3) 10 5 10 5 40. 3n 4 4 3n 3n 4 (3n 4) 10 5 3n 4 3n 4 10 5 3n 4 15 3n 4
1
x6 6 (x 6)(5x 1) 6 5x 1 (5x 1)2 (5x 1)2 (5x 1)2
9x x 9x(x 3) x(x 10) x 10 x 3 (x 10)(x 3) (x 3)(x 10)
468
2
5x2 x 30x 6 6 (5x 1)2 5x2 29x (5x 1)2 x(5x 29) (5x 1)2
4 50. 1 4 1 a 2 4 2a (2 a) 2(2 a) (1) 2 2a 2 a 1 2 2a 3 2a
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ISM: Prealgebra and Introductory Algebra
2
52.
x2 4x 4
54.
1 2 1(x 2) x 2 (x 2)2 (x 2)(x 2) 2 x2 (x 2)2 x4 (x 2)2
Chapter 12: Rational Expressions
60.
7 2 y2 3y 2 y 1 7 2( y 2) ( y 1)( y 2) ( y 1)( y 2) 7 2 y 4 ( y 1)( y 2) 3 2 y ( y 1)( y 2)
56.
9z 5 62.
27 3 y2 81 2( y 9) 27 2 3( y 9) ( y 9)( y 9)2 2( y 9)( y 9) 54 3y 27 2( y 9)( y 9) 3y 27 2( y 9)( y 9) 3( y 9) 2( y 9)( y 9) 3 2( y 9)
58.
6 2 5 y2 25 y 30 4 y2 8 y 64y 2( y 3) 5 5( y 3)( y 2)4 y 4 y( y 2)( y 3)5 24 y 10 y 30 20 y( y 3)( y 2) 14 y 30 20 y( y 3)( y 2) 2(7 y 15) 20 y( y 3)( y 2) 7 y 15 10 y( y 3)( y 2)
2
2
2
2
64. 2z z 2z 2z (z 2z ) 4z 1 4z 1 2 4z 1 2 2z z 2z 4z 1 2 4z z 4z 1 z(4z 1) 4z 1 z 66.
x4 x 1 x2 12x 20 x2 8x 20 (x 4)(x 2) (x 1)(x 2) (x 10)(x 2)(x 2) (x 10)(x 2)(x 2) x2 2x 4x 8 x2 2x x 2 (x 10)(x 2)(x 2) 2x2 5x 6 (x 10)(x 2)(x 2)
5z 9z 5 5z 15 81z2 25 3 5 (9z 5)(9z 5) z 3(9z 5)
68.
9 12 9 3x 3 2 x 1 3x 3 x 1 12 9 3(x 1) (x 1)(x 1) 3 4 9 4(x 1) 2
4 2 2x2 5x4 3 x 3 2(2x 1)
(2x 1)(x 3) 4 4x 2 (2x 1)(x 3) 4x 2 (2x 1)(x 3) 2(2x 1) (2x 1)(x 3)
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(2x 1)(x 3)
469
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions 70. 5x 1 8 5x 9 9 x 5 72.
4x2 9 0 (2x 3)(2x 3) 0 2x 3 0 or 2x 3 0 2x 3 2x 3 3 3 x x 2 2
74. 2(3x 1) 15 7 6x 2 15 7 6x 17 7 6x 24 x 4 76.
5 7x 11 5 7x 11 x 2 x2 4 x x 2 (x 2)(x 2) x 5x(x 2) 7x x 11(x 2)(x 2) (x 2)x(x 2) (x 2)(x 2) x x(x 2)(x 2) 5x2 10x 7x2 11x2 44 x(x 2)(x 2) 2 x 10x 44 x(x 2)(x 2)
8
78.
x 6x 5 2
80.
3x
2
8 3x 2 x 1 (x 5)(x 1) (x 5)(x 1) (x 1)(x 1) 8(x 1) 3x(x 1) 2(x 5) (x 5)(x 1)(x 1) (x 5)(x 1)(x 1) (x 1)(x 1)(x 5)
x 4x 5 2
2
8x 8 3x2 3x 2x 10 (x 5)(x 1)(x 1)
3x2 7x 2 (x 5)(x 1)(x 1)
x 10
8 9 x 10 8 9 2 2 x 3x 4 x 6x 5 x x 20 (x 4)(x 1) (x 1)(x 5) (x 4)(x 5) (x 10)(x 5) 8(x 4) 9(x 1) (x 4)(x 1)(x 5) (x 1)(x 5)(x 4) (x 4)(x 5)(x 1) 2
470
x2 5x 10x 50 8x 32 9x 9 (x 4)(x 1)(x 5)
x2 2x 73 (x 4)(x 1)(x 5)
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
82. Perimeter 2 3 2 2 y 5 y 6 4 y 5 y 6y 4( y 5) ( y 5) y y( y 5) 6 y 4 y 20 y( y 5) 10 y 20 y( y 5) 10( y 2) y( y 5) 3 2 6 6 Area y 5 y y( y 5) y2 5 y 10( y 2) The perimeter is feet and the area is y( y 5)
6 y2 5 y square feet.
84.
DA D( A 1) A 12 24 DA 24 D( A 1)( A 12) ( A 12)24 24( A 12)
24DA D( A2 12 A A 12) 24( A 12)
2
24DA DA 12DA DA 12D 24( A 12)
11DA DA 12D 24( A 12)
2
86. answers may vary x 2 180 x 2 88. 100 x 1 x 180x x 2 x
90. answers may vary
Section 12.5 Practice Exercises 1. The LCD is 20. x 4 1 4 5 20 x 4 1 20 20 4 5 x 420 1 20 20 20 4 5 5x 16 1 20 5x 15 x 3 x 4 1 Check: 4 5 20 3 4 1 0 4 5 20 15 16 1 0 20 20 20 1 1 True 20 20 The solution is 3. 2. The LCD is 15. x 2 x 1 1 3 5 15 x 2 x 1 1 15 3 5 15 15 x2 x 1 1 15 15 5 15 15 3 5(x 2) 3(x 1) 1 5x 10 3x 3 1 2x 13 1 2x 12 x 6 x 2 x 1 1 Check: 3 5 15 6 6 1
2
x
180x x 2 x 179x 2 x 179x 2 The supplement measures . x
3
0
1
5 15 4 7 1 0 3 5 15 20 21 1 0 15 15 1 15 1 15 15 True
The solution is 6.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
3. The LCD is x. 6 2 x7 6x x 2 x(x 7) x 6 x(2) x x x x7 x 2x 6 x2 7x 0 x2 5x 6 0 (x 6)(x 1) x6 0 or x 1 0 x 6 x 1 6 Check x = 6: 2 x 7 x 6 2 0 67 6 2 10 1 1 1 True 6 Check x = 1: 2 x 7 x 6 2 01 7 1 26 08 8 8 True Both 6 and 1 are solutions. 4. The LCD is (x + 3)(x 3). 2 3 2 x 3 x3 3 x2 9 2 (x 3)(x 3) (x 3)(x 3) 2 x 3 x 3 2 3 x 9 2 (x 3)(x 3) (x 3)(x 3) 2 (x 3)(x 3)
x3
x3 2(x 3) 3(x 3) 2 2x 6 3x 9 2 5x 3 2 5x 5 x 1
2 x 9
The solution is 1.
472
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
5. The LCD is x 1. 5x 5 3 x 1 x 1 5 5x (x 1) x 1 (x 1) x 1 3 5x 5 (x 1) (x 1) (x 1) 3 x 1 x 1 5x 5 3(x 1) 5x 5 3x 3 2x 2 x 1 1 makes the denominator 0 in the original equation. Therefore 1 is not a solution and this equation has no solution. 6. The LCD is x + 3. (x
x
6 x3 6
2x 2 x3 2x
3) x 6 (x 3) 2x 2 (x 3)(x) (x 3) x 3 (x 3) x 3 (x 3)(2) x3 x 3 (x 3)(x) 6 2x 2(x 3) x2 3x 6 2x 2x 6 x2 x 12 0 (x 4)(x 3) 0 x 4 0 or x 3 0 x4 x3 x = 3 can’t be a solution of the original equation. The only solution is 4. 7. The LCD is abx. 1 1 1 a b x 1 1 1 abx abx a b x 1 1 1 abx abx abx a b bx ax ab x bx ab ax bx a(b x) bx a(b x) b x b x bx a b x
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473
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
Vocabulary, Readiness & Video Check 12.5 1.
Exercise Set 12.5
3x 1 4 5 4 2 4 3x 1 4 2 4 5 4 4
2. The LCD is 5. x x 5 2 9 5 2 5(9) 5 x 10 45 x x 55 2 9 Check: 5
6x 20 1 The correct choice is c. 1 3 2. 5x 5x(2) x 5x 1 3 5x 5x 5x(2) x 5x 5 3 10x The correct choice is b. 3. The LCD of
4. The LCD of
5. The LCD of
8
1
, , 3x x 9
8
8 1
3 ,
7
7
9
and
x 1
x2
and
7 2
(x 1)
2 x 4
is 9x; d.
3 3 is (x 1)2; a. 3 (x 2)(x 2)
, and
is (x2 4); c.
7. These equations are solved in very different ways, so we need to determine the next correct step to make. For a linear equation, we first “move” variable terms to one side and numbers to the other; for a quadratic equation, we first set the equation equal to 0. 8. If there are variables in any denominators, we should first check to see if the proposed solutions make these denominators zero in the original equation, giving us an undefined rational expression. If so, that solution is rejected, and it is not a solution to the equation. 9. the steps for solving an equation containing rational expressions; as if it’s the only variable in the equation
474
209
5
9 3 1 1 is 12x; b. , , and x 4 12 3 4
1 6. The LCD of
55
11 0 9 9 9 True The solution is 55. 4. The LCD is 18. x 4x x 6 3 18 x 4x x 18 18 6 24x x 3x 3 18 27x x 26x 0 x0 x 4x x Check: 6 3 18 0 4(0) 0 0 6 3 18 00 0 0 0 0 True The solution is 0. 6. The LCD is x. 4 5 1 x 4 x 5 x(1) x 5x 4 x 4x 4 0 4x 4 x 1 4 5 1 Check: x 4 5 01 (1) 5 (4) 0 1 1 1 True The solution is 1.
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ISM: Prealgebra and Introductory Algebra
8. The LCD is y. 5 2 6 y y y 5 2 y 6 y y y y 2 6y 5 y 2
0 y2 6 y 7 0 ( y 7)( y 1) y 7 0 or y 1 0 y7 y 1 5 2 Check y = 7: 6 y y y 5 2 6 0 7 7 7 42 5 49 2 0 7 7 7 7 47 47 True 7 7 Check y = 1:
6
5
y
y 5
2
y 2 0 1 1 1 6 (5) 0 1 (2) 1 1 True 6
The solutions are 7 and 1. 10. The LCD is 5 6 = 30. b b2 5 6 b b 2 30 30 5 6 6b 5(b 2) 6b 5b 10 b 10 b b2 Check: 5 6 10 10 2 0 5 6 10 12 0 5 6 2 2 True The solution is 10.
Chapter 12: Rational Expressions 12. The LCD is 4 2 = 8. a5 a5 a 4 2 8 a 5 a 5 a 8 8 4 2 8 2(a 5) 4(a 5) a 2a 10 4a 20 a 6a 30 a 5a 30 a 6 a5 a5 a Check: 4 2 8 (6) 5 (6) 5 6 0 4 2 8 1 1 3 0 4 2 4 1 2 3 0 4 4 4 3 3 True 4 4 The solution is 6.
14. The LCD is 4 3x. 6 3 4 3x 6 (4 3x)(3) 4 3x 6 12 9x 18 9x 2x 6 Check: 3 46 3x 0 3 (4 3x)
4 3(2) 6 0 3 46 6 0 3 2 3 3 True The solution is 2.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
16. The LCD is a + 2. 2a 7a 5 a 2 a 2 7a 2a (a 2) a 2 5 (a 2) a 2 2a 5(a 2) 7a 2a 5a 10 7a 3a 10 7a 10 10a 1 a 7a 2a 5 Check: a2 a2 2(1) 7(1) 50 (1) 2 (1) 2 2 7 50 1 7 1 7 True The solution is 1. 18. The LCD is y 2. 2y 4 4 y2 y2 2y 4 ( y 2) ( y 2)(4) y 2 y 2 2y 4 4y 8 4 2y 2 y Since 2 makes a denominator 0 in the original equation, y = 2 does not check and the equation has no solution. 20. The LCD is (x 2)(x 2) x2 4. 1 4 2 1 x2 x 4 1 4 2 (x2 4) 1(x 4) x 2 x2 4 (x 2) 4 x2 4 x 2 x2 4
1
Check x = 3:
476
4
1
x2 1 4 3 2 32 4 01 1 4 01 5 5 5 1 True 5 The solution is 3. x2 4
22. The LCD is y + 1. 5y 3 4 5yy1 y3 1 ( y 1) ( y 1)(4) y 1 y 1 5y 3 4y 4 5y 4y 7 y7 5y
3
4 y 1 y 1 5(7) 3 04 7 1 7 1 35 3 04 8 8 32 04 8 4 4 True The solution is 7.
Check:
24. The LCD is ( y 3)( y 3) y2 9. 4y 3y 1 3 y3 y3 4y 3y 1 ( y2 9) 3 ( y2 9) y 3 y3 4 y( y 3) 3( y2 9) ( y 3)(3y 1) 4 y2 12 y 3y2 27 3y 2 y 9 y 3 y2 12 y 27 3y 2 10 y 3 0 2 y2 22 y 24
0 x2 x 6 0 (x 3)(x 2) x 3 0 or x 2 0 x3 x 2 Since 2 makes a denominator 0 in the original equation, x = 2 does not check.
y 12 0 or y 12
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0 2( y2 11y 12) 0 2( y 12)( y 1) y 1 0 y 1
ISM: Prealgebra and Introductory Algebra
Check y = 12:
4y
3y 1 y 3 y 3 4(12) 3(12) 1 12 3 3 0 12 3 48 27 36 1 0 9 9 15 21 35 0 9 15 7 7 True 3 3 3
Check y = 1:
3y 1
4y
3 y 3 y3 4(1) 3(1) 1
1 3 3 0 1 3 4 3 1 3 0 4 2 4 1 3 0 2 2 2 True The solutions are 12 and 1. 26. The LCD 6 is 33y. 3y y 1
Chapter 12: Rational Expressions 30. The LCD is 3 2x = 6x. 5 3 3 53 32x 2 3 6x 3 2x 6x 2 5(2x) 3(3) 3(3x) 10x 9 9x x9 The solution x = 9 checks.
32. The LCD is (x + 1)(x 1). 3 x 1 x 1 x 1 x (x 1)(x 1) (x 1)(x 3 1) 1 x 1 x 1 1(x 1)(x 1) 3(x 1) x(x 1)
x2 1 3x 3 x2 x 2x 4 x2 The solution x = 2 checks. 34. The LCD is 15. 3x x 6 2 5 2 3x5 x 36 15 15 5 5 3
6 3 3y 3y(1) 3y y 6 3(3) 3y 6 9 3y 15 3y 5 y The solution y = 5 checks.
3(3x) 5(x 6) 3(2) 9x 5x 30 6 4x 36 x 9 The solution x = 9 checks.
28. The LCD is (x 6)(x 2). 5 x x 6 x2 (x 6)(x 2)
5 x (x 6)(x 2) x6 x 2 5(x 2) x(x 6)
36. The LCD is x(x + 4). 15 x4 x4 x 15 x4 x(x 4) x 4 x(x 4) x 15x (x 4)(x 4) 15x x2 16 0 x2 15x 16 0 (x 16)(x 1)
5x 10 x 6x 2
0 x2 11x 10 0 (x 10)(x 1) x 10 0 or x 1 0 x 10 x 1 The solutions x = 10 and x = 1 check.
x 16 0 or x 1 0 x 16 x 1 The solutions x = 16 and x = 1 check.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
38. The LCD is (x 2)(x 2) x2 4. 1 4 1 2 x1 2 x 2 4 x 42 (x2 4) (x 4) 1 2 x 2 x 4 x 2 1(x 2) 4 1(x 2) x2 4 x2 2x 4 x2
The solution x = 2 makes the denominators x2 4 and x 2 equal to zero, so the equation has no solution. 40. x2 7x 12 (x 3)(x 4) The LCD is (x + 3)(x + 4). 3 12x 19 5 x3 3 x2 7x 12 x 4 (x 3)(x 4) (x 3)(x 4) 12x 19 5 x3 (x 3)(x 4) x 4 3(x 4) 12x 19 5(x 3) 3x 12 12x 19 5x 15 3x 12 7x 4 8 4x 2 x The solution x = 2 checks.
42. t 2 2t 3 (t 3)(t 1) The LCD is (t + 3)(t 1). 2t 3 2 5 6t t1 t2 3 t2 2t 3 2t 3 (t 3)(t 1) (t 3)(t 1) 5 6t t 1 t 3 2 t 2t 3 (2t 3)(t 3) 2(t 1) 5 6t
2t 2 6t 3t 9 2t 2 5 6t 2t 2 7t 11 5 6t
2t 1 0 2t 1 1 t 2
2t2 13t 6 0 (2t 1)(t 6) 0 or t 6 0 t 6
The solutions t
1 2
478
and t = 6 check.
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ISM: Prealgebra and Introductory Algebra
44. The LCD is Q. V T Q V Q(T ) Q Q QT V V Q T 46. The LCD is t + B. A i tB A (t B)i (t B) t B ti Bi A ti A Bi A Bi t i 48. The LCD is W. A L W A W W W (L) A WL A W L 50. The LCD is 24. D( A 1) 24 D( A 1) 24(C) 24 24 24C D( A 1) 24C DA D 24C D DA 24C D A D
Chapter 12: Rational Expressions
54. The LCD is 5xy. 1 2 1 1 5 2 y x 1 5xy 5xy 5 y x xy 2(5x) 5 y xy 10x 5 y x( y 10) 5 y 5y x y 10 1 56. The reciprocal of x + 1 is x 1 . 58. The reciprocal of x, subtracted from the 1 1 reciprocal of 5 is . 5 x 60.
1
of the beach is cleaned in 1 hour.
4 62. answers may vary 1 64.
C
66.
5
2
is an equation. The LCD is 9x. 3 1 5 2 x 9 3 1 5 2 9x x 9 9x 3 9 5x 6x 9 x The solution x = 9 checks. x
9
5 x 1 5 x 1
2 x 2 x
52. The LCD is 2. CE2 W 2 CE2 2(W ) 2 2 2W CE 2 2W C E2
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is an expression.
5 x
2 (x 1)
(x 1) x x (x 1) 5x 2(x 1) x(x 1) 5x 2x 2 x(x 1) 3x 2 x(x 1)
479
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
68.
70.
5x
25x
180 2 2 The5xLCD25x is 2. 2 2(180) 2 2 5x 25x 360 30x 360 x 12 5x 5(12) 60 30 2 2 2 25x 25(12) 300 150 2 2 2 The angles measure 30 and 150. 80 100 90 x x The80 LCD is x. 100 x x(90) x x 80 100 90x 180 90x 2x 80 80 40 x 2 100 100 50 x 2 The angles measure 40 and 50.
72. a2 2a 8 (a 4)(a 2) a2 9a 20 (a 4)(a 5) a2 3a 10 (a 5)(a 2) The LCD is (a + 4)(a + 5)(a 2). 4 1 4 2 2 2 a 42a 8 a 19a 20 a 3a 10 4 (a 4)(a 5)(a 2) a2 2a 8 a2 9a 20 (a 4)(a 5)(a 2) a2 3a 10 4(a 5) 1(a 2) 4(a 4) 4a 20 a 2 4a 16 3a 22 4a 16 a6 The solution a = 6 checks.
480
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ISM: Prealgebra and Introductory Algebra
Mid-Chapter Review 1.
1 x
2 3
1 3 x3
2 x 3 x
3 3x
2x
5a
3x
7.
3 2x 3x
Chapter 12: Rational Expressions
This is an expression. 2.
3
5
3 6
5 a
18
a 6 a 6 6 a 6a This is an expression. 1
3.
2
1 1 2x 1 1 x x(x 1) x(x 1)(1) x 1 x 2x (x 1) x(x 1) 2x x 1 x2 x
18 5a
6a
x 1 x2 x
6a
0 x2 1 1 x2 There are no solutions. This is an equation.
3
x 3 x 1 2 3 3x x 3 x 3 2x 9 2x 6 x3 This is an equation.
4 1 6 x 3 x x(x 3) 1 6 4 x(x 3) x(x 3) x 3 x x(x 3) 4x 1(x 3) 6 4x x 3 6 3x 3 x 1 This is an equation. 15x 2x 16 3 5x 2(x 8) 5 2 10 9. x8 3x x8 3x This is an expression.
3x
4.
2
8.
3
5 1 3 a 5 6 6a 6a(1) a 6 18 5a 6a 18 a This is an equation. 2 1 2 x 1(x 1) x 1 x (x 1) x x (x 1) 2x x 1 x(x 1) x(x 1) 2x x 1 x(x 1) x 1 x(x 1) This is an expression.
5z 9z 5 5z 10. 9z 5 15 81z 2 25 3 5 (9z 5)(9z 5) z 3(9z 5) This is an expression.
4 1 4 x 1 (x 3) 6. x 3 x (x 3) x x (x 3) 4x x3 x(x 3) x(x 3) 4x x 3 x(x 3) 3x 3 x(x 3) 3(x 1) x(x 3) This is an expression.
4 p 3 3p 8 4 p 3 3p 8 7 p 5 12. 2 p 7 2 p 7 2p 7 2p7 This is an expression.
5.
11.
2x 3 3x 6 2x 1 3x 6 5x 7 x 3 x3 x3 x3 This is an expression.
x5 13.
8
7 2 x5 8 14 7 14 2
2(x 5) 7(8) 2x 10 56 2x 46 x 23 This is an equation.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
14.
1 x 1 2 8 1 x 1 8 8 2 8 4 x 1 3 x This is an equation. 5a 10
15.
16.
19. x 7 x 2 (x 7) 5 x 2 x 5x x 5 5x 5x 35 x 2 5x 4x 37 5x This is an expression.
5a 10 10a 18 a2 4 18 10a 5(a 2) 2 5a 2 9 (a 2)(a 2) 5 5a 9(a 2) 25a 9(a 2) This is an expression.
a2 4
21.
x2 3x 1
5 2
(3x 1)
(x 2)(3x 1)
5
(3x 1)(3x 1) (3x 1)2 (x 2)(3x 1) 5
(3x 1)2 3x x 6x 2 2 5 (3x 1) 2
3x2 5x 3
(3x 1)2
This is an expression. 18.
4 (2x 5)
2
x 1 2x 5
4
(x 1)(2x 5)
(2x 5) (2x 5)(2x 5) 4 (x 1)(2x 5) 2
(2x 5)2 4 2x2 5x 2x 5 (2x 5)2
(10x 9) 3
x4
x 3 3x 3(10x 9) x 4 3x 30x 27 x 4 3x 29x 23 3x This is an expression. 3x
2 3 5 x 3 x2 9 x 3 5 2 3 2 2 9) 2 (x 9) (x x x 3 9 x3 3(x 3) 5 2(x 3) 3x 9 5 2x 6 3x 9 2x 1 5x 8 8 x 5 This is an equation.
22.
9 2 1 x2 4 x 2 x 2 9 2 1 2 (x2 4) 2 (x 4) x2 x2 x 4 9 2(x 2) 1(x 2) 9 2x 4 x 2 2x 5 x 2 3x 7 7 x 3 This is an equation.
23. answers may vary 24. answers may vary
2x2 3x 1 (2x 5)2
This is an expression.
482
x4
9 4(x 1) This is an expression. 17.
10x 9 x
9 12 9 3x 3 x2 1 3x 3 x2 1 12 9 3(x 1) (x 1)(x 1) 3 4
20.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
25. 414.92 feet per second 414.92 ft 3600 sec 1 mi 1 sec 1 hr 5280 ft 414.92 3600 mi 5280 hr 1, 493, 712 mi 5280 hr 282.9 mi/hr
1 1 1 2 3 x 1 1 1 6x 6x 6x 2 3 3x 2x 6 x 5x 6 6 1 x 1 5 5
26. a.
Together they can sort one batch in 1
b.
2920 miles per hour 2920 mi 1 hr 5280 ft 1 hr 3600 sec 1 mi 2920 5280 ft 3600 sec 15, 417, 600 ft 3600 sec 4282.7 ft/sec
3.
18 feet per second 18 ft 1 mi 3600 sec 1 sec 5280 ft 1 hr 18 3600 mi 5280 hr 64,800 mi 5280 hr 12.3 mi/hr
x
1
1
hours.
5 distance
=
rate
time
Car
600
x + 15
600 x15
Motorcycle
450
x
450
600
x
450
x 15 x 600x 450(x 15) 600x 450x 6750 150x 6750 x 45 x + 15 = 45 + 15 = 60 The speed of the motorcycle is 45 mph. The speed of the car is 60 mph.
Section 12.6 Practice Exercises 1.
hr
Vocabulary, Readiness & Video Check 12.6
x
1. The time to complete the job working together will be less than both of the individual times. The answer is c.
2 3 6 x 1 x 6 6 2 3 63x 2 x 2x 2 x1 The number is 1.
2. The time to fill the pond with both pipes on at the same time will be less than both of the individual times. The answer is a. 3. A number: x
Hours to Complete Total Job
Part of Job Completed in 1 Hour
Guillaume
2
1 2
Greg
3
1 3
Together
x
1 x
2.
The reciprocal of the number:
1
x The reciprocal of the number, decreased by 3: 1 3 x 4. A number: y The reciprocal of the number:
1
y The reciprocal of the number, increased by 2: 1 2 y
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483
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
5. A number: z The sum of the number and 5: z + 5 The reciprocal of the sum of the number and 5: 1 z5
4. Let n be the number. 1 6 n 5 n 6 n n 5(n) n 6 n2 5n
6. A number: x The difference of the number and 1: x 1 The reciprocal of the difference of the number 1 and 1: x 1 7. A number: y Twice the number: 2y Eleven divided by twice the number:
n2 5n 6 0 (n 3)(n 2) 0 n3 0 or n 2 0 n 3 n 2 The number is 3 or 2.
11 2y
8. A number: z Triple the number: 3z Negative 10 divided by triple the number: 10 3y 9. divided by, quotient 10. Two machines (or people) take different amounts of time to complete the task, one faster and one slower than the other. When working together, they will complete the task in less time than the faster machine, so our answer must be less than the time of the faster machine. 11.
d
t
325
x+7
325 x7
Motorcycle
290
x
290
Exercise Set 12.6
484
r
Car
325 290 x 7 x
2.
=
12
4 x2 x2 12(x 2) 4(x 2) 12x 24 4x 8 8x 32 x4
x
6. Let x be the time in hours that it takes them to complete the job working together. The 1 experienced bricklayer completes of the job in 3 1 1 hour. The apprentice completes of the job in 6 1 1 hour. Together, they complete of the job in x one hour. 1 1 1 3 6 x The LCD is 6x. 1 1 1 6x 6x 3 6 x 2x x 6 3x 6 x2 It takes them 2 hours to construct the wall. 8. Let t be the amount of time to move 1200 pounds. 300 1200 t 1 12 1200 1 1 300t 2
300t 1800 t6 It will take 6 minutes for the conveyor belts to move 1200 pounds of cans.
10. Let r be the speed of the boat in still water. distance
=
rate
Upstream
9
r3
Downstream
11
r+3
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time 9 r 3
11
r 3
ISM: Prealgebra and Introductory Algebra
Since the time is the same,
Chapter 12: Rational Expressions
9
11 . r 3 r 3
The LCD is (r 93)(r + 3). 11 (r 3)(r 3) (r 3)(r 3)
r 3 r 3 9(r 3) 11(r 3) 9r 27 11r 33 60 2r 30 r The speed of the boat in still water is 30 mph. 12. Let x be the speed of the semi-truck through the flatland. distance
=
rate
It will take Mr. Dodson and his son 2
2
days to
9 paint the house. 18. Let r be the rate in still water.
time
Flatland
300
x
300
Mountains
180
x 20
180 x20
distance
=
rate
time
x
300 180 x x 20 The LCD is300 x(x 20). x(x 20) x(x 20) 180 x x 20 300(x 20) 180x 300x 6000 180x 120x 6000 x 50 The speed through the flatland is 50 mph and the speed through the mountains is 300 mph.
1 1 1 4 5 x The LCD is 20x. 1 1 1 20x 20x 4 5 x 5x 4x 20 9x 20 20 2 x 2 9 9
14. Let x be the number. 4x 5 7 6 2 4x 5 7 6 6 6 2 4x 5 21 4x 16 x4 The number is 4.
Downstream
9
r+6
Upstream
3
r6
9
r 6
3
r 6
9 3 r6 r6 9(r 6) 3(r 6) 9r 54 3r 18 6r 72 r 12 9 9 9 1 r 6 12 6 18 2 It takes the same amount of time to travel the 9 miles downstream as it does the 3 miles upstream. If it takes half an hour to travel downstream then it would take 1 hour to travel the entire 12 miles. t
20. Let x be the number. 1 9x 7 15 x 2 9 x The LCD is x(x + 2). 15 9x 7 x(x 2) 9x(x 2) x x 15(x 2) 2 x(9x 7) 9x(x 2)
16. Let x be the time in days that it takes Mr. Dodson and his son to paint the house together. Mr. 1 Dodson can paint of the house in 1 day. His 4 1 son can paint of the house in 1 day. Together, 5 1 they paint of the house in 1 day. x
15x 30 9x2 7x 9x2 12x
9x2 8x 30 9x2 18x 30 10x 3 x The number is 3.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
22. Let r be the rate of the car. distance = rate Jet
1080
Car
240
time
6r
1080 6r
r
240
28. Let x be the amount of time the second proofreader takes to do the job alone. 1 1 1 4 x 5 2
1 1 2 1 4 1 x 5 2 20x 20x 4 x 5 5x 20 8x 20 3x 20 x 32
r
The car ride is one hour longer than the jet ride. 240
1080
1 r 6r The LCD is 6r. 240 1080 6r r 6r 6r 1
1440 1080 6r 360 6r 60 r 1080 1080 1080 3 6r 6(60) 360 240 240 4 r 60 It took the manager 3 hours to travel by jet and 4 hours to travel by car. 24. Let x be the number. x 7 1 5 5 x 7 5 5 1 5 5 x5 7 x 12 The number is 12.
6
3
x
It will take the second proofreader 6
2
hours to
3 do the job alone. 30.
r x3
Upstream
t
=
6
d 6
x3
Downstream
10
x+3
10
x3
6 10 x 3 x 3 6(x 3) 10(x 3) 6x 18 48 10x 4x 30 12 x The speed of the boat in still water is 12 mph. 32. Let x be the speed of the second car.
26. Let x be the time Gary traveled at 70 mph. Since Gary was stopped for 30 minutes, he spent a total of 4 hours driving. He drove at 60 mph for (4 x) hours. The distance he traveled at 700 mph is 70x, and the distance he traveled at 60 mph is 60(4 x). The total distance is 255 miles. 255 70x 60(4 x) 255 70x 240 60x 15 10x 15 x 10 3 x 2 1 Gary traveled at 70 mph for 1 hours before 2 getting stopped.
486
distance
=
rate
time
first car
224
x + 14
224 x14
second car
175
x
175 x
224 175 x x 14 224x 175(x 14) 224x 175x 2450 49x 2450 x 50 x + 14 = 50 + 14 = 64 The speed of the first car is 64 miles per hour and the speed of the second car is 50 miles per hour.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
34. Let x be the speed of the wind. distance
=
40. Let x be the speed traveled on the winding road.
rate
against wind
1365
490 x
1365 490 x
with wind
1575
490 + x
1575 490 x
1365 1575 490 x 490 x 1365(490 x) 1575(490 x) 668,850 1365x 771, 750 1575x 2940x 102, 900 x 35 The speed of the wind is 35 miles per hour. 36. Let t be the time it takes the bus to overtake the car. distance Bus
60t
=
distance
time
rate
time
60
t
Car 40(t + 1.5) 40 t + 1.5 60t 40(t 1.5) 60t 40t 60 20t 60 t 3 It will take the bus 3 hours to overtake the car. 38. Let x be the time it takes the faster pump to fill the tank. 1 1 1 x 2x 18 1 1 1 18x x 2x 18x 18 18 9 x 27 x 54 2x It will take the faster pump 27 minutes to fill the tank by itself and 54 minutes for the slower pump to fill the tank by itself.
=
rate
time
level road
3(x + 20)
x + 20
3
winding road
4x
x
4
The total distance is 305 miles. 305 3(x 20) 4x 305 3x 60 4x 245 7x 35 x 55 = x + 20 The average speed on the level road is 55 mph. 42. Let x be the amount of time it takes to balance the books working together. 1 1 1 8 12 x 1 1 1 24x 24x 8 12 x 3x 2x 24 5x 24 24 4 x 4 5 5 4 It will take 4 hours to balance the company’s 5 books together. 44. Let x be the speed of the slower car. distance
=
rate
time
Slower car
2.5x
x
2.5
Faster car
2.5(x + 8)
x+8
2.5
2.5x 2.5(x 8) 280 2.5x 2.5x 20 280 5x 20 280 5x 260 x 52 x + 8 = 52 + 8 = 60 The slower car’s speed was 52 mph and the faster car’s speed was 60 mph. 96 15 5 5 5 3 46. 17 7 24 4 6
6
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6
487
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
Section 12.7 Practice Exercises
48. 43 74 3 4 7 1
5
6
8
8
8 8 6 4 10 8
1.
3 7 3 5 3 9 27 5 9
7
3
3(3)
2
35
2(4)
2(4) 8 9 8 124 12 3 8 8 1 12 7 8
50. From the graph, the U.S. wind capacity in 2020 was approximately 122,000 megawatts.
1 8 12 7 1 2 4 3 4 7 2 21
52. answers may vary 54. Let x be the time it takes pump C to fill the tank. 1 1 1 1 1 9 115 1 x 5 1 45x 45x 9 15 x 5 5x 3x 45 9x 8x 45 9x 45 x It takes pump C 45 hours to fill the tank. 56. You can use cross products when each side of the equation contains a single fraction, as in case a.
2( x)
2 1 5 x
3.
2x 1 15 3
5( x)
1(5)
x(5)
2x 1(5) 15 3(5) 2x 5 5x 5x 2x 5 15 15 2x5 5x 2x5 15
2x 5 15 5x 2x 5 2x 15 3 5 5x 2x 15 3 x
D RT D RT
R R D D T or T R R 60. Let x be the top speed of the Bugatti Chiron Super Sport 300+. Then the top speed of the Koenigsegg Absolut is x + 26, Since d = rt, then d t r. 495 456 x 26 x 495 x 456(x 26) 495x 456x 11,856 39x 11,856 x 304 x + 26 = 304 + 26 = 330 The top speed of the Chiron is 304 miles per hour and the top speed of the Absolut is 330 miles per hour.
488
7 5
4(3) 3(4) 2. 41 33 1(4) 3 2 8
6 10 4 8 6 8 4 10 2 3 2 4 4 25 6 5
58.
9
4. The LCD is 24. 32 24 43 23 4 3 13 24 1 3
2 8 2 8 24 3 24 2 24 41 24 33 2 8 18 16 12 9 2 21
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ISM: Prealgebra and Introductory Algebra
y(1) y yx 2x 1 y y
5. The LCD is y. x 1 x y 1 y 2x1 y
Chapter 12: Rational Expressions
y
2x
2x1 y
y 5 y 6xy 65y x 6y x y y 6xy x x 3 3
6xy 6xy 6xy 6xy(x)
8. The LCD of
5x 6 y2 2
2
2xy 6x y 5x 6 y2 2xy( y 3x)
Vocabulary, Readiness & Video Check 12.7
2.
5xy
y 2 y2 2 5x 2 5x 2 2
3.
x 5 x2
4.
x2
2.
2a b
6.
a a 20 10 10 b b 20 20 20
1
4 22 2 2 3 = 12; c.
,
1 12 5
112
3 85 10
6x3 6x 3 2x 1 5x2 2 x1 10x 5x2 10x
6x 3 10x 5x2 2x 1 3(2x 1) 10x 2 5x (2x 1) 6 x
1 1 1 is , , and 2 3 2
3 2 1 1 1 1 , , , and is 5 3 10 2 5 6 23 2 3 5 = 30; b.
6. The LCD of
5
9
5 x2
1
1
8 5 8 12 8 12
y 6y 9 6 y 9 27 6 y 4y 4. 4611 y 11 9 11 4 y 11 4 y 22
5
5. The LCD of
11 11 10 7 7 , and 5 is , , 6 2 3 x2 9 3 3 x
10. In Method 2, we find the LCD of all fractions in the complex fraction, then multiply it by both the numerator and the denominator so that both will no longer contain fractions; in Method 1, we find the LCD of the fractions only in the numerator and then the LCD of the fractions only in the denominator in order to perform the addition and/or subtraction and get single fractions in the numerator and in the denominator.
1
10 z x 2 3x 3x
2222 x
Exercise Set 12.7
10 x 10x x z z x x x 3
16x
,
9. a single fraction in the numerator and in the denominator
y 3
1.
3
2 x2 18x2; c.
y x
5 6y
3
,
2 2 2 2 x2 16x2; a.
6. The LCD is 6xy.
2
x x 3 3 is 8 2 2 2 , and 4x 2 2 x
y
2x1 y
5
7. The LCD of
x
18 12 6 4 2 8. 43 12 8 6 24 38 16 9 4 13 11 4 11 12 4 48 11 37 10. 12 12 5 1 12 5 1 60 3 63 4 4 31
24 3 1
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489
10 7 3 7 6
73
12. 10 1 5
10
3
14.
16.
3
4
20.
87y
7
21
8y
315
4y
5
m 2 m2 4 m 2 2m 4 m 2 (m 2)(m 2)
45
m2 m2 2
74y 1 7 4y 8 y 21 8 y 21 6
x2
x 4 x4
y2 1
3 2
y2 3 6 y2 1 2
6 y2 1y 56
2(3 2 y2 )
6 y 5 y2
32.
x(2x 1) 1 (2x 1) x
36.
25 5 x5 x 35 5
25 5x 25 3 5x 25 5x 50 5x 22
24
26.
56 y 168 9 y 27 56( y 3)
28.
490
4 x2 4 x2 38. x2 2x 2x 2x x2 4 x2 or (2 x)(2 x) x 2 x 2 2x
56 9
2x 2 x
3 40 y 3 40 y 8 5y 8
5 y3 8
3 5y
5x 50 5x 22 5(x 10) 5x 22
3 3 y9 8
9( y 3)
3 1 43
5y 3 8 5y
53y
3(3)
3 1 43
9 3 4
25 5 x5 3 (x 5) x5 5 (x 5)
7 y21
24
2x 6x x2 9 2x(x 3) (x 3)(x 3) 2x x3
y(6 5 y)
(2x 1)(x 1) x 1 2x 1
x2 x2 1 9 2
1 9
3 3 y9 8
x 2 16
2x x 1 2x 1 x
24.
6 x2 2 6x 34. 2 xx
2
7 y21
2(m 2)
8 18 18 9
18 8 9
1 1 x 2x1 (2x 1) x 2x1 x 1 x (2x 1) 1 2x1 2x1
18 x2
2
6 4 y2
m2
m2
x 2 2 x 2 2 2x 2 2 x 2 2 2
1y 56
22.
5
3 15
2m 4 m2 30. 2m4 m 2 m2 4 m2 4
1
8 4 8 15 8 4 8 4 32 15
21 4y
18.
5
10 21
2
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
9 1
9
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ISM: Prealgebra and Introductory Algebra
40.
Chapter 12: Rational Expressions 11 1 1 54. 3 x 3 x 1 9 x2 19 12
4( x1) 4 x xx x(x1) ( x1)x x x1 1 1 1(x6) 12x 2x x6 2x(x6) 2 ( x6)2x 4x4 x
x 1x 13 x3 2 1x3x 19
x(x1)
x62x
9x2 x29 x3 2 3x x 9 9x2
2x(x6) x2 4x4 x(x1) 3x6 2x(x6)
( x2)2 x(x1) 3( x2) 2x(x6) 2
(x 2) 2x(x 6) x(x 2) 1) 3(x 2(x 2)(x 6) 3(x 1)
Venus Williams has earned $2 million more prize money than Simona Halep over her career. 44.
39
1 million or 1,000,000
39 Maria Sharapova earned $1,000,000 on average per tournament. 46. answers may vary 3 5 3n 5 3n 5 1 3n 5 n2 nn n2 2 2 n2 2 2
48. n
2n
50. The distance is the average of the two given distances. 7.2 10.3 17.5 3 8.75 or 8 2 2 3 4 He should drill 8.75 or 8 inches from the 4 left side of the board. J3 52.
7
J 2 3 2J 3 1 2J 3 2 2 2
7
2
7
14
x3
3x 3x
9x2 (x 3)(x 3)
x3
4x 1 56. 3 x1 3 1x x 3 1x 3x 1 x x 4 x1
42. 42 40 = 2
x 3 9x2 3x x2 9
58. w
4 1
x
4 1
A l
4x2 3 6x3 5
4x 2 5 3 6x 3 2(2x 1) 5 3 3(2x 1) 10 9 10 The width is meters. 9
Chapter 12 Vocabulary Check 1. A rational expressionP is an expression that can be written in the form
, where P and Q are Q polynomials and Q is not 0. 2. In a complex fraction, the numerator or denominator or both may contain fractions. 3. For a rational expression,
a b
a b
a
. b
4. A rational expression is undefined when the denominator is 0.
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491
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
5. The process of writing a rational expression in lowest terms is called simplifying. 6. The expressions
2x
and
7
7
6.
8. A unit fraction is a fraction that equals 1.
9.
x2 4 0 (x 2)(x 2) 0 x20 x 2 x5 The expression is undefined for x = 2 x2 4 and x = 2.
x 2 0 or x2
4x2 4x 15 0 (2x 5)(2x 3) 0 2x 5 0 or 2x 3 0 2x 5 2x 3 5 3 x x 2 2 5x 9 The expression is undefined for 2 4x 4x 15 5 3 x and x . 2 2
3. Replace z with 2. 2 z 2 (2) 4 z 5 2 5 3 4. Replace x with 5 and y with 7. x2 xy y2 (5)2 (5)(7) (7)2 57 x y 25 35 49 12 11 12
3(x 4) x(x 4)
x2 3x 10 x4
8.
Chapter 12 Review
5.
x 4x
x2 7.
7. The least common denominator of a list of rational expressions is a polynomial of least degree whose factors include all factors of the denominators in the list.
2.
2
are called
2x
reciprocals.
1.
3x 12
x2 5x 4 x3 4x
11.
x x2
(x 5)(x 2) x4 (x 4)(x 1)
1 x 5 1
x 1
x2 2x 15
x2 x 6 x2 3x 10
5(x2 25)
(x 5)(x 3) 5(x 5)(x 5) (x 5)(x 3) 5(x 5) x 3
(x 3)(x 2) (x 5)(x 2)
x3 x5
x2 2x x(x 2) x 12. x2 2x 8 (x 4)(x 2) x 4 2 2 13. x 2 xa xb ab (x 2 xa) (xb ab) x xc bx bc (x xc) (bx bc) x(x a) b(x a) x(x c) b(x c) (x b)(x a) (x b)(x c) xa xc
x2 5x 2x 10 14.
(x2 5x) (2x 10)
x2 3x 2x 6
2x 6 2(x 3) 2 x2 3x x(x 3) x
(x2 3x) (2x 6) x(x 5) 2(x 5) x(x 3) 2(x 3) (x 2)(x 5) (x 2)(x 3)
492
3
x(x2 4) x 3x 2 (x 2)(x 1) x(x 2)(x 2) (x 2)(x 1) x(x 2) x 1 2
5x2 125 10.
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x5 x3
ISM: Prealgebra and Introductory Algebra 15x3 y2 15.
16.
z
Chapter 12: Rational Expressions
3 5 x x2 y2 z 3x2 y 5xy3 z 5 x y y2
2x2 9x 9
z
23.
x2 y2 24.
18. 2x 5 2x 2x 5 2x x 6 x 6 x 6 (x 6) 2x(2x 5) (x 6)2 2x(2x 5) (x 6)2
2x
x2 5x 24 x2 x 6 x2 x 12 x2 10x 16 (x 8)(x 3) (x 3)(x 2) (x 4)(x 3) (x 8)(x 2) x3 x4
4x 4 y xy
2
3x 3y 2
x y
26.
4x 4 y x2 y 3x 3y xy2
4(x y) x2 y 3(x y) xy2 4x 3y
2
2x2 9x 9
2x 2 x 3x 8x 12 (2x 3)(x 3) 2x 4(2x 3) x(x 3) 1 2
3x2 2xy y2
3x2 6x x2 xy 2 x2 y2 3x 6x 2 3x2 2xy y2 x xy (x y)(x y) 3x(x 2) x(x y) (3x y)(x y) 3(x 2) 3x y
x7 x 7 25. x2 9x 14 x2 9x 14 x2 9x 14 x7 (x 7)(x 2) 1 x2
2 2 19. x 2 5x 24 x 210x 16 x x 12 x x6
2
x5 x 5 2 2 x 2x 15 x 2x 15 x 2x 15 x5 (x 5)(x 3) 1 x 3 2
4x 5 27.
3x
2
2x 5
3x2
2
21. x x 42 (x 3) (x 7)(x 6) (x 3) x3 x3 x7 x7 (x 6)(x 3) 22.
x2 3x
y3 9x2 y3 9x2 9x2 8 8 y3 8 y3
x2 9 x 2 (x 3)(x 3) x 2 x 3 17. x2 4 x 3 (x 2)(x 2) x 3 x 2
20.
8x 12
2a 2b a b 2(a b) ab 2 2 3 3 a b (a b)(a b) 2 3
4x 5 2x 5 3x2 2x 10 3x2 2(x 5) 3x2
28. 9x 7 3x 4 9x 7 3x 4 6x2 6x2 6x2 6x 3 6x2 3(2x 1) 6x2 2x 1 2x2 29. The LCD is 2 7 x or 14x.
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493
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
30.
x2 5x 24 (x 8)(x 3) x2 11x 24 (x 8)(x 3) The LCD is (x + 3)(x + 8)(x 8).
32.
33.
5 2x2 y
7x
10x2 y 7x 2x2 y 14x3 y
9
9 4 y2 x
5
31.
4y
4 y 4 y2 x
x2
x2 (x 2)(x 9) x 11x 18 (x 2)(x 5) (x 2)(x 9)(x 5) 3x 5 x 4x 4 2
4 35.
36.
5x2
2 x 3
37.
4 x3
494
16 y3 x
2
34.
36 y2 x
6 y
4
x2 3x 10 (x 2)(x 5)(x 9)
3x 5 (x 2)2 (3x 5)(x 3) (x 2)2 (x 3) 3x2 4x 15 (x 2)2 (x 3)
4y 5x2 y
6 5x2 y 5x2
4 y 30x2
5x2 y
2(2 y 15x2 )
5x2 y
2(x 1) 4(x 3) x 1 (x 3)(x 1) 2x 2 4x 12 (x 3)(x 1) 2x 10 (x 3)(x 1) 2(x 5) (x 3)(x 1)
2
4 2(x 3) x3
4 2x 6 x3
2x 2 x3
2(x 1) x3
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3
38.
x 2x 8 2
39.
2
Chapter 12: Rational Expressions
3 2 x 3x 2 (x 4)(x 2) (x 2)(x 1) 3(x 1) 2(x 4) (x 4)(x 2)(x 1) 3x 3 2x 8 (x 4)(x 2)(x 1) 5x 5 (x 4)(x 2)(x 1) 5(x 1) (x 4)(x 2)(x 1)
2
4 4 2x 5 2x 5 2 6x 9 2x 3x 3(2x 3) x(2x 3) x(2x 5) 4(3) 3x(2x 3) 2x2 5x 12 3x(2x 3) (2x 3)(x 4) 3x(2x 3) x4 3x
40.
x 1
x 1 x 1 x 1 x 2x 1 x 1 (x 1)(x 1) x 1 1 x 1 x 1 x 1 1 x 1 x 1 x x 1 2
41.
n
9
n
10 5 n n 10 10 9 10 5 n 90 2n 3n 90 n 30 The solution is 30.
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495
42.
2 1 1 x 1 x 2 2 1 2 1 2(x 1)(x 2) x 1 x 2 2(x 1)(x 2) 2 2 2(x 2) 2(x 1) 1(x 1)(x 2) 4x 8 2x 2 (x2 x 2) 2x 10 x2 x 2 x2 x 12 0 (x 4)(x 3) 0 x40 or x 3 0 x 4 x3 The solutions are 4 and 3.
43.
y 2 y 16 y 3 2y 2 4y 4 y 1 y 2 y 16 y 3 2( y 1) 4( y 1) y 1 4( y 1) y 2 y 16 y 3 4( y 1) 2( y 1) 4( y 1) y 1
2 y 2 y 16 4( y 3) 4 y 16 4 y 12 0y 4 0 4; no solution
44.
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
45.
2 4 8 x2 3 x4 3 x2 9 8 (x2 9) x 3 x 3 (x2 9) 2 x 9 2(x 3) 4(x 3) 8 2x 6 4x 12 8 2x 18 8 2x 10 x5 The solution is 5. x3 x6 0 3 x x65 x x 1 (x 1)(x 5) (x 1)(x 5)0 x 1 x 5 (x 5)(x 3) (x 1)(x 6) 0 x2 2x 15 x2 5x 6 0 7x 9 0 7x 9 9 x 7 9 The solution is . 7
496
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ISM: Prealgebra and Introductory Algebra
46.
x5
Chapter 12: Rational Expressions
6
50. Let x be the speed of the boat in still water.
x 6 x(x 5) x x x2 5x 6
distance
x2 5x 6 0 (x 6)(x 1) 0 x6 0 or x 1 0 x 6 x 1 The solutions are 6 and 1.
48
x4
Downstream
72
x+4
48 x4
72 x 4
51. Let x be the time for Maria alone. Together, 1 Mark and Maria complete 5
of the job in
1 hour. Individually, they complete
1 7
1
49. Let x be the speed of the faster car. =
rate
Faster car
90
x
Slower car
60
x 10
and
1
of
x
the job in 1 hour. 1 1 1 15 7 x1 1 35x 35x 5 7 x 7x 5x 35 2x 35 35 1 x 17 2 2 1 It will take Maria 17 hours to manicure Mr. 2 Sturgeon’s lawn alone.
distance
Upstream
time
The speed of the boat in still water is 20 mph.
n(4 n) n(4 n) n 4 n 4n n 4 2n 2n The number is 2.
rate
The times are equal. 48 72 x4 x4 48(x 4) 72(x 4) 48x 192 72x 288 480 24x 20 x
47. Let n be the number. 1 31 7 5 n 2 n 6 5 3 7 n 2n 6 5 3 7 6n n 6n 2n 6 30 9 7n 21 7n 3n The number is 3. 48. Let n be the number. 1 1 n 4n 1
=
time
The times are equal. 90 60 x x 10 90(x 10) 60x 90x 900 60x 30x 900 x 30 x 10 = 20 The faster car is traveling at 30 mph and the slower car is traveling at 20 mph.
90 x
60 x10
52. Let x be the time for the pipes to fill the pond 1 of the pond in working together. Pipe A fills 20 1 1 day, pipe B fills of the pond in 1 day, and 15 1 together they fill of the pond in one day. x
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
1
1 1 120 115 x 1 60x 60x 20 15 x 3x 4x 60 7x 60 60 4 x 8 7 7 4 It takes 8 days to fill the pond using both 7 pipes. 53.
5x 10xy 27 5x 27 21 10xy 21
x6 2 4 56. x2 8 8 4 x 2 4 (x 2) x2 6
6 4x 8 8 4x 8 4x 14 4x 2(2x 7) 2 2x 2x 7 2x 4x 12
57.
5x 21 27 10xy 5 x 3 7 3 9 2 5 x y 7 18 y 37
6
58.
31 31 35 30 31 30 35 31 31 5 6 5 7 31 6 7
55. 3y 1 2 y 2 2 y 1 y 1 y
498
8x(x 3)
1 2x
x(x2 6x 9)
(x 7)(x 3) x(x 3)(x 3) (x 7)(x 3) x(x 3) x7
x2 x 72 x2 6x 27 60. x2 x 30 x2 9x 18 x2 x 72 x2 9x 18 x2 x 30 x2 6x 27 (x 9)(x 8) (x 6)(x 3) (x 6)(x 5) (x 9)(x 3) (x 9)(x 8) (x 5)(x 9)
y 3 1
1y y
4(x 3)
2 2 59. x 29x 20 x 2 9x 20 x 25 x 8x 16 (x 4)(x 5) (x 4)(x 5) (x 5)(x 5) (x 4)(x 4) x4 x4
56 65 2110 35 625 30 31 35 31 30
x2 4x 21
25
3 1
8x2 24x
x3 6x2 9x
2 54. 5 7 57 75 1 5 16 55 5
6 4(x 2) 8 4(x 2)
3
(x 2)
4
x6 6 61. 2 2 2 x 36 x 36 x 36 x6 (x 6)(x 6) 1 x6 x
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ISM: Prealgebra and Introductory Algebra
62.
63.
5x 1 3x 2 5x 1 3x 2 2x 1 4x 4x 4x 4x 3x 6x
Chapter 12: Rational Expressions
66.
x2 9x 14 x2 4x 21 3x 6x (x 7)(x 2) (x 7)(x 3)
3x(x 3) 6x(x 2) (x 7)(x 2)(x 3) (x 7)(x 3)(x 2)
3x2 9x 6x2 12x (x 7)(x 2)(x 3) 3x2 21x (x 7)(x 2)(x 3)
3x(x 7) (x 7)(x 3x 2)(x 3)
(x 2)(x 3) 4
64.
65.
2 2 3x2 8x 3 3x 7x 2 4 2 (3x 1)(x 3) (3x 1)(x 2) 4(x 2) 2(x 3) (3x 1)(x 3)(x 2) (3x 1)(x 2)(x 3) 4x 8 2x 6 (3x 1)(x 3)(x 2) 6x 2 (3x 1)(x 3)(x 2) 2(3x 1) (3x 1)(x 3)(x 2) 2 (x 3)(x 2) 4
3 2 a 1 3 a 1 4 (a 1) a 1 2 (a 1) a 1 4 2(a 1) 3 4 2a 2 3 2 2a 3 2a 1 a1 2 1 The solution is . 2
x
x
xx 3 x 3 x (x 3) x 3 4 (x 3) x 3 x 4(x 3) x x 4x 12 x 5x 12 x 4x 12 0 4x 12 x 3 3 makes the denominator x + 3 zero. Therefore there is no solution.
67. Let n be the number. 2n 1 n 3 6 2 2n 1 n 6 3 6 6 2 4n 1 3n n 1 0 n1 The number is 1. 68. Let x be the time to paint the shed together. Mr. 1 Crocker can paint of the shed in one day, and 3 1 his daughter can paint . Together, they can 4 1 paint . x 1 1 1 3 4 x 1 1 1 12x 12x 3 4 x 4x 3x 12 7x 12 12 5 x 7 or 1 7 5
It will take 1
7 work together.
days to paint the shed when they
12 1
1
4
3 3 69. 1 4 1 12 1 4 1 4 6 10 3
2
4 2 70.
3 2 x 4 2 4x 2
x 6 3 x x
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x 3 6 x
6x 3
2(2x 1)
3(2x 1)
2
3
499
ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
71. 1.8 square yards 1.8 sq yd
32 sq ft
8.
1 cu yd 72. 135 cubic feet 135 cu ft 3 3 cu ft 135 cu yd 27 5 cu yd Chapter 12 Getting Ready for the Test
2.
x 8
x 8
x 8
8 x 1(8 x) Choice B is correct. 8
4
x2
x2
8 4 2
2
x x
x 8
1
32 x4
Choice C is correct. 8 3.
4.
5.
x2
4
8
x2
x2
x2
4 Choice D is correct. 8 x2
4
x2
8 4 x2
8x2 4x2
1 2 contains an equal sign, so it is an
x 3 x equation; B.
1 sq yd
1.8 9 sq ft 16.2 square feet
1.
5
8 x2 21 2 2 4 x
12
x2 Choice D is correct.
9.
a5 9 contains an equal sign, so it is an 11 equation; B.
10.
a5 9 does not contain an equal sign, so it is 11 an expression; A.
11. The LCD of 4, 6, and 1 is 12. x3 5 3 4 6 x3 5 3 12 12 4 6 1 x3 5 3 12 12 12 4 6 1 3(x 3) 2 5 12 3 The correct choice is C.
9
2. a.
C
b.
C
z ,
25x 10x3 2 3 3 25x 5 x and 10x 2 5 x LCD 2 52 x3 2 25 x3 50x3
Choice D is correct. 7.
500
5
1
does not contain an equal sign, so it is an x 3 expression; A.
x2 4x 3 0 (x 1)(x 3) 0 x 1 0 or x 3 0 x 1 x 3 x5 The expression is undefined for x2 4x 3 x = 1 and x = 3.
Choice A is correct. 6.
12. “The quotient of a number and 5 equals the sum x of that number and 12” translates to x 12; 5 A. Chapter 12 Test 1.
7x 5 2x 7x (5 2x) x 1 x 1 x 1 7x 5 2x x 1 5x 5 x 1 5(x 1) x 1 5
3.
100x 3000 100(200) 3000 $115 x 200
100x 3000 x 100(1000) 3000 1000 $103
3x 6 3(x 2) 3 5x 10 5(x 2) 5
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ISM: Prealgebra and Introductory Algebra
4.
x6 x6 1 x 12x 36 (x 6)(x 6) x 6
5.
7 x (x 7) 1 x7 x7
6.
Chapter 12: Rational Expressions
13.
2
yx x2 y2
(x y) 1 (x y)(x y) x y
3 2 2 7. 2m 2m 12m 2m(m m 6) m2 5m 6 (m 3)(m 2) 2m(m 3)(m 2) (m 3)(m 2) 2m(m 2) m2
14.
10.
3
(5x 5)
x 1 11.
3
x2 9 15.
5(x 1) 15
x 1
y2 5 y 6 y 2 ( y 3)( y 2) y 2 2y 4 2y 6 2( y 2) 2( y 3) y2 4 5
6
6 3 6 3 x 1 x 1 (x 1)(x 1) x 1 6 3(x 1) (x 1)(x 1) (x 1)(x 1) 6 3x 3 (x 1)(x 1) 3x 3 (x 1)(x 1) 3(x 1) (x 1)(x 1) 3 x 1
56 1 12. 2x 5 2x 5 2x 5 2x 5
2
ay 3a 2 y 6 (ay 3a) (2 y 6) 8. ay 3a 5 y 15 (ay 3a) (5 y 15) a( y 3) 2( y 3) a( y 3) 5( y 3) (a 2)( y 3) (a 5)( y 3) a2 a 5 x2 13x 42 x2 4 2 2 9. x 10x x 6 x2 13x 2142 xx2 x 6 2 x 10x 21 x2 4 (x 6)(x 7) (x 3)(x 2) (x 3)(x 7) (x 2)(x 2) (x 6)(x 7) (x 7)(x 2)
5a 2 a a 6 a 3 5a 2 (a 3)(a 2) a 3 5a 2(a 2) (a 3)(a 2) (a 3)(a 2) 5a 2(a 2) (a 3)(a 2) 5a 2a 4 (a 3)(a 2) 3a 4 (a 3)(a 2) 2
16.
2
x 3x
x2 4x 1 2x 10
2x 10 2 x 3x x 4x 1 (x 3)(x 3) 2(x 5) 2 x(x 3) x 4x 1 2(x 3)(x 5) x(x2 4x 1)
x2 9 2
5 x2 x2 11x 18 x2 3x 10 x2 5 (x 9)(x 2) (x 5)(x 2) (x 2)(x 5) 5(x 9) (x 9)(x 2)(x 5) (x 5)(x 2)(x 9)
x2 3x 10 5x 45 (x 9)(x 2)(x 5)
x2 2x 35 (x 9)(x 2)(x 5)
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
17.
3 4y y 6 y 5 y2 5 y 4
21. The LCD is x2 25 (x 5)(x 5). 10 3 1 x2 25 x 5 x 5 10 1 3 (x2 25) (x 5)(x 5) x 5 x 5 x 2 25
2
4y 3 ( y 5)( y 1) ( y 1)( y 4) 4 y( y 4) 3( y 5) ( y 1)( y 5)( y 4)
10 3(x 5) 1(x 5) 10 3x 15 x 5 10 4x 10 20 4x 5 x
4 y2 16 y 3y 15 ( y 1)( y 5)( y 4)
4 y2 13y 15 ( y 1)( y 5)( y 4)
Since x = 5 causes the denominators x2 25 and x 5 to be 0, the equation has no solution.
18. The LCD is 3 5 y = 15y. 4 5 1 y35 4 5 1 15 y 15 y y 3 5 60 25 y 3y 60 22 y 60 y 22 30 y 11 30 The solution is . 11 19.
5
22.
(x
14 x 1 14
4
2x x 1
2x
1) x (x 1) 4 1 4(x 1) 2xx 1 x(x 1)x 14
x2 x 14 4x 4 2x x2 x 14 2x 4 x2 3x 10 0 (x 5)(x 2) 0 x 5 0 or x 2 0 x5 x 2 The solutions are 5 and 2.
4
2
y 1 y 2 5( y 2) 4( y 1) 5 y 10 4 y 4 y 6
23.
5x 2 5x2 10x yz 2 3 10x yz z 3 z 2 3
The solution is 6. 20. The LCD is 2(a 3). a 3 3 a 3 a 3 2 a 3 3 2(a 3) 2(a 3) a 3 a 3 2 2a 6 3(a 3) 2a 6 3a 9 5a 15 a3 Since a = 3 causes the denominator a 3 to be 0, the equation has no solution.
502
x
5x
z
yz2 10x 5 x x z z2
y z2 2 5 x xz 2y
b a ba a b ab 24. a1 1b 1 1 ab b
a
b 2
a
b a2 ab (b a)(b a) ab ba
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
y2 5 1 2 2 y 5 y 1 y2 25. 1 2 2 2 1 2 y y y y 2 y 2 y 5 1
1 1 1 60x 12 15 x 5x 4x 60 9x 60 60 x 9 20 x 3 20 2 It takes both pipes 6 hours to fill the 3 3 tank. 60x
26. Let n be the number. 1 n5 6 n 5 n 6 n 5 n n n(6) n n2 5 6n 2 n 6n 5 0 (n 5)(n 1) 0 n 5 0 or n 1 0 n5 n 1 The number is 1 or 5.
Cumulative Review Chapters 112 7 1.
4
Downstream
14 16
5 3.
=
rate
time
x2
14
x+2
16
x2
2
1
28. Let x be the time in hours that it takes for both inlet pipes together to fill the tank. 1 The first pipe fills of the tank in 1 hour, the 12 1 second pipe fills of the tank in 1 hour, and 15 1 the two pipes fill of the tank in 1 hour. x 1 1 1 12 15 x The LCD is 60x.
6. 3
20
4
100%
100%
2
1
4
20
400
100%
200
100%
100
9
700
% 35%
% 80%
5
3
1
x 2
14 16 x2 x2 14(x 2) 16(x 2) 14x 28 16x 32 60 2x 30 x The speed of the boat in still water is 30 mph.
9 5. 2
7
5
3 4.
Upstream
20
2.
27. Let x be the speed of the boat in still water. Let x + 2 be the speed of the boat going downstream. Let x 2 be the speed of the boat going upstream. distance
9
4
9 4
2 % 66 % 3 3
1 % 11 % 9 9
100%
900 4
% 225%
3 15 15 1500 100% % 375% 4 4 4 4
7. 2 (z 5) = 2 (5 z) illustrates the commutative property of multiplication. 8. 3 + y = y + 3 illustrates the commutative property of addition. 9. (x + 7) + 9 = x + (7 + 9) illustrates the associative property of addition. 10. (x 7) 9 = x (7 9) illustrates the associative property of multiplication. 11. Let x be the length of the shorter piece. Then 3x is the length of the longer piece. x 3x 48 4x 48 x 12 3x = 3(12) = 36 The shorter piece is 12 inches. The longer piece is 36 inches.
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
12. Let x be the length of the longer pieces. Then x 3 is the length of the shorter piece. x x x 3 45 3x 3 45 3x 48 x 16 x 3 = 16 3 = 13 The longer pieces both have length 16 feet. The shorter piece has length 13 feet. 13.
y mx b y b mx yb x m
15.
x 4 6 x 4 4 6 4 x 10 –12–10 –8 –6 –4 –2 0
x2
19. 4 20.
3
2
4
6 8
3
1 16
1
1
(3)3
27
1 27
2
2 3 33. 4x 7 8x 4x 2 4x 6 11 2x 3 2x 3
4x2 4x 6 2x 3 8x33 4x22 0x1 7 8x 12x
4 256
52
(3)
3
(3)
8x2 0x1 8x2 12x 12x 7
27
12x 18 11
(3)
(4)9 (4) (4)7
97
2
(4)
16
5 2 5 2 4 23. 2x y 2 x y 2x51 y21 2x y xy 1 x y
504
4
32. 5y(7y 3y 1) 5y(7y2 ) (5y)(3y) (5y)(1) 35y3 15y2 5y
4
(3)5
22.
1
31. 3x2 (5x2 6x 1)
8 712 7124 7 5, 764,801 4 7
2
1 9 2 26. 9x 9 2 2 x x
15x4 18x3 3x2
73
21.
1 13b a 13ab
30. 3y(4y2 2) 3y(4y2 ) 3y(2) 12y3 6y
x
4
54
29. 5x(2x3 6) 5x(2x3 ) 5x(6) 10x4 30x
9 9 18. x x x91 x8 x x1
47
13b a
3x2 (5x2 ) (3x2 )(6x) (3x2 )(1)
52
x
a4
1 2 3 25. 2x 2 3 3 x x
28. (3)3
16. x 1 10 x 9 17.
a4
a5
13b
(2)
y mx b y mx b
x
13a5b
27. (2)4
14.
5
24.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 6 x2 7x 4 34.
2x 1
3x 5
Chapter 12: Rational Expressions 3x2 2x 10x 5 3x2 2x 10x 5 43. x 1 x 1 x 1 3x2 8x 5 x 1 (3x 5)(x 1) x 1 3x 5
9 2x 1
3x 5 2x 1 6x 2 7x 4 6x 3x 2
10 x 4 10x 5 9
44.
4x
8
x2
x2
6x
35. x2 7x 12 (x 4)(x 3) 45.
2
x 4
36. x 17x 70 (x 7)(x 10) 2
2 2 37. 25x 20xy 4 y (5x 2 y)(5x 2 y) (5x 2 y)2
38. 36a2 48ab 16b2 4(9a2 12ab 4b2 ) 4[(3a)2 2 3a 2b (2b)2 ] 4(3a 2b)2
2 46.
39.
x2 9x 22 0 (x 11)(x 2) 0 x 11 0 or x 2 0 x 11 x 2 The solutions are 11 and 2.
40.
x2 2x 15 0 (x 5)(x 3) 0 x5 0 or x 3 0
47.
x 5 x3 The solutions are 5 and 3. 2 41. x x 6 x(x 1) 2 3 2 3x 5x 5 3x 5(x 1) 5
42. 3x 12 5x 20 3x 12 4x 2 5x 20 2 4x 3(x 4) 2 2x 2 5(x 4) 3 2x 5 6x 5
48.
x 3
4x 8 x2
4(x 2)
4
x2
3 6x 3(x 2) x 2 (x 2)(x 2) (x 2)(x 2) 6x 3x 6 (x 2)(x 2) 3x 6 (x 2)(x 2) 3(x 2) (x 2)(x 2) 3 x2
5x x2 9
2(x 3)
(x 3)(x 3) 2x 6 5x (x 3)(x 3) 7x 6 2 x 9
5x (x 3)(x 3)
t 4 t 3 5 2 9 18 t 4 t 3 5 18 2 9 18 18 9(t 4) 2(t 3) 5 9t 36 2t 6 5 7t 30 5 7t 35 t 5 The solution is 5. y y 1 2 y y4 6 1 12 12 6 2 4 6y 3y 2 3y 2 2 y 3 2 The solution is . 3
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ISM: Prealgebra and Introductory Algebra
Chapter 12: Rational Expressions
49. Let x be the time in hours for Sam and Frank to complete the tour together. Sam completes 13 of a tour in 1 hour, and Frank completes
1 7
z 1 1 2 51. z1 2z 2z2 2z z 3 6 6 6 2z
2z
of a
2z
26
1
tour in 1 hour. Together, they complete x of a tour in 1 hour. 1 1 1 3 7 x 1 1 1 21x 21x 3 7 x 7x 3x 21 10x 21 21 1 x 2 10 10 Sam and Frank can complete a quality control tour together in 2
1
x 9 x1 52. 9 3x 9x 9x 1 x3 2
x
hours.
10 50. Let x be the time in hours for both machines to complete the task together. The first machine 1 completes of the task in 1 hour and the 18 1 second machine completes of the task in 1 12 1 hour. Together, the machines complete of the x task in 1 hour. 1 1 1 18 112 x 1 1 36x 36x 18 12 x 2x 3x 36 5x 36 36 x 5 Both machines can complete the task in 36 1 7 hours. 5 5
506
z 2z 2z 6 2z 6 2 2z z 2 3 (2 z) 2 z (2 z) 3 z
x x x2 9 9x x3 x 2
x 9 x3 9x x 2 x 9 x 9x x 3 x (x 3) (x 3) 9 x (x 3) x 3 9
Copyright © 2025 Pearson Education, Inc.
Chapter 13 Section 13.1 Practice Exercises 1. a.
b.
Point (4, 2) lies in quadrant I.
b.
Point (1, 3) lies in quadrant III.
c.
Point (2, 2) lies in quadrant IV.
d.
Point (5, 1) lies in quadrant II.
e.
Point (0, 3) lies on the y-axis.
f.
Point (3, 0) lies on the x-axis.
g.
Point (0, 4) lies on the y-axis.
h. i.
1 Point 2 , 0 lies on the x-axis. 2 3 Point 1, 3 lies in quadrant IV. 4
c. 3. a.
Let x = 3 and y = 1. x 3y 6 3 3(1) 0 6 33 0 6 6 6 True Yes, (3, 1) is a solution.
b.
Let x = 6 and y = 0. x 3y 6 6 3(0) 0 6 6006 6 6 True Yes, (6, 0) is a solution.
c.
Let x = 2 and y
y 6 5 4
(0, 3)
3
1 2 (–5, 1) 2 , 0 2 1 –6 –5 –4 –3 –2 –1–1
(–1, –3)
2. a.
–2
–3 –4 –5 –6
(3, 0)
(4, 2)
1 2 3 4 5 6
(2, –2)
The number of wildfires varies greatly from year to year.
x
3 1, 3 4 (0, –4)
2
.
3
The ordered pairs are (2016, 68), (2017, 71), (2018, 58), (2019, 50), (2020, 57), (2021, 59), and (2022, 69)
x 3y 6 2 2 3 06 3 2 2 0 6 0 6 False 2 No, 2, is not a solution. 3 4. a.
x + 2y = 8 In (0, ), the x-coordinate is 0. 0 2y 8 2y 8 y4 The ordered pair is (0, 4).
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
b.
x + 2y = 8 In ( , 3) the y-coordinate is 3. x 2(3) 8 x6 8 x2 The ordered pair is (2, 3).
c.
x + 2y = 8 In (4, ), the x-coordinate is 4. 4 2 y 8 2 y 12 y6
b.
Let x = 0. 1 y x 1 3 1 y (0) 1 3 y 0 1 y 1
c.
Let y = 0. 1 0 x 1 3 1 1 x 3 3(1) x 3 x
The ordered pair is (4, 6). 5. a.
b.
c.
Let x = 3 y = 2x y = 2(3) y=6
The ordered pairs are (3, 2), (0, 1), and (3, 0).
Let y = 0. y = 2x 0 = 2x 0=x Let y = 10. y 2x 10 2x 5 x
The ordered pairs are (3, 6), (0, 0), and (5, 10). x
y
a.
3
6
b.
0
0
c.
5
10
6. a.
Let x = 3. 1 y x 1 3 1 y (3) 1 3 y 11 y 2
x
y
a.
3
2
b.
0
1
c.
3
0
7. When x = 1, y = 50x + 250 y = 50(1) + 250 y = 50 + 250 y = 200 When x = 2, y = 50x + 250 y = 50(2) + 250 y = 100 + 250 y = 150 When x = 3, y = 50x + 250 y = 50(3) + 250 y = 150 + 250 y = 100 When x = 4, y = 50x + 250 y = 50(4) + 250 y = 200 + 250 y = 50 The completed table is shown.
508
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ISM: Prealgebra and Introductory Algebra
x
1
2
3
4
y
200
150
100
50
Chapter 13: Graphing Equations and Inequalities
Exercise Set 13.1 2.
Vocabulary, Readiness & Video Check 13.1 1. The horizontal axis is called the x-axis. 2. The vertical axis is called the y-axis. 3. The intersection of the horizontal axis and the vertical axis is a point called the origin. 4. The axes divide the plane into regions, called quadrants. There are four of these regions. 5. In the ordered pair of numbers (2, 5), the number 2 is called the x-coordinate and the number 5 is called the y-coordinate. 6. Each ordered pair of numbers corresponds to one point in the plane. 7. An ordered pair is a solution of an equation in two variables if replacing the variables by the coordinates of the ordered pair results in a true statement. 8. The graph of paired data as points in a rectangular coordinate system is called a scatter diagram. 9. origin; left or right; up or down 10. paired data, which can be written as ordered pairs and graphed 11. (7, 0) and (0, 7); since one of these points is a solution and one is not, it shows that it is very important to remember that the first number is the x-value and the second number is the y-value and not to mix them up. 12. Replace both values of the ordered pair in the linear equation and see if a true statement results.
(2, 4) is in quadrant I. (2, 1) and (3.4, 4.8) are in quadrant II. (3, 3) is in quadrant III. 1 (5, 4) and 3 , 5 are in quadrant IV. 3 3 , 0 lies on the x-axis. 4 (0, 2) lies on the y-axis. 4. Point B is 3 2
units right from the origin and is
1 on the x-axis, so its coordinates are 3 , 0 . 2 6. Point D is 1 unit left and 3 units up from the origin, so its coordinates are (1, 3). 8. Point F is on the y-axis and 1 unit down from the origin, so its coordinates are (0, 1). 10. Point A is 2 units right from the origin and is on the x-axis, so its coordinates are (2, 0). 12. Point C is 2 units left and 3 units up from the origin, so its coordinates are (2, 3). 14. Point E is 1 unit right and 1 unit down from the origin, so its coordinates are (1, 1). 16. Point G is 2 units left from the origin and is on the x-axis, so its coordinates are (2, 0). 18. a.
The ordered pairs are (2016, 66.8), (2017, 69.5), (2018, 90.5), (2019, 97.1), (2020, 103.6), and (2021, 123.6).
b. The ordered pair (2021, 123.6) indicates that in 2021, $123.6 billion was spent on petrelated expenditures.
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Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
c.
d. The scatter diagram shows that pet-related expenditures increased every year. 20. a.
The ordered pairs are (8.00, 1), (7.50, 10), (6.50, 25), (5.00, 50), and (2.00, 100).
b. The ordered pair (2.00, 100) indicates for the price of $2.00 per board, 100 boards were purchased. c.
d. answers may vary 22. a.
The ordered pairs are (2016, 439), (2017, 441), (2018, 443), (2019, 444), (2020, 444), (2021, 445), and (2022, 446).
b. We plot the ordered pairs. We label the horizontal axis “Year” and the vertical axis “Farm Size (in acres).”
510
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities 30. x 5y = 1 In ( , 2), the y-coordinate is 2. x 5(2) 1 x 10 1 x 11 In (4, ), the x-coordinate is 4. 4 5 y 1 5 y 5 y 1 The completed coordinates are (11, 2) and (4, 1).
24. For (2, 3), let x = 2 and y = 3. 3x y 8 3(2) 3 0 8 6308 9 8 False No, (2, 3) is not a solution. For (0, 8), let x = 0 and y = 8. 3x y 8 3(0) 8 0 8 080 8 8 8 True Yes, (0, 8) is a solution.
32. y
1
34.
y 9x
x2 5 In (10, ), the x-coordinate is 10. 1 y (10) 2 5 y 2 2 y 4 In ( , 1), the y-coordinate is 1. 1 1 x2 5 1 3 5x 15 x The completed coordinates are (10, 4) and (15, 1).
For (8, 0), let x = 8 and y = 0. 3x y 8 3(8) 0 0 8 24 0 0 8 24 8 False No, (8, 0) is not a solution. 26. For (0, 0), let x = 0 and y = 0. 1 y x 2 1 0 0 (0) 2 0 0 True Yes, (0, 0) is a solution. For (4, 2), let x = 4 and y = 2. 1 y x 2 1 2 0 (4) 2 2 2 False No, (4, 2) is not a solution.
1
yx 9
28. For (2, 2), let x = 2 and y = 2. y 2 2 2 False No, (2, 2) is not a solution.
For (0, 2), let x = 0 and y = 2. y 2 2 2 True Yes, (0, 2) is a solution.
y = 9x
19 (0) 0
0
3
9(3) = 27
1 (2) 2 9
36. For (2, 2), let x = 2 and y = 2. y 2 2 2 True Yes, (2, 2) is a solution.
x 19 y
9
2
x y 4 x 4 y x 4 y x = y + 4
y = x + 4
0 + 4 = 4
0
0
0 + 4 = 4
(3) + 4 = 3 + 4 = 7
3
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Chapter 13: Graphing Equations and Inequalities
38.
y
1
46. y = 3x x = 0: y = 3(0) = 0 x = 2: y = 3(2) = 6 y = 9: 9 3x 3 x The ordered pairs are (0, 0), (2, 6), and (3, 9).
x
3 3y x x = 3y
y1x
0
1 (0) 0 3
x
y
6
1 (6) 2 3
0
0
3(1) = 3
1
2
6
3
9
3
40. 2x y 4 2x y 4 1 x y 2 2
y = 2x + 4
y (–3, 9) 10 (–2, 6) 8
x 1 x2 2
y = 2x + 4
0
2(0) + 4 = 0 + 4 = 4
1 (0) 2 0 2 2
0
1 (2) 2 1 2 1
2
2
2
42.
6 4 2
y 5x 10 y 10 5x 1 y2 x 5
1
0
1 (5) 2 1 2 1 5
5
0
5(0)+10 = 0 + 10 = 10
44. x 6 y 3 x 6y 3
6y x 3 x 3 y 6
x = 6y + 3
y x3 6
0
03 3 1 6 6 2
1
13 2 1 6 6 3
6(1) + 3 = 6 + 3 = 3
1
x
x3
2 1
(0) 3 0 3 3
2
x = 4: y
1 (0) 2 0 2 2 5
2 4 6 8 10
–4 –6 –8 –10
x = 0: y
y = 5x + 10
(0, 0)
–10 –8 –6 –4 –2–2
48. y
x 1 y2 5
512
ISM: Prealgebra and Introductory Algebra
y = 0: 0 3
1 2 1
(4) 3 2 3 1 x3
2 1
x 2 6 x The ordered pairs are (0, 3), (4, 1), and (6, 0). x
y
0
3
4
1
6
0
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
y 5 4
3 2
(–4, 1)
(0, 3)
1
(–6, 0)
–5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
50. a.
x y = 0.25x + 9
0
1
5
10
0.25(0) + 9 =0+9 =9
0.25(1) + 9 = 0.25 + 9 = 9.25
0.25(5) + 9 = 1.25 + 9 = 10.25
0.25(10) + 9 = 2.5 + 9 = 11.50
b. Find x when y = 12.25. 12.25 0.25x 9 3.25 0.25x 13 x The employee must produce 13 units to earn $12.25 per hour. 52. a.
x
0
y = 3.37x + 88.18
3.37(0) + 88.18 = 0 + 88.18 = 88.18
6 3.37(6) + 88.18 = 20.22 + 88.18 = 108.40
b.
Find x when y = 105.00. 105.00 3.37x 88.18 16.82 3.37x 5 x 2015 + 5 = 2020 The average NFL ticket price was approximately $105.00 in 2020.
c.
Find x when y = 130.00. 130.00 3.37x 88.18 41.82 3.37x 12 x 2015 + 12 = 2027 The average NFL ticket price is predicted to be $130.00 in 2027.
9 3.37(9) + 88.18 = 30.33 + 88.18 = 118.51
54. The shortest bar on the graph corresponds to Germany, so Germany is predicted to be the least popular tourist destination. 56. The bars for Italy, Thailand, Spain, France, United States, and China extend beyond 70, so these six countries are predicted to have more than 70 million tourist visits per year. 58. The height of the bar for Mexico appears to be 58, so Mexico is predicted to have 58 million tourist arrivals per year. 60. x y 3 y x 3 y x 3 Copyright © 2025 Pearson Education, Inc.
513
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities 62. 5x 2 y 7 2 y 7 5x 7 5x y 2
Section 13.2 Practice Exercises 1. a.
3x + 2.7y = 5.3 is a linear equation in two variables because it is written in the form Ax + By = C with A = 3, B = 2.7, and C = 5.3.
b.
x2 y 8 is not a linear equation in two variables because x is squared.
c.
y = 12 is a linear equation in two variables because it can be written in the form Ax + By = C: 0x + y = 12.
d.
5x = 3y is a linear equation in two variables because it can be written in the form Ax + By = C: 5x + 3y = 0.
64. 4 y 8x y 2x 66. False; point (3, 0) lies on the x-axis. 68. False;
2x 3y 6 2 2(2) 3 0 6 3 420 6 26
70. Points in quadrant I are to the right and up from the origin, so the x- and y-coordinates are positive. (positive, positive) corresponds to quadrant I. 72. Points in quadrant II are to the left and up from the origin, so the x-coordinate is negative and the y-coordinate is positive. (negative, positive) corresponds to quadrant II. 74. If the y-coordinate of a point is 0, the point is neither up nor down from the origin, so it is on the x-axis. 76. answers may vary
2. x + 3y = 6 x = 0: 0 3y 6 3y 6 y2 x = 3: 3 3y 6 3y 3 y 1 y = 0: x 3(0) 6 x0 6 x6 The ordered pairs are (0, 2), (3, 1), and (6, 0). y
78. A point three units to the left of the origin has coordinates (3, 0). 80. The length of the rectangle is 2 (6) = 8 and the width of the rectangle is 5 0 = 5. Area = length width = 8 5 = 40 The area is 40 square units.
12 10 8 6 4
2 –12–10 –8 –6 –4 –2–2
86. no; answers may vary 88. answers may vary 90. answers may vary
514
(6, 0)
2 4 6 8 10 12
–4 –6 –8 –10 –12
82. Seven years after 2010, or in 2017, there were 1822 Target stores in the United States. 84. answers may vary
(3, 1) (0, 2)
3. 2x + 4y = 8 x = 2: 2(2) 4 y 8 4 4y 8 4y 4 y 1 x = 0: 2(0) 4 y 8 0 4y 8 4y 8 y2
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x
ISM: Prealgebra and Introductory Algebra x = 2: 2(2) 4 y 8 4 4 y 8 4 y 12 y3 The ordered pairs are (2, 1), (0, 2), and (2, 3).
Chapter 13: Graphing Equations and Inequalities
y 12 10 8
(–6, 7)
4 2
y
–12 –10 –8 –6 –4 –2–2
6 5 4 3 2
(–2, 1)
1
–6 –5 –4 –3 –2 –1–1
6
(4, 2) 2 4 6 8 10 12
x
–4 –6 –8 –10
(2, 3) (0, 2) 1 2 3 4 5 6
(0, 4)
–12
x
6. The equation x = 3 can be written in standard form as x + 0y = 3. No matter what value replaces y, x is always 3. It is a vertical line. Plot points (3, 2), (3, 0), and (3, 4), for example.
–2 –3 –4 –5 –6
y
4. y = 2x x = 2: y = 2(2) = 4 x = 0: y = 2(0) = 0 x = 3: y = 2(3) = 6 The ordered pairs are (2, 4), (0, 0), and (3, 6).
6 5 4 3 2 1 –6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6
(3, 2) (3, 0) 1 2 3 4 5 6
x
(3, –4)
7. y = 2x + 3 x = 2: y = 2(2) + 3 = 4 + 3 = 7 x = 0: y = 2(0) + 3 = 0 + 3 = 3 x = 3: y = 2(3) + 3 = 6 + 3 = 3 The ordered pairs are (2, 7), (0, 3), and (3, 3). 5. y
1
x4
2 1 x = 6: y (6) 4 3 4 7 2 1 x = 0: (0) 4 0 4 4 2 1 x = 4: y (4) 4 2 4 2 2 The ordered pairs are (6, 7), (0, 4), and (4, 2).
The graph of y = 2x + 3 is the same as the graph of y = 2x except that the graph of y = 2x + 3 is moved three units upward.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
8. a.
y = 430x + 7000 x = 0: y = 430(0) + 7000 = 7000 x = 5: y = 430(5) + 7000 = 9150 x = 10: y = 430(10) + 7000 = 11,300 The ordered pairs are (0, 7000), (5, 9150), and (10, 11,300).
5.
y
3
x
10
32 5
10
–10
10
–10
6. y
2
x
9
22 3 10
–10
10
–10
b. 2028 2019 = 9 The graph shows that we predict approximately 10,900 wind turbine service technicians in 2028.
Vocabulary, Readiness & Video Check 13.2 1. In the definition, x and y both have an understood power of 1. Video Notebook Number 3 shows an equation where y has a power of 2, so it is not a linear equation in two variables.
Calculator Explorations 2. It is always good practice to use a third point as a check to see that our points lie along a straight line.
1. y = 3x + 7 10
–10
10
–10
3. An infinite number of points make up the line, and every point corresponds to an ordered pair that is a solution of the linear equation in two variables. Exercise Set 13.2
2. y = x + 5 10
–10
2. Yes; it can be written in the form Ax + By = C. 10
4. No; y is cubed. –10
6. Yes; it can be written in the form Ax + By = C.
3. y = 2.5x 7.9
8. Yes; it can be written in the form Ax + By = C.
10
–10
10
–10
4. y = 1.3x + 5.2 10
–10
10
–10
516
10. x y = 4 x = 0: 0 y 4 y 4 y = 2: x 2 4 x6 x = 1: 1 y 4 y 5 y 5 The ordered pairs are (0, 4), (6, 2), and (1, 5).
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ISM: Prealgebra and Introductory Algebra
16. y = 5x + 2 x = 0: y = 5(0) + 2 = 0 + 2 = 2 x = 1: y = 5(1) + 2 = 5 + 2 = 3 x = 2: y = 5(2) + 2 = 10 + 2 = 8 The ordered pairs are (0, 2), (1, 3), and (2, 8).
y 10 8 6 4 2
(6, 2)
–10 –8 –6 –4 –2–2
2 4 6 8 10
Chapter 13: Graphing Equations and Inequalities
y
x
10 8 6
(–1, –5)–4 (0, –4) –6 –8 –10
`
4
(0, 2)
2
12. y = 5x x = 1: y = 5(1) = 5 x = 0: y = 5(0) = 0 x = 1: y = 5(1) = 5 The ordered pairs are (1, 5), (0, 0), and (1, 5). y
–10 –8 –6 –4 –2–2 –4 –6 –8 –10
2 4 6 8 10
(1, –3)
x
(2, –8)
18. x + y = 7
(–1, 5) 5
4 3 2
1
–5 –4 –3 –2 –1–1 –2 –3 –4 –5
14. y
1
(0, 0) 1 2 3 4 5
x
(1, –5)
x
20. x + y = 6
2
x = 0: y
1
(0) 0 2 1 x = 4: y (4) 2 2 1 x = 2: y (2) 1 2 The ordered pairs are (0, 0), (4, 2), and (2, 1). y 5 4 3 2
(0, 0)1 –5 –4 –3 –2 –1–1
(–4, –2)
22. x + 5y = 5 (2, 1) 1 2 3 4 5
x
–2 –3 –4 –5
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517
Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
24. y = 2x + 7
32. x = 4y
26. y = 5
34. 2x + y = 2
28. x = 1
36. y
1
x3
4
30. y = x 38. 2x 7y = 14
518
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
40. y = 1.5x 3
42. Comparing y = 5x 4 to y = mx + b, we see that b = 4. We see that graph B crosses the y-axis at (0, 4). 44. Comparing y = 5x + 2 to y = mx + b, we see that b = 2. We see that graph A crosses the y-axis at (0, 2). 46. a.
b. x = 4: y = 4.1(4) + 10 = 16.4 + 10 = 26.4 The ordered pair is (4, 26.4). c. 48. a.
2016 + 4 = 2020 In 2020, U.S. food delivery app revenue was $26.4 billion. y = 1.1x + 86 x = 6: y = 1.1(6) + 86 = 92.6 The ordered pair in (6, 92.6).
b. 2015 + 6 = 2021 Six years after 2015, in 2021, 92.6% of U.S. adults say they use the Internet. c.
2026 is 11 years after 2015, so let x = 11. y = 1.1x + 86 x = 11: y = 1.1(11) + 86 = 98.1 If the trend continues, 98.1% of U.S. adults will use the Internet in 2026.
d.
answer may vary
50. The length of the side of the square is 2 (3) = 2 + 3 = 5. Therefore the remaining vertices could be 5 units up from the given two or 5 units down from the given two. Five units up from y = 1 is y = 1 + 5 = 4. Five units down from y = 1 is y = 1 5 = 6. The remaining vertices could be (3, 4), (2, 4) or (3, 6), (2, 6).
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519
Chapter 13: Graphing Equations and Inequalities 52. y x = 5 x = 0: y 0 5 y 5 y = 0: 0 x 5 x 5 x
y
0
5
5
0
ISM: Prealgebra and Introductory Algebra
60. y = |x| x = 0: y = |0| = 0 x = 1: y = |1| = 1 x = 1: y = |1| = 1 x = 2: y = |2| = 2 x = 2: y = |2| = 2
54. x = 3y x = 0: 0 3y 0 y y = 0: x 3(0) x0
x
y
0
0
1
1
1
1
2
2
2
2 y
x
y
0
0
5 4 3 2 1
0
0
–5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
56. y = 2x y = 2x + 5 y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
62. The perimeter is the distance around. x y x y 50 2x 2 y 50 x y 25 y = 20: x 20 25 x5 If y is 20 miles, then x is 5 miles. 64. no; answers may vary
58. y = x y=x7
Section 13.3 Practice Exercises y
1. x-intercept: (2, 0) y-intercept: (0, 4)
10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10
2. x-intercepts: (4, 0), (2, 0) y-intercept: (0, 2) 2 4 6 8 10
x
3. x-intercept and y-intercept: (0, 0) 4. no x-intercept y-intercept: (0, 3) 5. x-intercepts: (1, 0), (5, 0) y-intercepts: (0, 2), (0, 2)
520
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ISM: Prealgebra and Introductory Algebra 6. 2x y = 4 y = 0: 2x 0 4 2x 4 x2 x-intercept: (2, 0) x = 0: 2(0) y 4 0 y 4 y 4 y 4 y-intercept: (0, 4) x = 3: 2(3) y 4 6 y 4 y 2 y2 Third point: (3, 2)
Chapter 13: Graphing Equations and Inequalities
8. 3x = 2y + 4 y = 0: 3x 2(0) 4 3x 0 4 3x 4 4 1 x 1 3 3 1 x-intercept: 1 , 0 3 x = 0: 3(0) 2 y 4 0 2y 4 4 2 y 2 y y-intercept: (0, 2) y = 1: 3x 2(1) 4 3x 2 4 3x 6 x2 Third point: (2, 1)
7. y = 3x y = 0: 0 3x 0x x-intercept: (0, 0) x = 0: y 3(0) y0 y-intercept: (0, 0) Let x = 1 to find a second point. x = 1: y = 3 (1, 3) is another point on the line. Let x = 1 to find a third point. x = 1: y = 3 (1, 3) is a third point on the line.
9. x = 3 The graph of x = 3 is a vertical line with x-intercept (3, 0). y
(–3, 0)
6 5 4 3 2 1
–6 –5 –4 –3 –2 –1–1
1 2 3 4 5 6
x
–2 –3 –4 –5 –6
Copyright © 2025 Pearson Education, Inc.
521
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
10. y = 4 The graph of y = 4 is a horizontal line with y-intercept (0, 4). y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1–1
6. 5.9x 0.8y = 10.4 10
–10
(0, 4)
10
–10
Vocabulary, Readiness & Video Check 13.3 1 2 3 4 5 6
–2 –3 –4 –5 –6
3. The graph of the equation y = 1 is a horizontal line.
1. x = 3.78y
4. The graph of the equation x = 5 is a vertical line.
10
10
5. A point where a graph crosses the y-axis is called a y-intercept. 6. A point where a graph crosses the x-axis is called an x-intercept.
–10
2. 2.61y = x
7. Given an equation of a line, to find the x-intercept (if there is one), let y = 0 and solve for x.
10
–10
1. An equation that can be written in the form Ax + By = C is called a linear equation in two variables. 2. The form Ax + By = C is called standard form.
Calculator Explorations
–10
x
10
8. Given an equation of a line, to find the y-intercept (if there is one), let x = 0 and solve for y.
–10
3. 3x + 7y = 21 10
–10
10
9. False; a horizontal line (other than y = 0) has no x-intercept and a vertical line (other than x = 0) has no y-intercept. 10. True
–10
11. True
4. 4x + 6y = 12
12. False; x = 5: y = 5(5) = 25, so the point (5, 1) is not on the graph of y = 5x.
10
–10
10
–10
14. It is always good practice to use a third point as a check to see that your points lie along a straight line.
5. 2.2x + 6.8y = 15.5 10
–10
13. because x-intercepts lie on the x-axis; because y-intercepts lie on the y-axis
10
15. For a horizontal line, the coefficient of x will be 0; for a vertical line, the coefficient of y will be 0.
–10
522
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
Exercise Set 13.3
14. x 2y = 8 y = 0: x 2(0) 8 x 0 8 x 8 x-intercept: (8, 0) x = 0: 0 2 y 8 2 y 8 y4 y-intercept: (0, 4)
2. x-intercept: (4, 0) y-intercept: (0, 3) 4. x-intercepts: (3, 0), (3, 0) y-intercept: (0, 5) 6. x-intercepts: (4, 0), (1, 0), (1, 0) y-intercept: (0, 4)
y
8. x-intercepts: (3, 0), (3, 0) y-intercepts: (0, 2), (0, 2)
10 8 6
10. x y = 4 y = 0: x 0 4 x 4 x-intercept: (4, 0) x = 0: 0 y 4 y 4 y4 y-intercept: (0, 4)
4
(–8, 0)
–10 –8 –6 –4 –2–2
(–4, 0)
–5 –4 –3 –2 –1–1
(0, 4)
1 2 3 4 5
2 4 6 8 10
x
16. 2x + 3y = 6 y = 0: 2x 3(0) 6 2x 0 6 2x 6 x3 x-intercept: (3, 0) x = 0: 2(0) 3y 6 0 3y 6 y2 y-intercept: (0, 2)
5 3 2 1
(0, 4)
–4 –6 –8 –10
y 4
2
x
–2 –3 –4 –5
y
12. x = 2y y = 0: x = 2(0) = 0 x-intercept: (0, 0) x = 0: 0 2 y 0 y y-intercept: (0, 0) Let y = 4 to find a second point. y = 4: x = 2(4) = 8 (8, 4) is another point on the line.
5 4 3
2 1 –5 –4 –3 –2 –1–1
(0, 2) (3, 0) 1 2 3 4 5
x
–2 –3 –4 –5
18. y = 2x y = 0: 0 2x 0x x-intercept: (0, 0) (0, 0) is also the y-intercept. Let x = 4 to find a second point. x = 4: y 2(4) 8 (4, 8) is another point on the line.
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523
Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
24. The graph of x = 0 is a vertical line with x-intercept (0, 0). y 5 4 3 2 1 –5 –4 –3 –2 –1–1
20. y = 2x + 10 y = 0: 0 2x 10 2x 10 x 5 x-intercept: (5, 0) x = 0: y 2(0) 10 y 0 10 y 10 y-intercept: (0, 10)
(–5, 0)
8 6 4 2
–10 –8 –6 –4 –2–2
y 5 4 3 2 1
(0, 10)
2 4 6 8 10
–5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
x
–4 –6 –8 –10
22. The graph of y = 5 is a horizontal line with y-intercept (0, 5).
28. y 6 0 y6 The graph of y = 6 is a horizontal line with y- intercept (0, 6). y
y
10 8 6 4 2
5 4 3 2 1 –5 –4 –3 –2 –1–1
x
26. x 2 0 x2 The graph of x = 2 is a vertical line with x-intercept (0, 2).
y 10
1 2 3 4 5
–2 –3 –4 –5
–10 –8 –6 –4 –2–2 1 2 3 4 5
2 4 6 8 10
x
–4 –6 –8 –10
x
–2 –3 –4 –5
30. x = y y = 0: x = 0 = 0 x-intercept: (0, 0) (0, 0) is also the y-intercept. Let x = 3 to find a second point. x = 3: 3 y 3 y (3, 3) is another point on the line.
524
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities 36. x + 9y = 10 y = 0: x 9(0) 10 x 0 10 x 10 x 10 x-intercept: (10, 0) x = 0: 0 9 y 10 9 y 10 10 y 9 10 y-intercept: 0, 9
32. x + 3y = 9 y = 0: x 3(0) 9 x0 9 x9 x-intercept: (9, 0) x = 0: 0 3y 9 3y 9 y3 y-intercept: (0, 3) y 10 8 6 4 2 –10 –8 –6 –4 –2–2
(0, 3)
38. x 1
(9, 0) 2 4 6 8 10
x
3 4
3 This is a vertical line with x-intercept 1 , 0 . 4
–4 –6 –8 –10
y
34. 4 = x 3y y = 0: 4 x 3(0) 4 x0 4x x-intercept: (4, 0) x = 0: 4 0 3y 4 3y 4 y 3 4 y-intercept: 0, 3
5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
1 40. y 2 2 This is a horizontal line with y-intercept 1 0, 2 2 .
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525
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
3 42. y x 3 5
3
3
x3
5
48.
x3
5
–5 –4 –3 –2 –1–1
(0, 3) (5, 0) 1 2 3 4 5
–2 –3 –4 –5
44. 9x 6y + 3 = 0 y = 0: 9x 6(0) 3 0 9x 0 3 0 9x 3 1 x 3 1 x-intercept: , 0 3 x = 0: 9(0) 6 y 3 0 0 6y 3 0 6 y 3 1 y 2 1 y-intercept: 0, 2 526
1 1
1
12 3 9 9 10 9 1
3 54. The graph of y = 2x + 3 has x-intercept , 0 2 and y-intercept (0, 3). This is graph b.
5 3 2 1
1 0
52. The graph of y = 2x + 2 has x-intercept (1, 0) and y-intercept (0, 2). This is graph d.
y 4
45
50. 2 2 0 0 3 5 2
x5 x-intercept: (5, 0) 3 x = 0: y (0) 3 5 y 03 y3 y-intercept: (0, 3)
0
y = 0:
46.
56. The smallest number of x- and y-intercepts occurs when the line is vertical or horizontal. Then it has 1 x- or y-intercept.
x
58. The circle can have a maximum of 4 x- and y-intercepts because it could cross each axis twice; for example a circle centered at the origin. 60. answers may vary 62. 3x + 6y = 1200 y = 0: 3x 6(0) 1200 3x 1200 x 400 The ordered pair (400, 0) corresponds to manufacturing 400 chairs and 0 desks. 64. Manufacturing 50 chairs corresponds to x = 50. 3x + 6y = 1200 x = 50: 3(50) 6 y 1200 150 6 y 1200 6 y 1050 y 175 When 50 chairs are manufactured, 175 desks can be manufactured.
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c.
8.
ISM: Prealgebra and Introductory Algebra 66.
Chapter 13: Graphing Equations and Inequalities y y 5 1 4 2. m 2 1 x2 x1 3 (2) 5
y
5 4 3 2 1 –1–1 –2 –3 –4 –5
y (1, 0) 1 2 3 4 5 6 7 8 9
x (–2, 1)
x=5
b. The y-intercept of (0, 650) means that there were 650 stores in 2015 (0 years after 2015).
0 1.6x 47.4 1.6x 47.4 x 29.6 The x-intercept is (29.6, 0).
a.
y = 0:
b.
About 29.6 years after 2013, there should be no more desktop Internet use.
answers may vary Section 13.4 Practice Exercises y y1 1 3 4 2 1. m 2 x2 x1 4 (2) 6 3
3. y
2
x 2 is in slope-intercept form,
3 y = mx + b. The slope is m
2
4. y 2x 7 y 2x 7 1 1 1 y 2x 7 The slope is m = 2 and the y-intercept is (0, 7). 5. 5x 2 y 8 2 y 5x 8 2 y 5x 8 2 2 2 5 y x4 2 The slope is m
y
5 2
6 5 4 3 2 1
–2 –3 –4 –5 –6
x
1 2 3 4 5 6
, and the 3 y-intercept is (0, b), which is (0, 2).
70. y = 1.6x + 47.4
–6 –5 –4 –3 –2 –1–1
(3, 5)
–2 –3 –4 –5 –6
y = 7.6x + 650 x = 0: y 7.6(0) 650 650 The y-intercept is (0, 650).
(–2, 3)
6 5 4 3 2 1
–6 –5 –4 –3 –2 –1–1
The equation of the line is x = 1. 68. a.
and the y-intercept is
(0, 4). 6. y = 3 is a horizontal line. Horizontal lines have a slope of m = 0. 1 2 3 4 5 6
(4, –1)
x
7. x = 2 is a vertical line. Vertical lines have undefined slopes. a.
x y 5 2x y 5 y x 5 y 2x 5 slope = 1 slope = 2 The slopes are not the same, so the lines are not parallel. The product, (1)(2) = 2, is not 1, so the lines are not perpendicular.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
b.
5 y 2x 3 2 3 y x 5 5 slope
5x 2 y 1 2 y 5x 1 5 1 y x 2 2 5 slope
2
–10
y = 2x + 1
–10
4x 2 y 8 2 y 4x 8
4x 8 2 2 y 2x 4 slope = 2 slope = 2 The slopes are the same, so the lines are parallel. rise 3 9. grade 0.15 15% run 20 The grade is 15%. y
12.15 11.61 0.54 0.09 2023 2017 6 Each year, the number of workers employed in the U.S. restaurant industry increases by 0.09 million, or 90,000, workers per year. Calculator Explorations 10. m
10
The lines are parallel since they all have a slope 1 1 of 4 . The graph of y 4 x 5 is the graph of 1 y x moved 5 units up with a y-intercept of 5. 4 1 1 The graph of y x 8 is the graph of y x 4 4 moved 8 units down with a y-intercept of 8.
5 2 The slopes are not the same, so the 2 5 lines are not parallel. The product, 1, is 5 2 1, so the lines are perpendicular. c.
10
3.
10
4. –10
10
–10
The lines are parallel since they all have a slope 3 3 of . The graph of y x 5 is the graph 4 4 3 of y x moved 5 units down with a 4 3 y- intercept of 5. The graph of y x 6 is 4 3 the graph of y x moved 6 units up with a 4 y-intercept of 6.
10
1.
Vocabulary, Readiness & Video Check 13.4 –10
10
1. The measure of the steepness or tilt of a line is called slope. –10
The lines are parallel since they all have a slope of 3.8. The graph of y = 3.8x 3 is the graph of y = 3.8x moved 3 units down with a y-intercept of 3. The graph of y = 3.8x + 9 is the graph of y = 3.8x moved 9 units up with a y-intercept of 9. 10
5. If the graph of a line moves upward from left to right, the line has positive slope.
10
–10
The lines are parallel since they all have a slope of 4.9. The graph of y = 4.9x + 1 is the graph of y = 4.9x moved 1 unit up with a y-intercept of 1. The graph of y = 4.9x + 8 is the graph of y = 4.9x moved 8 units up with a y-intercept of 8. 528
3. The slope of a horizontal line is 0. 4. The slope of a vertical line is undefined.
2. –10
2. If an equation is written in the form y = mx + b, the value of the letter m is the value of the slope of the graph.
6. If the graph of a line moves downward from left to right, the line has negative slope. 7. Given two points of a line, slope
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change in y change in x
.
ISM: Prealgebra and Introductory Algebra
8. The line slants downward, so its slope is negative. 9. The line slants upward, so its slope is positive. 10. The line is vertical, so its slope is undefined. 11. The line is horizontal, so its slope is 0. 12. Since m
7
Chapter 13: Graphing Equations and Inequalities 10. (x1, y1) (3, 2), (x2 , y2 ) (1, 3) y2 y1 3 (2) 3 2 5 m x2 x1 1 (3) 1 3 2 12. (x1, y1) (3, 3), (x2 , y2 ) (1, 4) y y1 4 3 4 3 7 7 m 2 x2 x1 1 (3) 1 3 4 4
is positive, the line slants upward.
6
14. The line goes up. The slope is positive.
13. Since m = 3 is negative, the line slants downward. 14. Since m = 0, the line is horizontal. 15. Since m is undefined, the line is vertical. 16. Whatever y-value we decide to start with in the numerator, we must start with the corresponding x-value in the denominator. 17. Solve the equation for y; the slope is the coefficient of x. 18. Zero slope indicates m = 0 and a horizontal line; undefined slope indicates m is undefined and a vertical line; “no slope” refers to an undefined slope.
16. The line is horizontal. The slope is 0. 18. The slope is negative. The line is “downward.” 20. The slope is undefined. The line is vertical. 22. Line 1 has a positive slope and line 2 has a negative slope, so line 1 has the greater slope. 24. Line 1 has a positive slope and line 2 has a slope of 0, so line 1 has the greater slope. 26. y = 2x + 6 The slope is m = 2 and the y-intercept is (0, 6). 28. y = 7.6x 0.1 The slope is m = 7.6 and the y-intercept is (0, 0.1).
19. Slope-intercept form; this form makes the slope easy to see, and we need to compare slopes to determine if two lines are parallel or perpendicular.
30. 5x y 10 y 5x 10 The slope is m = 5 and the y-intercept is (0, 10).
20. Step 4: INTERPRET the results.
32. The line is horizontal, so it has a slope of 0.
Exercise Set 13.4 y y1 4 16 12 2. m 2 3 x2 x1 3 (1) 4 y y1 6 1 5 5 4. m 2 x2 x1 2 3 1
y y1 33 0 6. m 2 0 x2 x1 2 (8) 6
34. 3x 5 y 1 5 y 3x 1 3 1 y x 35 1 5 y x 5 5 3 The slope is m . 5 36. y = 2 is a horizontal line, so its slope is 0.
y y1 5 (3) 53 8 8. m 2 x2 x1 2 (2) 2 2 0 The slope is undefined.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
38.
x 4 y x y 4 1 y x 4
50. y
40. x = 5 is a vertical line, so its slope is undefined. 42. 4x 7 y 9 7 y 4x 9 4x 9 y 7 7 4 9 y x 7 7 4 The slope is m . 7
x2 x1
4 (2) 1 6
y
1
1
5 1
x
, so the lines are not parallel. 5 5 1 1, so the lines are not 1 1 5 5 25 perpendicular. The lines are neither parallel nor perpendicular. 52. y = 4x 2 4x y 5 y 4x 5 4 4, so the lines are not parallel. (4)(4) = 16 1, so the lines are not perpendicular. The lines are neither parallel nor perpendicular.
44. 24x 3y 5.7 3y 24x 5.7 24x 5.7 y 3 3 y 8x 1.9 The slope is m = 8. y2 y1
x 20
5
1 The slope is m . 4
46. m
1
54. x 2 y 2 2 y x1 2 y x 1
42 5
2 2x 4 y 3 4 y 2x 3 1 3 y x 2 4
6 5
1
and the y-intercepts are 2 different, so they are parallel.
Both lines have slope 6
a.
The slope of a parallel line is . 5
b.
The slope of a perpendicular line is 1 5 6 . 6 5
48. m y2 y1 10 (1) 10 1 9 9 4 6 4 6 10 10 x2 x1 9 a.
The slope of a parallel line is
.
56. 10 + 3x = 5y 3 3 2 x y or y x 2 5 5 5x 3y 1 3y 5x 1 5 1 y x 3 3 3 5 so the lines are perpendicular. 1, 5 3
10
b.
The slope of a perpendicular line is
58. slope
rise
run 1
10 9 . 9
5 10
1 2
The pitch of the roof is
1
.
2
10
60. slope
rise
16
0.16 16% run 100 The grade of the road is 16%.
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ISM: Prealgebra and Introductory Algebra
62. slope
rise
0.25
0.02 2%
run 12 The grade is 2%. rise 1 foot 0.083 8.3% run 12 feet The grade is 8.3%.
64. grade
y y1 66. m 2 x2 x1 37 49 2021 2013 12 8 1.5 or 1.5 1 Every 1 year the average daily desktop Internet use per person worldwide decreases by 1.5 minutes.
Chapter 13: Graphing Equations and Inequalities
82. answers may vary 84. For 2018, the average fuel economy was 24.4 miles per gallon and for 2019 it was 24.1 miles per gallon. This is a decrease of 24.4 24.1 = 0.3 mile per gallon. 86. The average fuel economy was the highest for 2020, the high point on the graph. The average fuel economy for automobiles produced during 2020 was 25.3 miles per gallon. 88. The points are (2020, 25.3) and0.3 (2021, 25.0). y2 y1 25.0 25.3 0.3 m x2 x1 2021 2020 1 0.3 The slope is or 0.3. 1 90. pitch
run 1 x 3 18 3x 18 x6
68. m y2 y1 121.2 115.6 5.6 0.8 x2 x1 2021 2014 7 Every year, the number of U.S. households with televisions increases by 0.8 million. 70. y 7 9(x 6) y 7 9x 54 y 9x 61
92. a.
b.
72. y (3) 4(x (5)) y 3 4(x 5) y 3 4x 20 y 4x 17
c.
74. The line is horizontal, so its slope is 0; A. 76. (x1, y1) (1, 3),(x2 , y2 ) (0, 0) y y1 0 3 3 m 2 3 x2 x1 0 1 1 The slope is m = 3; C. 78. (x1, y1) (2, 1), (x2 , y2 ) (2, 2) y y1 2 1 3 m 2 x2 x1 2 (2) 4 3 The slope is m ; F. 4 80. no; answers may vary
rise
The ordered pairs are (2020, 270) and (2022, 419). y y1 419 270 149 m 2 74.5 x2 x1 2022 2020 2 The slope is 74.5 For the years 2020 through 2022, attendance at the National Air and Space Museum increased at a rate of 74.5 thousand per year.
94. Slope through (3, 0) and (1, 1): 1 0 1 m 1 (3) 4 Slope through (3,4 0) and (4, 4): 40 4 m 4 (3) 1 1 Their product is (4) 1, so the sides are 4 perpendicular. y2 y1 9.3 6.7 2.6 96. m x x 8.3 2.1 10.4 0.25 2
1
98. m y2 y1 5.1 0.2 4.9 0.875 x2 x1 7.9 2.3 5.6
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Chapter 13: Graphing Equations and Inequalities
10
100.
4.
–10
10 –10
As the absolute value of the slope becomes larger, the line becomes steeper. Section 13.5 Practice Exercises 1. From y
2
x 4, the y-intercept is (0, 4). The
2 3 5 5 1 4 y y1 m(x x1) 5 y 3 (x 1) 4 5 4( y 3) 4 (x 1) 4
5. m
3
2 slope is , so another point on the graph is 3 (0 + 3, 4 + 2) or (3, 2). y
4( y 3) 5(x 1) 4 y 12 5x 5 5x 4 y 17
12 10 8 6 4 2 –12 –10 –8 –6 –4 –2–2 –4
–6 –8 –10 –12
y y1 m(x x1) y (4) 3(x 2) y 4 3(x 2) y 4 3x 6 y 3x 2 3x y 2 The slope-intercept form is y = 3x + 2 and the standard form is 3x + y = 2.
2 4 6 8 10 12
x
(0, –4)
6. The equation of a vertical line can be written in the form x = c, so an equation of the vertical line through (3, 2) is x = 3. 7. Since the graph of y = 2 is a horizontal line, any line parallel to it is also horizontal. The equation of a horizontal line can be written in the form y = c, so an equation of the horizontal line through (4, 3) is y = 3.
2. 3x y 2 y 3x 2 3 and the y-intercept is 1 (0, 2). Another point on the graph is (0 + 1, 2 3) or (1, 1). The slope is 3
8. a.
y y1 m(x x1) y 200 50(x 10) y 200 50x 500 y 50x 700
y 6 5 4 3
2 1 –6 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 –6
3. y mx b 3 y x (2) 5 3 y x2 5 532
b. Let x = 7.50. y = 50x + 700 y = 50(7.50) + 700 y = 375 + 700 y = 325 The predicted weekly sales is 325.
(0, 2)
1 2 3 4 5 6
(10, 200), (9, 250) y y1 250 200 50 m 2 50 x2 x1 9 10 1
x
Calculator Explorations 10
1. –10
10
–10
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ISM: Prealgebra and Introductory Algebra
10
2. –10
11. Write the equation with x- and y-terms on one side of the equal sign and a constant on the other side.
10
12. Yes, if one of the points given is the y-intercept. We will still need to use the slope formula to find the slope, but then we’ll have the slope and y-intercept for the slope-intercept form.
–10 10
3. –10
13. Video Notebook Number 6: y = 3; Video Notebook Number 7: x = 2
10
–10
14. We need to know what our variables stand for in order to solve part b., and that depends on how we set up our ordered pairs in part a.
10
4. –10
Chapter 13: Graphing Equations and Inequalities
10
Exercise Set 13.5
–10
Vocabulary, Readiness & Video Check 13.5 1. The form y = mx + b is called slope-intercept form. When a linear equation in two variables is written in this form, m is the slope of its graph and (0, b) is its y-intercept. 2. The form y y1 m(x x1) is called pointslope form. When a linear equation in two variables is written in this form, m is the slope of its graph and (x1, y1) is a point on the graph.
2. From y = 4x 1, the y-intercept is (0, 1). The 4 , so another point on the graph slope is 4 or 1 is (0 + 1, 1 4) or (1, 5). y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
3. y 7 = 4(x + 3); point-slope form 4. 5x 9y = 11; standard form 5. y
4. From y
3
1 x ; slope-intercept form 4 3
6. y 2
1
slope is
1 4
1 2 3 4 5
(0, –1)
x
x 3, the y-intercept is (0, 3). The
1
, so another point on the graph is 4 (0 + 4, 3 + 1) or (4, 2).
(x 2); point-slope form
y
3
10 8 6 4 2
1 7. y ; horizontal line 2
–10 –8 –6 –4 –2–2
8. x = 17; vertical line 9. Start by graphing the y-intercept. From this point, find another point by applying the slopeif necessary, rewrite the slope as a fraction. 1 10. 0, 6
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2 4 6 8 10
x
–4 (0, –3) –6 –8 –10
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Chapter 13: Graphing Equations and Inequalities 6. From y = 6x, the y-intercept is (0, 0). The slope 6 , so another point on the graph is is 6 or 1 (0 + 1, 0 + (6)) or (1, 6).
x
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
3
y 2 2 x y 3 2 y x0 3
12.
and the y-intercept is (0, 0). 3 Another point on the graph is (0 + 3, 0 + 2) or (3, 2).
(0, 0) 1 2 3 4 5
x
y
–2 –3 –4 –5
5 4 3 2 (0, 0)1
8. 3x y 2 y 3x 2 The slope is 3
–5 –4 –3 –2 –1–1
3
14. y = mx + b y = 3x + (3) y = 3x 3
y 5 4 3 1 –5 –4 –3 –2 –1–1
16. y mx b 3 y 2x 4
(0, 2) 1 2 3 4 5
x
–2 –3 –4 –5
10. 3x 4 y 4 4 y 3x 4 3 y x 1 4 3 The slope is and the y-intercept is (0, 1). 4 Another point on the graph is (0 + 4, 1 + 3) or (4, 2). y
–2 –3 –4 –5
534
18. y mx b 4 y x0 5 4 y x 5 20. y = mx + b y = 0x + (2) y = 2 22. y mx 1 b 1 y x 3 2 1 1 y x 2 3
5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
and the y-intercept is (0, 2). 1 Another point on the graph is (0 + 1, 2 + 3) or (1, 5).
2
2
The slope is
1 2 3 4 5
(0, –1)
x
24.
y y1 m(x x1) y 3 4(x 1) y 3 4x 4 y 4x 1 4x y 1
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ISM: Prealgebra and Introductory Algebra
26.
28.
y y1 m(x x1) y (12) 2[x (11)] y 12 2(x 11) y 12 2x 22 y 2x 34 2x y 34 y y m(x x ) 1
1
2
y 9 [x (8)] 3 2 y 9 (x 8) 3 2 3( y 9) 3 (x 8) 3 3y 27 2x 16 3y 2x 43 2x 3y 43 or 2x 3y 43 30.
y y m(x x ) 1
1
1 y 0 (x 4) 5 1 y (x 4) 5 5 y 1(x 4) 5 y x 4 x 5y 4 32. m y2 y1 8 2 6 3 x2 x1 8 6 2 y y1 m(x x1) y 2 3(x 6) y 2 3x 18 y 16 3x 16 3x y 3x y 16 y y1 1 0 1 34. m 2 x2 x1 6 (4) 10 y y1 m(x x1) 1 y 0 [x (4)] 10 1 y (x 4) 10 10 y x 4 4 x 10 y x 10 y 4
Chapter 13: Graphing Equations and Inequalities 36. m y2 y1 110 11 11 x2 x1 1 7 8 8 y y1 m(x x1) 11 y 10 (x 7) 8 8( y 10) 11(x 7) 8 y 80 11x 77 8 y 3 11x 3 11x 8 y 11x 8 y 3 y y
1 0
1
2
2
1 2 2 3 1 38. m 2 1 31 3 1 3 x2 x1 0
y y1 m(x x1) 2 y 0 (x 0) 23 y x 3 3y 2x 2x 3y 0 40. The horizontal line through (1, 4) has equation y = 4. 42. The vertical line through (1, 3) has equation x = 1. 2
44. The horizontal line through , 0 has equation 7 y = 0. 46. The line y = 5 is horizontal, so a line perpendicular to y = 5 is a vertical line. The vertical line through (1, 2) has equation x = 1. 48. The line y = 4 is horizontal, so a line parallel to y = 4 is also a horizontal line. The horizontal line through (0, 3) has equation y = 3. 50. The line x = 7 is vertical, so a line perpendicular to x = 7 is a horizontal line. The horizontal line through (5, 0) has equation y = 0. 52. y mx b 5 y x 3 7
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities 54. m y2 y1 5 (6) 11 1 x2 x1 6 5 11 y y1 m(x x1) y (6) 1(x 5) y 6 x 5 y x 1 56. A line with a slope of 0 is a horizontal line. this one has a y-intercept of (0, 12.1). The equation is y = 12.1. 58.
70. a. b.
66. A line perpendicular to the x-axis is a vertical line. x=0 68. y y1 m(x x1) 3 y 4 (x 4) 5 3 12 y4 x 5 5 3 12 20 y x 5 5 5 3 32 y x 5 5
1
y y1 m(x x1) y 4.3 1.5(x 0) y 4.3 1.5x y 1.5x 4.3 c.
72. a.
62. A line parallel to the y-axis is a vertical line. x=1 y2 y1 7 0 7 64. m x2 x1 4 0 4 y y1 m(x x1) 7 y 0 (x 0) 4 7 y x 4
y2 y1 7.3 4.3 3 m x x 2 0 2 1.5 2
y y1 m(x x1) y (8) 5(x 6) y 8 5x 30 y 5x 38
60. y = mx + b y = 2x + (4) y = 2x 4
The ordered pairs are (0, 4.3) and (2, 7.3).
2028 corresponds to x = 9. x = 9: y 1.5(9) 4.3 13.5 4.3 17.8 If the trend were to continue, there would be 17.8 million digital magazine subscriptions in 2028. The ordered pairs are (1, 30,000) and (4, 66,000). p p1 66, 000 30, 000 m 2 t2 t1 4 1 36, 000 3 12, 000 p p1 m(t t1) p 30, 000 12, 000(t 1) p 30, 000 12, 000t 12, 000 p 12, 000t 18, 000
b. t = 7 p 12, 000(7) 18, 000 84, 000 18, 000 102, 000 The profit is predicted to be $102,000 after 7 years. 74. a.
The ordered pairs are (0, 2.29) and (8, 3.77). y y1 3.77 2.29 1.48 m 2 0.185 x2 x1 80 8 y y1 m(x x1) y 2.29 0.185(x 0) y 2.29 0.185x y 0.185x 2.29
b. The year 2030 is 12 years after 2018. x = 12: y 0.185(12) 2.29 2.22 2.29 4.51 The global value of the hot sauce market is predicted to be $4.51 billion in 2030.
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ISM: Prealgebra and Introductory Algebra
76. a.
The ordered pairs are (0, 10,884) and (7, 12,697). y y1 12, 697 10,884 m 2 70 x2 x1 1813 7 259 y y1 m(x x1) y 10,884 259(x 0) y 10,884 259x y 259x 10,884
86. The graph of y
The ordered pairs are (7, 225,000) and (12, 195,000). y y1 195, 000 225, 000 m 2 x2 x1 12 7 30, 000 5 6000 y y1 m(x x1) y 225, 000 6000(x 7) y 225, 000 6000x 42, 000 y 6000x 267, 000 or V = 6000x + 267,000
b. x = 2018 2000 = 18 V = 6000(18) + 267,000 V = 108,000 + 267,000 V = 159,000 The depreciated value in 2018 was $159,000.
5
x 2 has slope m
3 y-intercept (0, 2). This is graph A.
5 3
and
88. The slope of the line y = 3x 1 is m = 3. y y1 m(x x1) y 0 3(x 3) y 3x 9 3x y 9 90. a.
b. The year 2026 is 12 years after 2014. x = 12: y 259(12) 10,884 3108 10,884 13, 992 The number of Dunkin’ stores worldwide is predicted to be 13,992 in 2026. 78. a.
Chapter 13: Graphing Equations and Inequalities
b.
A line parallel to the line y = 2x + 3 will have slope 2. y y1 m(x x1) y 0 2(x 4) y 2x 8 2x y 8 A line perpendicular to the line y = 2x + 3 1 will have slope . 2 y y1 m(x x1) 1 y 0 (x 4) 2 1 y x2 2 2y x 4 4 x 2y x 2y 4
Mid-Chapter Review 1. Select two points on the line, such as (0, 0) and (1, 2). y y1 2 0 2 m 2 2 x2 x1 1 0 1 2. Horizontal lines have slopes of m = 0. 3. Select two points on the line, such as (0, 1) and (3, 3). y y1 3 1 2 2 m 2 x2 x1 3 0 3 3
80. x = 5: x2 3x 1 52 3(5) 1 25 15 1 10 1 11
4. Vertical lines have undefined slopes. 82. x = 3: x2 3x 1 (3)2 3(3) 1 9 9 1 19 84. The graph of y = x + 1 has slope m = 1 and y-intercept (0, 1). This is graph C. Copyright © 2025 Pearson Education, Inc.
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Chapter 13: Graphing Equations and Inequalities 5. y = 2x y = 0: 0 2x 0x The x-intercept is (0, 0). x = 0: y = 2(0) = 0 The y-intercept is (0, 0). Find another point, for example let x = 1. y = 2(1) = 2 Another point on the line is (1, 2).
ISM: Prealgebra and Introductory Algebra
8. The graph of y = 4 is a horizontal line with a y-intercept of (0, 4). y 5 4 3 2 1 –5 –4 –3 –2 –1–1
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
6. x + y = 3 y = 0: x 0 3 x3 The x-intercept is (3, 0). x = 0: 0 y 3 y 3 The y-intercept is (0, 3).
The y-intercept is (0, 3). y 10 8 6 4 2
y 5 4 3 2 1
–10 –8 –6 –4 –2–2
1 2 3 4 5
x
x
7. The graph of x = 1 is a vertical line with an x-intercept of (1, 0). y 5 4 3 2 1 1 2 3 4 5
x
10. y = 3x + 2 y = 0: 0 3x 2 2 3x 2 x 3 2 The x-intercept is , 0 . 3 x = 0: y 3(0) 2 y 02 y2 The y-intercept is (0, 2).
–2 –3 –4 –5
538
2 4 6 8 10
–4 –6 –8 –10
–2 –3 –4 –5
–5 –4 –3 –2 –1–1
x
9. x 2y = 6 y = 0: x 2(0) 6 x0 6 x6 The x-intercept is (6, 0). x = 0: 0 2 y 6 2 y 6 y 3
–2 –3 –4 –5
–5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
3 11. y x 3 4 y = 0: 0
3 4 3
13. y = mx + b y = 3x 1 The slope is m = 3. x3
14. y = mx + b y = 6x + 2 The slope is m = 6.
3 x 4 4x The x-intercept is (4, 0). 3 x = 0: y (0) 3 4 y 03 y3 The y-intercept is (0, 3).
15. y = mx + b 7x 2 y 11 2 y 7x 11 7 11 y x 2 2 7 The slope is m . 2
y 5 4
3 2 1 –5 –4 –3 –2 –1–1
16.
(0, 3) (4, 0) 1 2 3 4 5
–2 –3 –4 –5
12. 5x 2y = 8 y = 0: 5x 2(0) 8 5x 0 8 5x 8 8 x 5 8 The x-intercept is , 0 . 5 x = 0: 5(0) 2 y 8 0 2y 8 2 y 8 y 4 The y-intercept is (0, 4).
x
y = mx + b 2x y 0 y 2x y 2x y 2x 0 The slope is m = 2. 17. The graph of x = 2 is a vertical line. Vertical lines have undefined slopes. 18. The graph of y = 4 is a horizontal line. Horizontal lines have slopes of m = 0. 19. y mx b 1 y 2x 3 1 y 2x 3 20. y y1 m(x x1) y 3 4[x (1)] y 3 4(x 1) y 3 4x 4 y 4x 1
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities y y1 3 0 3 1 21. m 2 x2 x1 1 2 3 y y1 m(x x1) y 0 1(x 2) y x2 x y 2 or x y 2 22. 6x y 7 y 6x 7 y 6x 7 m=6
2x 3y 4 3y 2x 4 2 4 y x 3 3 2 m 3
The x-value 1 is paired with two y-values, 4 and 3, so this set of ordered pairs is not a function.
3. a.
This is the graph of the relation {(3, 2), (1, 1), (0, 0), (1, 1)}. Each x-coordinate has exactly one y-value, so this is the graph of a function.
b. This is the graph of the relation {(1, 1), (1, 2), (1, 0), (3, 1)}. The x-value 1 is paired with two y-values, 1 and 2, so this is not the graph of a function.
2 Since 6 , the lines are not parallel. Since 3 2 6 4 1, the lines are not 3 perpendicular. The lines are neither parallel nor perpendicular.
4. a.
No vertical line will intersect the graph more than once, so the graph is the graph of a function.
b.
No vertical line will intersect the graph more than once, so the graph is the graph of a function.
23. 3x 6 y 4 y 2x 6 y 3x 4 m 2 1 2 y x 2 3 1 m 2 1 Since (2) 1, the lines are perpendicular. 2
c.
Vertical lines can be drawn that intersect the graph in two points, so the graph is not the graph of a function.
d.
A vertical line can be drawn that intersects this line at every point, so the graph is not the graph of a function.
24. a.
b. c.
The ordered pairs are (2019, 4732) and (2021, 5041).
For the years 2019 through 2021, the amount of yogurt produced increased at a rate of 154.5 million pounds per year.
1. The domain is the set of x-coordinates: {3, 4, 7}. The range is the set of y-coordinates: {0, 1, 5, 6}. a.
540
5. a, b, and c are functions because their graphs are nonvertical lines. d is not a function because its graph is a vertical line. 6. a.
y y1 5041 4732 309 m 2 154.5 x2 x1 2021 2019 2
Section 13.6 Practice Exercises
2.
b.
Each x-value is only assigned to one y-value, so the relation is a function.
b.
7.
According to the graph, the time of sunrise on March 1st is 6:30 a.m. According to the graph, the sun rises at 6 a.m. in the middle of March and the middle of October.
f (x) x2 1 a.
f (1) 12 1 11 2 Ordered pair: (1, 2)
b.
f (3) (3)2 1 9 1 10 (3, 10)
c.
f (0) 02 1 0 1 1 (0, 1)
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ISM: Prealgebra and Introductory Algebra
8. a.
In this function, x can be any real number. The domain of h(x) is the set of all real numbers.
b. Since we cannot divide by 0, the domain of f(x) is the set of all real numbers except 0. 9. a.
b.
The x-values go from 4 to 6, so the domain is 4 x 6. The y-values go from 2 to 3, so the range is 2 y 3. There are no restrictions on the x-values, so the domain is all real numbers. The y-values are all less than or equal to 3, so the range is y 3.
Vocabulary, Readiness & Video Check 13.6 1. A set of ordered pairs is called a relation. 2. A set of ordered pairs that assigns to each x-value exactly one y-value is called a function. 3. The set of all y-coordinates of a relation is called the range. 4. The set of all x-coordinates of a relation is called the domain. 5. All linear equations are functions except those whose graphs are vertical lines. 6. All linear equations are functions except those whose equations are of the form x = c.
Chapter 13: Graphing Equations and Inequalities
13. f(2) = 6 corresponds to (2, 6) and f(3) = 11 corresponds to (3, 11). Exercise Set 13.6 2. The domain is the set of x-coordinates: {2, 1, 3} The range is the set of y-coordinates: {6, 2, 4} 4. The domain is the set of x-coordinates: {5} The range is the set of y-coordinates: {3, 0, 3 , 4} 6. Each x-value is only assigned to one y-value, so the relation is a function. 8. The x-value 1 is paired with two y-values, 2 and 4, so the relation is not a function. 10. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 12. The vertical line x = 2 will intersect the graph in two points, so the graph is not the graph of a function. 14. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 16. Vertical lines can be drawn that intersect the graph in two points, so the graph is not the graph of a function. 18. If x = 5, the relation is not also a function; d.
7. If f(3) = 7, the corresponding ordered pair is (3, 7). 8. The domain of f(x) = x + 5 is all real numbers. 9. For the function y = mx + b, the dependent variable is y and the independent variable is x. 10. A relation is a set of ordered pairs and an equation in two variables defines a set of ordered pairs. Therefore, an equation in two variables can also define a relation. 11. Yes, this is a function. The definition restricts x-values to be assigned to exactly one y-value, but it makes no such restriction on the y-values. 12. A vertical line represents one x-value paired with many y-values. A function only allows an x-value paired with exactly one y-value, so if a vertical line intersects a graph more than once, there’s an x-value paired with more than one y-value and we don’t have a function.
20. 2x 3y = 9 The graph of this linear equation is not a vertical line, so the equation describes a function. 22. x = 3 The graph of this linear equation is a vertical line, so the equation does not describe a function. 24. y = 9 The graph of this linear equation is not a vertical line, so the equation describes a function. 2 26. y x 3 Each x-value will generate exactly one y-value, so the equation describes a function.
28. On November 1, the graph shows sunset to be at approximately 3:45 p.m. 30. At 9 p.m., the graph shows this happens in midMay and in mid-July.
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Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
32. Yes; every location has exactly 1 sunset time per day.
60. f(x) = 3x f(2) = 3(2) = 6 f(0) = 3(0) = 0 f(3) = 3(3) = 9 The ordered pairs are (2, 6), (0, 0), and (3, 9).
34. For 2006, the graph shows the minimum wage was $5.15 per hour. 36. According to the graph, the minimum wage that was in effect for the greatest number of years was $7.25 per hour. 38. answers may vary 40. According to the graph, the postage would be $2.94. 42. From the graph, the heaviest large envelope that can be mailed for $4.00 weighs 12 ounces. 44. yes; answers may vary 46. h(x) = 3x h(1) = 3(1) = 3 h(0) = 3(0) = 0 h(4) = 3(4) = 12
62. f(x) = |2 x| f(2) = |2 (2)| = |2 + 2| = |4| = 4 f(0) = |2 0| = |2| = 2 f(3) = |2 3| = |1| = 1 The ordered pairs are (2, 4), (0, 2), and (3, 1). 64. The domain of f(x) is all real numbers. 66. x 6 cannot be 0. x6 0 x6 The domain of f(x) is all real numbers except 6. 68. The domain is all real numbers. The range is y 5. 70. The domain is all real numbers. The range is all real numbers.
48. h(x) 3x2
72. The domain is {3}. The range is all real numbers.
h(1) 3(1)2 3(1) 3 h(0) 3(0)2 3(0) 0
74. When x = 3, y = 2, so the ordered pair solution is (3, 2).
h(4) 3(4)2 3(16) 48 50. The ordered pair solution corresponding to f(7) = 2 is (7, 2).
76. When x = 3, y = 2, so f(3) = 2.
52. The ordered 7 pair 7 corresponding to solution g(0) is 0, . 8 8
78. When y = 1, x = 0 and x = 4. 80. 3x 1 11 3x 12 x4
54. The ordered pair solution corresponding to h(10) = 1 is (10, 1). 56. f(x) = 3 7x f(2) = 3 7(2) = 3 + 14 = 17 f(0) = 3 7(0) = 3 0 = 3 f(3) = 3 7(3) = 3 21 = 18 The ordered pairs are (2, 17), (0, 3), and (3, 18). 58.
f (x) x2 4
82. 2x 3 3 2x 0 x0 3
5
6
10
16
84. 2 2 x2 x 2 x 2 x 2 x 2 16 The perimeter is inches. x2
f (2) (2)2 4 4 4 0
86. (9, 20) written in function notation is g(9) = 20.
f (0) 02 4 0 4 4
88. (3, 4) written in function notation is f(3) = 4.
f (3) 3 4 9 4 5 The ordered pairs are (2, 0), (0, 4), and (3, 5).
90. The other x-intercept of the graph is (3, 0). In function notation, this is f(3) = 0.
2
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ISM: Prealgebra and Introductory Algebra
92. answers may vary 94. y = 3 written in function notation is f(x) = 3. 96.
f (x)
136
x
25 a.
f (35)
136
(35)
4760
952
190.4 25 25 5 The proper dosage for a 35-pound dog is 190.4 milligrams.
b.
f (70)
136
(70)
9520
1904
Chapter 13: Graphing Equations and Inequalities 3. Graph the boundary line, x 4y = 4, with a solid line. Test (0, 0): x 4 y 4 0 4(0) 4 00 4 0 4 True Shade the half-plane containing (0, 0).
380.8 25 25 5 The proper dosage for a 70-pound dog is 380.8 milligrams.
Section 13.7 Practice Exercises 1. x 4y > 8 a.
(3, 2): 3 4(2) 8 3 8 8 11 8 False (3, 2) is not a solution of the inequality.
b.
4. Graph the boundary line, y = 3x, with a dashed line. Test (1, 1): y 3x 1 3(1) 1 3 True Shade the half-plane containing (1, 1).
(9, 0): 9 4(0) 8 90 8 9 8 True (9, 0) is a solution of the inequality.
2. Graph the boundary line, x y = 3, with a dashed line. Test (0, 0): x y 3 00 3 0 3 False Shade the half-plane not containing (0, 0). 5. Graph the boundary line, 3x + 2y = 12, with a solid line. Test (0, 0): 3x 2 y 12 3(0) 2(0) 12 0 0 12 0 12 False Shade the half-plane not containing (0, 0).
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Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
Vocabulary, Readiness & Video Check 13.7 1. The statement 5x 6y < 7 is an example of a linear inequality in two variables. 2. A boundary line divides a plane into two regions called half-planes. 3. The graph of 5x 6y < 7 includes its corresponding boundary line. false
6. Graph the boundary line, x = 2, with a dashed line. Test (0, 0): x 2 0 2 True Shade the half-plane containing (0, 0).
4. When graphing a linear inequality, to determine which side of the boundary line to shade, choose a point not on the boundary line. true 5. The boundary line for the inequality 5x 6y < 7 is the graph of 5x 6y = 7. true 6. The graph shown is y < 2. 7. An ordered pair is a solution of an inequality if replacing the variables with the coordinates of the ordered pair results in a true statement. 8. We find the boundary line equation by replacing the inequality symbol with =. The points on this line are solutions (line is solid) if the inequality is or ; they are not solutions (line is dashed) if the inequality is > or <. Exercise Set 13.7
7. Graph the boundary line, y
1 4
x 3, with a
solid line. Test (0, 0): y 0
1 4 1
x3
(0) 3 4 0 03 0 3 False Shade the half-plane not containing (0, 0).
2. y x < 2 (2, 1): 1 2 2 1 2 False (2, 1) is not a solution of the inequality. (5, 1): 1 5 2 6 2 True (5, 1) is a solution of the inequality. 4. 2x + y 10 (0, 11): 2(0) 11 10 0 11 10 11 10 True (0, 11) is a solution of the inequality. (5, 0): 2(5) 0 10 10 0 10 10 10 True (5, 0) is a solution of the inequality. 6. y > 3x (0, 0): 0 3(0) 0 0 False (0, 0) is not a solution to the inequality.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
(1, 4): 4 3(1) 4 3 True (1, 4) is a solution to the inequality.
14. Graph the boundary line, x = 2y, with a dashed line. Test (1, 1): x 2 y 1 2(1) 1 2 True Shade the half-plane containing (1, 1).
8. Graph the boundary line, x + y = 2, with a solid line. Test (0, 0): x y 2 0 0 2 0 2 True Shade the half-plane containing (0, 0).
10. Graph the boundary line, x 3y = 3, with a dashed line. Test (0, 0): x 3y 3 0 3(0) 3 0 3 True Shade the half-plane containing (0, 0).
12. Graph the boundary line, y = 3x, with a solid line. Test (1, 1): y 3x 1 3(1) 1 3 True Shade the half-plane containing (1, 1).
16. Graph the boundary line, y = x + 1, with a solid line. Test (0, 0): y x 1 0 0 1 0 1 True Shade the half-plane containing (0, 0).
18. Graph the boundary line, y = 2, with as dashed line. Test (0, 0): y 2 0 2 False Shade the half-plane not containing (0, 0).
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Chapter 13: Graphing Equations and Inequalities 20. Graph the boundary line, x = 1, with a solid line. Test (0, 0): x 1 0 1 False Shade the half-plane not containing (0, 0).
22. Graph the boundary line, 4x + 3y = 12, with a solid line. Test (0, 0): 4x 3y 12 4(0) 3(0) 12 0 0 12 0 12 False Shade the half-plane not containing (0, 0).
24. Graph the boundary line, x = y, with a solid line. Test (1, 1): x y 1 1 False Shade the half-plane not containing (1, 1).
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ISM: Prealgebra and Introductory Algebra 26. Graph the boundary line, x y = 10, with a dashed line. Test (0, 0): x y 10 0 0 10 0 10 False Shade the half-plane not containing (0, 0).
28. Graph the boundary line, y = 0, with a solid line. Test (1, 1): y 0 1 0 False Shade the half-plane not containing (1, 1).
30. Graph the boundary line, 3x + 5y = 2, with a solid line. Test (0, 0): 3x 5 y 2 3(0) 5(0) 2 0 0 2 0 2 False Shade the half-plane not containing (0, 0).
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ISM: Prealgebra and Introductory Algebra
32. Graph the boundary line, y Test (0, 0): y 0
2
2 5
Chapter 13: Graphing Equations and Inequalities
x 3, with a dashed line.
x3
5 2
(0) 3 5 0 03 0 3 False Shade the half-plane not containing (0, 0).
1 34. Graph the boundary line, y x 4, with a dashed line. 3 1 Test (0, 0): y x 4 3 1 0 (0) 4 3 0 04 0 4 False Shade the half-plane not containing (0, 0).
36. The point of intersection appears to be (3, 0). 38. The point of intersection appears to be (3, 3). 40. The graph is the half-plane with the solid boundary line y = 2x. The choice is C. 42. The graph is the half-plane with the solid boundary line y = 3x. The choice is D. 44. The inequality is x + y 13.
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Chapter 13: Graphing Equations and Inequalities
ISM: Prealgebra and Introductory Algebra
46. Test (1, 1): y 5x 1 5(1) 1 5 False (1, 1) is not included in the graph of y > 5x. 48. Test (1, 1): x 3 1 3 False (1, 1) is not included in the graph of x > 3. 50. a.
The sum of x and y must be at most $5000. x + y 5000
b.
c.
answers may vary
Section 13.8 Practice Exercises 1. Use (4, 8). y kx 8 k 4 8 k 4 4 4 2k Since k = 2, the equation is y = 2x. 2. Let y = 15 and x = 45. y kx 15 k(45) 15 k (45) 45 45 1 k 3 1 The equation is y x. 3
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ISM: Prealgebra and Introductory Algebra
Let x = 3. 1 y x 3 1 y 3 3 y 1 Thus, when x is 3, y is 1.
Let r = 4. A r2 A 42 A 16 The area is 16 square feet. 7.
3. Use (1, 2) and (0, 0). 0 (2) 2 slope 2 0 (1) 1 Thus, k = 2 and the variation equation is y = 2x. 4. Use (4, 5). k y x k 5 4 k 45 4 4 20 k
d kt2 144 k (3)2 144 9k 16 k The equation is d 16t 2. Let t = 5. d 16t 2 d 16 52 d 16 25 d 400 The object will fall 400 feet in 5 seconds.
Vocabulary, Readiness & Video Check 13.8
Since k = 20, the equation is y
20
.
x
5. Let y = 4 and x = 0.8. k y x k 4 0.8 k 0.8(4) 0.8
1. y
k
, where k is a constant. inverse
x 2. y = kx, where k is a constant. direct 3. y = 5x direct 5 4. y x inverse
0.8
3.2 k
7 5. y 2 inverse x
The equation is y 3.2 . x Let x = 20. 3.2 y 20 y 0.16 Thus, when x is 20, y is 0.16. 6.
Chapter 13: Graphing Equations and Inequalities
6. y 6.5x4 direct 7. y
11
8. y = 18x direct
A kr
2
49 k (7) 49 49k k
inverse
x
9. y 12x
2
The formula for the area of a circle is A r 2 .
2
direct
20 10. y 3 inverse x 11. linear; slope
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
12. With inverse variation, we know that y
k
or
x yx = k, so we just need to multiply the values of x and y together to find k.
13. No; the direct relationship is the power of x times a constant, and the inverse relationship is the reciprocal of the power of x times a constant. 14. This is a direct variation problem, y = kx (with k positive), so as either amount (y or x) increases, the other amount also increases. We’re asked to find the new distance given a weight increase, so we know that our answer will also show a distance increase.
12.
y 0.1
k x k
4 0.4 k 0.4 y x 14. a varies directly as b is written as a = kb. k 16. s varies inversely as t is written as s . t 18. p varies inversely as x2 is written as p
Exercise Set 13.8 2.
y kx 14 k(2) 7k y = 7x
22. y varies directly as d 2 is written as y kd 2 . 24. y = kx y = 27 when x = 3: 27 k (3) 9k y = 9x x = 2: y = 9 2 = 18 y = 18 when x = 2. 26. y
y y1 1 0 1 6. m 2 40 4 x2 x1 1 y x 4 y2 y1 5 0 5 8. m x x 2 0 2 y
5
x
1
k x
y = 200 when x = 5: 200
k
5 1000 k y
1000 x
x = 4: y
1000
250
4 y = 250 when x = 4.
2 10.
y 11
k x k
2 22 k 22 y x
550
x2
20. x varies directly as y4 is written as x ky4 .
4. y kx 3 k (9) 1 k 3 1 y x 3
2
k
28. s kt 3 s = 270 when t = 3: 270 k (3)3 270 27k 10 k s 10t3
t = 1: s 10(1)3 10(1) 10 s = 10 when t = 1.
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.
ISM: Prealgebra and Introductory Algebra
30.
k p q2 5 5 k p when q = 8: 16 16 82 5 k 16 64 20 k 20 p 2 q 1 20 q : p 20 4 80 2 2 1
Chapter 13: Graphing Equations and Inequalities
w = 20: d
1
(20) 6
3 The spring stretches 6
2 3 2
3 is attached to the spring.
2
1 p = 80 when q . 2 32. Let p be the paycheck amount when h hours are worked. p = kh p = 304.50 when h = 30: 304.50 k (30) 10.15 k p = 10.15h h = 34: p = 10.15(34) = 345.10 The pay is $345.10 for 34 hours. 34. Let c be the cost per notebook when n notebooks are manufactured. k c n k c = 0.50 when n = 10,000: 0.50 10, 000 5000 k 5000 c n 5000 n = 18,000: c 0.28 18, 000 The cost to manufacture 18,000 notebooks is $0.28 per notebook. 36. Let d be the distance when a weight of w is attached. d = kw d = 10 when w = 30: 10 k (30) 1 k 3 1 d w 3
inches when 20 pounds
38. Let t be the time and r be the rate. k r t k r = 55 when t = 4: 55 4 220 k 220 r t 220 r = 60: 60 t 2 t3 3 2 The trip will take 3 hours at 60 mph. 3 40. d kr 2 r = 60 when d = 300: 300 k (60)2 300 k (3600) 1 k 12 1 d r2 12 1 1 r = 30: d (30)2 (900) 75 12 12 It takes 75 feet to stop a car traveling 30 mph. 42.
x y 9 x y 14 2 y 23
44.
1.9x 2 y 5.7 1.9x 0.1y 2.3 2.1y 8.0
46. If y varies directly as x2 , then y kx2. If x is tripled, to become 3x, then y k (3x)2 9(kx), and y is multiplied by 9. 48. 6 min
6 min
1 hr
1
hr 1 60 min 10 4 min 1 hr 1 4 min hr 1 60 min 15
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
r
k
16. The slope of a line measures the steepness or tilt of a line.
t r = 100 when t
1 10
: 100
k
17. The equation y = kx is an example of direct variation.
1 10
100 10k 10 k r t
10 t 1 15
18. The equation y
k x
is an example of inverse
variation. : r
Chapter 13 Review
10
15 10 150 1
16.
15
The rate must be 150 mph. Chapter 13 Vocabulary Check 1. An ordered pair is a solution of an equation in two variables if replacing the variables by the coordinates of the ordered pair results in a true statement. 2. The vertical number line in the rectangular coordinate system is called the y-axis.
7. 2 + y = 6x In (7, ), the x-coordinate is 7. 2 y 6(7) 2 y 42 y 44 The ordered pair solution is (7, 44).
3. A linear equation can be written in the form Ax + By = C. 4. A(n) x-intercept is a point of the graph where the graph crosses the x-axis. 5. The form Ax + By = C is called standard form. 6. A(n) y-intercept is a point of the graph where the graph crosses the y-axis. 7. A set of ordered pairs that assigns to each x-value exactly one y-value is called a function. 8. The equation y = 7x 5 is written in slopeintercept form. 9. The set of all x-coordinates of a relation is called the domain of the relation. 10. The set of all y-coordinates of a relation is called the range of the relation. 11. The set of ordered pairs is called a relation. 12. The equation y + 1 = 7(x 2) is written in point-slope form. 13. To find an x-intercept of a graph, let y = 0. 14. The horizontal number line in the rectangular coordinate system is called the x-axis.
8. y = 3x + 5 In ( , 8), the y-coordinate is 8. 8 3x 5 13 3x 13 x 3 13 The ordered pair solution is , 8 . 3
9. 9 = 3x + 4y y = 0: 9 3x 4(0) 9 3x 0 9 3x 3 x y = 3: 9 3x 4(3) 9 3x 12 3 3x 1 x x = 9: 9 3(9) 4 y 9 27 4 y 36 4 y 9 y
15. To find a y-intercept of a graph, let x = 0. 552
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ISM: Prealgebra and Introductory Algebra
x
y
3
0
1
3
9
9
13. x y = 1 y 5 4 3 2 1 –5 –4 –3 –2 –1–1
10. y = 5 for each value of x. x
y
7
5
7
5
0
5
Chapter 13: Graphing Equations and Inequalities
1 2 3 4 5
x
2 4 6 8 10
x
2 4 6 8 10
x
2 4 6 8 10
x
–2 –3 –4 –5
14. x + y = 6 y 10 8 6 4 2
11. x = 2y y = 0: x 2(0) x0
–10 –8 –6 –4 –2–2 –4 –6 –8 –10
y = 5: x 2(5) x 10 y = 5: x 2(5) x 10
15. x 3y = 12 y
12. a.
x
y
0
0
10
5
10
5
10 8 6 4 2 –10 –8 –6 –4 –2–2
y = 5x + 2000 x = 1: y = 5(1) + 2000 = 5 + 2000 = 2005
16. 5x y = 8
x = 100: y 5(100) 2000 500 2000 2500
y
x = 1000: y 5(1000) 2000 5000 2000 7000
b.
–4 –6 –8 –10
x
1
100
1000
y
2005
2500
7000
10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10
Let y = 6430 and solve for x. 6430 5x 2000 4430 5x 886 x 886 compact disc holders can be produced for $6430. Copyright © 2025 Pearson Education, Inc.
553
Chapter 13: Graphing Equations and Inequalities
23. x 3y = 12 y = 0: x 3(0) 12 x 0 12 x 12 x-intercept: (12, 0) x = 0: 0 3y 12 3y 12 y 4 y-intercept: (0, 4)
17. x = 3y y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
24. 4x + y = 8 y = 0: 4x 0 8 4x 8 x 2 x-intercept: (2, 0) x = 0: 4(0) y 8 0 y 8 y8 y-intercept: (0, 8)
18. y = 2x y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
ISM: Prealgebra and Introductory Algebra
x
–2 –3 –4 –5
25. (x1, y1) (1, 2), (x2 , y2 ) (3, 1) y y 1 2 3 3 m 2 1 x2 x1 3 (1) 3 1 4
19. The x-intercept is (4, 0). The y-intercept is (0, 2). 20. The x-intercepts are (2, 0) and (2, 0). The y-intercepts are (0, 2) and (0, 2).
26. (x1, y1) (2, 2), (x2 , y2 ) (3, 1) y y1 1 (2) 1 2 1 m 2 x2 x1 3 (2) 3 2 5
21. y = 3 is a horizontal line with y-intercept (0, 3).
27. When m = 0, the line is horizontal. The choice is d.
y
28. The slope is m = 1. The choice is b.
5 4 3 2 1 –5 –4 –3 –2 –1–1
29. When the slope is undefined, the line is vertical. The choice is c. 1 2 3 4 5
30. The slope is m = 4. The choice is a.
x
–2 –3 –4 –5
31. m y2 y1 8 5 3 x2 x1 6 2 4
22. x = 5 is a vertical line with x-intercept (5, 0). y 10 8 6 4 2 –10 –8 –6 –4 –2–2 –4 –6 –8 –10
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32. m y2 y1 2 7 5 5 x2 x1 1 4 3 3 33. m y2 y1 9 3 12 4 x2 x1 2 1 3
2 4 6 8 10
x
34. m y2 y1 6 1 6 1 7 1 x2 x1 3 (4) 3 4 7
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
y y1 7416 7293 123 24.6 43. m 2 x2 x1 2020 2015 5 The total number of U.S. magazines in print increases by 24.6 magazines per year.
35. y = mx + b y = 3x + 7 The slope is m = 3. 36.
y = mx + b x 2y 4 2 y x 4 1 y x2 2
44. m y2 y1 2693 2315 378 63 x2 x1 2022 2016 6 The number of U.S. lung transplants increases by about 63 transplants per year.
1 The slope is m . 2 37. y = mx + b y = 2 y = 0x 2 The slope is m = 0. 38. x = 0 is a vertical line. The slope is undefined. x y 3 39. x y 6 y x 6 y x 3 y x6 m=1 m = 1 Since (1)(1) = 1, the lines are perpendicular. 3x y 10 y 3x 10 y 3x 10 m = 3 m = 3 Since the slopes are equal, the lines are parallel.
40. 3x y 7 y 3x 7
41. y 4x
1
4x 2 y 1 2 y 4x 1 1 y 2x 2 m=4 m = 2 Since 4 2 and (4)(2) = 8 1, the lines are neither parallel nor perpendicular. 2
1 42. y 6x 3
m=6
Chapter 13: Graphing Equations and Inequalities
x 6y 6 6 y x 6 1 y x 1 6 1 m 6
1 Since (6) 1, the lines are perpendicular. 6
45. y = mx + b x 6 y 1 6 y x 1 1 1 y x 6 6 1 1 m 6 ; y-intercept 0, 6 46. y = mx + b 3x y 7 y 3x 7 m = 3; y-intercept (0, 7) 47. y mx b 1 y 5x 2 48. y mx b 2 y x6 3 49. y = mx + b y = 2x + 1 m = 2, y-intercept (0, 1) The choice is d. 50. y = mx + b y = 4x y = 4x + 0 m = 4; y-intercept (0, 0) The choice is c. 51. y = mx + b y = 2x y = 2x + 0 m = 2; y-intercept (0, 0) The choice is a. 52. y = mx + b y = 2x 1 m = 2; y-intercept (0, 1) The choice is b.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities 53. y y1 m(x x1) y 0 4(x 2) y 4x 8 4x + y = 8 54.
55.
56.
59. The x-value 7 is paired with two y-values, 1 and 5, so the relation is not a function. 60. Each x-value is only assigned to one y-value, so the relation is a function.
y y1 m(x x1) y (5) 3(x 0) y 5 3x y 3x 5 3x y 5
61. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 62. No vertical line will intersect the graph more than once, so the graph is the graph of a function.
y y1 m(x x1) 3 y 4 (x 1) 5 3 5( y 4) 5 (x 1) 5 5 y 20 3(x 1) 5 y 20 3x 3 5 y 3x 17 3x 5 y 17
63. The vertical line x = 3 will intersect the graph at more than one point, so the graph is not the graph of a function.
y y1 m(x x1) 1 y 3 [x (3)] 3y 9 3 (3)] 1[x 3y 9 (x 3) 3y 9 x 3 3y x 6 x 3y 6
66. f(x) = 2x + 6 f(2) = 2(2) + 6 = 4 + 6 = 10
64. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 65. f(x) = 2x + 6 f(0) = 2(0) + 6 = 0 + 6 = 6
67. f(x) = 2x + 6
y y 7 7 14 14 57. m 2 1 x2 x1 2 1 y y1 m(x x1) y 7 14(x 1)
1
y 7 14x 14 y 14x 21 58. m y2 y1 6 5 6 5 1 1 x x 4 (2) 4 2 2 2 2
1 1 2 6 1 6 5 2 2 68. f(x)= 2x 1 +6 1 f 2 6 1 6 7 2 2 f
69. Graph the boundary line, x + 6y = 6, with a dashed line. Test (0, 0): x 6 y 6 0 6(0) 6 00 6 0 6 True Shade the half-plane containing (0, 0).
1
y y1 m(x x1) 1 y 5 [x (2)] 2 1 y 5 (x 2) 2 1 2( y 5) 2 (x 2) 2 2 y 10 1(x 2) 2 y 10 x 2 2 y x 8 1 y x4 2 556
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ISM: Prealgebra and Introductory Algebra 70. Graph the boundary line, x + y = 2, with a dashed line. Test (0, 0): x y 2 0 0 2 0 2 True Shade the half-plane containing (0, 0).
71. Graph the boundary line, y = 7, with a solid line. Test (0, 0): y 7 0 7 True Shade the half-plane containing (0, 0).
72. Graph the boundary line, y = 4, as a solid line. Test (0, 0): y 4 0 4 False Shade the half-plane not containing (0, 0).
Chapter 13: Graphing Equations and Inequalities 73. Graph the boundary line, x = y, as a solid line. Test (1, 1): x y 1 1 True Shade the half-plane containing (1, 1).
74. Graph the boundary line, x = y, as a solid line. Test (1, 1): x y 1 1 True Shade the half-plane containing (1, 1).
75. y = kx y = 40 when x = 4: 40 k (4) 10 k y = 10x x = 11: y = 10(11) = 110 y = 110 when x = 11. 76. y
k x
y = 4 when x = 6: 4
k
6 24 k y
24 x
x = 48: y y
24 1 48 2
1 2
when x = 48.
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557
Chapter 13: Graphing Equations and Inequalities
77. y
81. 2x 5y = 9 y = 1: 2x 5(1) 9 2x 5 9 2x 14 x7
k x3
y = 12.5 when x = 2: 12.5 12.5
k 23 k
8 100 k y
x = 2: 2(2) 5 y 9 4 5y 9 5 y 5 y 1
100 x3
x = 3: y 100 100 27 33 100 y when x = 3. 27
y = 3: 2x 5(3) 9 2x 15 9 2x 6 x 3
78. y kx2 y = 175 when x = 5: 175 k(5)2 175 25k 7k 2
y 7x
x = 10: y 7(10)2 7(100) 700 y = 700 when x = 10. 79. Let c be the cost for manufacturing m milliliters. k c m k c = 6600 when m = 3000: 6600 3000 19,800, 000 k 19,800, 000 c m 19,800, 000 Let m = 5000: c 3960 5000 It costs $3960 to manufacture 5000 milliliters. 80. Let d be the distance when a weight of w is attached. d = kw d = 8 when w = 150: 8 k(150) 4 k 75 4 d w 75 4 4 Let w = 90: d 90 4 75 5 4 The spring stretches 4 inches when 90 pounds 5 is attached. 558
ISM: Prealgebra and Introductory Algebra
x
y
7
1
2
1
3
3
82. x = 3y x = 0: 0 3y 0 y y = 1: x 3(1) x 3 x = 6:
6 3y 2 y
x
y
0
0
3
1
6
2
83. 2x 3y = 6 y = 0: 2x 3(0) 6 2x 0 6 2x 6 x3 x-intercept: (3, 0) x = 0: 2(0) 3y 6 0 3y 6 3y 6 y 2 y-intercept: (0, 2)
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ISM: Prealgebra and Introductory Algebra
87. y = 4x y = 0: 0 4x 0x x-intercept: (0, 0) The y-intercept is also (0, 0). Find another point. Let x = 1. y = 4(1) = 4 Another point is (1, 4).
84. 5x + y = 10 y = 0: 5x 0 10 5x 10 x 2 x-intercept: (2, 0) x = 0: 5(0) y 10 0 y 10 y 10 y-intercept: (0, 10)
y 5 4 3 2
85. x 5y = 10 y = 0: x 5(0) 10 x 0 10 x 10 x-intercept: (10, 0) x = 0: 0 5 y 10 5 y 10 y 2
1
–5 –4 –3 –2 –1–1
y
–10 –8 –6 –4 –2–2 –4 –6 –8 –10
(0, 0) 1 2 3 4 5
x
–2 –3 –4 –5
y-intercept: (0, 2) 10 8 6 4 2
Chapter 13: Graphing Equations and Inequalities
(10, 0) 2 4 6 8 10
x
(0, –2)
88. 2x + 3y = 6 y = 0: 2x 3(0) 6 2x 0 6 2x 6 x 3 x-intercept: (3, 0) x = 0: 2(0) 3y 6 0 3y 6 y 2 y-intercept: (0, 2) y
86. x + y = 4 y = 0: x 0 4 x4 x-intercept: (4, 0) x = 0: 0 y 4 y4 y-intercept: (0, 4)
5 4 3 2 (–3, 0) 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5
(0, –2)
x
(0, 4)
89. x = 3 is a vertical line with x-intercept (3, 0). y (4, 0) 1 2 3 4 5
5 4 3 2 1
x
–5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
Copyright © 2025 Pearson Education, Inc.
559
Chapter 13: Graphing Equations and Inequalities 90. y = 2 is a horizontal line with y-intercept (0, 2).
y y1 2 (5) 25 7 91. m 2 1 x2 x1 4 3 4 3 7 92. m y2 y1 8 3 11 11 x2 x1 6 1 7 7 93. (x1, y1) (0, 4), (x2 , y2 ) (2, 0) y y1 0 (4) 4 m 2 2 x2 x1 20 2 94. (x1, y1) (0, 2), (x2 , y2 ) (6, 0) y2 y1 0 2 2 1 m x2 x1 6 0 6 3 95. y = mx + b 2x 3y 15 3y 2x 15 2 y x5 3 2 m ; y-intercept: (0, 5) 3 96. y = mx + b 6x y 2 0 6x y 2 y 6x 2 m = 6; y-intercept: (0, 2) 97.
560
y y1 m(x x1) y (7) 5(x 3) y 7 5x 15 y 5x 8 5x y 8
ISM: Prealgebra and Introductory Algebra 98. y y1 m(x x1) y 6 3(x 0) y 6 3x 6 3x y 3x y 6 99. m y2 y1 5 9 5 9 4 4 1 x2 x1 2 (3) 2 3 y y1 m(x x1) y 9 4[x (3)] y 9 4(x 3) y 9 4x 12 y 4x 3 4x y 3 100. m y2 y1 9 1 10 5 x2 x1 5 3 2 y y1 m(x x1) y 1 5(x 3) y 1 5x 15 y 5x 16 5x y 16 Chapter 13 Getting Ready for the Test 1. A. x y = 7 2 = 5 B. x y = 0 (5) = 0 + 5 = 5 C. x y = 2 3 = 5 5 D. x y = 2 (7) = 2 + 7 = 5 Choice C is correct. 2. All ordered-pair solutions of y = 4 are of the form (x, 4) where x can be any real number. (4, 0) is not of this form, so choice A is correct. 3. The lines x = 0 (the y-axis) and y = 0 (the x-axis) each have an infinite number of intercepts. Any other line must intercept at least one of the axes, so the fewest intercepts a line may have is 1; B. 4. A is not a linear equation because it involves x. B is a linear equation since 2x 62 is 2x = 36. C is not a linear equation since both x and y are raised to the third power. D is not a linear equation because it involves |x|. Choice B is correct.
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ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
5. The line has negative slope and goes downward more steeply than the other line with negative slope, so the slope is m = 10; B.
4. The slope of a horizontal line is m = 0. y y1 2 (5) 7 1 5. m 2 x2 x1 1 6 7
6. The line has positive slope and goes upward more steeply than the other line with positive slope, so the slope is m = 5; A.
6. m y2 y1 1 (8) 1 8 7 7 1 0 1 1 x2 x1
7. The line has negative slope, but does not go downward as steeply as the other line with 4 negative slope, so the slope is m ; D. 7
7. y = mx + b 3x y 5 y 3x 5 m=3
8. The line has positive slope, but does not go upward as steeply as the other line with positive 1 slope, so the slope is m ; C. 2
8. x = 6 is a vertical line. The slope of a vertical line is undefined. 9. 2x + y = 8 y = 0: 2x 0 8 2x 8 x4 x-intercept: (4, 0) x = 0: 2(0) y 8 y 8
9. (0, 5) in function notation is f(0) = 5; C. 10. Let (x1, y1) (2, 3) and (x2 , y2 ) (0, 9). y y1 9 3 6 m 2 3 x2 x1 0 2 2 y y1 m(x x1) y 3 3(x 2) y 3 3x 6 y 3x 9 This is the slope-intercept form for the equation of the line through the two points ; C.
y-intercept: (0, 8) y 10 8 6 4 2 –10 –8 –6 –4 –2–2
11. f(0) = 2; B
2 4 6 8 10
x
–4 –6 –8 –10
12. f(4) = 0; D 13. If f(x) = 0 then x = 2 or x = 4; A, C. 14. f(1) = 3; E Chapter 13 Test 1. 12y 7x = 5 x = 1: 12 y 7(1) 5 12 y 7 5 12 y 12 y 1 The ordered pair is (1, 1). 2. y = 17 for each value of x. The ordered pair is (4, 17).
10. x + 4y = 5 y = 0: x 4(0) 5 x 0 5 x 5 x 5 x-intercept: (5, 0) x = 0: 0 4 y 5 4y 5 5 y 54 y-intercept: 0, 4
3. (x1, y1) (1, 1), (x2 , y2 ) (4, 1) y y1 1 (1) 11 2 m 2 x2 x1 4 (1) 4 1 5 Copyright © 2025 Pearson Education, Inc.
561
Chapter 13: Graphing Equations and Inequalities
y 10 8 6 4 2 –10 –8 –6 –4 –2–2
2 4 6 8 10
x
–4 –6 –8 –10
11. Graph the boundary line, x y = 2, with a solid line. Test (0, 0): x y 2 0 0 2 0 2 True Shade the half-plane containing (0, 0).
ISM: Prealgebra and Introductory Algebra 13. 5x 7y = 10 y = 0: 5x 7(0) 10 5x 0 10 5x 10 x2 x-intercept: (2, 0) x = 0: 5(0) 7 y 10 0 7 y 10 7 y 10 10 y 7 10 y-intercept: 0, 7 y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
12. Graph the boundary line, y = 4x, with a solid line. Test (1, 1): y 4x 1 4(1) 1 4 True Shade the half-plane containing (1, 1).
562
14. Graph the boundary line, 2x 3y = 6, with a dashed line. Test (0, 0): 2x 3y 6 2(0) 3(0) 6 0 0 6 0 6 True Shade the half-plane containing (0, 0).
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ISM: Prealgebra and Introductory Algebra 15. Graph the boundary line, 6x + y = 1, with a dashed line. Test (0, 0): 6x y 1 6(0) 0 1 0 0 1 0 1 True Shade the half-plane containing (0, 0).
Chapter 13: Graphing Equations and Inequalities y y1 7 0 7 7 19. m 2 x2 x1 60 6 6
16. The graph of y = 1 is a horizontal line with a y-intercept of (0, 1). y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
17. y = 2x 6 The slope is m = 2 and the y-intercept is (0, 6). 4x 2 y 2x y y 2x 0 The slope is m = 2 and the y-intercept is (0, 0). Since the slopes are different, the lines are not parallel. Since 2(2) = 4 1, the lines are not perpendicular. The lines are neither parallel nor perpendicular. y y1 m(x x1) 1 y 2 (x 2) 4 1 4( y 2) 4 (x 2) 4 4 y 8 1(x 2) 4 y 8 x 2 4 y x 10 x 4 y 10
20. m y2 y1 3 (5) 3 5 8 x2 x1 1 2 1 y y1 m(x x1) y (5) 8(x 2) y 5 8x 16 y 8x 11 8x y 11 1 21. m ; b = 12 8 1 y x 12 8 8 y x 8(12) 8 y x 96 x 8 y 96
–2 –3 –4 –5
18.
y y1 m(x x1) 7 y 0 (x 0) 6 7 y x 6 7 6y 6 x 6 6 y 7x 7x 6 y 0
22. Each x-value is only assigned to one y-value, so the relation is a function. 23. The x-value 3 is assigned to two y-values, 3 and 2, so the relation is not a function. Note that the x-value 0 is also assigned to two y-values, 5 and 0. 24. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 25. No vertical line will intersect the graph more than once, so the graph is the graph of a function. 26. f(x) = 2x 4 a. f(2) = 2(2) 4 = 4 4 = 8 b. f(0.2) = 2(0.2) 4 = 0.4 4 = 3.6 c.
f(0) = 2(0) 4 = 0 4 = 4
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563
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
27.
f (x) x3 x a.
f (1) (1) (1) 11 0
b.
f (0) 0 0 0 0 0
c.
x2
3
y = 8 when x = 5:
k
8 8
3
52 k
25 200 k
f (4) 43 4 64 4 60
28. 2x 2(2 y) 42 2(x 2 y) 2(21) x 2 y 21 Let y = 8: x 2(8) 21 x 16 21 x5 When y = 8 meters, x = 5 meters. 29. a.
k
32. y
The ordered pairs are (2018, 7.3), (2019, 8.9), (2020, 10.1), (2021, 12.4), and (2022, 13.3).
b.
200
y
x2
Let x = 15: y 200 200 8 152 225 9 8 When x = 15, y . 9 Cumulative Review Chapters 113 1. Area length width (380 miles) (280 miles) 106, 400 square miles The area of Colorado is about 106,400 square miles. 2. 21 7 = 147 There are 147 trees in the orchard. 3. 1 (10) (8) 9 9 (8) 9 17 9 8 4. 2 (7) 3 (4) 9 3 (4) 6 (4) 10 5.
8
8 8 8 8 64 3x 8 3x 8 24x
30. m y2 y1 7.37 2.12 5.25 2.625 x2 x1 2022 2020 2 For every 1 year, gross box office sales increase by $2.625 billion. 31. y = kx y = 10 when x = 15: 10 k (15) 2 k 3 2 y x 3 2 Let x = 42: y (42) 28 3 When x = 42, y = 28.
564
3x 6.
3
2c 7.
14 8
3 4 3 4 12 2c 4 2c 4 8c
13 3
7
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7
7 3 8 7 4 5 7
ISM: Prealgebra and Introductory Algebra
8.
15 4
14
Chapter 13: Graphing Equations and Inequalities 14. 3x 9 3x 9 3 3 x 3 {x|x 3}
5
5 2 4 5 3 10 5
2 5
–5 –4 –3 –2 –1
9. 2x + 5 = 2(3.8) + 5 = 7.6 + 5 = 2.6
15. a.
10. 6x 1 = 6(2.1) 1 = 12.6 1 = 13.6 3.142 3.14 11. 7 22.000 21 10 7 30 28 20 14 6 7 3.14 22
2
3
4 5
The degree of the trinomial 2t 2 3t 6 is 2, the greatest degree of any of its terms.
The degree of the polynomial 7 x 3x3 2x2 1 is 3. It is not a monomial,
binomial, nor a trinomial, so the answer is none of these. 16. a.
The degree of the binomial 7y + 2 or 7 y1 2 is 1.
b.
The degree of the trinomial 8x x2 1 is 2, the greatest degree of any of its terms.
c.
The degree of the polynomial 9 y3 6 y 2 y2 is 3. It is not a monomial, binomial, or trinomial, so the answer is none of these.
17. (2x2 5x 1) (2x2 x 3) 2x2 5x 1 2x2 x 3 (2x2 2x2 ) (5x x) (1 3) 4x2 6x 2 2 18. (9x 5) (x 6x 5)
9x 5 x2 6x 5 x2 (9x 6x) (5 5) x2 3x
2x 4 2x 4 2 2 x 2 –5 –4 –3 –2 –1
1
b. The degree of the binomial 15x 10 or 15x1 10 is 1. c.
1.9473 1.947 12. 19 37.0000 19 18 0 17 1 90 76 140 133 70 57 13 37 1.947 19 13.
0
19. (3y 1)2 (3y 1)(3y 1) (3y)(3y) (3y)(1) 1(3y) 1(1) 9 y2 3y 3y 1 0
1
2
3
9 y2 6 y 1
4 5
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565
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
20. (2x 5)2 (2x 5)(2x 5) (2x)(2x) (2x)(5) (5)(2x) (5)(5)
x2 9x 14 0 (x 2)(x 7) 0 x 2 0 or x 7 0 x2 x7 The solutions are 2 and 7.
4x2 10x 10x 25 4x2 20x 25 21. 9a5 18a2 3a 3a(3a4 ) 3a(6a) 3a(1) 3a(3a4 6a 1) 22. 2x5 x3 x3 2x2 x3 1 x3 (2x2 1) 2
23. x 4x 12 Look for two numbers whose product is 12 and whose sum is 4.
2 2 29. 2x 11x 5 4x 2 2x 11x 5 10 5x 25 10 5x 25 4x 2 (2x 1)(x 5) 2 5 5(x 5) 2(2x 1) 1 or 1 1
3x2 17x 6
30.
x 4x 12 (x 2)(x 6) 2
24. x2 4x 21 Look for two numbers whose product is 21 and whose sum is 4. x2 4x 21 (x 3)(x 7) 25. Factors of 8x2: 8x2 8x x, 8x2 4x 2x Factors of 5: 5 = 1 5 8x 22x 5 (4x 1)(2x 5) 26. Factors of 15x2: 15x2 15x x, 15x2 5x 3x Factors of 2: 2 = 2 1, 2 = 2 1 15x2 x 2 (5x 2)(3x 1) 27.
4x2 28x 49
2x 2 5x 5 4x 24 (3x 1) (x 6) 2 (x 1) 5 (x 1) 2 2(x 6) 3x 1 5 2 3x 1 10 4b
31. 2
x2 9x 14
28.
9a
4b 9a
(2x 7)2 0 2x 7 0 2x 7 7 x 2 7 The solution is . 2
566
4b 3ab 4b(3ab) 12ab2 9a 3ab 9a(3ab) 27a2b
7x 7x 32. 11y 1 11y
4x2 28x 49 0 (2x)2 2 2x 7 72 0
1
7x 9x2 y 11y 9x2 y 7x(9x2 y) 11y(9x2 y) 63x3 y 99x2 y2
33. 1 m 1 m m 1 1 m 1 1(m 1) m 1(m 1) m 1 m 1 m m 1 2m 1 m 1
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ISM: Prealgebra and Introductory Algebra
34. 1
35.
Chapter 13: Graphing Equations and Inequalities
m
1 m m 1 1 m 1 1(m 1) m 1(m 1) m 1 m 1 m m 1 1 m 1 3
x 8 x 6 x 3 x(x 8) x 3x 6 x2 8x
2
In ( , 6), the y-coordinate is 6. y = 6: 3x 6 12 3x 6 x2 The ordered pair solution is (2, 6).
c.
In (1, ), the x-coordinate is 1. x = 1: 3(1) y 12 3 y 12 y 15 The ordered pair solution is (1, 15).
6
40. x + 4y = 20 a.
0 x2 5x 6 0 (x 3)(x 2) x3 0 or x 2 0 x 3 x 2 The solutions are 3 and 2. 36.
b.
In (0, ), the x-coordinate is 0. x = 0: 0 4 y 20 4 y 20 y 5 The ordered pair solution is (0, 5).
b.
10
x5 x 10 x 2 x(x 5) x 2x 10 x2 5x 0 x2 3x 10 0 (x 5)(x 2) x5 0 or x 2 0 x 5 x2 The solutions are 5 and 2.
In ( , 0), the y-coordinate is 0. y = 0: x 4(0) 20 x 0 20 x 20 x 20
x1 y
The ordered pair solution is (20, 0). c.
x 12 x 12 The ordered pair solution is (12, 2).
x1 y x1 x 1 y y y y x 2 y x y(2) 37. x x 2y 2 y
y
y
x 4 38. 2x 2x 2 2 2 2 2 x 4 x 2
2 x2
39. 3x + y = 12 a.
In ( , 2), the y-coordinate is 2. y = 2: x 4(2) 20 x 8 20
In (0, ), the x-coordinate is 0. x = 0: 3(0) y 12 0 y 12 y 12 The ordered pair solution is (0, 12).
41. 2x + y = 5 y = 0: 2x 0 5 2x 5 5 x 2 The x-intercept is 5 , 0 . 2 x = 0: 2(0) y 5 0 y 5 y5 The y-intercept is (0, 5).
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567
ISM: Prealgebra and Introductory Algebra
Chapter 13: Graphing Equations and Inequalities
y
45.
y y1 m(x x1) y 5 2[x (1)] y 5 2(x 1) y 5 2x 2 y 2x 3 2x y 3
46.
y y1 m(x x1) y (7) 5(x 2) y 7 5x 10 y 5x 3 5x y 3
5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5
1 2 3 4 5
x
2x + y = 5
42. y = 2x y
2
5 4 3 2 1
47. g(x) x 3 2
–5 –4 –3 –2 –1–1
1 2 3 4 5
–2 –3 –4 –5
y = –2x
b.
g(2) (2)2 3 4 3 1 The ordered pair is (2, 1).
c.
g(0) 02 3 0 3 3 The ordered pair is (0, 3).
48. 3 4
and the y-intercept is (0, 1).
44. y = mx + b 7x 4 y 10 4 y 7x 10 7 5 y x 4 2 7 The slope is m and the y-intercept is 4 5 0, 2 .
568
g(2) 2 3 4 3 1 The ordered pair is (2, 1).
x
43. y = mx + b 3x 4 y 4 4 y 3x 4 3 y x 1 4 The slope is m
a.
f (x) 3x2 2 a.
f (0) 3(0)2 2 3 0 2 0 2 2 The ordered pair is (0, 2).
b.
f (4) 3(4)2 2 316 2 48 2 50 The ordered pair is (4, 50).
c.
f (1) 3(1)2 2 31 2 3 2 5 The ordered pair is (1, 5).
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Chapter 14 Section 14.1 Practice Exercises
5.
1.
5x 2 y 3 y 3x 9 0 3(3) 5(3) 2(9) 0 3 99 15 18 0 3 3 3 (3, 9) is a solution of the system.
2.
2x y 8 x 3y 4 2(3) (2) 0 8 3 3(2) 0 4 6 2 08 3 6 0 4 88 3 4 (3, 2) is not a solution of the system. y
3.
The system has no solution because the lines are parallel. 6.
12 10 8 6 4 2 –12 –10 –8 –6 –4 –2–2
x–y=6
–4 –6 –8 –10 –12
2 4 6 8 10 12
x
(2, –4) –3x + y = –10
(2, 4) is a solution of the system. 4.
There are an infinite number of solutions because it is the same line. 7. 5x 4 y 6 4 y 5x 6 5 3 y x 4 2
x y 3 y x 3 y x 3
5 The slope of the first line is , while the slope 4 of the second line is 1. Since the slopes are not equal, the system has one solution. 8. (4, 1) is a solution of the system.
2
x y 6
3 y
2 3
x6
3y 2x 5 2 5 y x 3 3
2 Both lines have slope , but the y-intercepts are 3 different. The lines are parallel, so the system has no solution.
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569
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
4. A solution of a system of two equations in two variables is an ordered pair of numbers that is a solution of both equations in the system.
Calculator Explorations 1. y 2.68x 1.21 y 5.22x 1.68
5. A system of equations that has no solution is called an inconsistent system.
10 –10
10
6. In a system of linear equations in two variables, if the graphs of the equations are different, the equations are independent equations.
–10
(0.37, 0.23) is the approximate point of intersection. 2. y = 4.25x + 3.89 y = 1.88x + 3.21
8. Since the lines are parallel and do not intersect, there is no solution.
10 –10
9. Since the lines are the same, there is an infinite number of solutions.
10 –10
(0.11, 3.42) is the approximate point of intersection. 3. 4.3x 2.9 y 5.6 y (5.6 4.3x) / 2.9 8.1x 7.6 y 14.1 y (14.1 8.1x) / 7.6 10 –10
10. The lines intersect at (3, 4); therefore there is one solution. 11. The ordered pair must satisfy all equations of the system in order to be a solution of the system, so we must check that the ordered pair is a solution of both equations. 12. Graphing is not the most accurate method, especially if your graph is off just slightly, or the point of intersection does not have integer coordinates.
10 –10
(0.03, 1.89) is the approximate point of intersection. 4. 3.6x 8.6 y 10 y (10 3.6x) / 8.6 4.5x 9.6 y 7.7 y (7.7 4.5x) / 9.6 10 –10
7. The lines intersect at (1, 3); therefore there is one solution.
13. Writing the equations of a system in slopeintercept form lets us see and compare their slopes and y-intercepts. Different slopes mean one solution; same slopes with different y-intercepts mean no solution; same slopes with same y-intercepts mean an infinite number of solutions.
10
Exercise Set 14.1 –10
(0.41, 0.99) is the approximate point of intersection.
2. a.
Vocabulary, Readiness & Video Check 14.1 1. In a system of linear equations in two variables, if the graphs of the equations are the same, the equations are dependent equations. 2. Two or more linear equations are called a system of linear equations. 3. A system of equations that has at least one solution is called a consistent system. 570
b.
First equation: 2x y 5 2(5) 0 0 5 10 0 0 5 10 5 False (5, 0) is not a solution of the first equation, so it is not a solution of the system. First equation: 2x y 5 2(2) 1 5 4 1 0 5 5 5 True
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ISM: Prealgebra and Introductory Algebra
Second equation: x 3y 5 2 3(1) 0 5 2305 5 5 True Since (2, 1) is a solution of both equations, it is a solution of the system. 4. a.
First equation: 2x 3y 8 2(2) 3(4) 0 8 4 12 0 8 8 8 True Second equation: x 2y 6 2 2(4) 0 6 2 8 0 6 6 6 True Since (2, 4) is a solution of both equations, it is a solution of the system.
b. First equation: 2x 3y 8 2(7) 3(2) 0 8 14 60 8 8 8 True Second equation: x 2y 6 7 2(2) 0 6 7406 3 6 False (7, 2) is not a solution of the second equation, so it is not a solution of the system. 6. a.
First equation: x 5 y 4 4 5(0) 0 4 4 0 0 4 4 4 True Second equation: 2x 10 y 8 2(4) 0 10(0) 8 80 08 8 8 True Since (4, 0) is a solution of both equations, it is a solution of the system.
Chapter 14: Systems of Equations
b. First equation: x 5 y 4 6 5(2) 0 4 6 (10) 0 4 4 4 True Second equation: 2x 10 y 8 2(6) 010(2) 8 12 0 20 8 12 12 True Since (6, 2) is a solution of both equations, it is a solution of the system. 8. a.
First equation: 4x 1 y 4(0) 0 11 0 0 True Second equation: x 3y 8 0 3(1) 0 8 030 8 3 8 False (0, 1) is not a solution of the second equation, so it is not a solution of the system.
b. First equation: 4x 1 y 1 1 4 0 1 6 3 2 2 True 3 3 Second equation: 1 x 13y 8 3 0 8 6 3 1 10 8 6 5 8 False 6 1 1 , is not a solution of the second 6 3 equation, so it is not a solution of the system.
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
10.
14.
First equation: x y 3 4 (1) 0 3 4 1 0 3 3 3 True Second equation: x y 5 4 (1) 0 5 4 1 0 5 5 5 True The solution of the system is (4, 1).
First equation: y 3x 3 0 3(1) 3 3 True Second equation: 2x y 5 2(1) 3 0 5 2 3 0 5 5 5 True The solution of the system is (1, 3). 16.
12.
First equation: y 3x 4 5 0 3(3) 4 5094 5 5 True Second equation: y x2 5 0 3 2 5 5 True The solution of the system is (3, 5).
First equation: x y 1 2 (1) 01 2 1 0 1 1 1 True Second equation: x y 3 2 (1) 0 3 3 3 True The solution of the system is (2, 1). 18.
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ISM: Prealgebra and Introductory Algebra
First equation: 2x y 1 2(1) 3 01 2 3 0 1 1 1 True Second equation: 3x y 0 3(1) 3 0 0 3 3 0 0 0 0 True The solution of the system is (1, 3).
Chapter 14: Systems of Equations
24.
First equation: x y 5 4 1 0 5 5 5 True Second equation: x4 4 4 True The solution of the system is (4, 1).
20.
26.
First equation: y x 1 2 0 1 1 2 2 True Second equation: y 3x 5 2 0 3(1) 5 2 0 3 5 2 2 True The solution of the system is (1, 2).
First equation: x 3y 7 1 3(2) 0 7 1 6 0 7 7 7 True Second equation: 2x 3y 4 2(1) 3(2) 0 4 26 0 4 4 4 The solution of the system is (1, 2).
22.
28. First equation: x y 4 x4 y Second equation: x y 1 x 1 y The lines have the same slope, but different y-intercepts, so they are parallel. The system has no solution. Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
First equation: y 2x 3 y 2x 3 Second equation: 4x 2 2 y 2 y 4x 2 y 2x 1 The lines have the same slope, but different y-intercepts, so they are parallel. The system has no solution.
34.
First equation: y x5 2 0 35 2 2 True Second equation: y 2x 4 2 0 2(3) 4 20 64 2 2 True The solution of the system is (3, 2).
30.
First equation: x 2 y 6 2 y x 6 x y 3 2 Second equation: 2x 4 y 12 4 y 2x 12 x y 3 2 The graphs of the equations are the same line, so the system has an infinite number of solutions. 32.
36.
First equation: 4x y 7 4(3) 5 0 7 12 5 0 7 7 7 True Second equation: 2x 3y 9 2(3) 3(5) 0 9 6 15 0 9 9 9 True The solution of the system is (3, 5).
First equation: x 5 5 5 True Second equation: y 3 3 3 True The solution of the system is (5, 3).
574
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
46. y = 3 horizontal line
38.
First equation: 3x y 6 y 3x 6 y 3x 6 Second equation: 1 y 2 x 3 1 y x2 3 y 3x 6 The graphs of the equations are the same line, so the system has an infinite number of solutions. 40. 3x y 1 y 3x 1
3x 2 y 6 2 y 3x 6 3 y x3 2
x = 4 vertical line
a.
The slopes of the lines are different, so the lines intersect.
b.
The system has one solution.
48. 2 y x 2 1 y x 1 2
y 2x 3 y 2x 3
a.
The slopes of the lines are different, so the lines intersect.
b.
The system has one solution.
50. 8 y 6x 4 8 y 6x 4 3 1 y x 4 2
4 y 2 3x 4 y 3x 2 3 1 y x 4 2
a.
The slopes of the lines are different, so the lines intersect.
b.
The system has one solution.
52. 2x y 0 y 2x
y = 2x + 1
a.
The slopes of the lines are different, so the lines intersect.
a.
The lines have the same slope, but different y-intercepts, so the lines are parallel.
b.
The system has one solution.
b.
The system has no solution.
42. 3x y 0 y 3x
2 y 6x y 3x
a.
The lines are identical.
b.
The system has an infinite number of solutions.
44. 3x y 2 3x 2 y
1 3
y 2 3x y 9x 6
a.
The slopes of the lines are different, so the lines intersect.
b.
The system has one solution.
54. 2x 3(x 6) 17 2x 3x 18 17 x 1 y 1 3 4 y 3( y 1) 3 y 3y 3 3 2y 6 y3
56. y 12
58. 3z (4z 2) 9 3z 4z 2 9 z 7 z 7
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
60. Answers may vary. Possible answer:
62. Answers may vary. Possible answer:
64. The per person availability of corn sweeteners was greater than the per person availability of refined sugar for the years 19862009. 66. The per person availability of chicken was closer to the per person availability of beef in 2005 than in 2021. 68. answers may vary 70. answers may vary 72. a.
(1, 1) appears in both tables, so it is a solution of the system.
b.
Section 14.2 Practice Exercises 1. 2x 3y 13 x y 4 Substitute y + 4 for x in the first equation and solve for y. 2( y 4) 3y 13 2 y 8 3y 13 5y 5 y 1 Solve for x. x=y+4=1+4=5 The solution of the system is (5, 1). 2. 4x y 2 y 5x Substitute 5x for y in the first equation and solve for x. 4x 5x 2 x 2 x 2 Solve for y. y = 5x = 5(2) = 10 The solution of the system is (2, 10). 3. 3x y 5 3x 2 y 7 Solve the first equation for y. 3x y 5 y 3x 5 Substitute 3x + 5 for y in the second equation. 3x 2 y 7 3x 2(3x 5) 7 3x 6x 10 7 9x 3 3 1 x 9 3 Solve for y. 1 y 3x 5 3 5 1 5 4 3 1 The solution of the system is , 4 . 3
c.
Yes; the two lines intersect at (1, 1).
4. 5x 2 y 6 3x y 3 Solve the second equation for y. 3x y 3 y 3x 3 Substitute 3x 3 for y in the first equation.
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ISM: Prealgebra and Introductory Algebra 5x 2 y 6 5x 2(3x 3) 6 5x 6x 6 6 x 0 x0 Solve for y. y = 3x 3 = 3(0) 3 = 0 3 = 3 The solution of the system is (0, 3). 5. x 3y 6 1 y x2 3 1 Substitute x 2 for y in the first equation. 3 x 3y 6 1 x 3 x 2 6 3 x x 6 6 0x 0 00 The statement 0 = 0 indicates that this system has an infinite number of solutions. It is a graph of the same line. 6. 2x 3y 6 4x 6 y 12 2x 3y 6 3y 2x 6 2 y x2 3 2 Substitute 3 x 2 for y in the second equation. 4x 6 y 12 2 4x 6 x 2 12 3 4x 4x 12 12 12 12 The statement 12 = 12 indicates that this system has no solution. It is a graph of parallel lines. Vocabulary, Readiness & Video Check 14.2 1. x = 1: y = 4(1) = 4 The solution is (1, 4). 2. 0 = 34 is a false statement. The system has no solution. 3. 0 = 0 is a true statement. The system has an infinite number of solutions.
Chapter 14: Systems of Equations
4. y = 0: x = 0 + 5 = 5 The solution is (5, 0). 5. x = 0: 0 y 0 y0 The solution is (0, 0). 6. 0 = 0 is a true statement. The system has an infinite number of solutions. 7. We solved one equation for a variable. Next, be sure to substitute this expression for the variable into the other equation. Exercise Set 14.2 2. x y 20 x 3y Substitute 3y for x in the first equation and solve for y. x y 20 3y y 20 4 y 20 y5 Now solve for x. x = 3y = 3(5) = 15 The solution of the system is (15, 5). 4. x y 6 y 4x Substitute 4x for y in the first equation and solve for x. x y 6 x (4x) 6 3x 6 x 2 Now solve for y. y = 4x = 4(2) = 8 The solution of the system is (2, 8). 6. y 2x 3 5 y 7x 18 Substitute 2x + 3 for y in the second equation and solve for x. 5 y 7x 18 5(2x 3) 7x 18 10x 15 7x 18 3x 3 x1 Now solve for y. y = 2x + 3 = 2(1) + 3 = 5 The solution of the system is (1, 5).
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 8. y 5x 3 y 8x 4 Substitute 5x 3 for y in the second equation and solve for x. y 8x 4 5x 3 8x 4 3x 7 7 x 3 Now solve fory. 7 35 9 44 y 5x 3 5 3 3 3 3 3 44 7 The solution of the system is , . 3 3
10. 4x 3y 10 y x 5 Substitute x 5 for y in the first equation and solve for x. 4x 3y 10 4x 3(x 5) 10 4x 3x 15 10 x 5 Now solve for y. y = x 5 = 5 5 = 10 The solution of the system is (5, 10). 12. x 3y 5 2x 2 y 6 Solve the first equation for x. x 3y 5 x 3y 5 Substitute 3y 5 for x in the second equation and solve for y. 2x 2 y 6 2(3y 5) 2 y 6 6 y 10 2 y 6 4 y 16 y 4 Now solve for x. x = 3y 5 = 3(4) 5 = 12 5 = 7 The solution of the system is (7, 4).
578
14. 2x 3y 18 x 2 y 5 Substitute 2y 5 for x in the first equation and solve for y. 2x 3y 18 2(2 y 5) 3y 18 4 y 10 3y 18 7 y 28 y4 Now solve for x. x = 2y 5 = 2(4) 5system = 8 is 5 =(3,3 4). The solution of the
16. 3y x 6 4x 12 y 0 Solve the first equation for x. 3y x 6 3y 6 x Substitute 3y 6 for x in the second equation and solve for y. 4x 12 y 0 4(3y 6) 12 y 0 12 y 24 12 y 0 24 y 24 y 1 Now solve for x. x = 3y 6 = 3(1) 6 = 3 6 = 3 The solution of the system is (3, 1). 18. 2 y x 2 6x 12 y 0 Solve the first equation for x. 2y x 2 2y 2 x Substitute 2y 2 for x in the second equation and solve for x. 6x 12 y 0 6(2 y 2) 12 y 0 12 y 12 12 y 0 12 0 False Since the statement 12 = 0 is false, the system has no solution. 20. 3x y 14 4x 3y 22 Solve the first equation for y. 3x y 14 y 3x 14 Substitute 3x 14 for y in the second equation and solve for x.
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ISM: Prealgebra and Introductory Algebra 4x 3y 22 4x 3(3x 14) 22 4x 9x 42 22 5x 20 x 4 y = 3x 14 = 3(4) 14 = 12 14 = 2 The solution of the system is (4, 2). 22. 5x 4 y 2 6 7 y 3x 3x 4x y 3 Simplify each equation. 8x 4 3y 7x y 3 Solve the second simplified equation for y. 7x y 3 7x 3 y Substitute 7x 3 for y in the first simplified equation and solve for x. 8x 4 3y 8x 4 3(7x 3) 8x 4 21x 9 13 13x 1 x Now solve for y. y = 7x 3 = 7(1) 3 = 7 3 = 4 The solution of the system is (1, 4). 24. 10x 5 y 21 x 3y 0 Solve the second equation for x. x 3y 0 x 3y Substitute 3y for x in the first equation and solve for y. 10x 5 y 21 10(3y) 5 y 21 30 y 5 y 21 35 y 21 21 3 y 35 5 Now solve for x. 3 9 x 3y 3 5 5 9 3 The solution of the system is , . 5 5
Chapter 14: Systems of Equations 26. 2x y 7 4x 3y 11 Solve the first equation for y. 2x y 7 2x 7 y Substitute 2x + 7 for y in the second equation and solve for x. 4x 3y 11 4x 3(2x 7) 11 4x 6x 21 11 2x 10 x 5 Now solve for y. y = 2x + 7 = 2(5) + 7 = 10 + 7 = 3 The solution of the system is (5, 3). 28. x 3y 18 3x 2 y 19 Solve the first equation for x. x 3y 18 3y 18 x Substitute 3y 18 for x in the second equation and solve for y. 3x 2 y 19 3(3y 18) 2 y 19 9 y 54 2 y 19 7 y 54 19 35 7 y 5 y Now solve for x. x = 3y 18 = 3(5) 18 = 15 18 = 3 The solution of the system is (3, 5). 30. 6x 3y 12 9x 6 y 15 Solve the first equation for y. 6x 3y 12 3y 6x 12 y 2x 4 Substitute 2x + 4 for y in the second equation and solve for x. 9x 6 y 15 9x 6(2x 4) 15 9x 12x 24 15 3x 9 x3 Now solve for y. y = 2x + 4 = 2(3) + 4 = 6 + 4 = 2 The solution of the system is (3, 2).
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 32. 2x 4 y 6 5x 10 y 16 Solve the first equation for x. 2x 4 y 6 2x 4 y 6 x 2 y 3 Substitute 2y + 3 for x in the second equation and solve for y. 5x 10 y 16 5(2 y 3) 10 y 16 10 y 15 10 y 16 0 1 Since the statement 0 = 1 is false, the system has no solution.
1 34. x 2 y 1 4 x 8 y 4 Solve the second equation for x. x 8y 4 x 8y 4 Substitute 8y + 4 for x in the first equation and solve for y. 1 x 2y 1 4 1 (8 y 4) 2 y 1 4 2 y 1 2 y 1 00 Since 0 = 0 is a true statement, the two equations in the original system are equivalent. The system has an infinite number of solutions. 5 36. x y 2 6 12x 5 y 9 5 Substitute y 2 for x in the second equation 6 and solve for y. 12x 5 y 9 5 12 y 2 5 y 9 6 10 y 24 5 y 9 5 y 15 y3 Now solve for x. 5 5 5 1 x y 2 (3) 2 2 6 6 2 2 1 The solution of the system is , 3. 2 580
38.
x y 10 5(x y) 5(10) 5x 5 y 50
40.
5a 7b 4 4(5a 7b) 4(4) 20a 28b 16
42. 2x 5y 2x 11y 16 y 44.
9q p 9q p 0
46. 5x 2 y 4x 2 y 2(2 y 6) 7 3(2x y) 4x 1 9 Simplify each equation. x 4y 5 2x 3y 10 Substitute 4y + 5 for x in the second simplified equation and solve for y. 2x 3y 10 2(4 y 5) 3y 10 8 y 10 3y 10 5y 0 y0 Now solve for x. x = 4y + 5 = 4(0) + 5 = 0 + 5 = 5 The solution of the system is (5, 0). 48. answers may vary 50. The solution of the system is (0, 0). 52. a.
There is 1 solution for a system with lines intersecting in one point.
b.
There are 0 solutions for a system with parallel lines.
c.
There are an infinite number of solutions for a system with the same line.
54. Using a graphing calculator, the solution of the system is (3.5, 5.5). 56. Using a graphing calculator, the solution of the system is (8.1, 7.1).
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ISM: Prealgebra and Introductory Algebra
58. a.
Chapter 14: Systems of Equations
y 849x 60, 028 y 2255x 34, 720 Substitute 2255x + 34,720 for y in the first equation and solve for x. y 849x 60, 028 2255x 34, 720 849x 60, 028 2255x 849x 25, 308 1406x 25, 308 x 18 Now solve for y. y 849x 60, 028 849(18) 60, 028 15, 282 60, 028 75, 310 The solution is (18, 75,310).
b. In 1990 + 18 = 2008, the number of men and women receiving doctoral degrees was the same, about 75,310. c.
answers may vary Section 14.3 Practice Exercises 1. x y 13 x y 5 Add the equations to eliminate y; then solve for x. x y 13 x y 5 2x 18 x 9 Now solve for y. x y 13 9 y 13 y4 The solution of the system is (9, 4).
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 2. 2x y 6 x 4 y 17 Multiply the first equation by 4. 4(2x y) 4(6) 8x 4 y 24 x 4 y 17 x 4 y 17 Add the equations to eliminate y; then solve for x. 8x 4 y 24 x 4 y 17 7x 7 x 1 Now solve for y. 2x y 6 2(1) y 6 2 y 6 y 4 y4 The solution of the system is (1, 4). 3. x 3y 2 3x 9 y 5 Multiply the first equation by 3. 3(x 3y) 3(2) 3x 9 y 6 3x 9 y 5 3x 9 y 5 Add the equations to eliminate y; then solve for x. 3x 9 y 6 3x 9 y 5 0 1 Since the statement 0 = 1 is false, there is no solution to the system. 4. 2x 5 y 1 4x 10 y 2 Multiply the first equation by 2. 2(2x 5 y) 2(1) 4x 10 y 2 4x 10 y 2 4x 10 y 2 Add the equations to eliminate y; then solve for x. 4x 10 y 2 4x 10 y 2 00 Since the statement 0 = 0 is true, there are an infinite number of solutions. 5. 4x 5 y 14 3x 2 y 1 Multiply the first equation by 2 and multiply the second equation by 5.
582
5 y) 2(14) 8x 10 y 28 2(4x 5(3x 2 y) 5(1) 15x 10 y 5 Add the equations to eliminate y; then solve for x. 8x 10 y 28 15x 10 y 5 23 x 1
23x
Now solve for y. 4x 5 y 14 4(1) 5 y 14 5 y 10 y2 The solution of the system is (1, 2). 4 x 6. y 3 3 x 5 1 y 2 2 2 Multiply the first equation by 3 and the second equation by 2. x 4 3 y 3 x 3y 4 3 3 x 5 1 x 5 y 1 2 y 2 2 2 2 Add the equations to eliminate x; then solve for y. x 3y 4 x 5 y 1 2 y 3 3 y 2 Now solve for y through elimination. Multiply the first equation by 15 and the second equation by 6. x 4 5x 15 y 20 15 y 15 3 3 3x 15 y 3 5 1 6 x y 6 2 2 2 Add the equations to eliminate y; then solve for x. 5x 15 y 20 3x 15 y 3 17 2x 17 x 2 3 17 The solution to the system is , . 2 2
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ISM: Prealgebra and Introductory Algebra Vocabulary, Readiness & Video Check 14.3 1. 3x 2 y 9 x 5 y 14 Multiply the second equation by 3, then add the resulting equations. 3x 2 y 9 3x 15 y 42 17 y 51 The y’s are not eliminated; the statement is false. 2. 3x 2 y 9 5 y 14 x Multiply the second equation by 3, then add the resulting equations. 3x 2 y 9 3x 15 y 42 17 y 51 The statement is true. 3. 3x 2 y 9 x 5 y 14 Multiply the first equation by 5 and the second equation by 2, then add the two new equations. 15x 10 y 45 2x 10 y 28 17x 17 The statement is true. 4. 3x 2 y 9 x 5 y 14 Multiply the first equation by 5 and the second equation by 2, then add the two new equations. 15x 10 y 45 2x 10 y 28 13x 20 y 73 The y’s are not eliminated; the statement is false. 5. The multiplication property of equality; be sure to multiply both sides of the equation by the nonzero number chosen. Exercise Set 14.3 2. 4x y 13 2x y 5 Add the equations to eliminate y; then solve for x.
Chapter 14: Systems of Equations 4x y 13 2x y 5 6x 18 x3 Now solve for y. 4x y 13 4(3) y 13 12 y 13 y 1 The solution of the system is (3, 1). 4. x 2 y 11 x 5 y 23 Add the equations to eliminate x; then solve for y. x 2 y 11 x 5 y 23 3y 12 y4 Now solve for x. x 2(4) 11 x 8 11 x 3 The solution of the system is (3, 4). 6. 4x y 13 6x 3y 15 Multiply the first equation by 3. 3(4x y) 3(13) 12x 3y 39 6x 3y 15 6x 3y 15 Add the equations to eliminate y; then solve for x. 12x 3y 39 6x 3y 15 54 18x x 3 Now solve for y. 4x y 13 4(3) y 13 12 y 13 y 1 The solution of the system is (3, 1). 8. 4x 2 y 2 3x 2 y 12 Add the equations to eliminate y; then solve for x. 4x 2 y 2 3x 2 y 12 7x 14 x2
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
16. 2x y 6 4x 2 y 12 Multiply the first equation by 2. y) 2(6) 4x 2 y 12 2(2x 4x 2 y 12 4x 2 y 12
Now solve for y. 4x 2 y 2 4(2) 2 y 2 8 22 yy 26 y 3 The solution of the system is (2, 3). 10. x 4 y 14 5x 3y 2 Multiply the first equation by 5. 5(x 4 y) 5(14) 5x 20 y 70 5x 3y 2 5x 3y 2 Add the equations to eliminate x; then solve for y. 5x 20 y 70 5x 3y 2 17 y 68 y4 Now solve for x. x 4 y 14 x 4(4) 14 x 16 14 x 2 The solution of the system is (2, 4). 12. x y 1 x 2 y 0 Add the equations to eliminate x; then solve for y. x x y 1 2y 0 y 1 Now solve for x. xx1y 11 x2 The solution of the system is (2, 1). 14. 3x y 4 9x 3y 6 Multiply the first equation by 3. 3(3x y) 3(4) 9x 3y 12 9x 3y 6 9x 3y 6 Add the equations to eliminate y. 9x 3y 12 9x 3y 6 0 6 Since the statement 0 = 6 is false, the system has no solution. 584
Add the equations to eliminate y. 4x 2 y 12 4x 2 y 12 00 Since the statement 0 = 0 is true, the system has an infinite number of solutions. 18. 6x 5 y 25 4x 15 y 13 Multiply the first equation by 3. 5 y) 3(25) 18x 15 y 75 3(6x 4x 15 y 13 4x 15 y 13 Add the equations to eliminate y; then solve for x. 18x 15 y 75 4x 15 y 13 88 22x x4 Now solve for y. 6x 5 y 25 6(4) 5 y 25 24 5 y 25 5 y 1 y
1 5
1
The solution of the system is 4, . 5 10x 3y 12 10x 3y 12 20. 5x 4 y 16 5x 4 y 16 Multiply the second equation by 2. 10x 3y 12 10x 3y 12 2(5x 4 y) 2(16) 10x 8 y 32 Add the equations to eliminate x; then solve for y. 10x 3y 12 10x 8 y 32 5 y 20 y 4 Now solve for x.
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ISM: Prealgebra and Introductory Algebra 10x 3y 12 10x 3(4) 12 10x 12 12 10x 0 x0 The solution of the system is (0, 4). 3y 10 22. 2x 3x 4 y 2 Multiply the first equation by 3 and the second equation by 2. 3y) 3(10) 6x 9 y 30 3(2x 6x 8 y 4 2(3x 4 y) 2(2) Add the equation to eliminate x; then solve for y. 6x 9 y 30 6x 8 y 4 17 y 34 y2 Now solve for x. 2x 3y 10 2x 3(2) 10 2x 6 10 2x 4 x 2 The solution of the system is (2, 2). 24. 9x 3y 12 12x 4 y 18
Chapter 14: Systems of Equations 12x 10 y 14 12x 18 y 21 8 y 7 7 y 8 Multiply the first original equation by 6 and the second original equation by 5. 5 y) 6(7) 30 y 42 6(6x 36x 5(4x 6 y) 5(7) 20x 30 y 35 Add the equations to eliminate y; then solve for x. 36x 30 y 42 20x 30 y 35 16x 7 7 x 16 7 7 The solution of the system is , . 8 16 x y 28. 2 8 3 y x 0 4 Multiply the first equation by 8 and the second by 4. x y 8(3) 4x y 24 8 2 8 y 4x y 0 4 x 4(0) 4 Add the equations to eliminate y; then solve for x.
Multiplying the first equation by 4 and the second equation by 3. 3y) 4(12) 36x 12 y 48 4(9x 3(12x 4 y) 3(18) 36x 12 y 54 Add the equations to eliminate y. 36x 12 y 48 36x 12 y 54 06 Since the statement 0 = 6 is false, the system has no solution. 26. 6x 5 y 7 4x 6 y 7 Multiply the first equation by 2 and the second equation by 3. 2(6x 5 y) 2(7) 12x 10 y 14 12x 18 y 21 3(4x 6 y) 3(7) Add the equations to eliminate x; then solve for y.
4x y 24 4x y 0 8x 24 x3 Now solve for y. y x 0 4 y 3 0 4 y 3 4 12 y The solution of the system is (3, 12). 10 x 4 y 4 30. 3 5x 6 y 6 Multiply the first equation by 3 and the second equation by 2.
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585
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 10 10x 12 y 12 3 x 4 y 3( 4) 3 2(5x 6 y) 2(6) 10x 12 y 12 Add the equations to eliminate x. 10x 12 y 12 10x 12 y 12 00
Add the equations to eliminate x; then solve for y. 9x 8 y 27 9x 9 y 24 17 y 51 y 3 Now solve for x.
586
3x 9 8 3x 1 1 x 3 1 The solution of the system is , 3 . 3 x y 1 36. 2 4 x y 1 4 8 Multiply the first equation by 4 and the second by 8. x y 4 4(1) 2x y 4 8 x2 y4 8(1) 2x y 8 4 8 Add the equations to eliminate y. 2x y 4 2x y 8 0 12 Since the statement 0 = 12 is false, the system has no solutions.
Since the statement 0 = 0 is true, the system has an infinite number of solutions. 3y 32. 2x 3 4 x y 13 9 3 Multiply the first equation by 4 and the second equation by 27. 3y 8x 3y 12 4 2x 4( 3) 27x 4 3y 117 y 13 27 x 27 9 3 Add the equations to eliminate y; then solve for x. 8x 3y 12 27x 3y 117 35x 105 x3 Now solve for y. y 13 x 9 3 y 13 3 9 3 9 3 y 9 13 9 3 27 y 39 y 12 The solution of the system is (3, 12). 34. 9(x 3) 8 y 3x 3y 8 Multiply the second equation by 3. 9x 8 y 27 9(x 3) 8 y 9x 9 y 24 3(3x 3y) 3(8)
3x 3y 8 3x 3(3) 8
7 3 3x y 38. x 25 45 y 2 3 4 Multiply the first equation by 4 and the second equation by 24.
7 3 4 3x y 4 x 52 4 24 y 24 5 4 2 3 12x 14 y 3 12x 40 y 30 Add the equations to eliminate x; then solve for y. 12x 14 y 3
12x 40 y 30 54 y 27 27 1 y 54 2 Now solve for x.
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ISM: Prealgebra and Introductory Algebra 3x
7
y
3
7 1 3 3x 2 22 44 7 3 3x 7 4 4 3 4 3x 4 4 4 12x 7 3 12x 10 10 5 x 12 6 The solution of the system is
Chapter 14: Systems of Equations 40x 50 y 105 2(20x 60 y) 2(105) 40x 50 y 105 40x 120 y 210 Add the equations to eliminate x; then solve for y. 40x 50 y 105 40x 120 y 210
1 5 , . 6 2
40. 2.5x 6.5 y 47 0.5x 4.5 y 37 Multiply the first equation by 10 and the second equation by 50. 10(2.5x 6.5 y) 10(47) 50(0.5x 4.5 y) 50(37) 25x 65 y 470 25x 225 y 1850 Add the equations to eliminate x; then solve for y. 25x 65 y 470 25x 225 y 1850 290 y 2320 y 8 Now solve for x. 0.5x 4.5 y 37 0.5x 4.5(8) 37 0.5x 36 37 0.5x 1 x2 The solution of the system is (2, 8). 42. 0.04x 0.05 y 0.105 0.2x 0.6 y 1.05 Multiply the first equation by 1000 and the second equation by 100 to eliminate decimals. 1000(0.04x 0.05y) 1000(0.105) 100(0.2x 0.6 y) 100(1.05) 40x 50 y 105 20x 60 y 105 Multiply the second equation by 2.
70 y 105 y 1.5 Now solve for x. 0.2x 0.6 y 1.05 0.2x 0.6(1.5) 1.05 0.2x 0.9 1.05 0.2x 0.15 x 0.75 The solution of the system is (0.75, 1.5). 44. The sum of three consecutive integers is 66 is written as x + (x + 1) + (x + 2) = 66. 46. Twice the sum of 8 and a number is the difference of the number and 20 is written as 2(8 + x) = x 20. 48. The quotient of twice a number and 7 is subtracted from the reciprocal of the number and 1 2x the result is 2 is written as 2. x 50. a.
7
To eliminate the variable x, multiply the first equation by 5 and the second equation by 3. 5(3x y) 5(8) 3(5x 3y) 3(2) 15x 5 y 40 15x 9 y 6 Add the equations to eliminate x; then solve for y. 15x 5 y 40 15x 9 y 6 14 y 46 y
46 23 14 7
b. To eliminate the variable y, multiply the first equation by 3. y) 3(8) 9x 3y 24 3(3x 5x 3y 2 5x 3y 2 Add the equations to eliminate y; then solve for x.
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Chapter 14: Systems of Equations 9x 3y 24 5x 3y 2 22 14x 22 11 x 14 7 11 23 The solution of the system is , . 7 7 52. a; answers may vary
c.
54. answers may vary 56. a.
When b = 2, the system has an infinite number of solutions.
b.
When b is any real number except 2, the system has a single solution. 5x 2 y 19.8 58. 3x 5 y 3.7 Multiply the first equation by 5 and the second equation by 2. 5(5x 2 y) 5(19.8) 2(3x 5 y) 2(3.7) 25x 10 y 99 6x 10 y 7.4 Add the equations to eliminate y; then solve for x. 25x 10 y 99 6x 10 y 7.4 106.4 19x x 5.6 Now solve for y. 5x 2 y 19.8 5(5.6) 2 y 19.8 28 2 y 19.8 2 y 8.2 y 4.1 The solution of the system is (5.6, 4.1). 60. a.
588
b.
0.42x y 13.71 0.15x y 15.77 Add the equations to eliminate y. 0.42x y 13.71 0.15x y 15.77 0.57x 2.06 x4 Let x = 4 in the first equation.
0.42(4) y 13.71 1.68 y 13.71 y 15.39 y 15 The solution is approximately (4, 15). In 2003 (1999 + 4), the percent of workers age 55 and older was the same as the percent of workers ages 16 to 24. In 2003, each of these age groups composed 15% of the work force.
Mid-Chapter Review 1. 2x 3y 11 y 4x 3 Substitute 4x 3 for y in the first equation and solve for x. 2x 3y 11 2x 3(4x 3) 11 2x 12x 9 11 10xx 220 Now solve for y. y = 4x 3 = 4(2) 3 = 8 3 = 5 The solution to the system is (2, 5). 2. 4x 5 y 6 y 3x 10 Substitute 3x 10 for y in the first equation and solve for x. 4x 5 y 6 4x 5(3x 10) 6 4x 15x 50 6 11x 44 x4 Now solve for y. y = 3x 10 = 3(4) 10 = 12 10 = 2 The solution of the system is (4, 2). 3. x y 3 x y 7 Add the equations to eliminate y; then solve for x. x y 3 xy7 2x 10 x5 Now solve for y.
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ISM: Prealgebra and Introductory Algebra x y 3 5 y 3 y 2 The solution of the system is (5, 2). 4. x y 20 x y 8 Add the equations to eliminate y; then solve for x. x y 20 x y 8 2x 12 x6 Now solve for y. x y 8 6 y 8 y 14 The solution of the system is (6, 14). 5. x 2 y 1 3x 4 y 1 Solve the first equation for x. x 2y 1 x 2 y 1 Substitute 2y + 1 for x in the second equation and solve for y. 3x 4 y 1 3(2 y 1) 4 y 1 6 y 3 4 y 1 2 y 4 y2 Now solve for x. x = 2y + 1 = 2(2) + 1 = 4 + 1 = 3 The solution of the system is (3, 2). 6. x 3y 5 5x 6 y 2 Solve the first equation for x. x 3y 5 x 3y 5 Substitute 3y + 5 for x in the second equation and solve for y. 5x 6 y 2 5(3y 5) 6 y 2 15 y 25 6 y 2 9 y 27 y3 Now solve for x. x = 3y + 5 = 3(3) + 5 = 9 + 5 = 4 The solution of the system is (4, 3).
Chapter 14: Systems of Equations 7. y x 3 3x 2 y 6 Substitute x + 3 for y in the second equation and solve for x. 3x 2 y 6 3x 2(x 3) 6 3x 2x 6 6 x0 Now solve for y. y=x+3=0+3=3 The solution of the system is (0, 3). 8. y 2x 2x 3y 16 Substitute 2x for y in the second equation and solve for x. 2x 3y 16 2x 3(2x) 16 2x 6x 16 8x 16 x 2 Now solve for y. y = 2x = 2(2) = 4 The solution of the system is (2, 4). 9. y 2x 3 y 5x 18 Substitute 2x 3 for y in the second equation and solve for x. y 5x 18 2x 3 5x 18 15 3x 5 x Now solve for y. y = 2x 3 = 2(5) 3 = 10 3 = 7 The solution of the system is (5, 7). 10. y 6x 5 y 4x 11 Substitute 6x 5 for y in the second equation and solve for x. y 4x 11 6x 5 4x 11 2x 6 x 3 Now solve for y. y = 6x 5 = 6(3) 5 = 18 5 = 23 The solution of the system is (3, 23).
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Chapter 14: Systems of Equations 1 1 11. x y 6 2 3x 2 y 3
12.
590
Multiply the first equation by 6 to eliminate fractions. 1 1 6 x y 6 2 6 6x y 3 Now solve for y. y = 6x + 3 Substitute 6x + 3 for y in the second equation and solve for x. 3x 2 y 3 3x 2(6x 3) 3 3x 12x 6 3 9x 3 3 1 x 9 3 Now solve for y. 1 y 6x 3 6 3 2 3 1 3 1 The solution of the system is , 1. 3 x
1
y
5
3 12 8x 3y 4 Multiply the first equation by 12 to eliminate fractions. 1 5 12 x y 12 12 3 12x 4 y 5 Multiply the revised first equation by 3 and the second original equation by 4. 3(12x 4 y) 3(5) 36x 12 y 15 32x 12 y 16 4(8x 3y) 4(4) Add the equations to eliminate y; then solve for x. 36x 12 y 15 32x 12 y 16 1 4x 1 x 4 Now solve for y.
8x 3y 4 1 8 3y 4 4 2 3y 4 3y 6 y2 1 The solution of the system is , 2 . 4
x 5y 1 13. 2x 10 y 3 Solve the first equation for x. x 5y 1 x 5y 1 Substitute 5y + 1 for x in the second equation and then solve for y. 2x 10 y 3 2(5 y 1) 10 y 3 10 y 2 10 y 3 2 3 Since the statement is false, the system has no solution. 14. x 2 y 3 3x 6 y 9 Solve the first equation for x. x 2 y 3 x 2y 3 Substitute 2y 3 for x in the second equation and solve for y. 3x 6 y 9 3(2 y 3) 6 y 9 6 y 9 6 y 9 00 Since the statement 0 = 0 is true, the system has an infinite number of solutions. 15. 0.2x 0.3y 0.95 0.4x 0.1y 0.55 Multiply both equations by 100 to eliminate decimals. 100(0.2x 0.3y) 100(0.95) 100(0.4x 0.1y) 100(0.55) 20x 30 y 95 40x 10 y 55 Multiply the second revised equation by 3. 20x 30 y 95 20x 30 y 95 3(40x 10 y) 3(55) 120x 30 y 165
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ISM: Prealgebra and Introductory Algebra Add the equations to eliminate y; then solve for x. 20x 30 y 95 120x 30 y 165 70 140x x 0.5 Now solve for y. 0.4x 0.1y 0.55 0.4(0.5) 0.1y 0.55 0.2 0.1y 0.55 0.1y 0.35 y 3.5 The solution of the system is (0.5, 3.5). 16. 0.08x 0.04 y 0.11 0.02x 0.06 y 0.09 Multiply both equations by 100 to eliminate decimals. 0.04 y) 100(0.11) 100(0.08x 100(0.02x 0.06 y 100(0.09) 8x 4 y 11 2x 6 y 9 Multiply the second revised equation by 4. 8x 4 y 11 8x 4 y 11 4(2x 6 y) 4(9) 8x 24 y 36 Add the equations to eliminate x; then solve for y. 8x 4 y 11 8x 24 y 36 20 y 25 y 1.25 Now solve for x. 0.08x 0.04 y 0.11 0.08x 0.04(1.25) 0.11 0.08x 0.05 0.11 0.08x 0.06 x 0.75 The solution of the system is (0.75, 1.25). 17. x 3y 7 2x 6 y 14 Substitute 3y 7 for x in the second equation and solve for y. 2x 6 y 14 2(3y 7) 6 y 14 6 y 14 6 y 14 00 Since the statement 0 = 0 is true, the system has an infinite number of solutions.
Chapter 14: Systems of Equations x 18. y 3 2 2x 4 y 0 x Substitute 3 for y in the second equation and 2 solve for x. 2x 4 y 0 x 2x 4 3 0 2 2x 2x 12 0 12 0 Since the statement 12 = 0 is false, the system has no solution. 19. 2x 5 y 1 3x 4 y 33 Multiply the first equation by 3 and the second equation by 2. 3(2x 5 y) 3(1) 6x 15 y 3 2(3x 4 y) 2(33) 6x 8 y 66
Add the equations to eliminate x; then solve for y. 6x 15 y 3 6x 8 y 66 23y 69 y 3 Now solve for x. 2x 5 y 1 2x 5(3) 1 2x 15 1 2x 14 x7 The solution of the system is (7, 3).
20. 7x 3y 2 6x 5 y 21 Multiply the first equation by 5 and the second equation by 3. 5(7x 3y) 5(2) 35x 15 y 10 3(6x 5 y) 3(21) 18x 15 y 63 Add the equations to eliminate y; then solve for x. 35x 15 y 10 18x 15 y 63 53 53x x 1 Now solve for y.
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 7x 3y 2 7(1) 3y 2 7 3y 2 3y 9 y 3 The solution of the system is (1, 3). 21. answers may vary 22. answers may vary Section 14.4 Practice Exercises 1. x y 50 x y 22 Add the equations to eliminate y; then solve for x. x y 50 x y 22 2x 72 x 36 Now solve for y. x y 50 36 y 50 y 14 The two numbers are 36 and 14. 2. A C 587 7 A 5C 3379 Multiply the first equation by 5. 5( A C) 5(587) 5A 5C 2935 7 A 5C 3379 7 A 5C 3379 Add the equations to eliminate C; then solve for A. 5A 5C 2935 7 A 5C 3379 2A 444 A 222 Now solve for C. A C 587 222 C 587 C 365 There were 222 adults and 365 children. r
t =
d
Faster car
x
6
6x
Slower car
y
6
6y
3.
6x 6 y 780 x y 5
592
Substitute y + 5 for x in the first equation and solve for y. 6x 6 y 780 6( y 5) 6 y 780 6 y 30 6 y 780 12 y 750 y 62.5 Now solve for x. x y 5 62.5 5 67.5 One car’s speed is 62.5 mph and the other car’s speed is 67.5 mph. 4. Let x be the liters of 20% solution. Let y be the liters of 70% solution. x y 50 0.2x 0.7 y 0.6(50) Multiply the first equation by 2 and the second equation by 10. 2(x y) 2(50) 10(0.2x 0.7 y) 10(30) 2x 2 y 100 2x 7 y 300 Add the equations to eliminate x; then solve for y. 2x 2 y 100 2x 7 y 300 5 y 200 y 40 Now solve for x. x y 50 x 40 50 x 10 10 liters of the 20% alcohol solution and 40 liters of the 70% alcohol solution make 50 liters of the 60% alcohol solution. Vocabulary, Readiness & Video Check 14.4 1. Up to now we’ve been working with one variable/unknown and one equation. Because systems involve two equations with two unknowns, for these applications we need to choose two variables to represent two unknowns and translate the problem into two equations. Exercise Set 14.4 2. In choice a, the perimeter is 6 + 6 + 7 = 19, not 20 feet. In choice c, no 2 sides have the same length. Choice b gives the solution, since 7 + 7 + 6 = 20.
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ISM: Prealgebra and Introductory Algebra
4. In choice a, the total cost is 2(12) + 4(4) = 24 + 16 = $40 but the second part is 3(12) + 5(4) = 36 + 20 = $56, not $55. In choice b, the total cost is 2(15) + 4(2) = 30 + 8 = $38, not $40. Choice c gives the solution since 2(10) + 4(5) = 20 + 20 = $40 and 3(10) + 5(5) = 30 + 25 = $55. 6. In choice a, the total number of gallons is 15 + 5 = 20, not 28. In choice b, 3 times 8 is 24, not 20. Choice c gives the solution, since 21 + 7 = 28 and 3 times 7 is 21. 8. Let x be the first number and y the second. x y 16 x 3y 2 18 10. Let x be the amount in his checking account and y be the amount in his savings account. x y 2300 y 4x 12. Let x be the first number and y be the second. x y 76 x y 52 Add the equations to eliminate y; then solve for x. xx 52 76 yy 2x
128 x 64
Now solve for y. x y 76 64 y 76 y 12 The numbers are 64 and 12. 14. Let x be the first number and y the second. xx y2 y254 Substitute 2y + 4 for x in the second equation and solve for y. x y 25 2 y 4 y 25 3y 21 y7 Now solve for x. x y 25 x 7 25 x 18 The numbers are 18 and 7.
Chapter 14: Systems of Equations
16. Let x be the number of points that Breanna Stewart scored and y be the number that Kelsey Plum scored. x y 15 x y 1467 Substitute y + 15 for x in the second equation and solve for y. x y 1467 y 15 y 1467 2 y 15 1467 2 y 1452 y 726 Now solve for x. x = y + 15 = 726 + 15 = 741 Breanna Stewart scored 741 points and Kelsey Plum scored 726 points. 18. Let x be the cost of a DVD and y be the cost of a CD. 5x 2 y 65 3x 4 y 81 Multiply the first equation by 2. 2(5x 2 y) 2(65) 10x 4 y 130 3x 4 y 81 3x 4 y 81 Add the equations to eliminate y and then solve for x. 10x 4 y 130 3x 4 y 81 7x x 49 7 Now solve for y. 5x 2 y 65 5(7) 2 y 65 35 2 y 65 2 y 30 y 15 DVDs cost $7 and CDs cost $15 each. 20. Let x be 47¢ stamps and y be 34¢ stamps. x y 40 0.47x 0.34 y 18.15 Solve the first equation for x. x y 40 x y 40 Substitute y + 40 for x in the second equation and solve for y. 0.47 x 0.34 y 18.15 0.47( y 40) 0.34 y 18.15 0.47 y 18.8 0.34 y 18.15 0.13 y 0.65 y5
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Chapter 14: Systems of Equations
Now solve for x. x = y + 40 = 5 + 40 = 35 They purchased 35 47¢ stamps and 5 34¢ stamps. 22. Let x be the number of shares of Meta Platform, Inc. stock and y be the number of shares of Alphabet, Inc. stock. y x 15 288.70x 120.76 y 10, 410.06 Substitute x + 15 for y in the second equation and solve for x. 288.70x 120.76 y 10, 410.06 288.7x 120.76(x 15) 10, 410.06 288.7x 120.76x 1811.4 10, 410.06 409.46x 8598.66 x 21 Now solve for y. y = x + 15 = 21 + 15 = 36 The couple owned 21 shares of Meta Platform, Inc. and 36 shares of Alphabet, Inc. 24. Let x be the daily fee and y be the mileage charge. 300 y 178 5x 4x 500 y 197 Multiply the first equation by 4 and the second by 5. 4(5x 300 y) 4(178) 5(4x 500 y) 5(197) 20x 1200 y 712 20x 2500 y 985 Add the equations to eliminate x; then solve for y. 20x 1200 y 712 20x 2500 y 985 1300 y 273 y 0.21 Now solve for x. 5x 300 y 178 5x 300(0.21) 178 5x 63 178 5x 115 x 23 There is a $23 daily fee and a $0.21 per mile mileage charge.
26. 10 5 (x y) 4 10 4(x y) Multiply the first equation by
4
equation by 4
. 4 4 5
(x y) 58 x y 5 51 4 x y 1 (10) [4(x y)] 2 4 4 Add the equations to eliminate y; then solve for x. 8 x y 2.5 x y 10.5 2x 5.25 x Now solve for y. 8 x y 8 5.25 y 2.75 y The rate of the current is 2.75 mph. (10)
28. Let x be the rate of the plane in still air; then let y be the rate of the wind. 19 2400 (x y) 4 2400 6(x y) 4 Multiply the first equation by and the second 19 1 equation by . 6 4 4 19 (2400) (x y) 19 19 505.26 1 1 4 400 xx yy (2400) [6(x y)]
6
6
Add the equations to eliminate y; then solve for x. 505.26 x y 400 x y 905.26 2x 452.6 x Now solve for y. 2400 6(x y) 2400 6(452.6) 6 y 2400 2715.6 6 y 315.6 6 y 52.6 y The speed of the plane in still air is 452.6 mph and the rate of the wind is 52.6 mph.
and the second
5 594
1
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ISM: Prealgebra and Introductory Algebra
30. Let x be the speed of the eastbound train, and let y be the speed of the westbound train. 1.25x 1.25 y 150 y 2x Substitute 2x for y in the first equation. 1.25x 1.25 y 150 1.25x 1.25(2x) 150 1.25x 2.5x 150 3.75x 150 x 40 Now solve for y. y = 2x = 2(40) = 80 The eastbound train is traveling at 40 mph and the westbound train is traveling at 80 mph. 32. Let x be liters of the 40% solution, and let y be liters of the 10% solution. x y 15 0.40x 0.10 y 0.25(15) Solve the first equation in terms of x. x y 15 x y 15 Substitute y + 15 for x in the second equation and solve for x. 0.40x 0.10 y 3.75 0.40( y 15) 0.10 y 3.75 0.40 y 6 0.10 y 3.75 0.30 y 2.25 y 7.5 Now solve for x. x = y + 15 = 7.5 + 15 = 7.5 The technician will need 7.5 liters of the 10% solution and 7.5 liters of the 40% solution to make 15 liters of the 25% solution. 34. Let x be the number of pounds of macadamias, and let y be the number of pounds of standard mix. x y 40 16.50x 9.25 y 10(40) Solve the first equation in terms of x. x y 40 x y 40 Substitute y + 40 for x in the second equation and solve for y. 16.50x 9.25 y 10(40) 16.50x 9.25 y 400 16.50( y 40) 9.25 y 400 16.50 y 660 9.25 y 400 7.25x 260 x 35.9
Chapter 14: Systems of Equations
Now solve for y. x = y + 40 = 35.9 + 40 = 4.1 Combining 4.1 pounds of macadamias and 35.9 pounds of standard mix will produce 40 pounds, which costs $10 per pound. 36. Let x be the measure of one angle, and let y be the measure of the other angle. x + y = 180 y = 4x + 20 Substitute 4x + 20 for y in the first equation and solve for x. x y 180 x 4x 20 180 5x 160 x 32 Now solve for y. y = 4x + 20 = 4(32) + 20 = 148 One angle is 32 and the other is 148. 38. Let x be the measure of one angle and y be the measure of the other. x y 90 x 10 3y Substitute 10 + 3y for x in the first equation and solve for y. x y 90 10 3y y 90 10 4 y 90 4 y 80 y 20 x = 10 + 3y = 10 + 3(20) = 10 + 60 = 70 The angles measure 20 and 70. 40. Let x be the number of adults, and let y be the number of children. x y 387 6.80x 3.40 y 2444.60 Solve the first equation in terms of x. x y 387 x 387 y Substitute 387 y for x in the second equation and solve for y. 6.80x 3.40 y 2444.60 6.80(387 y) 3.40 y 2444.60 2631.60 6.80 y 3.40 y 2444.60 3.4 y 187 y 55 Now solve for x. x = 387 y = 387 55 = 332 332 adults and 55 children attended the supper.
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
42. Let x be the rate of the faster car and y be the rate of the slower car. 3x + 3y = 123 x=y+1 Substitute y + 1 for x in the first equation and solve for y. 3x 3y 123 3( y 1) 3y 123 3y 3 3y 123 6 y 120 y 20 Now solve for x. x = y + 1 = 20 + 1 = 21 One car traveled at 21 mph and the other traveled at 20 mph. 44. Let x be the number of ounces of the 10% solution and y be the number of ounces of the 50% solution. x y 120 0.10x 0.50 y 0.20(120) Solve the first equation in terms of x. x y 120 x 120 y Substitute 120 y for x in the second equation and solve for y. 0.10x 0.50 y 0.20(120) 0.10x 0.50 y 24 0.10(120 y) 0.50 y 24 12 0.10 y 0.50 y 24 0.40 y 12 y 30 Now solve for x. x = 120 y = 120 30 = 90 Combining 90 ounces of 10% solution with 30 ounces of 50% solution will make 120 ounces of 20% solution. 46. Let x be the length and y the width. 2(x y) 60 x y6 Substitute y + 6 for x in the first equation and solve for y. 2(x y) 60 2( y 6 y) 60 2(2 y 6) 60 4 y 12 60 4 y 48 y 12 Now solve for x. x = y + 6 = 12 + 6 = 18 The length is 18 inches and the width is 12 inches. 596
48. 32 3 3 9 50. (11y)2 (11y)(11y) 121y2 52. (8x5 )2 (8x5 )(8x5 ) 64x10 54. The strength of the acid mixture must be between 50% and 100%, so choices c and d are the only possibilities. 56. 2x y 152 1 x y4 2 1 Substitute y 4 for x in the first equation and 2 solve for y. 2x y 152 1 2 y 4 y 152 2 y 8 y 152 2 y 144 y 72 Now solve for x. 1 1 x y 4 (72) 4 36 4 40 2 2 The width is 40 feet and the length is 72 feet.
Chapter 14 Vocabulary Check 1. In a system of linear equations in two variables, if the graphs of the equations are the same, the equations are dependent equations. 2. Two or more linear equations are called a system of linear equations. 3. A system of equations that has at least one solution is called a(n) consistent system. 4. A solution of a system of two equations in two variables is an ordered pair of numbers that is a solution of both equations in the system. 5. Two algebraic methods for solving systems of equations are addition and substitution. 6. A system of equations that has no solution is called a(n) inconsistent system. 7. In a system of linear equations in two variables, if the graphs of the equations are different, the equations are independent equations.
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ISM: Prealgebra and Introductory Algebra
Chapter 14 Review 1. a.
3. a.
2x 3y 12 3x 4 y 1 First equation: 2x 3y 12 2(12) 3(4) 012 24 12 0 12 12 12 True Second equation: 3x 4 y 1 3(12) 4(4) 01 36 16 0 1 20 1 False (12, 4) is not a solution of the system.
b. First equation: 2x 3y 12 2(3) 3(2) 012 6 6 0 12 12 12 True Second equation: 3x 4 y 1 3(3) 4(2) 01 98 01 1 1 True (3, 2) is a solution of the system. 2. a.
Chapter 14: Systems of Equations
2x 3y 1 3y x 4 First equation: 2x 3y 1 2(2) 3(2) 01 4 6 01 10 1 False (2, 2) is not a solution of the system.
b. First equation: 2x 3y 1 2(1) 3(1) 01 2 3 0 1 1 1 True Second equation: 3y x 4 3(1) (1) 0 4 3 1 0 4 4 4 True (1, 1) is a solution of the system.
5x 6 y 18 2 y x 4 First equation: 5x 6 y 18 5(6) 6(8) 018 30 48 0 18 18 18 True Second equation: 2 y x 4 2(8) (6) 0 4 16 6 0 4 10 4 False (6, 8) is not a solution of the system.
b. First equation: 5x 6 y 18 5 5(3) 6 018 2 15 15 0 18 0 18 False 5 3, is not a solution of the system. 2 4. a.
4x y 0 8x 5 y 9 First equation: 4x y 0 3 4 (3) 0 0 4 3 3 0 0 0 0 True Second equation: 8x 5 y 9 3 8 5(3) 0 9 4 6 15 0 9 9 9 True 3 , 3 is a solution of the system. 4
b. First equation: 4x y 0 4(2) 8 0 0 8 8 0 0 0 0 True
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Chapter 14: Systems of Equations
Second equation: 8x 5 y 9 8(2) 5(8) 0 9 16 40 0 9 24 9 False (2, 8) is not a solution of the system.
8. x 3 y 2 The solution of the system is (3, 2).
5. x y 5 x y 1 The solution of the system is (3, 2).
9. 2x y 5 x 3y The solution of the system is (3, 1).
6. x y 3 x y 1 The solution of the system is (1, 2).
10. 3x y 2 y 5x The solution of the system is (1, 5).
7. x 5 y 1 The solution of the system is (5, 1).
11. y 3x 6x 2 y 6 There are no solutions to the system because the lines are parallel.
598
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 15. x 3y 3 2x y 4 Solve the second equation in terms of y. 2x y 4 y 2x 4
12. x 2 y 2 2x 4 y 4 Since they are the same lines, there are an infinite number of solutions.
13. y 2x 6 3x 2 y 11 Substitute 2x + 6 for y in the second equation and solve for x. 3x 2 y 11 3x 2(2x 6) 11 3x 4x 12 11 x 1 x 1 Now solve for y. y = 2x + 6 = 2(1) + 6 = 2 + 6 = 4 The solution of the system is (1, 4). 14. y 3x 7 2x 3y 7 Substitute 3x 7 for y in the second equation and solve for x. 2x 3y 7 2x 3(3x 7) 7 2x 9x 21 7 7x 14 x2 Now solve for y. y = 3x 7 = 3(2) 7 = 6 7 = 1 The solution of the system is (2, 1).
Substitute 2x + 4 for y in the first equation and solve for x. x 3y 3 x 3(2x 4) 3 x 6x 12 3 5x 15 x3 Now solve for y. y = 2x + 4 = 2(3) + 4 = 6 + 4 = 2 The solution of the system is (3, 2). 16. 3x y 11 x 2 y 12 Solve the first equation in terms of y. 3x y 11 y 3x 11 Substitute 3x + 11 for y in the second equation and solve for x. x 2 y 12 x 2(3x 11) 12 x 6x 22 12 5x 10 x2 Now solve for y. y = 3x + 11 = 3(2) + 11 = 6 + 11 = 5 The solution of the system is (2, 5). 17. 4 y 2x 6 x 2 y 3 Solve the second equation in terms of x. x 2 y 3 x 2y 3 Substitute 2y 3 for x in the first equation and solve for y. 4 y 2x 6 4 y 2(2 y 3) 6 4y 4y 6 6 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 18. 9x 6 y 3 6x 4 y 2 Solve the first equation in terms of x.
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 9x 6 y 3 6 3 x y 9 9 2 1 x y 3 3 2 1 Substitute y for x in the second equation 3 3 and solve for y. 6x 4 y 2 1 2 6 y 4y 2 3 3 4 y 2 4 y 2 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 19. xy yx 6 4 Substitute x 4 for y in the first equation and solve for x. x y 6 x ( x 4) 6 0 10 False Since the statement 0 = 10 is false, there is no solution for the system. 20. 3x y 6 y 3x 2 Substitute 3x + 2 for y in the first equation and solve for x. 3x y 6 3x 3x 2 6 2 6 False Since the statement 2 = 6 is false, there is no solution for the system. 21. 2x 3y 6 x 3y 12 Add the equations to eliminate y and solve for x. 2x 3y 6 x 3y 12 3x 18 x 6 Now solve for y. x 3y 12 6 3y 12 3y 6 y2 The solution for the system is (6, 2).
600
22. 4x y 15 4x 3y 19 Add the equations to eliminate x and solve for y. 4x y 15 4x 3y 19 4 y 4 y 1 Now solve for x. 4x y 15 4x 1 15 4x 16 x4 The solution for the system is (4, 1). 23. 2x 3y 15 x 4 y 31 Multiply the second equation by2x 2. 3y 15 2x 3y 15 2(x 4 y) 2(31) 2x 8 y 62
Add the equations to eliminate x and solve for y. 2x 3y 15 2x 8 y 62 11y 77 y7 Now solve for x. x 4 y 31 x 4(7) 31 x 28 31 x3 The solution of the system is (3, 7).
24. x 5 y 22 4x 3y 4 Multiply the first equation by 4. 5 y) 4(22) 4x 20 y 88 4(x 4x 3y 4 4x 3y 4 Add the equations to eliminate x and solve for y. 4x 20 y 88 4x 3y 4 23y 92 y4 Now solve for x. x 5 y 22 x 5(4) 22 x 20 22 x 2 The solution of the system is (2, 4).
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ISM: Prealgebra and Introductory Algebra
25. 2x 6 y 1 x 3y 1 2 Multiply the second equation by 2. 2x 6 y 1 1 2x 6 y 1 2(x 3y) 2 2x 6 y 1 2 Add the equations to eliminate x. 2x 6 y 1 2x 6 y 1 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 26. 0.6x 0.3y 1.5 0.04x 0.02 y 0.1 Multiply the first equation by 10 and the second by 100 to eliminate decimals. 10(0.6x 0.3y) 10(1.5) 100(0.04x 0.02 y) 100(0.1) 6x 3y 15 4x 2 y 10 Multiply the first revised equation by 2 and the second revised equation by 3. 2(6x 3y) 2(15) 12x 6 y 30 3(4x 2 y) 3(10) 12x 6 y 30 Add the equations to eliminate y. 12x 6 y 30 12x 6 y 30 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 2 3 27.
x y 2 4 3y x 6 3 Multiply the first equation by 12 and the second by 3 to eliminate fractions. 3 2 9x 8 y 24 12 x y 12(2) 3 4 3x y 18 y 3 x 3(6) 3 Multiply the second revised equation by 3. 9x 8 y 24 9x 8 y 24 3(3x y) 3(18) 9x 3y 54 Add the equations to eliminate x and solve for y.
Chapter 14: Systems of Equations 9x 8 y 24 9x 3y 54 5 y 30 y 6 Now solve y for x. x 6 63 6 3 x (2) 6 x8 The solution of the system is (8, 6). x
2y 0 28. 10x 3x 5 y 33 Multiply the first equation by 5 and the second by 2. 2 y) 5(0) 50x 10 y 0 5(10x 2(3x 5 y) 2(33) 6x 10 y 66 Add the equations to eliminate y and solve for x. 50x 10 y 0 6x 10 y 66 66 44x 66 3 x 44 2 Now solve for y. 10x 2 y 0 3 10 2 y 0 2 15 2 y 0 2 y 15 15 y 2 3 15 The solution of the system is , . 2 2 29. Let x be the smaller number and y be the larger. x y 16 3y x 72 Solve the first equation in terms of y. x y 16 y x 16 Substitute x + 16 for y in the second equation and solve for x. 3y x 72 3(x 16) x 72 3x 48 x 72
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4x 24 x 6
601
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
Now solve for y. y = x + 16 = (6) + 16 = 6 + 16 = 22 The two numbers are 6 and 22. 30. Let x be the number of orchestra seats and y be the number of balcony seats. x y 360 45x 35 y 15,150 Solve the first equation in terms of x. x = y + 360 Substitute y + 360 for x in the second equation and solve for y. 45x 35 y 15,150 45( y 360) 35 y 15,150 45 y 16, 200 35 y 15,150 10 y 1050 y 105 Now solve for x. x y 360 x 105 360 x 255 There are 255 orchestra seats and 105 balcony seats. 31. Let x be the rate of the current and y be the speed in still water. 19(x y) 340 14(x y) 340 Multiply the first equation by
1 19
by
1
and the second
. 14 1 1 [19(x y)] (340) 19 x y 17.9 19 1 1 x y 24.3 [14(x y)] 14 14 (340) Add the equations to eliminate y and solve for x. x y 17.9 x y 24.3 2x 42.2 x 21.1 Now solve for x. x y 24.3 21.1 y 24.3 y 3.2 The speed in still water is 21.1 mph and the current of the river is 3.2 mph.
32. Let x = number of cc of 6% acid solution and y = number of cc of 14% acid solution. x y 50 0.06x 0.14 y 0.12(50) 0.06x 0.14 y 6 Solve the first equation in terms of x. x y 50 x y 50 Substitute y + 50 for x in the second equation and solve for y. 0.06x 0.14 y 6 0.06( y 50) 0.14 y 6 0.06 y 3 0.14 y 6 0.08 y 3 y 37.5 Now solve for x. x = y + 50 = 37.5 + 50 = 12.5 12.5 cc of the 6% solution and 37.5 cc of the 14% solution are needed to make 50 cc of the 12% solution. 33. Let x be the cost of an egg and y be the cost of a strip of bacon. 3x 4 y 3.80 2x 3y 2.75 Multiply the first equation by 2 and the second by 3. 2(3x 4 y) 2(3.80) 6x 8 y 7.60 3(2x 3y) 3(2.75) 6x 9 y 8.25 Add the equations to eliminate x and solve for y. 6x 8 y 7.60 6x 9 y 8.25 y 0.65 y 0.65 Now solve for x. 2x 3y 2.75 2x 3(0.65) 2.75 2x 1.95 2.75 2x 0.80 x 0.40 Each egg costs $0.40 and each strip of bacon costs $0.65. 34. Let x be time spent jogging, and let y be the time spent walking. x y 3 7.5x 4 y 15 Solve the first equation in terms of x. x y 3 x y 3 Substitute y + 3 for x in the second equation and solve for y.
602
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 7.5x 4 y 15 7.5( y 3) 4 y 15 7.5 y 22.5 4 y 15 3.5 y 7.5 y 2.14 Now solve for x. x y 3 x 2.14 3 x 0.86 He spent 0.86 hour jogging and 2.14 hours walking. 35. x 2 y 1 2x 3y 12 The solution to the system is (3, 2).
36. 3x y 4 6x 2 y 8 The system has an infinite number of solutions because it is the same line.
37. x 4 y 11 5x 9 y 3 Solve the first equation in terms of x. x 4 y 11 x 4 y 11 Substitute 4y + 11 for x in the second equation and solve for y.
Chapter 14: Systems of Equations 5x 9 y 3 5(4 y 11) 9 y 3 20 y 55 9 y 3 29 y 58 y2 Now solve for x. x = 4y + 11 = 4(2) + 11 = 8 + 11 = 3 The solution of the system is (3, 2). 38. x 9 y 16 3x 8 y 13 Solve the first equation in terms of x. x 9 y 16 x 9 y 16 Substitute 9y + 16 for x in the second equation and solve for y. 3x 8 y 13 3(9 y 16) 8 y 13 27 y 48 8 y 13 35 y 35 y 1 Now solve for x. x = 9y + 16 = 9(1) + 16 = 9 + 16 = 7 The solution of the system is (7, 1). 39. y 2x 4x 7 y 15 Substitute 2x for y in the second equation and solve for x. 4x 7 y 15 4x 7(2x) 15 4x 14x 15 10x 15 15 3 x 10 2 Now solve for y. 3 y 2x 2 3 2 3 The solution of the system is 2 , 3.
3y 2x 15 2x 3y 15 2x 3y 21 40. 2x 3y 21 Add the equations to eliminate x. 2x 3 y 15 2x 3 y 21 06 False Since the statement 0 = 6 in false, there is no solution for the system.
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603
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 41. 3x y 4 4 y 12x 16 Solve the first equation in terms of y. 3x y 4 3x 4 y Substitute 3x 4 for y in the second equation and solve for x. 4 y 12x 16 4(3x 4) 12x 16 12x 16 12x 16 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 42. x y 19 x y 3 Add the equations to eliminate y and solve for x. x y 19 x y 3 2x 16 x8 Now solve for y. x y 19 8 y 19 y 11 The solution of the system is (8, 11). 43. x 3y 11 4x 5 y 10 Solve the first equation in terms of x. x 3y 11 x 3y 11 Substitute 3y 11 for x in the second equation and solve for y. 4x 5 y 10 4(3y 11) 5 y 10 12 y 44 5 y 10 17 y 34 y2 Now solve for x. x = 3y 11 = 3(2) 11 = 6 11 = 5 The solution of the system is (5, 2). 44. x 15y 44 2x 3y 20 Solve the first equation in terms of x. x 15 y 44 15 y 44 x Substitute 15y 44 for x in the second equation and solve for y. 604
2x 3 y 20 2(15 y 44) 3 y 20 30 y 88 3 y 20 27 y 108 y 4 Now solve for x. x = 15y 44 = 15(4) 44 = 60 44 = 16 The solution of the system is (16, 4). 45. 2x y 3 6x 3y 9 Solve the first equation in terms of y. 2x y 3 y 2x 3 Substitute 2x + 3 for y in the second equation and solve for x. 6x 3y 9 6x 3(2x 3) 9 6x 6x 9 9 00 Since the statement 0 = 0 is true, there are an infinite number of solutions for the system. 46. 3x y 5 3x y 2 Solve the first equation in terms of y. 3x y 5 y 3x 5 Substitute 3x + 5 for y in the second equation and solve for x. 3x y 2 3x 3x 5 2 5 2 False Since the statement 5 = 2 is false, there is no solution for the system. 47. Let x be the smaller number and y be the larger number. x y 12 3x y 20 Solve the first equation in terms of y. x y 12 y x 12 Substitute x + 12 for y in the second equation and solve for x. 3x y 20 3x (x 12) 20 2x 12 20 2x 8 x4
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 1.05x 0.63y 20.58 1.05( y 26) 0.63y 20.58 1.05 y 27.3 0.63y 20.58 0.42 y 6.72 y 16 Now solve for x. x = y + 26 = 16 + 26 = 10 The purchase included 10 $1.05 stamps and 16 63¢ stamps.
Now solve for y. y = x + 12 = 4 + 12 = 8 The two numbers are 4 and 8. 48. Let x be the smaller number and y be the larger number. x y 18 2x y 23 Solve the first equation in terms of x. x y 18 x y 18 Substitute y 18 for x in the second equation and solve for y. 2x y 23 2( y 18) y 23 2 y 36 y 23 y 13 Now solve for x. x = y 18 = 13 18 = 5 The two numbers are 5 and 13. 49. Let x be nickels and y be dimes. x y 65 0.05x 0.1y 5.30 Solve the first equation in terms of x. x y 65 x y 65 Substitute y + 65 for x in the second equation and solve for y. 0.05x 0.1y 5.30 0.05( y 65) 0.1y 5.30 0.05 y 3.25 0.1y 5.30 0.05 y 2.05 y 41 Now solve for x. x = y + 65 = 41 + 65 = 24. There are 24 nickels and 41 dimes. 50. Let x be the number of $1.05 stamps and y be the number of 63¢ stamps. x y 26 1.05x 0.63y 20.58 Solve the first equation for x. x y 26 x y 26 Substitute y + 26 for x in the second equation and solve for y.
Chapter 14 Getting Ready for the Test 1. A.
5x y 7 5(1) 2 0 7 5 2 0 7 7 7 True
x y 3 1 2 0 3 1 2 0 3 3 3 False
B.
3x y 5 3(1) 2 0 5 3 2 0 5 5 5 True
x y 1 1 2 0 1 1 1 True
C.
x2 1 2 False
x y 1 1 2 0 1 1 1 True
D.
y 1 2 1 False
x y 3 1 2 0 3 1 3 False
(1, 2) is a solution to system B.
x 3 2. Graph D is the graph of . y 3 x 3 3. Graph A is the graph of . y 3
4. Graph B is the graph of
x 3 . y 3
x 3 5. Graph C is the graph of . y 3 6. When solving a system of two linear equations in two variables, if all variables subtract out and the resulting equation is 0 = 5, this means that the system has no solution; C. 7. the slopes of the two lines are different, so the system has one solution; B.
Copyright © 2025 Pearson Education, Inc.
605
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
8. The two lines have the same slope, but different y-intercepts, so the graph of the system is two parallel lines. The system has no solution; A. 9. Solve the second equation for y. 8x 2 y 4 2 y 8x 4 y 4x 2 The two lines have the same slope and the same y-intercept, so the graph of the system is one line. The system has an infinite number of solutions; D. 10. the slopes of the two lines are different, so the system has one solution; B. 11.
7. x y 2 3x y 2
3(5x y) 3(8) 3(5x) 3( y) 3(8) 15x 3y 24 Choice D is correct.
12.
6. 4x 3y 24 4x 5 y 8 First equation: 4x 3y 24 4(3) 3(4) 0 24 12 12 0 24 24 24 True Second equation: 4x 5 y 8 4(3) 5(4) 0 8 12 20 0 8 8 8 True (3, 4) is a solution of the system.
5(2x 3y) 5(1) 5(2x) (5)(3y) 5(1) 10x 15 y 5 Choice C is correct.
Chapter 14 Test 1. False; a system of two linear equations can have no solutions, exactly one solution, or infinitely many solutions.
8. y 3x 3x y 6 y
2. False; a solution has to be a solution to both equations to be a solution of the system.
5 4 3 2 1
3. True; when the resulting statement is false the system has no solutions. 4. False; when 3x = 0 x = 0; the system does have a solution. 5. First equation: 2x 3 y 5 2(1) 3(1) 0 5 230 5 5 5 True Second equation: 6x y 1 6(1) (1) 01 6 1 0 1 5 1 False Since the statement 5 = 1 is false, (1, 1) is not a solution of the system.
606
–5 –4 –3 –2 –1–1
no solution
1 2 3 4 5
x
–2 –3 –4 –5
9. 3x 2 y 14 y x 5 Substitute x + 5 for y in the first equation and solve for x. 3x 2 y 14 3x 2(x 5) 14 3x 2x 10 14 x 10 14 x 4 y = x + 5 = 4 + 5 = 1 The solution of the system is (4, 1).
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
15 1 10. x 2 y 2 4 4x y Multiply the second equation by 1. (1)4x (1)( y) 4x y Substitute 4x for y in the first equation. 1 15 x 2y 2 4 1 15 x 2(4x) 2 4 1 15 x 8x 2 4 1 15 4 2 x 8x 4 4 2x 32x 15 30x 15 15 1 x 30 2 Now solve for y. 1 y 4x 4 2 2 1 The solution of the system is 2 , 2 . 11. x y 28 x y 12 Add the equations to eliminate y. x y 28 x y 12 2x x 40 20
13. 3x y 7 4x 3y 1 Solve the first equation for y. 3x y 7 y 7 3x Substitute 7 3x for y in the second equation and solve for x. 4x 3y 1 4x 3(7 3x) 1 4x 21 9x 1 21 5x 1 5x 20 x4 y = 7 3x = 7 3(4) = 7 12 = 5 The solution of the system is (4, 5).
Now solve for y. x y 28 20 y 28 y 8 The solution of the system is (20, 8). 12. 4x 6 y 7 2x 3y 0 Multiply the second equation by 2. 4x 6 y 7 4x 6 y 7 2(2x 3y) 2(0) 4x 6 y 0 Add the equations to eliminate x. 4x 6 y 7 4x 6 y 0 07 Since the statement 0 = 7 is false, the system has no solution.
14. 3(2x y) 4x 20 x 2 y 3 Simplify the first equation. 3(2x y) 4x 20 6x 3y 4x 20 2x 3y 20 2x 3y 20 x 2y 3 Multiply the second equation by 2. 2x 3y 20 2x 3y 20 2(x 2x 4 y 6 2 y) 2(3) Add the equations to eliminate x; then solve for y. 2x 3y 20 2x 4 y 6 7 y 14 y2 Now solve for x. x 2y 3 x 2(2) 3 x4 3 x7 The solution of the system is (7, 2). x3 2 y 15. 2 4 7 2x y 3 2 Multiply the first equation by 4 and the second equation by 6 to eliminate fractions and simplify.
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607
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
x3 2 y 4 4 2x 6 2 y 2 4 14 4x 3y 7 2x 3 6 y2 6 2x y 8 4x 3y 14 Multiply the first revised equation by 3. 3(2x y) 3(8) 6x 3y 24 4x 3y 14 4x 3y 14 Add the equations to eliminate y. 6x 3y 24 4x 3y 14 10 2x x5 Now solve for y. 2x y 8 2(5) y 8 10 y 8 y 2 The solution of the system is (5, 2).
16. 8x 4 y 12 y 2x 3 Substitute 2x 3 for y in the first equation and solve for x. 8x 4 y 12 8x 4(2x 3) 12 8x 8x 12 12 00 Since the statement 0 = 0 is true, the system has an infinite number of solutions. 17. 0.01x 0.06 y 0.23 0.2x 0.4 y 0.2 Multiply the first equation by 100 and the second equation by 10 to eliminate decimals. 100(0.01x 0.06 y) 100(0.23) 10(0.2x 0.4 y) 10(0.2) x 6 y 23 2x 4 y 2 Multiply the first equation by 2. 2(x 6 y) 2(23) 2x 12 y 46 2x 4 y 2 2x 4 y 2 Add the equations to eliminate x; then solve for y. 2x 12 y 46 2x 4 y 2 16 y 48 y3 608
Now solve for x. x 6 y 23 x 6(3) 23 x 18 23 x 5 The solution of the system is (5, 3). 2 18. x 3 y 3 2x 3y 10 Solve the first equation in terms of x. 2 x y 3 3 2 x y3 3 2 Substitute y 3 for x in the second equation 3 and solve for y. 2x 3y 10 2 2 y 3 3y 10 3 4 y 6 3y 10 3 4 y 18 9 y 30 5 y 48 48 y 5 Now solve for x. 2 2 48 32 15 47 x y 3 3 3 3 5 5 5 5 47 48 The solution of the system is , . 5 5 19. Let x be the first number and the y be the second. x y 124 x y 32 Add the equations to eliminate y then solve for x. x y 124 x y 32 2x 156 x 78 Now solve for y. x y 124 78 y 124 y 46 The numbers are 78 and 46.
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ISM: Prealgebra and Introductory Algebra
20. Let x = number of cc of 12% solution and y = number of cc of 16% solution. x 80 y 0.12x 0.22(80) 0.16 y Substitute x + 80 for y in the second equation; then solve for x. 0.12x 17.6 0.16 y 0.12x 17.6 0.16(x 80) 0.12x 17.6 0.16x 12.8 4.8 0.04x 120 x 120 cc of the 12% saline is needed to make the 16% solution. 21. Let t be the number of farms in Texas (in thousands) and let m be the number of farms in Missouri (in thousands). t m 342 t m 152 Substitute m + 152 for t in the first equation and solve for m. t m 342 m 152 m 342 2m 152 342 2m 190 m 95 t = m + 152 = 95 + 152 = 247 There were 247 thousand farms in Texas and 95 thousand farms in Missouri. 22. Let x be speed of one hiker and y be the speed of the other hiker. 4x 4 y 36 y 2x Substitute 2x for y in the first equation; then solve for x. 4x 4 y 36 4x 4(2x) 36 4x 8x 36 12x 36 x3 Now solve for y. y = 2x = 2(3) = 6 One hiker hikes at 3 mph and the other hikes at 6 mph. Cumulative Review Chapters 114 1. 8 15 = 8 + (15) = 7 2. 4 7 = 4 + (7) = 3
Chapter 14: Systems of Equations 4. 3 (2) = 3 + 2 = 5 5. x = 60 + 4 + 10 x = 56 + 10 x = 46 6.
4x 2 3x 4x 3x 2 3x 3x x 2
7. “1.2 is 30% of what number?” translates as 1.2 = 30% x. 8. “9 is 45% of what number?” translates as 9 = 45% x. 9. x 50 8 50 x 8 50 50 x 0.16 16% 16% of 50 is 8. 10. x 16 4 16x 4 16 16 x 0.25 25% 25% of 16 is 4. 11. Let x be the number of freshmen at Slidell High. 31 is 4% of what number? 31 4% x 31 0.04x 31 0.04x 0.04 0.04 775 x There are 775 freshmen at Slidell High. 12. Let x be the number of apples in the shipment. 29 is 2% of what number? 29 2% x 29 0.02x 29 0.02x 0.02 0.02 1450 x There are 1450 apples in the shipment. 13. 2(x 5) 10 3(x 2) x 2x 10 10 3x 6 x 2x 20 2x 6 20 6 Since the statement 20 = 6 is false, the equation has no solution.
3. 4 (5) = 4 + 5 = 1 Copyright © 2025 Pearson Education, Inc.
609
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 14. 4(4y 2) 2(1 6y) 8 16y 8 2 12y 8 16y 8 12y 10 4y 8 10 4y 2 4y 2 4 4 1 y 2
2x 4 23. 3x 1 6x 10x 5 2
6x2 2x 12x 5 12x 4 1 6x2 10x 1 1 2x 4 3x 1 3x 1 3x 2
15. 5x 7 2(x 3) 5x 7 2x 6 13 7x 13 x 7 13 x x 7
24. x 1 3x2 x 4 3x2 3x 2x 4 2x 2 2 3x2 x 4 13 7
–5
–4
–3
–2
–1
0
1
x 1
2
3
4
5
–4
–3
–2
2x2 7x 4 0 (2x 1)(x 4) 0 2x 1 0 or x 4 0 2x 1 x4 1 x 2 1 The solutions are x or x = 4. 2
–1
0
1
2
3
4
5
7
m m7 where n 0 17. n7 n
x(x 5) 24
26. 3
x 1
2x2 7x 4
2 x {x|x 2} –5
2
x(2x 7) 4
25.
16. 7x 4 3(4 x) 7x 4 12 3x 4 12 4x 8 4x
3x 2
4
3
4
34
18. (5x )(7x ) 5(7)x x 35x
7
35x
x2 5x 24 x 5x 24 0 (x 3)(x 8) 0 2
2x4 19.
3y5 5x2
20.
4
2
4y3
24 x44 34 y54 52 x22 42 y32
16x16 81y20 25x4
where y 0
where y 0
16y6
x3 0 or x 8 0 x 3 x 8 The solutions are 3 and 8. 27. x = one leg x + 2 = other leg x + 4 = hypotenuse
3 2 3 2 21. (2x 8x 6x) (2x x 1) 3
2
3
2
2x 8x 6x 2x x 1
x x2 4x 4 x2 8x 16
9x2 6x 1
x2 4x 12 0 (x 6)(x 2) 0 x 6 0 or x 2 0 x6 x 2
22. (7x 1) (x 3) (7x 1) (x 3) 7x x 1 3 8x 4
610
x2 (x 2)2 (x 4)2 2
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ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations
Discard x = 2 because length cannot be negative. one leg: x = 6 other leg: x + 2 = 6 + 2 = 8 hypotenuse: x + 4 = 6 + 4 = 10 The lengths of the sides of the right triangle are 6, 8, and 10 units. 28. Let x be the number. x x2 132
2y
2y 7 30.
7 2y 7
36. {(2, 3), (2, 0), (2, 2), (2, 4)} Domain: {2} Range: {2, 0, 3, 4}
2y 7 1 2y 7
37. x 2 y 7 2x 2 y 13 Multiply the first equation by 1. 1(x 2 y) 1(7) x 2 y 7 2x 2 y 13 2x 2 y 13 Add the equations to eliminate y; then solve for x.
x2 3 9x 3 x2 3 9x 3 x9 x9 x9
x2 9x x9 x(x 9) x9 x
x 2 y 7 2x 2 y 13 x 6 Now solve for y. x 2y 7 6 2y 7 2y 1 1 y 1 2 The solution of the system is 6, . 2
31. y = mx + b If y = 1, then m = 0. 32. Pick any 2 points on the line x = 2. Let (x1, y1) (2, 0) and (x2 , y2 ) (2, 5). y y 50 5 m x22 x11 2 2 0 undefined
33. m y2 y1 4 5 1 1 x2 x1 3 2 5 5 y y1 m(x x1) 1 y 5 (x 2) 5 1 2 y5 x 5 5 2 1 5( y 5) 5 x 5 5 y 25 x 52 x 5 y 23
y y1 m(x x1) y (6) 1(x 5) y 6 x 5 y x 1 x y 1 35. {(0, 2), (3, 3), (1, 0), (3, 2)} Domain: {1, 0, 3} Range: {2, 0, 2, 3}
x2 x 132 0 (x 12)(x 11) 0 x 12 0 or x 11 0 x 12 x 11 The number is 12 or 11. 29.
y y1 5 (6) 11 1 34. m 2 x2 x1 6 5 11
3y x 6 38. 4x 12y 0 Solve the first equation for x. 3y 6 = x Substitute 3y 6 for x in the second equation and solve for y. 4x 12y 0 4(3y 6) 12y 0 12y 24 12y 0 24y 24 0 24y 24 y1 Now solve for x.
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611
ISM: Prealgebra and Introductory Algebra
Chapter 14: Systems of Equations 3y x 6 3(1) x 6 3 x6 3 x The solution of the system is (3, 1). y 5 39. x x 2y 2 0 6 2 Multiply the first equation by 6 and the second equation by 6. y 5 6 x 6 6x 3y 15 2 2 x y x 3y 0 6 6(0) 6 2 Add the equations to eliminate y; then solve for x. 6x 3y 15 x 3y 0 7x 15 15 x 7 Now solve for y y.5 x 2 2 15 y 5 7 2 2 15 y 5 14 14 7 230 7 y 35 2 7 y 5 5 y 7 5 15 The solution of the system is , . 7 7 3 3y x 8 2 40. x y 13 9 3 Multiply the first equation by 8 and the second equation by 9. 3y 3 8 x 8 8x 3y 12 8 2 y 13 9x y 39 9 x 9 3 9 Multiply the second equation by 3.
612
8x 3y 12 8x 3y 12 3(9x y) 3(39) 27x 3y 117 Add the equations to eliminate y, then solve for x. 8x 3y 12 27x 3y 117 35x 105 x 3 Now solve for y. 8x 3y 12 8(3) 3y 12 24 3y 12 3yy 12 36 The solution of the system is (3, 12). 41. Let x be the first number and y be the second. x y 37 x y 21 Add the equations to eliminate y; then solve for x. x y 37 x y 21 2x 58 x 29 Now solve for y. x y 37 29 y 37 y 8 The two numbers are 29 and 8. 42. Let x be the first number and y be the second. x y 75 x y 9 Add the equations to eliminate y, then solve for x. x y 75 xy9 2x 84 x 42 Now x ysolve 75for y. 42 y 75 y 33 The two numbers are 42 and 33.
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Chapter 15 Section 15.1 Practice Exercises
18. 3 64x9 y24 4x3 y8 because (4x3 y8 )3 64x9 y24 .
100 10, because 102 100 and 10 is
1.
positive. 2. 81 9. The negative sign in front of the radical indicates the negative square root of 81. 2
25 5 5 5 25 3. , because and is 81 9 9 81 9 positive.
19. 3 64x9 y24 4x3 y8 because (4x3 y8 )3 64x9 y24 . Calculator Explorations 6 2.449; since 6 is between perfect squares 4
1.
and 9, 2 1 1, because 1 1 and 1 is positive.
4. 5.
6.
3
27 3 because 33 27.
7.
3
8 2 because (2)3 8.
squares 196 and 225, 196 14 and
5
5 1 1 because (1) 1.
11.
4
256 4 because 44 256 and 4 is positive.
12.
6
1 is not a real number because the index 6 is
22 4.690.
z8 z4 because (z 4 )2 z8.
15.
x20 x
16.
4x 2x because (2x ) 4x .
17.
3 8 y12
20
because (x ) x .
3
3 2
4
4 3
82 is between
81 9
and 100 10.
14.
6
200 is between
82 9.055; since 82 is between perfect
5.
46 6.782; since 46 is between perfect
6.
squares 36 and 49, and
even and the radicand 1 is negative.
10 2
9 3 and
225 15.
squares 81 and 100,
10.
10
11 is between
200 14.142; since 200 is between perfect
4.
16 is not a real number since the index 4 is even and the radicand 16 is negative.
9 3 and
11 3.317; since 11 is between perfect
4
13. To three decimal places,
14 is between
squares 9 and 16, 16 4.
3
9 3.
16 4. 3.
1 1 1 8. 3 1 because . 64 4 4 64 9.
squares 9 and 16,
2
4 2 and
14 3.742; since 14 is between perfect
2.
0.81 0.9 because (0.9) 0.81 and 0.9 is positive.
6 is between
46 is between
36 6
49 7.
7.
3
40 3.420
8.
3
71 4.141
9.
4
20 2.115
10.
4
15 1.968
11.
5
18 1.783
12.
62
6
12
2 y because (2 y ) 8 y .
1.122
Copyright © 2025 Pearson Education, Inc.
613
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
Vocabulary, Readiness & Video Check 15.1 1. The symbol is used to denote the positive, or principal, square root. 2. In the expression 4 16, the number 4 is called the index, the number 16 is called the radicand, and is called the radical sign. 3. The reverse operation of squaring a number is finding a square root of a number. 4. For a positive number a, a is the negative square root of a and a is the positive square root of a. 5. An nth root of a number a is a number whose nth power is a.
2
1 1 1 1 1 4. , because and is 64 8 8 64 8 positive. 6. 36 6. The negative sign indicates the negative square root of 36. 8.
25 is not a real number, because there is no real number whose square is 25.
10. 49 7. The negative sign indicates the negative square root of 49. 2
4 2 2 2 4 and is , because 12. 81 9 9 9 81 positive. 400 20, because 202 400 and 20 is
14. 6.
4 2; the statement is false.
7.
9 is not a real number; the statement is false.
8.
1000 31.623; the statement is false.
positive.
positive. 18.
9. True
169 13, because 132 169 and 13 is
16.
10. True 11. The radical sign, , indicates a positive square root only. A negative sign before the , indicates a negative square radical sign, root.
20.
12. The square root of a negative number is not a real number; the cube root of a negative number is a real number.
24.
2 1 1 , because 1 1 and 1 is 121 11 11 11 121 positive.
0.49 0.7, because 0.72 0.49 and 0.7 is
positive. 22. 3 64 4, because 43 64. 3
27 3, because (3)3 27.
26. 3 27 3, because 33 27.
13. an odd-numbered index
3
1 1 , because . 64 4 64 4
3 1
1
14. Take the two integers that your answer falls between and square them; then check to make sure that the radicand falls between these two squares.
28.
15. Divide the index into each exponent in the radicandbut still check by raising your answer to a power equal to the index.
32. 4 81 3, because 34 81 and 3 is positive.
30.
3
1 1, because (1)3 1.
34.
49 7, because 72 49 and 7 is positive.
36.
9 is not a real number, because there is no real number whose square is 9.
Exercise Set 15.1 2.
614
64 8, because 82 64 and 8 is positive.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra 3
8 2 2 8 , because . 38. 3 27 27 3 3 40. 5 32 2, because 25 32. 51
42.
Chapter 15: Roots and Radicals 78. 90 = 9 10 1 is not a real number because the index is even and the radicand is negative.
80. a.
3 1
b.
1, because 1 1. 5
44.
10 3.162
46.
27 5.196
4 1
c.
is not a real number because the index is even and the radicand is negative.
d.
5 1
8 2.828
48.
82. The length of a side is 2
1 1 , 9 81
5 5 2 10 y10 y , because ( y ) y .
52.
is a real number because the index is
odd.
29 5.39 480 29 480 5.39 2587.2 The area is approximately 2587 square feet.
50.
is a real number because the index is
odd.
x6 x3 , because (x3 )2 x6 .
54.
1 1 . The length of a side of 81 9 1
the square is
meter.
9
84. The length of a side is 12
6 2
56.
36x
58.
100z 4 10z2 , because (10z2 )2 100z4 .
60.
x12 y20 x6 y10 , because (x6 y10 )2 x12 y20 .
2
86. 3 3 1 1 since 13 1. 88.
4m14n2 2m7 n, because (2m7 n)2 4m14n2 .
62.
64. 3 x12 y18 x4 y6 , because (x4 y6 )3 x12 y18.
67.24 feet. Since
8.2 67.24, 67.24 8.2. The length of a side is 8.2 feet.
12
6x , because (6x ) 36x . 6
1 meter. Since 81
1, 600, 000, 000 200 since 2002 40, 000 and 40, 0002 1, 600, 000, 000
90. Since
28 is between
25 and
36, then
28 is between 5 and 6. 3
66.
27a6b30 3a2b10 , because
92. Since
(3a2b10 )3 27a6b30 . 2
70.
2
2
4x 2x 4x , because . 81 9 81 9 2
2x
81 and 100, then
98 is between 9 and 10.
y4 y8 y4 y8 , because . 7 49 7 49
68.
98 is between
94. 3 195,112 58, because 583 195,112. One side of the cube would be 58 feet. 96. answers may vary
72. 8 = 4 2 74. 75 = 25 3 76. 44 = 4 11 Copyright © 2025 Pearson Education, Inc.
615
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
98.
x
y3x
8
3 8 2
2
3 2 1.3
1
3 1 1
0
30 0
1
21 1
2
3 2 1.3
8
38 2
5. 7 75 7 25 3 7 25 3 75 3 35 3 6.
16 16 4 81 81 9
7.
2 25 8. 45 49
2
25
5
45 49
9 5 7
3 5 7
x11 x10 x x10 x x5 x
9.
10.
2
18x4 9 2 x4 9x4 2 9x4 2 3x2 2
11. 2
100.
4x (2x)2 2x 2 x
102.
x2 6x 9 (x 3)2 x 3 12.
104. The graph of y x 3 ‘starts’ at (3, 0). 106. The graph of y x 5 ‘starts’ at (5, 0). Section 15.2 Practice Exercises 1.
40 4 10 4 10 2 10
2.
18 9 2 9 2 3 2
3.
500 100 5 100 5 10 5
4.
27
27 9 3
x8
x8
x4
9
3 3 3
x4
x4
7 y7 7 y7 25 25
y6 7 y 5 6
y 7 5 y3 7 y 5 13. 3 88 3 8 11 3 8 3 11 23 11
15 The radicand 15 contains no perfect square factors other than 1. Thus 15 is in simplest form.
14. 3 50 The number 50 contains no perfect cube factors, so 3 50 cannot be simplified further. 10 3 10 3 10 15. 3 27 3 27 3 81 3 81 3 27 3 3 27 3 3 33 3 16. 3 38 8 2 2 2
616
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Vocabulary, Readiness & Video Check 15.2 1. If
a and
b are real numbers, then
a b a b. 2.
If
a and
b are real numbers, then
a a . b b 3.
16 25 16 25 4 5 20
4.
36 3 36 3 6 3 6 3
5.
48 2 12 2 43
Chapter 15: Roots and Radicals
10.
90 9 10 9 10 3 10
12.
150 25 6 25 6 5 6
14.
75 25 3 25 3 5 3
16. 9 36 9 6 54 18. 11 99 11 9 11 11 9 11 11 3 11 33 11 20. 6 75 6 25 3 6 5 3 30 3
2 4 3 22 3 4 3 The statement is false.
22.
3 7 63 63 9 7 9 7 16 4 4 4 16
24.
46 24 24 4 6 2 6 13 13 13 169 169
26.
100 100 10 7 49 49
28.
27 100
6. True; 6 has no perfect cube factors. 7. Factor until we have a product of primes. A repeated prime factor means a perfect squareif more than one factor is repeated, we can multiply all the repeated factors together to get one larger perfect square factor. 8. In words, the quotient rule for square roots says that the square root of a quotient is equal to the square root of the numerator over the square root of the denominator. 9. The power must be 1. Any even power is a perfect square and can be simplified; any higher odd power is the product of an even power times the variable with a power of 1. 10. If a factor is repeated the same number of times as the index, then we have a perfect root and the product rule can be applied.
30.
27
100
93 10
9 3 3 3 10 10
30 30 30 49 49 7
84 84 32. 121 121 4 21 11 4 21 11 2 21 11
Exercise Set 15.2 2.
44 4 11 4 11 2 11
34.
y3
y2 y
y16 y
y2
yy y
y16 y y8 y
4.
28 4 7 4 7 2 7
36.
y17
6.
21 is in simplest form.
38.
81b5 81b4 b 81b4 b 9b2 b
8.
125 25 5 25 5 5 5
40.
40 y10 4 y10 10 4 y10 10 2 y5 10
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617
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
42.
44.
46.
63 p2
34 34 4 68. 3 27 3 27 3
9 7 9 7 3 7 p p p p2
63
2 6 y2 6 y
70. 3 108 3 27 4 3 27 3 4 33 4
y2 6 y 6 z8 z8
z16
z16
500
500 100 5 100 5 10 5 y11 y11 y11 y22
y22
72. (6x)(8x) 48x2 74. (2x + 3) + (x 5) = 3x 2 76. (9 y2 )(8 y2 ) 72 y4
48. 6 49 6 7 42
a12b14 a a12b14 a
a13b14
78. 50.
25 25 5 144 144 12
a6b7 a
52.
700 100 7 100 7 10 7
54.
45 45 9 5 9 5 3 5 64 8 8 8 64
56.
50n13 25n12 2n
3x4 y5 3y
82.
25n12 2n 84.
5n6 2n
8 10 27x8 y11 9x y 3y 9x8 y10 3y
80.
3
27x12 3 27 3 x12 3x4 11, 600 400 29 400 29 20 29
The distance is 20 29 feet. 58.
41x5 9 y8
41x5
86. answers may vary; possible answer: let a = 1 and b = 1 so a b 2 and a b 2.
9 y8 x4 41x 3y 4
88.
6
x4 41x 3y 4 x2 41x 3y 4
60. 3 81 3 27 3 3 27 3 3 33 3 62. 3 56 3 8 7 3 8 3 7 23 7 3 32 383 4 3 8 4 32 23 4 5 64. 3 5 5 125 3 125
618
6 594 6 3564 6 324 11 6 324 11 6 18 11 6 3 11
The length of a side is 3 11 feet. 90.
37 3 37 3 37 66. 3 27 3 27 3
6A
6A 6 6144 36,864 192 32 6 6 6 6 The length of a side is 32 feet.
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ISM: Prealgebra and Introductory Algebra
92. C 1003 n 700 1003 500 700 1493.70 The cost is approximately $1493.70. 94. B
18385 3600 18385 3600 15, 555 60 2.1 The surface area is approximately 2.1 square meters.
9.
36 48 4 3 9 6 16 3 4 3 3 6 16 3 4 3 3 6 4 3 4 3 3 38 3
10.
9x4 36x3 x3
hw 3600
Chapter 15: Roots and Radicals
3x2 36x2 x x2 x 3x2 36x2 x x2 x 3x2 6x x x x 3x2 5x x 11. 103 81p6 3 24 p6 103 27 p6 3 3 8 p6 3 103 27 p6 3 3 3 8 p6 3 3
Section 15.3 Practice Exercises
10 3 p2 3 3 2 p2 3 3
1. 6 11 9 11 (6 9) 11 15 11 2.
7 3 7 1 7 3 7 (1 3) 7 2 7
3.
2 2 15 1 2 1 2 15 (1 1) 2 15 2 2 15
4. 3 3 3 2 cannot be simplified further since the radicands are not the same. 5. 63 5 23 5 113 7 (6 2)3 5 113 7 43 5 113 7 6. 4 13 53 13 cannot be simplified further since
30 p2 3 3 2 p2 3 3 28 p2 3 3 Vocabulary, Readiness & Video Check 15.3 1. Radicals that have the same index and same radicand are called like radicals. 2. The expressions 73 2 x and 3 2 x are called like radicals. 3. 11 2 6 2 17 2, choice b. 4.
5 is the same as 1 5, choice b.
5.
5 5 2 5, choice c.
the indices are not the same. 7.
6. 9 7 7 8 7, choice a.
27 75 9 3 25 3 9 3 25 3 3 3 5 3 8 3
7. Both like terms and like radicals are combined using the distributive property; also, only like (vs. unlike) terms can be combined, as with like radicals (same index and same radicand).
8. 3 20 7 45 3 4 5 7 9 5 3 4 5 7 9 5 3 2 5 7 3 5 6 5 21 5 15 5
8. Sometimes we can’t see until we simplify that there are like radicals to combine, so we may incorrectly think we cannot add or subtract unless we simplify first. 9. the product rule for radicals Exercise Set 15.3 2.
5 9 5 1 5 9 5 (1 9) 5 8 5
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619
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
4. 12 2 3 2 8 2 10 (12 3 8) 2 10 17 2 10
22.
6. 4 3 3 3 5 (4 3) 3 5 1 3 5 3 5 8.
13 13 13 1 13 1 13 13 (11) 13 13 2 13 13
10. 5 7 2 11 7 2 (5 11) 7 2 6 7 12.
50 18 25 2 9 2 25 2 9 2 5 2 3 2 (5 3) 2 8 2
24.
26.
20.
620
2
2
6 3 6 6 (1 3 1) 6 5 6
28. 2 80 45 2 16 5 9 5 2 16 5 9 5 24 5 3 5 8 5 3 5 (8 3) 5 5 5 30.
18. x 6 x2 2 x x 6x 2 x 5x 2 x
2 2 25 9
25 9 2 2 5 3 3 2 5 2 15 15 3 5 15 15 2 8 2 15 8 2 15
14. 5 32 72 5 16 2 36 2 5 16 2 36 2 5 4 2 6 2 20 2 6 2 (20 6) 2 14 2 16. 2 8 128 48 18 2 4 2 64 2 16 3 9 2 2 4 2 64 2 16 3 9 2 22 2 8 2 4 3 3 2 4 2 8 2 4 3 3 2 (4 8 3) 2 4 3 1 2 4 3 2 4 3
3 3 3 3 64 16 64 16 3 3 8 4 3 2 3 8 8 1 2 3 8 8 3 3 8 33 8
5 10 5 10
32. x 16x x3 x 16 x x2 x
9x 16x 2 x 9 x 16 x 2 x 3 x 4 x 2 x (3 4 2) x x
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x 16 x x2 x 4x x x x (4x x) x 3x x
ISM: Prealgebra and Introductory Algebra
34.
6 16 24 25 6 4 4 6 5 6 4 6 9 6 2 6 9 (1 2) 6 9 3 6 9
Chapter 15: Roots and Radicals
52.
x 3 36x2 2 9 6x 25 2 2x 6x 2 3 6x 5x 2 3 2x (6x 5x) 2 3 6x 3 2x x 2 3 6x 3 2x
36. 11 5 7 8 5 7 3 38. 5 18 2 32 5 9 2 2 16 2 5 9 2 2 16 2 53 2 24 2 15 2 8 2 (15 8) 2 23 2 40.
15 135 15 9 15 15 9 15 15 3 15 (1 3) 15 2 15
72x2 54x x 50 3 2x 36x2 2 9 6x x 25 2 3 2x
54. 83 4 23 4 3 49 (8 2)3 4 3 49 103 4 3 49 56. 53 9 2 113 9 2 53 9 113 9 2 (5 11)3 9 2 63 9 58. 3 32 3 4 3 8 4 3 4 3 8 2 4 3 4 23 4 3 4 (2 1)3 4 33 4
42. 8 2 5 2 8 (1 5) 2 8 6 2
60.
3 64
3 3 3 14 9 4 14 9 5 14
44. 5 2x 98x 5 2x 49 2x 5 2x 7 2x (5 7) 2x 12 2x 46.
3
3
9z 3 3z 3 3 9z 9z 3 3 3
3
3 2
3
z 5z3 z2
3
3
19z3 z2 5z3 z2 3
(19z3 5z3 ) z2 3
14z3 z2 66.
2
9x 2 4x 6x x 2 2
3
19 z9
x
18x2 24x3 2x2 2
3
3 3 2 9 2 3 2 64. 19 z11 z 125z 19 z z z 3 125 z
48. x 4x 9x3 x 4 x 9x2 x
50.
3
9z 3 27z3 3 3
100x2 3 x 36x2 10x 3 x 6x 4x 3 x
x 4 x 9x2 2x x 3x x (2x 3x) x 5x x
3
62. 3 27z3 3 81z3 3 3z 3 27z3 3
9x2 2 4x2 6x x2 2 3x 2 2x 6x x 2 (3x x) 2 2x 6x 4x 2 2x 6x
72x2 3 54 x 50 33 2 36x2 2 3 27 2 x 25 2 33 2 36x2 2 3 27 3 2 x 25 2 33 2 6x 2 3 3 2 5x 2 3 3 2 (6x 5x) 2 (3 3)3 2 x 2
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
68.
(3x 2)2 (3x)2 2(3x)(2) (2)2 9x2 12x 4
70. (x 5)2 (x)2 2(x)(5) (5)2 x2 10x 25 72. answers may vary 74. 15 6 15 6 20 6 30 6 (15 15 20 30) 6 80 6 The perimeter is 80 6 feet. 76. 4 52 4 80 4 4 13 4 16 5 4 4 13 4 16 5 4 2 13 4 4 5 8 13 16 5
The total length is 8 13 16 5
feet.
78. The expression 3 7 3 6 cannot be simplified. 80. The expression can be simplified. 5x 2 8x 2 (5x 8x) 2 13x 2 82. The expression can be simplified. 6 5 5 (6 1) 5 5 5 84. 7 x11 y7 x2 y 25x7 y5 8x8 y2 7 x10 y6 xy x2 y 25x6 y4 xy 4x8 y2 2 7 x10 y6 xy x2 y 25x6 y4 5 3
xy x y 5x y
5 3
5 3
7x y 7x y
2
3 2
xy 5x y
xy 4x8 y2
xy 2x y 2 4
xy 2x4 y 2
(7x5 y3 5x5 y3 ) xy 2x4 y 2 2x5 y3 xy 2x4 y 2 Section 15.4 Practice Exercises
2.
5 2 5 2 10 7 7 7 7 49 7
3.
6 3 18 9 2 9 2 3 2
4.
10x 2x 10x 2x
1.
20x2 4x2 5 4x2 5 2x 5 622
Copyright © 2025 Pearson Education, Inc.
2
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
7 3 7 7 7 3 7 21 5x x 3 5 5x x 5x 3 5
5. a.
7
b.
12.
2 45x
9 5x 2 5x 3 5x 5x 2 5x 3 5x 5x 10x 3 5x
5x x 3 5x 5 5 x2 3 25 x 5 x2 3 25 x x 5 3 5 x x 5 15 x
x 5 x 3
c.
6. a.
3 8
3 8
2
13.
2
3 364 8
3 2 7
5x 4 5x 2 5x (4) (4) 2
2
2
5x 8 5x 16 7. 8.
9.
10.
11.
21 3
90 90 45 2 2 125x3
5x 5 3
5
7 20
3
95
9
3
7 45 7 5 2 5 5 7 5 2 5 5 35 25 35 10
22 7 3 2 7 2
47 3 2 7
3 3 2 7
1 2 7 2 7
14.
5 3 5 3 3 3 3
2 5 2 1
3
5
3 2 7
3 5
125x3 2 25x 5x 5x 3
21 7 3
3 2 7 2 7 2 7
61 b.
10x 15x
x x x 3 5 x 5 3 x 3x 5x 15
2 45x 2
2 5 2 1 2 1 2 1
2 2 5 2 5 2 1 76 2 1 76 2
7
15.
7 2
2 x 2 x
Copyright © 2025 Pearson Education, Inc.
x
7 2
2 x 4 x
x
623
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
Vocabulary, Readiness & Video Check 15.4 1.
3.
8 y 8 y 8 y 64 y 64 y 2
2
7 3 7 3 21
2.
12.
10 10 10 10 100 or 10
14.
7 21y7 3y 21y 3y
63y8 9 y8 7
15 or 5 3 3
15
8
4. The process of eliminating the radical in the denominator of a radical expression is called rationalizing the denominator.
9 y 7 3y4 7 16.
2 2 2 18x2 y2 2x2 y 18x y 2x y
36x4 y3 36x4 y2 y
5. The conjugate of 2 3 is 2 3.
6. In each Video Notebook, the product rule is first used to multiply the radicals and then later used to simplify the radical. 7. The square root of a positive number times the square root of the same positive number (or the square root of a positive number squared) is that positive number. 8. if we notice that some simplifying can be done to the fraction if both radicands are under one radical
10
30 70
20.
6
3 2 6 3 6 2 18 12 9 2 4 3 9 2 4 3 3 22 3
22.
2 5b b 5b 5
2 b2 5 25 b
3 12 36 6
4.
2 14 28 4 7 4 7 2 7
8.
5 y 5 y
10.
2 b2 5 25 b 2b 5 5 b 24.
5 2 5 2 5 22 5 4 1
26.
7 5 2 5
2
5y 5y 2
3 10 3 10 3 10 2 9 10 2
2
7 2 7 5 5 2 5 5 14 35 10 25 14 35 10 5
9 10 90
624
5b 2 b 5 5b 2 b 5b 5 2 5b2 25b
2.
10 10
3 7 10 3 10 7
Exercise Set 15.4
6.
6x2 y y
18.
9. to write an equivalent expression without a radical in the denominator 10. Using the FOIL order to multiply, the Outer product and the Inner product are the only terms with radicals and they will subtract out.
36x4 y2 y
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ISM: Prealgebra and Introductory Algebra
28.
Chapter 15: Roots and Radicals
5 3 2 3 1 5 3 3 5 3 1 2 3 2 1 53 5 3 2 3 2 15 3 3 2 13 3 3
3x2 y
y 5 y 5 y (5)2 y 25
32.
x 4 x 2 x (4) (4) 2
16x2 y2 2x
2
4xy 2x 2
50.
34.
3y 2 3y 2 3y (2) (2) 2
2
3
x 8 x 16 2
3x2 y
32x3 y2
2
2
96x5 y3
16x2 y2 2x
30.
96x5 y3
48.
2
3
3
3
6 3
8 8 11 8 11 11 11 11 11
52.
3y 4 3y 4 1
54. 36.
40 10
40 42 10 56.
38.
40.
42.
44.
46.
55 5
55 11 5
96 96 12 4 3 4 3 2 3 8 8 24x7 6x 120 3
24x7 6 4x 2x3 6x 120 40 4 10 4 10 2 10 3
54x3
54x3 2x 2x 27x2 2
9x 3 9x2 3 3x 3
58.
60.
10z
1
10z
10z 10z
10z 10z
7 7 12 12 7 43 7 3 2 3 3 7 3 2 3 3 21 23 21 6 5 x
5
5
x
x
x 5x x x
1 1 27 27 1 93 1 33 1 3 3 33 3 3 3 3 9
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625
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
62.
11 11 11 14 154 14 14 14 14 14
72.
x x 20 y 20 y
64.
66.
3 50 50 3 25 2 3 5 2 3 2 5 2 2 6 5 2 6 10
68.
70.
4 5y
7x 2
3y 7x
2 5y
2
2
2 5y
74. 5y
3y 5 3y 3y 3
5y 5y
52
6 52 5 2 5 2 6 52
5 22 2
z 49 y
6
6 5 12 54 6 5 12 1 6 5 12
7x 2 2
76.
2 2 10 3 10 3 10 3 10 3 2 10 3 2 10 32
z 7 y
5xy 2 5y 5xy 10 y
z z 49 y 49 y
x
3y 2
x
5 y 5 y 3y 3y
x
3
z y 7 y y
2 10 6 10 9 2 10 6 1 2 10 6
yz 7 y yz 7y 78.
3 1 3 1 3 2 3 2 3 2 3 2
3 2 3 1 2 2 3 2
3 3 3 2 1 3 1 2 32 3 6 3 2 1 3 6 3 2
626
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
80.
2 2
2 3
2 2 2 3 2 3 2 3 22 2 3
2
2
2
3
2 2 2 3 22 2 3 43 2 2 6 42 3 1
3 x
5 3 5 3 15
3x 3x (11) 3x 2 3x The statement is true.
100.
11 2 11 2 22 The statement is true.
3 x
102. answers may vary
3 x 3 x 9 3 x
104. answers may vary
9
2 2 3 x
2 22 2 2 106. 2 2 2 3 2 3 2 2 2 2 2 2 3 2 2
27 9x 9 x 84.
96.
98.
2 2 6 42 3 9
S 4
The statement is false.
82.
S S S S 4 2 2 2
94.
4 x 1
x1 x 1
4 x 1 4
2
x 1
x 1 2
2
4 x 4 x 1
2
24 22 3 2 2 2
3
2 2 4 6 2 3
Mid-Chapter Review 1.
36 6, because 62 36 and 6 is positive.
2 y 9 1 2 y 10 y 5
2.
48 16 3 16 3 4 3
3.
x4 x2 , because (x2 )2 x4 .
88. 16x2 x 9 (4x)2
4.
y7 y6 y
5.
16x2 4x, because (4x)2 16x2 .
6.
18x11 9x10 2x 9x10 2x 3x5 2x
86. 2 y 1 32
16x2 x 9 16x2 x9 0 x 9 90. x2 3x 4 (x 2)2 x2 3x 4 x2 4x 4 3x 4x 0x
y6
y y3 y
7. 3 8 2, because 23 8. 8. 4 81 3, because 34 81.
92. Volume = (length)(width)(height) 3 2 2 32 2 3 The volume is 2 3 cubic feet.
9. 3 27 3, because (3)3 27. 10.
4 is not a real number.
Copyright © 2025 Pearson Education, Inc.
627
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
11.
12.
2 7 2 7 2 7 4 7 4 7 28 2 2 28. 3 5 3 5 3 5 9 5 9 5 45
11 11 11 9 3 9 3 7
64
37 3 64
3
2
27.
7 4
13. 16 4. The negative sign indicates the negative square root of 16.
2
29.
3
11 1 3 11 3 1 33 3
30.
6
3 2 6 3 6 2 18 2 6 92 2 6 3 2 2 6
14. 25 5. The negative sign indicates the negative square root of 25. 15.
16.
9 49
9 3 49 7
1 1 1 64 64 8
31.
8y 2 y 8y 2 y 16 y2 4 y
32.
15x2 3x2 15x2 3x2 45x
17.
a8a2 a8 b2 a4b
18.
x10 y20 x10 y20 x y
19.
25m 25
20.
9n16
9x4 5 3x2 5
5 10
x 5 x 2 x x 2 x 5 x 5 2
33. 6
9
m6 5m3
n16
x 3 x 10
3n
2
34.
26.
628
8
35.
2
37.
45
36.
23. 5 2 5 3 cannot be simplified.
25.
2
96 2 2 11 6 2
50 8 25 2 4 2 25 2 4 2 5 2 2 2 (5 2) 2 3 2
24. 2 x 25x 36x 3x 2 x 25 x 36 x 3x 2 x 5 x 6 x 3x (2 5 6) x 3x x 3x
3 2 (3)2 2(3) 2 2
8
21. 5 7 7 (5 1) 7 6 7 22.
4
15
8 42 2
45 3 15
24x5
2 24x5 12x4 4x4 3 2x 3 2x
2x 38.
75a4b5 5ab
75a4b5 5ab
15a3b4 a2b4 15a
2 15 2 15 30
ab2 15a
3 3 3 3 9 3
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
39.
1 1 1 1 6 6 6 6 6 6 6 6
40.
x
Chapter 15: Roots and Radicals
4.
x 5 x 5 x 5 10 4 5 2 5 2 5 5 25 x
20 41.
4 6 1
x
2 y 10 0 2 y 10 y 5
4 6 1 6 1 6 1
6 1 6 2 12 4
5.
4 64 6 1 4 6 4 5
42.
9 y2 2 y 10 3y 9 y2 2 y 10 2 (3y)2 9 y2 2 y 10 9 y2
2 1 2 1 5 x 5 x
x 5 x 5
2 12 x 5 2 x 5
52 1x 1 5 x 25 2x 5 2 x 5 x 25 2 x
x 1 x 5 x5 x 12
x 1
2
(x 5)
x 1 x2 10x 25 0 x2 11x 24 0 (x 8)(x 3) 0 x 8 or 0 x 3 8 x 3 x Replacing x with 3 results in a false statement. 3 is an extraneous solution. The only solution is 8. 6. 3 x x 15 2 2 x 3 x 15
x 6 x 9 x 15 6 x 6 x 1 x 1
Section 15.5 Practice Exercises 1.
Vocabulary, Readiness & Video Check 15.5
x2 7
1. The squaring property can result in extraneous solutions, so we need to check our solutions in the original equationbefore the squaring property was appliedto make sure they are actual solutions.
x 2 72 2
x 2 49 x 51
2.
6x 1 x
6x 1 2
x
2. No; if the first squaring leaves a radical term, this new equation can be thought of as an equation that needs the property applied once.
2
6x 1 x 5x 1 0 5x 1 1 x 5
3.
Exercise Set 15.5 2.
x 4
x 42 2
x 16
x 9 2 x 7 x cannot equal 7. Thus, the equation has no solution.
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
4.
x 12 3
16.
x 12 32
2
x 12 9 x 3
x x2 4x 4 0 x2 5x 4 0 (x 4)(x 1) 0 x 4 or 0 x 1 4x 1 x x = 1 does not check, so the solution is x = 4.
6. 4 x 7 5 4 x 12 x 3
x 32 2
18.
x9
2
x 5 x2 2x 1 0 x2 3x 4 0 (x 4)(x 1) x 4 0 or x 1 0 x4 x 1 x = 1 does not check, so the solution is x = 4.
x cannot equal 1. Thus, the equation has no solution. x 4x 3 2 2 x 4x 3
(x 1) 2
2
x2 6x 8 0 (x 4)(x 2) 0
5x 42 x 8 2 5x 4 x 8 4x 4 8 4x 12 x3
4x2 3x 9 2x 2 2 2 4x 3x 9 (2x) 4x2 3x 9 4x2 3x 9 0 3x 9 x3
630
4x 7
x2 2x 1 4x 7
5x 4 x 8
14.
x 4x 7 1 x 1 4x 7
20.
x 4x 3 3x 3 x 1 12.
x 5 x 1
x 5 (x 1)2
8. 3 x 8 5 3 x 3 x 1
10.
x x2
x (x 2)2
2
x 4 0 or x4 22.
x2 0 x2
4x2 3x 2 2x 4x2 3x 2 2 2 (2x) 2 4x 3x 2 4x2 3x 2 0 3x 2 2 x 3 2 x does not check, so the equation has no 3 solution.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
24.
3x2 6x 4 2 3x2 6x 4 2 22 3x2 6x 4 4
3x 0 x0 26.
Chapter 15: Roots and Radicals
34.
x 5 32 2
3x2 6x 0 3x(x 2) 0 or x 2 0 x 2
x5 9 x4 36.
x8 x 2
x 8 x 2 2
2x 1 7 1 2x 1 6 2x 1 cannot equal 6. Thus, the equation has no solution.
2
x8 x 4 x 4
2
9x2 2x 8 (3x)2 9x2 2x 8 9x2 2x 8 0 2x 8 x4
2
32 x 9 x
x 25 x 55 2
x 5 x 55
9x2 2x 8 3x
38.
12 4 x 3 x
28.
x 5 2 5 x 5 3
40.
y
8
y 8
y 82 2
x 10 x 25 x x 55
y 64
10 x 30 3
x 32 2
30.
x 6 x 36 2
x 1 x 15
2
x 2 x 1 x 15 2 x 14 x 7
x 02 2
x0 x = 0 does not check, so the equation has no solution.
x 72 2
x 49
3x 7 5
3x 1 1 4 3x 1 3
3x 7 52
3x 1 32
44. 32.
2
3x 7 25 3x 18 x6
2
x 12 x 36 x 36 12 x 0 x 0
x 1 x 15 2
x 6 x 36
42.
x9
2
3x 1 9 3x 10 10 x 3
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
46.
56. 2x (x 3) 11 2x x 3 11 x 3 11 x 14
x 5 10 x 5
x 52 2
x 25
48.
58. Let x be the width of the rectangle, then the length is x + 2. 2x 2(x 2) 24 2x 2x 4 24 4x 4 24 4x 20 x5 x+2=5+2=7 The length of the rectangle is 7 inches.
5x 6 8x
5x 6 8x 2
2
5x 6 8x 6 3x 2x
50.
2x 5 x 5
2x 5 (x 5)2 2
2x 3
60.
2x 3 x 2 2
2x 5 x2 10x 25 0 x2 12x 20 0 (x 10)(x 2) x 10 0 or x 2 0 x 10 x2 x = 2 does not check, so the only solution is x = 10. 52.
2x 4 2 x 2x 4 x 2 2
2x 4 x2 4x 4
2x 3 x 2 4 x 2 4 x 1 4 x 2
(x 1)2 4 x 2
2
x2 2x 1 16(x 2) x2 2x 1 16x 32
64. a.
x2 0 x2
2x 5 1 x 2x 5 x 1
For V = 1000, r
2x 5 (x 1)2 2x 5 x2 2x 1 0 x2 4 0 (x 2)(x 2) x20 or x 2 0 x 2 x2 x = 2 does not check, so the only solution is x = 2.
10
1.3. 2 100 For V = 100, r 4.0. 2 For V = 10, r
2
632
62. no; answers may vary
0 x2 6x 8 0 (x 4)(x 2)
54.
2
x2 14x 33 0 (x 11)(x 3) 0 x 11 0 or x 3 0 x 11 x3
2x 4 (x 2)2
x 4 0 or x4
x2 2
2
1000 12.6. 2
V
10
100
1000
r
1.3
4.0
12.6
b. No, the volume increases by a factor of 10.
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
66. Use r
3V 3 4
with V = 212,067.
3 212, 067 3 636, 201 r3 37.0 4 4 The radius of the baseball is approximately 37.0 millimeters. 68. Use r 3
3V
with V = 435.
3 435 3 1305 4.7 4 4 The radius of the basketball is approximately 4.7 inches. r3
10 –10
10
–10
The solution of
x 1 2x 3 is x 2.43.
1600 b2 4225
The distance across the pond is 5 105 feet or approximately 51.2 feet. 4. v 2gh 2 32 20 1280 16 5 The velocity of the object after falling 20 feet is exactly 16 5 feet per second or approximately 35.8 feet per second. Vocabulary, Readiness & Video Check 15.6
10
72.
402 b2 652 b2 2625 b 2625 5 105
4
70.
3. The property owner is using a right triangle with one leg measuring 40 feet and hypotenuse measuring 65 feet to find the unknown distance. a2 b2 c2
1. The Pythagorean theorem applies to right –10
triangles only, and in the formula a2 b2 c2 , c is the length of the hypotenuse.
10 –10
The solution of x 5 7x 1 is x 0.48. Section 15.6 Practice Exercises 1. a2 b2 c2 32 42 c2 9 16 c2 25 c2 25 c 5c The hypotenuse has a length of 5 centimeters.
2. Our answer is a number that when squared equals another value. We’re looking for a distance, which must be positive, so our answer must be positive. 3. This Video Notebook asks for an answer rounded to a given place, meaning an estimated answer is expected rather than an exact answer. An exact answer would be given in radical form. Exercise Set 15.6 2. a2 b2 c2 52 b2 102
2. a2 b2 c2 52 32 c2 25 9 c
2
34 c2 34 c 5.83 c
25 b2 100 b2 75 b 75 5 3 8.66 The length of the leg is exactly 5 3 miles or approximately 8.66 miles.
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633
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
4. a2 b2 c2
16.
42 b2 82 16 b2 64 b2 48
a 2 b 2 c2 102 242 c2 100 576 c2
7 b 35 b2 28 2
b 28 b 2 7 5.29 18. The length of the shortest pipe is the length of the hypotenuse of a right triangle with legs both measuring 3 feet. a2 b2 c2
676 c2
32 32 c2
676 c2 26 c 8.
9 9 c2 18 c2
a2 b2 c2
18 c2 c 18 4.24 The length of the shortest pipe is 4.24 feet.
5 b 2 62 2
5 b2 36 b2 31 b2 31 b 31 5.57
20. The height of the kite is the length of one leg of a right triangle where the other leg is 32 feet and the hypotenuse is 80 feet. a2 b2 c2 322 b2 802
2 2 2 10. a b c
2
1024 b 6400
52 b2 92 25 b2 81
b2 5376 2
b 5376 b 5376 73 The height of the kite is 73 feet.
b2 56 b2 56 b 2 14 7.48 12. a2 b2 c2 22 72 c2 4 49 c22 53 c 53 c2 c 53 7.28
14. a2 b2 c2
d 22. t 4 1730 t 4 t 10.40 It takes 10.40 seconds. 24.
s 30 fd 60 30 0.95d 60 28.5d
28.5d
2
a2 12 52
602
a2 1 25
3600 28.5d 126 d The car will skid 126 feet.
a2 24 a2
24 a 2 6 4.90
634
7 b22 35 2 2
b2 48 b 4 3 6.93 6.
a2 b2 c2
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
26.
v 2.5r 30 2.5r
2
302 2.5r 900 2.5r 360 r The radius of curvature is 360 feet.
Chapter 15: Roots and Radicals
40. The height is the length of one leg of a right 1 triangle with other leg (1200) 600 inches 2 1 and hypotenuse (1201) 600.5 inches. 2 a2 b2 c2 6002 b2 600.52
28. d 3.5 h d 3.5 202.7 d 49.8 You can see a distance of 49.8 kilometers. 30. d 3.5 h d 3.5 386.1 d 68.8 You can see a distance of 68.8 kilometers. 32.
25 5 and 25 5, so 5 and 5 are numbers whose square is 25.
34.
49 7 and 49 7, so 7 and 7 are numbers whose square is 49.
36.
121 11 and 121 11, so 11 and 11 are numbers whose square is 121.
38. y is one leg of a right triangle with other leg 5 and hypotenuse 15.
360, 000 b2 360, 600.25 b2 600.25 b2 600.25 b 24.5 The height of the bulge is 24.5 inches. Chapter 15 Vocabulary Check 1. The expressions 5 x and 7 x are examples of like radicals. 2. In the expression 3 45 the number 3 is the index, the number 45 is the radicand, and called the radical sign.
is
3. The conjugate of a + b is a b. 4. The principal square root of 25 is 5.
a2 b2 c2
5. The process eliminating the radical in the denominator of a radical expression is called rationalizing the denominator.
52 y2 152
6. The Pythagorean theorem states that for a right
25 y 225
triangle, (leg)2 (leg)2 (hypotenuse)2 .
2
y2 200 Chapter 15 Review
y2 200 y 10 2 We now find the length of the second leg in the right triangle with leg 5 and hypotenuse 13.
2. 49 7. The negative indicates the negative
a 2 b 2 c2
square root of 49.
52 b2 132 25 b2 169 b2 144 b2 144 b 12 x y b 10 2 12
81 9, because 92 81 and 9 is positive.
1.
3.
3
27 3, because 33 27. 4
4. 4 81 3, because 3 81. 2
9 3 3 . 5. 9 because 64 8 8 64
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals 2
36 6 6 2 6 36 because , and . 81 9 9 3 9 81
6. 7.
4
16 2 because 24 16.
8.
3
8 2 because (2)3 8.
10. a, c; 5 and 4 5 are not real numbers because the radicands are negatives and the indexes are even. 6 2
11.
x
12.
4 4 2 8 x8 x , because (x ) x .
13.
25 3 25 3 5 3 8 8 8
28. 12 12 49 49 29.
30.
12
x , because (x ) x . 6
75 75 64 64
27. 50 50 9 9
9. c; 4 is not a real number because the radicand is negative and the index is even.
12
26.
11
11 11 x x2
7
7
x2
y4
y4
25 2 3 43 7
25 2 5 2 3 3
4 3 2 3 7 7
7 2 y
31.
y5 y5 100 100 x3 81
y4
y4 y 10
y2 y
10
x3
x2 x 9
10
x2 x x x 9 9
14.
25x4 5x2 , because (5x2 )2 25x4 .
15.
40 4 10 4 10 2 10
33. 5 8 8 2 (5 8) 2 3 2
16.
24 4 6 4 6 2 6
34.
17.
54 9 6 9 6 3 6
18.
88 4 22 4 22 2 22
19.
x5 x4 x x4 x x
20.
y7 y6 y
21.
20x2 4x2 5 4x2 5 2x 5
22.
50 y4 25 y4 2 25 y4 2 5 y
y6
81
3 6 3 (1 6) 3 5 3
35. 6 5 3 6 2 5 6 (6 2) 5 (3 1) 6 4 54 6 36. 7 8 2 7 6 2 (1 1) 7 (8 6) 2 2 7 2 2
2
x
y y3 y
37.
2
28 63 56 4 7 9 7 4 14 4 7 9 7 4 14 2 7 3 7 2 14 (2 3) 7 2 14
5 7 2 14
2
23. 3 54 3 27 2 3 27 3 2 33 2
38.
23 11 24. 3 88 3 8 11 3 8 3 11
636
9 y2 3y, because (3y)2 9 y2 . 32.
25.
y
18 18 9 2 9 2 3 2 25 5 5 5 25
75 48 16 25 3 16 3 4 25 3 16 3 4 5 34 34 (5 4) 3 4
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ISM: Prealgebra and Introductory Algebra
39.
40.
5 5 5 5 9 36 9 36 5 5 3 6 2 5 5 6 6 5 6 11
11 11 25 16
47.
41.
3 2 5 3 2 6 10 48.
11
2
49.
50.
44.
51.
2
50x 9 2x 72x 3x 25 2x 9 2x 36 2x 3x 25 2x 9 2x 36 2x 3x 5 2x 9 2x 6 2x 3x (5 9 6) 2x 3x 2 2x 3x
46.
52.
5
2
2
27 3
27 9 3 3
20
20 4 2 5
5
53.
160 160 20 4 5 4 5 2 5 8 8
54.
96 96 32 16 2 16 2 4 2 3 3
55.
30x6 30x6 2x3
2x3
15x3
3 6 3 6 18 9 2 9 2 3 2 5 15 5 15 75 25 3
2
y 4 y 2 y (4) 42
2
x2 15x
x2 15x x 15x
56.
25 3 5 3 45.
x 2 x 2 x (2) 22
y 8 y 16
9x2 5 3 x2 5 7x 5 10 3x 5 3x 5 7x 5 10 (3x 3x 7x) 5 10 x 5 10 or 10 x 5
43.
2
x4 x 4
9x 5 3 x 5 7x 5 10
42.
5 1 5 3 5 5 3 5 1 5 1 3 5 (3 1) 5 3 22 5
45x 3 5x 7x 5 10 2
3 2 6 5 3 6 5 3 2 6 2 5 18 5 3 2 6 10 9 2 5 3 2 6 10
25 16 11 11 5 4 4 11 5 11 20 20 9 11 20
2
Chapter 15: Roots and Radicals
54x5 y2 3xy2
5 7 2 5 2 7 10 14 11 3 5 11 5 3 55 15 Copyright © 2025 Pearson Education, Inc.
54x5 y2 3xy2
18x4 9x4 2 9x4 2 3x2 2
637
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
57.
2
2
11 58.
3
11
2 11 22 11 11
313 39 13 13
11
3
13 59.
11
13
13
13
5 5 5 6 6 6 6 6
56 6
60.
7 7 7 10 10 10 10 10
61.
5x 1 1 5x 5x 5x 5x 5x
62.
63.
64.
65.
5 3y
5
3 x
3 x
x
6 y
6 y
68.
y
69.
3
3 8 10 3
3 1
3 2 5 2
32 52 5 2 52 3 5 2 3 2 5 22 54 15 2 3 2 5 4 1 15 2 3 2 5 4
10
10 x 5 x 5 x 5
x 5
10
x 5
x 25 10 x 50 x 25
6y y
8
70.
8
x 1
3 52 52 52 52
3 1 3 1 2 3 1 2 1 3 11 3 1 6 2 3 1 2
3 1
3
2 1
7 10 70 10 10
5 3y 3y
y
y
3
2 1
x 3x x x
6
3y 3
66.
3y
3y
30 6
67.
71.
1 8 x 1 8 8 x x x 1 x 1 x 1 x 1
2x 6
2x2x 6362
52 5 4
2
5 2
x 18
1
5 2 or 3 5 6 8
10 3
10 3 10 3
10 3 10 9 8 10 3 8
72.
x 3 2 42 x 3 16 x 13
73.
x 38 x2 5
x 52
x3 4
1
10 3 or 8 10 24
x 25
8
74.
x 8 3 x 5 x cannot equal 5. Thus, the equation has no solution.
638
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
75.
Chapter 15: Roots and Radicals
2x 1 x 7
2 2 2 80. a b c
2x 1 (x 7)2 2
6 2 9 2 c2 36 81 c2
2x 1 x2 14x 49 0 x2 16x 48 0 (x 12)(x 4) x 12 0 or x 4 0 x 12 x4 x = 4 does not check, so the solution is x = 12. 76.
3x 1 x 1
117 c2 117 c2 c 3 13 10.82 81. The distance between Romeo and Juliet is the length of the hypotenuse of a right triangle with legs of length 20 feet and 12 feet. a2 b2 c2
3x 1 (x 1)2 2
202 122 c2
3x 1 x 2x 1 2
0 x 5x 0 x(x 5) x = 0 or x 5 0 x5 x = 0 does not check, so the solution is x = 5. 2
77.
x 3 x 15
x 3 x 15 2
2
400 144 c2 544 c2 544 c2 c 4 34 23.32 The distance is exactly 4 34 feet or approximately 23.32 feet. 82. The diagonal of a rectangle forms right triangles with the sides of the rectangle.
x 6 x 9 x 15 6 x6 x 1
a2 b2 c2 52 b2 102 25 b2 100
x 12 2
b2 75
x 1
78.
b2 75 b 5 3 8.66 The length of the rectangle is exactly
x 52 x 1 2
x 5 x 1
5 3 inches or approximately 8.66 inches.
x 5 x 2 x 1 6 2 x 3 x 2 2
3 x 9 x
83.
r
S r
4 72
4 r 2.4 The radius is 2.4 inches.
2 2 2 79. a b c
52 b2 92 25 b2 81 b2 56 b2 56 b 2 14 7.48
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
84.
S 4 S 4
r 6
95.
2
62
S 4 S 36 4 144 S The surface area is 144 square inches.
96.
2
85.
144 12, because 12 144 and 12 is
86. 3 64 4, because 43 64. 8 8 2 16 16x16 4x , because (4x ) 16x .
88.
4x24
89.
18x7 9x6 2x1 9x6
90.
48 y6 y4 81
92.
x9 9
94.
12 2
16 y6 3 y4 y 81
x9 9
2x 3x
16 y6 3
7 14 7 14 98 49 2 49 2 7 2
3
4 y3
98.
3
2x 3
9
9 2 3 3 2
2
x8 x 3
x 8 x x4 x 3 3
99.
3 3 3 2 3 3 6
2 4 5 1 2 5 1 2 4 5 4 1 10 2 4 5 4
12 75 4 3 25 3 4 3 25 3 2 3 5 3 (2 5) 3 7 3
100.
x 3 x 2 x (3) (3) 2
2
x6 x 9 101.
63 28 9 9 7 4 7 3 9 7 4 7 3 3 7 2 7 3 (3 2) 7 3 5 7 3
640
97. 24
2x , because (2x ) 4x .
91.
93.
12
45x3 x 20x 5x3 9x2 5x x 4 5x x2 5x 9x2 5x x 4 5x x2 5x 3x 5x 2x 5x x 5x (3x 2x x) 5x 4x 5x
positive.
87.
3 3 3 3 16 4 16 4 3 3 4 2 3 2 3 4 4 3 4
102.
120 5
60x9 15x
4
120 24 4 6 2 6 5
60x9 15x7
4x2 2x
103.
2 2 2 7 14 7 7 7 7 7
104.
2x 3 2x 3 3 2x 2x 2x 2x
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ISM: Prealgebra and Introductory Algebra
105.
3
3 x 6 x 6 x 6
x 6
Chapter 15: Roots and Radicals
2 2 2 111. a b c 2 2 3 7 c2
x 6
9 49 c2
3
x 36 3 x 18 x 36
106.
107.
7 5
7 5
5 3
58 c2 2 58 c c 58 7.62
5 3
5 3 5 3 7 5 3 7 5 5 53 59 35 3 7 5 5 15 4
112. The diagonal is the hypotenuse of a right triangle with the sides of the rectangle as its legs. a 2 b 2 c2 22 b2 62 4 b2 36 b2 32 b2 32 b 4 2 5.66 The length of the rectangle is exactly
4x 2
4x 22 2
4x 4 x 1
108.
4 2 inches or approximately 5.66 inches. Chapter 15 Getting Ready for the Test
x4 3
x 4 32 2
1. 16 4 while 38
x4 9 x 13 109.
4x 8 6 x 4x 8 x 6
4x 8 x2 12x 36 0 x2 16x 28 0 (x 14)(x 2) x 14 0 or x 2 0 x 14 x2 x = 2 does not check, so the solution is x = 14.
4.
2
x 8 x 4 x 4 12 4 x 3 x
x 32 9x
4 5 4 2 5 16 5 80 2
2
Choice B is correct. 5.
x8 x 2 2
8 2; B.
3. 7 3 2 7 3 2 7 6 Choice B is correct.
2
x 8 x 2
16 is not a real number,
2. 7 3 3 7 3 1 3 (7 1) 3 6 3 Choice C is correct.
4x 8 (x 6)2
110.
2, and
3
28
28 4 7 5 25 25 Choice C is correct.
4 7 2 7 5 5
6.
18x16 9x16 2 9x16 2 3x8 2 Choice D is correct.
7.
64 3 43 4 Choice B is correct. 3
2
8.
3 27 x
3
x39 3 (x9 )3 x9 Choice B is correct.
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ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
2 2 2, to rationalize the
9. Since
5
denominator of
, we multiply by
2 10.
2
13 13 4 13 (11 4) 13 2 13
16.
18 75 7 3 8 9 2 25 3 7 3
3 x 2 10x 9 2 2 3 x 10x 9 2
25
x 1 9x 9 2 (x 1) 2 9x 9
17.
x 2x 1 9x 9 Choice A is correct.
3 3 4 25
16 4, because 42 16 and 4 is positive.
2. 3 125 5, because 53 125. 3. 4 81 3, because 34 81. 2
9 3 3 9 , because . 16 4 16 4
18.
5. 4 81 is not a real number since the index 4 is even and the radicand 81 is negative. 6.
x10 x5 , because (x5 )2 x10 .
7. 8.
54 9 6 9 6 3 6 92 4 23 4 23 2 23
9.
y7 y6 y
10.
24x 4x 6 4x
y6
yy
3
19.
8
8
8
3
5
y 4
6 2x
12. 3 16 3 8 2 3 8 3 2 23 2
3
7 14 7 14 98 49 2 49 2 7 2 2
6 5 2 6 2 5
x 2 x 3 x x 3 x 2 x 2 3
6
x3 x 2 x 6 x x 6
11. 3 27 3 21.
50 10
5
642
y y
12 10 4 3 10 2 3 10 20.
13.
4 25 3 3 2 5 5 3 2 3 10 10 5 32 3 10 (5 2) 3 10 7 3 10
Chapter 15 Test
y
5
42 9 2 25 3 7 3 4 2 3 2 5 3 7 3 2 2 (3 2) 2 (5 7) 3 22 3
2
3 x 10x 9
11.
y2
15.
; C.
9x 10x 9 Choice C is correct.
4.
y2 y 5
y3 25
2
2
1.
y3
14.
5 5 16 4 16
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50 5 10
ISM: Prealgebra and Introductory Algebra
22.
40x4 2x
2x 2 x 5
29.
40x4 2x
Chapter 15: Roots and Radicals
2x 2 (x 5)2 2
20x3
2x 2 x2 10x 25
4x2 5x 4x2 5x 2x 5x 23.
24.
25.
2 3
2
8
8
2
6 3 3 3 3
3
5y
5y
5y
8 62
5y 8
0 x2 12x 27 0 (x 3)(x 9) x 3 0 or x 9 0 x3 x9 x = 3 does not check, so the solution is x = 9.
6 2
2 2 2 30. a b c 2 2 8 b 122
8 5y 5y
64 b2 144 b2 80 b2 80
6 2
b4 5
6 2
6 2 2 6 22 8 6 2
The length is 4 5 inches.
8
31. r
A 15 r r 2.19 The radius is 2.19 meters.
64
6 2
8
2
4
6 2
Cumulative Review Chapters 115
4 6 8
26.
1 3
x
27.
28.
3 x 2 3 x x 3 3
1
x 8 11 x 3 x
1 3 x
2
2 39 xx
x
1. 736.2359 rounded to the nearest tenth is 736.2. 2. 328.174 rounded to the nearest tenth is 328.2. 3.
23.850 1.604 25.454
4.
12.762 4.290 17.052
5.
3.7 y 3.33 3.7(9) 0 3.33
32 x9
3x 62 x 4 2
3x 6 x 4 3x 6 x 4 2x 6 4 2x 10 x5
33.3 3.33 False No, 9 is not a solution. 6.
2.8x 16.8 2.8(6) 0 16.8 16.8 16.8 True Yes, 6 is a solution.
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643
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals 2
1 1 1 1 1 1 1 and is 7. because 36 6 6 6 6 6 36 positive. 2
8.
2 2 2 2 4 2 4 and is because 25 5 5 5 5 5 25 positive.
15. xy 2x 3y 6 x( y 2) 3( y 2) ( y 2)(x 3) 16. 16x3 28x2 12x 21 4x 2 (4x 7) 3(4x 7) (4x 7)(4x 2 3) 17. 3x2 11x 6 (3x 2)(x 3) 18. 9x2 5x 4 9x2 9x 4x 4 9x(x 1) 4(x 1) (x 1)(9x 4)
9. 4(2x 3) 7 3x 5 8x 12 7 3x 5 8x 5 3x 5 5x 5 5 5x 10 x2
19. a.
The expression is undefined for x = 3, since the denominator, x 3, is 0 when x = 3.
b.
The expression is undefined for x = 2 and x = 1, since the denominator, x2 3x 2 (x 2)(x 1), is 0 when x = 2 or x = 1.
c.
There are no values for which the expression is undefined, since the denominator, 3, is never 0.
20. a.
The expression is undefined for x = 0, since the denominator, x, is 0 when x = 0.
b.
There are no values for which the expression is undefined, since the denominator, 5, is never 0.
c.
The expression is undefined for x = 2 and x = 2, since the denominator,
10. 3(2 5x) 24x 12 6 15x 24x 12 6 9x 12 9x 6 6 2 x 9 3 11. a.
12.
1.02 105 102, 000
b.
7.358103 0.007358
c.
8.4 107 84, 000, 000
d.
3.007 105 0.00003007
a.
8.26 104 82, 600
b.
9.9 102 0.099
x2 4 (x 2)(x 2), is 0 when x = 2 or x = 2. x2 4x 4 21.
c.
1.002 105 100, 200
d.
8.039 103 0.008039
13. (3x 2)(2x 5) (3x)(2x) (3x)(5) 2(2x) 2(5) 6x2 15x 4x 10
x2 2x
(x 2)(x 2) x(x 2)
x2 x
2 2 2 2 22. 16x 4y 4(4x y ) 4x 2y 2(2x y) 2 2(2x y)(2x y) 2(2x y) 2(2x y)
6x2 11x 10 14. (5x 1)(4x 1) (5x)(4x) (5x)(1) (1)(4x) (1)(1)
23. a.
20 x2 5x 4x 1 20 x2 x 1
644
b.
a 2a a a 0 4 8 4 4 3 10x
2
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7 15 14x 15 14x 25x 50x2 50x2 50x2
ISM: Prealgebra and Introductory Algebra
24. a.
b.
x 3x 2x 3x x 5 10 10 10 10 9 2
12a
25.
26.
Chapter 15: Roots and Radicals
5
36 15a 2 16a 48a2 48a 36 15a 48a2 3(12 5a) 3 16a2 12 5a 16a2
2 1 x 4x x 30 x2 5 x 6 1 (x 5)(x 6) (x 5)(x 6) x 5 (x 5)(x 6) x 6 4x 2(x 6) x 5 4x 2x 12 x 5 6x 12 x 5 5x 12 5 5x 17 17 x 5 4x
2
3 12x 19 5 2 x 3 3 x 7x 12 x 4 (x 3)(x 4) (x 3)(x 4) 12x 19 5 x 3 (x 3)(x 4) x 4 3(x 4) 12x 19 5(x 3) 3x 12 12x 19 5x 15 3x 12 7x 4 4x 12 4 4x 8 x2
27.
Copyright © 2025 Pearson Education, Inc.
645
ISM: Prealgebra and Introductory Algebra
Chapter 15: Roots and Radicals
y
28.
5 4 3 2 1 –5 –4 –3 –2 –1–1
41. 2 x2 25x5 x5 2x 25x4 x x4 x x=2
2x 25x4 2x 5x
x
1 2 3 4 5
x x
xx
2
2
4
x
x
2x (5x x ) x 2
–2 –3 –4 –5
2
2x 4x2 x 42. 5 x2 36x 49x2 5x 36 x (7x)2
29. 3 1 1, because 13 1.
5x 36 x 7x 5x 6 x 7x
30. 3 8 2, because 23 8.
12x 6 x
31.
3
32.
3
27 3, because (3)3 27. 8 2, because (2)3 8. 3
33.
3
1 . 1 , because 1 1 125 5 5 125 3
3 27 27 34. 3 , because . 64 4 64 4 3
35.
54 9 6 9 6 3 6
36.
40 4 10 4 10 2 10
37.
200 100 2 100 2 10 2
38.
646
44.
45.
7
2
7 2 7 7 7 7
4 4 54 5 5 5 5 5 x 5x 2
x 2 5x 2 2 x 5x 2 4x 2 2 x 4 1 x 2 The solution is
125 25 5 25 5 5 5
75 48 25 3 16 3 25 3 16 3 5 34 3 9 3
1
.
2
39. 7 12 2 75 7 4 3 2 25 3 7 4 3 2 25 3 7 2 3 25 3 14 3 10 3 (14 10) 3 4 3 40.
2 43.
46.
x 5 x 1
x 5 (x 1)2 2
x 5 x2 2x 1 0 x2 3x 4 0 (x 4)(x 1) x 4 0 or x 1 0 x 1 x4 The possible solution x = 1 does not check. The solution is 4.
Copyright © 2025 Pearson Education, Inc.
Chapter 16 Section 16.1 Practice Exercises
x
1. x2 16 0
The solutions are
x2 16 x 16 or x 16 x4 x 4 The solutions are 4 and 4.
7.
x
11 3 11 3
or x
3 3 33 x 3 The solutions are
11
3 11 3 x 3 3 33 x 3 33 33 . and 3 3
x 4 49 or x 4 49 x4 7 x 4 7 x 11 x 3 The solutions are 11 and 3. 4. (x 5)2 18
Vocabulary, Readiness & Video Check 16.1 1. To solve, a becomes the radicand and the square root of a negative number is not a real number. 2. A quadratic equation must be in the form of a variable (or polynomial) squared equal to some nonnegative number in order for the property to be used. For Video Notebook 6, we use h 16t2, so this equation can easily be placed in this form. The negative value is rejected because it will not be used in the context of the application. Exercise Set 16.1 2.
x 5 18 or x 5 18 x 5 3 2 x 5 3 2 x 53 2 x 53 2 x 53 2 The solutions are 5 3 2.
4.
5. (x 3) 5 This equation has no real solution because the square root of 5 is not a real number. 2
4x 1 15 4x 1 15 1 15 x 4
or
h 16t 2 759 t 2 27.5 t or 27.5 t 27.5 is rejected because time cannot be negative. The free-falling dive of 12,144 feet lasted approximately 27.5 seconds.
3. (x 4)2 49
6. (4x 1)2 15
1 15 . 4
12,144 16t 2
2. 3x2 11 11 x2 3 x
1 15 4
k2 9 0 (k 3)(k 3) 0 k 3 0 or k 3 0 k 3 k 3 The solutions are k = 3 and k = 3. m2 6m 7 m2 6m 7 0 (m 7)(m 1) 0 m 7 0 or m 1 0 m 7 m 1 The solutions are m = 7 and m = 1.
4x 1 15 4x 1 15 1 15 x 4
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647
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
2x2 98 0
6.
2(x2 49) 0 2(x 7)(x 7) 0 x 7 0 or x 7 0 x7 x 7 The solutions are x = 7 and x = 7. 7a2 175 0
8.
7(a2 25) 0 7(a 5)(a 5) 0 a 5 0 or a 5 0 a5 a 5 The solutions are a = 5 and a = 5. x2 10x 24
10.
x2 10x 24 0 (x 6)(x 4) 0 x6 0 or x 4 0 x 6 x 4 The solutions are x = 6 and x = 4. 12. x2 121 x 121 or x 121 x 11 x 11 The solutions are x = 11. 14. x2 22 x 22
or x
16. x x x
1 4
x x
5 2
22
5 10
5 5
10 . 5
9 2 x 9 2 2 2 3 2 x 2
or x
3 2 . 2
x2 15 x 15
or
x 15
The solutions are x 15. or x
16 x
26. (x 2)2 25
1 16 1
x 2 25 or x 2 25 x2 5 x 2 5 x3 x 7 The solutions are x = 3 and x = 7.
4
1 The solutions are x . 4
28. (x 7)2 2
18. x2 25 has no real solution because the square root of 25 is not a real number.
x7 2 x 7 2
or
x 7 2 x 7 2
The solutions are x 7 2.
648
5
The solutions are 22. 2x2 9 9 x2 2 9 x 2 9 2 x 2 2 3 2 x 2
2
24. 3x2 45 0 3x2 45
1 16 1
or x
The solutions are x
The solutions are x 22. 2
20. 5x2 2 2 x2 5 2 x 5 2 5 x 5 5 10 x 5
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ISM: Prealgebra and Introductory Algebra 2
1 1 30. m 3 9 1 1 m or 3 9 1 1 m 3 3 m0
Chapter 16: Quadratic Equations 40. (5x 11)2 54
m m
1 3 1
5x 11 54 5x 11 3 6
1 1
5x 11 3 6 11 36 x 5
5x 11 3 6 11 3 6 x 5 11 3 6 . The solutions are x 5
3
3 3 2 m 3
2 The solutions are m = 0 or m . 3
or
p 7 13
x2 35 x 35 or x 35
p 7 13
The solutions are x 35.
p 7 13
The solutions are p 7 13.
44. (x 5)2 20 x 5 20 x5 2 5 x 5 2 5
34. (4 y 3)2 81 4 y 3 81 4y 3 9
or
42. x2 35 0
32. ( p 7)2 13 p 7 13
or 5x 11 54 5x 11 3 6
4 y 3 81 4 y 3 9
4 y 6 6 3 y 4 2 3 The solutions are y = 3 and y . 2 36. (z 7) 20 has no real solution because the
x 5 20 x 5 2 5 x 5 2 5
The solutions are x 5 2 5.
4 y 12 y3
2
or
46.
1 2 y 2 5 y2 10 y 10 or
y 10
The solutions are y 10.
square root of 20 is not a real number. 38. (3x 17)2 28 3x 17 28 3x 17 2 7 3x 17 2 7 17 27 x 3
or 3x 17 28 3x 17 2 7 3x 17 2 7 17 2 7 x 3 17 2 7 The solutions are x . 3
48. (7x 2)2 11 7x 2 11 7x 2 11 x
or 7x 2 11 7x 2 11
2 11 7
The solutions are x
x 2 11
2 11 7
.
7 50. 3z 2 24 z2 8 z 8 z2 2
or z 8 z 2 2
The solutions are z 2 2.
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649
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
52. (10 9x)2 75 0
62.
(10 9x)2 75
10 r2
10 9x 75 10 9x 5 5 9x 10 5 5 10 5 5 x 9
10 r2
or 10 9x 75 10 9x 5 5 9x 10 5 5 10 5 5 x 9 10 5 5 The solutions are x . 9 54.
A r 2
10 r or 10 r Since the radius does not have a negative length, reject the solution 10. The radius is 10 units. 64. y2 10 y 25 y2 2( y)(5) (5)2 ( y 5)2 66. x2 20x 100 x2 2(x)(10) (10)2
A s2
(x 10)2
32 s2 32 s or 32 s 4 2s 4 2 s Since the length of a side is not a negative number, reject the solution 4 2. The sides have length 4 2 5.66 meters. 56.
y 5 11 y 5 11 y 5 11
h 16t
h 16t 2 400 16t 2 25 t2 25 t or 25 t 5t 5 t Since the time of a fall is not a negative number, reject the solution 5. It would take 5 seconds for the goggles to hit the water.
or
y 5 11 y 5 11
The solutions are y 5 11. 72. (x 1.37)2 5.71 x 1.37 5.71 or
x 1.37 5.71
x 1.37 2.39 x 1.37 2.39 x 3.76 x 1.02 The solutions are x = 3.76 and x = 1.02.
2
696 t2 696 t or 696 t 26.4 t 26.4 t Since the time would not be a negative number, reject the solution 26.4. The free-fall lasted approximately 26.4 seconds.
650
( y 5)2 11
As
11,136 16t2
60.
70. y2 10 y 25 11
2
3039 s2 3039 s or 3039 s 55.13 s 55.13 s Since the length of a side is not a negative number, reject the solution 3039. The sides have length 3039 feet 55.13 feet. 58.
68. answers may vary
74.
y 760(x 12)2 95, 628 400, 000 760x 21)2 95, 628 495, 628 760(x 12)2 495, 628 (x 12)2 760 495, 628 x 12 760 495, 628 12 x 760 14 x or 495, 628 x 12 760 495, 628 12 x 760 38 x If the trend continues, the first year after 2015 in
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ISM: Prealgebra and Introductory Algebra
which there will be 400,000 thousand paying Spotify subscribers worldwide will be 2029 (2015 + 14).
Chapter 16: Quadratic Equations
4.
2x2 10x 13 13 x2 5x 2 25 13 25 x2 5x 42 2 4 5 1 x 2 4 There is no real solution to this equation since the square root of a negative number is not a real number.
5.
2x2 6x 5 5 x2 3x 2 5 x2 3x 2 9 5 9 x2 3x 4 2 4 2 3 19 x 2 4 3 19 3 19 or x x 2 4 2 4 3 19 3 19 x x 2 2 2 2 3 19 3 19 x x 2 2 2 2 3 19 The solutions are . 2
Section 16.2 Practice Exercises 1.
x2 8x 1 0 x2 8x 1 x2 8x 16 1 16 (x 4)2 15 x 4 15 x 4 15
or x 4 15 x 4 15
The solutions are 4 15. 2.
x2 14x 32 x 14x 49 32 49 2
(x 7)2 17 x7 or 17 x 7 17
x 7
17 x 7 17
The solutions are 7 17. 3. 4x2 16x 9 0 9 x2 4x 0 4 9 x2 4x 4 9 x2 4x 4 4 4 25 2 (x 2) 4 25 25 x2 or x 2 4 4 5 5 x2 x2 2 2 5 5 x 2 x 2 2 2 9 1 x x 2 2 9 1 The solutions are and . 2 2
Vocabulary, Readiness & Video Check 16.2 1. By the zero-factor property, if the product of two numbers is zero, then at least one of these two numbers must be zero. 2. If a is a positive number, and if x2 a, then x a. 3. An equation that can be written in the form ax2 bx c 0 where a, b, and c are real numbers and a is not zero is called a quadratic equation. 4. The process of solving a quadratic equation by writing it in the form (x a)2 c is called completing the square. 5. To complete the square on x2 6x, add 9.
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651
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations 2
b 6. To complete the square on x bx, add . 2
4.
2
z2 6z 9 2 2 6 6 2 z 6z 9 2 2 z2 6z 9 9 9 (z 3)2 18
2
8 7. 42 16; p2 8 p 16 2 2
6 2 2 8. 3 9; p 6 p 9 2
z 3 18 z3 3 2 z 3 3
2
20 2 2 9. 10 100; x 20x 100 2 18
2
2
2
14
2
6.
2
y2 4 y 0 2 4 2 4 y2 4 y 0 2 2 2 y 4y 4 0 4 ( y 2)2 4 y 2 4 or y 2 4 y 2 2 y 2 2 y0 y 4
2
2 2 2 12. 1 1; y 2 y 1 2 13. When working with equations, whatever is added to one side must also be added to the other side to keep equality. 2
14. The coefficient of y is 2. The method of completing the square works only when the coefficient of the squared variable is 1, so we must first divide through by 2. Exercise Set 16.2 x2 10x2 24 2 10 10 2 x 10x 24 2 2 x2 10x 52 24 52 x2 10x 25 24 25
The solutions are y = 0 and y = 4. 8.
y2 5 y 6 0 y2 5 y2 6 2 5 5 y2 5 y 6 2 2 2 5 1 y 4 2 5 1 5 1 y or y 2 4 2 4 5 1 5 1 y y 2 2 2 2 6 4 y 3 y 2 2 2 The solutions are y = 3 and y = 2.
(x 5)2 1 x 5 1 or x 5 1 x 5 1 x 5 1 x6 x4 The solutions are x = 6 and x = 4.
652
2
z 3 18 z 3 3 2 z 3 3
11. 7 49; y 14 y 49 2
2.
or
The solutions are z 3 3 2.
2
10. 9 81; x 18x 81 2
z2 6z 9 0
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2
ISM: Prealgebra and Introductory Algebra
10.
x2 4x 2 0
Chapter 16: Quadratic Equations
x2 4x 2 2 2 4 4 2 x 4x 2 2 2 2 x 4x 42 2 4 (x 2) 2 x2 2 x 2 2
or
3 2 2 14 4 4 x 4x 2 3 2 2 2 (x 2) 3 2
x 2 2 x 2 2
The solutions are x 2 2. 12.
14.
3x2 12x 14 0 14 x2 4x 0 3 14 x2 4x
16.
This has no real solution because
x2 7x 5 2 2 x 2 7x 7 5 7 2 2 49 69 2 x 7x 4 4 2 7 69 x 2 4 7 69 7 69 x or x 2 4 2 4 7 69 7 69 x x 2 2 2 2 7 69 7 69 x x 2 2 2 2 7 69 The solutions are . 2 2x2 18x 40 x2 9x 20 2 2 x 2 9x 9 20 9 2 2 2 9 1 x 2 4 9 9 x 1 or x 1 4 4 2 2 9 1 9 1 x x 2 2 2 2 8 10 x 4 x 5 2 2 The solutions are x = 4 and x = 5.
2 is not a 3
real number. 4x2 20x 3
18.
4x2 20x 3 3 x2 5x 4 5 2 3 5 2 2 5x x 2 2 4 2 5 28 x 2 4
5 28 2 4 5 2 7 x 2 2 5 2 7 x 2 2 x
or x x
5 2 5
28
4 2 7
2
2
2 7 x 2 2 5 2 7 The solutions are x or 2 5 x 7. 2
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5
653
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
20.
4z 2 8z 1 0 1 z2 2z 0 4 1 z2 2z 4 2 2 1 2 2 2 z 2z 3 2 (z 21)2 4 4 3 3 z 1 or z 1 4 z 1
26.
4 3
z 1
2
3 2
The solutions are z 1
3
.
2 x2 6x 7 0
22.
x2 6x2 7 2 6 6 2 x 6x 7 2 2 (x 3)2 2 x3 2 x 3 2
or
3 2 3 2 x2 3x 18 2 2 2 3 81 x 4 2
x 3 2 x 3 2
3 81 3 81 or x 2 4 2 4 3 9 3 9 x x 2 2 26 9 12 x 6 x 3 2 2 The solutions are x = 6 or x = 3. x
x2 9x 3 0 x2 9x 3 92 9 2 x 9x 3 2 2 2 9 69 x 2 4 9 69 9 69 or x x 2 4 2 4 9 69 9 69 x x 2 2 2 2 2
The solutions are x 9 2 69 .
654
x(x 3) 18 x2 3x 18
28.
The solutions are x 3 2. 24.
2 y2 y 1 0 y 1 y2 0 2 2 y 1 y2 2 2 2 2 1 1 2 y 1 y 2 41 2 29 4 y 4 16 1 9 1 9 or y y 4 16 4 16 1 3 1 3 y y 4 4 4 4 4 2 1 y 1 y 4 4 2 1 The solutions are y = 1 and y . 2
30.
2 y2 8 y 9 0 9 y2 4 y 0 2 9 y2 4 y 2 2 2 9 4 4 2 y 4 y 2 2 12 ( y 2)2 2 The equation has no real solution, since the 1 square root of is not a real number. 2
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ISM: Prealgebra and Introductory Algebra
6x2 30x 36
32.
1786 20x2 147x 1487 299 20x2 147x 299 2 147 x x 20 20 2 147 299 147 2 2 147 20 x 20 x2 40 40 45, 529 147 x 1600 40 45, 529 147 x 1600 40 147 45, 529 x 40 40 9x or 45, 529 147 x 1600 40 147 45, 529 x 40 40 2 x Since the time will not be a negative number in this context, reject the solution x 2. The solution is x 9, so the model predicts that the year-opening price of gold was $1786 per ounce in 2022 (2013 + 9).
16 3 4 7 3 25 5 5 5 5
9 36. 10
y 20x2 147x 1487
48.
x2 5x 6 2 2 5 5 2 x 5x 6 2 2 2 1 x 5 2 4 x 52 1 or x 52 1 54 1 5 41 x x 2 2 2 2 6 4 x 3 x 2 2 2 The solutions are x = 3 and x = 2. 34.
Chapter 16: Quadratic Equations
9 7 2 1 49 5 100 10 10 10
38.
10 20 3 10 20 3 5 10 3 2 2 2
40.
12 8 7 12 8 7 3 2 7 3 2 7 16 16 16 4 4 4
50. The solutions are x = 4 and x = 6.
42. answers may vary
52. The solutions are x 0.65 and x 7.65.
44. answers may vary
Section 16.3 Practice Exercises 2
46. x kx k is a perfect square trinomial, 2 then 2 k 2
25 2 k k 25 or 25 2 2 k 25 k 2(5) k 10 k 10 2 x kx 25 is a perfect square trinomial when k = 10 or k = 10.
1.
2x2 x 5 0
a = 2, b = 1, c = 5 x
b b2 4ac 2a
x
(1) (1)2 4(2)(5) 2(2)
1 1 40 4 1 41 4
The solutions are
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1 41 . 4
655
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
3x2 8x 3
2.
1
5.
3x2 8x 3 0 a = 3, b = 8, c = 3
2 1 2 2 x 5x 11 2(0) 2 x2 10x 22 0 a = 1, b = 10, c = 22
b b2 4ac x 2a x
x
(10) (10)2 4(1)(22) 2(1) 10 12 2 10 2 3 2
5x 2 0 a = 5, b = 0, c = 2 6.
0 0 4(5)(2) 2(5) 40 10 2 10 10 10 5 10 . The solutions are 5 2
1. The quadratic formula is x
b b2 4ac . 2a
2. 5x2 7x 1 0; a = 5, b = 7, c = 1 3. x2 3x 7 0; a = 1, b = 3, c = 7 4. x2 6 0; a = 1, b = 0, c = 6 5. x2 x 1 0; a = 1, b = 1, c = 1
b b 4ac 2a 2
2 22 4(1)(3) 2(1)
6. 9x2 4 0; a = 9, b = 0, c = 4 2 8 2
There is no real number solution because not a real number.
656
1.4
Vocabulary, Readiness & Video Check 16.3
x 2x 3 0 a = 1, b = 2, c = 3
x
4 41
1.9
4
2
x
1 41
1
x2 2x 3
4.
The solutions are 5 3.
b b2 4ac x 2a
2 5 3 2 5 3
2
x
b b2 4ac 2a
x
5x2 2
3.
2 x2 5x 11 0
1
8 82 4(3)(3) 2(3) 8 64 36 6 8 100 6 8 10 6 8 10 1 8 10 x or 3 6 6 3 1 The solutions are and 3. 3
x2 5x 11
8 is
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ISM: Prealgebra and Introductory Algebra
7.
1 12 4(1)(2) 2(1)
Chapter 16: Quadratic Equations
1 1 8 2 1 9 2 1 3 2
x
2(1) 5 25 24 2 5 49 2 57 2 5 7 12 57 2 x 6 or x 1 2 2 2 2 The solutions are x = 6 and x = 1.
1 3 2 1 2 2 1 3 4 2 2 2
(5) (5)2 4(2)(3) 5 25 24 8. 4 2(2) 5 1 6 3 4 4 2
4. 7k 2 3k 1 0 a = 7, b = 3, c = 1
5 1 4 5 1 4
k
9.
7 72 4(2)(1) 7 49 8 7 41 4 4 2(2)
10.
5 52 4(1)(2) 5 25 8 5 17 2(1) 2 2
3 32 4(7)(1) 2(7) 3 9 28 14 3 37 14 3 37 The solutions are k . 14
12. The exact solution values are keyed into a calculator, and the decimal approximations are then rounded to the requested place value. Exercise Set 16.3 2. x2 5x 6 0 a = 1, b = 5, c = 6 x b b 4ac 2a 2
25x2 15 0
6.
25x2 0x 15 0 a = 25, b = 0, c = 15
yes, in order to make sure you have correct values for a, b, and c
b. No; it simplifies calculations, but you would still get a correct answer using fraction values in the formula.
b b2 4ac 2a
k
5 1 4 1 4 4
11. a.
(5) (5)2 4(1)(6)
x
b b2 4ac 2a
2 x 0 0 4(25)(15) 2(25) 1500 50 10 15 50 15 5
The solutions are x
Copyright © 2025 Pearson Education, Inc.
15 . 5
657
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
8. 3x2 2x 1 0 a = 3, b = 2, c = 1 x
b b 4ac 2a
x
2 22 4(3)(1) 2(3)
m2 14 5m
14.
m2 5m 14 0 a = 1, b = 5, c = 14
2
2 m b b 4ac 2a (5) (5)2 4(1)(14) m 2(1) 5 25 56 2 5 81 2 59 2 5 9 4 5 9 14 2 m 7 or m 2 2 2 2 The solutions are m = 7 and m = 2.
2 4 12 6 2 8 6 The equation has no solution, since the square root of 8 is not a real number.
y2 5 y 36
10.
y2 5 y 36 0 a = 1, b = 5, c = 36 b b2 4ac
y y
10 x 2 2x 16. 0 x2 2x 10 a = 1, b = 2, c = 10
2a (5) (5)2 4(1)(36)
2(1) 5 25 144 2 5 169 2 5 13 2 5 13 y 9 or y 5 13 4 2 2 The solutions are y = 9 and y = 4.
x
x
5x2 15 0 a = 5, b = 0, c = 15 b b 4ac x 2a 2
x
0 02 4(5)(15) 2(5)
300 10 10 3 10 3
2 22 4(1)(10)
2(1) 2 4 40 2 2 44 2 2 2 11 2 1 11 The solutions are x 1 11.
5x2 15
12.
b b2 4ac 2a
18.
3x2 9x 8 3x2 9x 8 0 a = 3, b = 9, c = 8 2 x b b 4ac 2a
The solutions are x 3.
658
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
x
Chapter 16: Quadratic Equations
24. 5x2 8x 2 0 a = 5, b = 8, c = 2
(9) (9)2 4(3)(8) 2(3)
2 b b 4ac x 2a
9 81 96 6 9 177 6
The solutions are x
y 9 177
8 64 40 10 8 24 10 8 2 6 10 4 6 5
.
6 11p2 2 10 p
20.
11p2 10 p 2 0 a = 11, b = 10, c = 2 p
b b2 4ac 2a
(10) (10)2 4(11)(2) p 2(11) 10 100 88 22 10 12 22 10 2 3 22 5 3 11 The solutions are p
5 3 . 11
22. x2 10x 19 0 a = 1, b = 10, c = 19 x x
b b2 4ac 2a (10) (10)2 4(1)(19) 2(1)
10 100 76 2 10 24 2 10 2 6 2 5 6
(8) (8)2 4(5)(2) 2(5)
The solutions are x
4 6
.
5 5 y2 4 y
26.
5 y2 y 4 0 a = 5, b = 1, c = 4 2 y b b 4ac 2a
y
1 12 4(5)(4) 2(5)
1 1 80 10 1 81 10 1 9 10 1 9 8 4 or y 1 9 10 1 y 10 10 5 10 10 4 The solutions are y and y = 1. 5
The solutions are x 5 6.
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659
6z 2 2 3z
28.
34. x2 2x 5 0 a = 1, b = 2, c = 5
6z2 3z 2 0 a = 6, b = 3, c = 2
z
2 x b b 4ac 2a
b b 4ac 2a 2
(2) (2)2 4(1)(5) 2(1) 2 4 20 2 2 24 2 22 6 2
x
3 32 4(6)(2) 2(6) 3 9 48 12 3 57 12 3 57 The solutions are z . z
1 6 The solutions are x 1 6.
12 30. k 2 2k 5 0 a = 1, b = 2, c = 5 k
36. 5x2 13x 6 0 a = 5, b = 13, c = 6
b b2 4ac 2a
2 22 4(1)(5) 2 4 20 2 16 2 2 2(1) The equation has no solution, since the square root of 16 is not a real number. 2z 2 z 3
32.
2z2 z 3 0 a = 2, b = 1, c = 3 b b2 4ac 2a (1) (1)2 4(2)(3)
z z
b b2 4ac 2a
x
k
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
2(2)
1 1 24 4
2 x (13) (13) 4(5)(6) 2(5) 13 169 120 10 13 289 10 13 17 10 13 17 30 13 17 4 2 x 3 or x 10 10 10 10 5 2 The solutions are x = 3 and x . 5 2 m 38. 3m 1 2 m2 3m 1 0 2 m2 6m 2 0 a = 1, b = 6, c = 2
1 25 4 1 5 4 1 5 6 3 1 5 4 z or z 1 4 4 2 4 4 3 The solutions are z and z = 1. 2
660
m
b b2 4ac 2a
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
m
Chapter 16: Quadratic Equations
10 2 12 3 10 2 8 1 or p 16 16 4 16 16 2 3 1 The solutions are p and p . 4 2
(6) (6)2 4(1)(2) 2(1)
p
6 36 8 2 6 28 2 62 7 2 3 7
2x2
44. 2
2x
x
a = 5, b = 2, c = 1
p
b b2 4ac 2a (2) (2)2 4(5)(1)
4 p2 2
4p 5p
3
5p
2 3
0 2 8 p2 10 p 3 0 a = 8, b = 10, c = 3 2 p b b 4ac 2a
p
(10) (10)2 4(8)(3) 2(8)
10 100 96 16 10 4 16 10 2 16
2 0
2
(5) (5)2 4(4)(7)
2(4) 5 25 112 8 5 137 8
2(5) 2 4 20 10 2 16 10 The equation has no solution, since the square root of 16 is not a real number. 42.
7
2 x b b 4ac 2a
5 2 1 p p 0 40. 2 2 5 p2 2 p 1 0
7
x
x 2 2 4x2 5x 7 0 a = 4, b = 5, c = 7
The solutions are m 3 7.
p
5
5
The solutions are x
46.
5 137 . 8
2 2 2 x 2x 0 3 3 2x2 6x 2 0 a = 2, b = 6, c = 2 2 x b b 4ac 2a
x
(6) (6)2 4(2)(2)
2(2) 6 36 16 4 6 52 4 6 2 13 4 3 13 2 The solutions are x
3 13
.
2
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661
48.
9z 2 12z 1
x
9z2 12z 1 0 a = 9, b = 12, c = 1
b b2 4ac z 2a
x2 7x 5
54.
x 7x 5 0 a = 1, b = 7, c = 5 2
2 x b b 4ac 2a
(7) (7)2 4(1)(5) x 2(1) 7 49 20 2 7 69 2 7 69 The solutions are x or x 7.7 or 2 x 0.7.
2x2 26 2x2 0x 26 0 a = 2, b = 0, c = 26 x x
b b2 4ac 2a 0 02 4(2)(26) 2(2)
208 4 4 13 4 13
56. 5x2 3x 1 0 a = 5, b = 3, c = 1 x
The solutions are x 13 or x 3.6 and x 3.6. 52. x2 4x 2 0 a = 1, b = 4, c = 2 b b2 4ac x
4 16 8 2
4 8 2 4 2 2 2 2 2 The solutions are x 2 2 or x 3.4 and x 0.6.
12 12 4(9)(1) 2(9) 12 144 36 18 12 108 18 12 6 3 18 2 3 3 2 3 The solutions are z . 3 50.
4 42 4(1)(2) 2(1)
2
z
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
2a
b b2 4ac 2a
2 x (3) (3) 4(5)(1) 2(5) 3 9 20 10 3 29
10 The solutions are x x 0.2.
662
Copyright © 2025 Pearson Education, Inc.
3 29 or x 0.8 and 10
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
58. x = 4 is a vertical line with x-intercept (4, 0). x
6 62 4(1)(95)
2(1) 6 36 380 2 6 416 2 6 4 26 2 3 2 26 x 3 2 26 7.2 or x 3 2 26 13.2 Since the width is not a negative number, discard the solution x = 13.2. x + 6 = 7.2 + 6 = 13.2 The width was approximately 7.2 feet and the length was approximately 13.2 feet.
60. y = 2x + 3 is a line with slope of 2 and y-intercept (3, 0).
2 68. y 2 5 y 1 0
a = 1, b 2 5, c = 1 y
2
62.
5x 2 x 5x2 x 2 0 c = 2, which is choice d.
y
0 7 y2 3y 0 c = 0, which is choice c. 66. Let x be the width of the table. Then the length is x + 6. Area (length)(width) 95 (x 6)(x) 0 x2 6x 95 a = 1, b = 6, c = 95 x
b b2 4ac 2a
2 5 4(1)(1) 2
2 5
2(1)
2 5 20 4 2 2 5 24 2 2 52 6 2 5 6
64. 7 y2 3y
95 x2 6x
b b2 4ac 2a
The solutions are y 5 6. 70. answers may vary 72. 1.2x2 5.2x 3.9 0 c = 1.2, b = 5.2, c = 3.9 x
b b2 4ac 2a
(5.2) (5.2)2 4(1.2)(3.9) 2(1.2) 5.2 27.04 18.72 2.4 5.2 45.76 2.4
x
Copyright © 2025 Pearson Education, Inc.
663
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x
5.2
Mid-Chapter Review Practice Exercises
45.76 5.0 or 2.4
5.2 45.76 x 0.7 2.4 The solutions are x 5.0 and x 0.7.
1. y2 4 y 6 0 Use the quadratic formula with a = 1, b = 4, and c = 6. 2 y b b 4ac 2a
74. h 16t 2 120t 80
0 16t2 120t 80 a = 16, b = 120, c = 80 t
b b2 4ac 2a
t
120 1202 4(16)(80) 2(16)
120 14, 400 5120 32
t
120 19, 520 32 120 19, 520 32
(4) (4)2 4(1)(6) 2(1) 4 16 24 2 4 40 2 4 2 10 2 2 10
y
The solutions are y 2 10. 2. (2x 5)2 45
0.6 or
Use the square root property.
120 19, 520 t 8.1 32 Since the time of the flight is not a negative number, discard the solution 0.6. The rocket will strike the ground approximately 8.1 seconds after it is launched. 76.
y 0.2x2 0.9x 15.4
(2x 5)2 45 2x 5 45 2x 5 3 5 2x 5 3 5 5 3 5 x 2
22 0.2x2 0.9x 15.4
The solutions are x
0 0.2x 0.9x 6.6 a = 0.2, b = 0.9, c = 6.6
3.
2 x b b 4ac 2a
(0.9) (0.9) 4(0.2)(6.6) x 2(0.2) 0.9 0.81 5.28 0.4 0.9 6.09 0.4 0.9 6.09 0.9 6.09 x 8 or x 4 0.4 0.4 Since the time is not negative in this context, discard the solution x 4. The solution is x 8, so the model predicts that the number of restaurant and food service jobs will reach 22 million in 2027 (2019 + 8). 664
.
2
2
2
5 3 5
5 3 x 2 2 3 2 5 x x 0 2 2 x2
2x2 5x 3 0 (2x 3)(x 1) 0 2x 3 0 or x 1 0 2x 3 x 1 3 x 1 x 2 3 The solutions are x and x = 1. 2
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
Mid-Chapter Review 1.
5x2 11x 2 0 (5x 1)(x 2) 0 5x 1 0 or x 2 0 5x 1 x2 1 x 5 1 The solutions are x and x = 2. 5
2.
5x2 13x 6 0 (5x 2)(x 3) 0
2 x (6) (6) 4(1)(7) 2(1) 6 36 28 2 6 8 2 62 2 2 3 2
The solutions are x 3 2.
5x 2 0 5x 2 2 x 5
or
5. a2 20 a 20 a 2 5
x3 0 x 3
The solutions are x
The solutions are a 2 5. 2 5
6. a2 72 and x = 3.
a 72 a 6 2 The solutions are a 6 2.
x2 1 2x
3.
x 2x 1 0 a = 1, b = 2, c = 1
7. x2 x 4 0 a = 1, b = 1, c = 4
2 x b b 4ac 2a
2 x b b 4ac 2a
(2) (2)2 4(1)(1) x 2(1)
(1) (1)2 4(1)(4) 2(1) 1 116 2 1 15 2 The equation has no solution, since the square root of 15 is not a real number.
2
x
2 44 2 2 8 2 22 2 2 1 2
8. x2 2x 7 0 a = 1, b = 2, c = 7
The solutions are x 1 2. x2 7 6x
4.
x2 6x 7 0 a = 1, b = 6, c = 7 x
b b2 4ac 2a
x
b b2 4ac 2a
2 x (2) (2) 4(1)(7) 2(1)
2 4 28 2 2 24 2 The equation has no solution, since the square root of 24 is not a real number.
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665
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
9. 3x2 12x 12 0
15.
x2 3x 2 0 (x 2)(x 1) 0 x 2 0 or x 1 0 x2 x 1 The solutions are x = 1 and x = 2.
16.
x2 7x 12 0 (x 3)(x 4) 0 x 3 0 or x 4 0 x 3 x 4 The solutions are x = 3 and x = 4.
3(x2 4x 4) 0 3(x 2)(x 2) 0 x 2 0 or x 2 0 x2 x2 The solution is x = 2. 10. 5x2 30x 45 0 5(x2 6x 9) 0 5(x 3)(x 3) 0 x 3 0 or x 3 0 x3 x3 The solution is x = 3. 11.
12.
13.
9 6 p p2 0 (3 p)(3 p) 0 3 p 0 or 3 p 0 3 p 3 p The solution is p = 3. 49 28 p 4 p2 0 (7 2 p)(7 2 p) 0 7 2p 0 or 7 2 p 0 7 2p 7 2p 7 7 p p 2 2 7 The solution is p . 2 4 y 16 0 2
4( y 4) 0 4( y 2)( y 2) 0 y20 y 2 0 or y 2 y2 2
The solutions are y = 2. 14.
3y2 27 0 3( y2 9) 0 3( y 3)( y 3) 0 y 3 0 or y3
y3 0 y 3
The solutions are y = 3.
666
17. (2z 5)2 25 2z 5 25 or 2z 5 25 2z 5 5 2z 5 5 2z 0 2z 10 z0 z 5 The solutions are z = 0 and z = 5. 18. (3z 4)2 16 3z 4 16 3z 4 4 3z 8 8 z 3
or 3z 4 16 3z 4 4 3z 0 z0
The solutions are z
8 3
and z = 0.
19. 30x 25x2 2 0 25x2 30x 2 a = 25, b = 30, c = 2 b b2 4ac 2a 2 ( 30) (30) 4(25)(2) x 2(25) 30 900 200 50 30 700 50 30 10 7 50 3 7 5 3 7 The solutions are x . 5 x
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
20. 12x 4x2 4
x
0 4x 12x 4 2
0 4(x2 3x 1)
5 25 80 20 5 105 20
a = 1, b = 3, c = 1 x
x
(5) (5)2 4(10)(2) 2(10)
b b2 4ac 2a
The solutions are x
(3) (3)2 4(1)(1)
5 105
.
20
2(1) 3 9 4 2 3 5 2 3 5 The solutions are x .
2 24. x
2
x
5 2 m 1 0 m 8 2 5m2 8m 4 0 (5m 2)(m 2) 0 5m 2 0 or m 2 0 5m 2 2 m 5
2 5
23.
x2
1 1 x 0 2 5
2
10x 5x 2 0 a = 10, b = 5, c = 2 2 x b b 4ac 2a
1
b b2 4ac 2a
x
The solutions are
m 2
The solutions are m
x
4 42 4(8)(1) 2(8) 4 16 32 16 4 48 16 4 4 3 16 1 3 4
2 2 1 m 1 0 m 21. 3 3 2m2 m 3 0 (2m 3)(m 1) 0 2m 3 0 or m 1 0 2m 3 m 1 3 m 2 3 The solutions are m and m = 1. 2 22.
1
0 2 8 8x2 4x 1 0 a = 8, b = 4, c = 1
and m = 2.
1 3 . 4
25. 4x2 27x 35 0 (4x 7)(x 5) 0 4x 7 0 or x 5 0 4x 7 x5 7 x 4 7 The solutions are x and x = 5. 4 26.
9x2 16x 7 0 (9x 7)(x 1) 0 9x 7 0 9x 7 7 x 9
or
x 1 0 x 1
The solutions are x
7 9
Copyright © 2025 Pearson Education, Inc.
and x = 1.
667
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
27. (7 5x)2 18 7 5x 18 5x 7 3 2 x
2 5x 7 3 2 7 3 73 2 x 5 5 7 3 2 73 2 The solutions are x or . 5 5
5 4x 75 4x 5 5 3 5 53 5 5 3 x 4 4
31. x x2 110
5 5 3
or
4
55 3 . 4
32. x 56 x2 0 56 x x2 0 (8 x)(7 x) 8 x 0 or 7 x 0 x 8 x7 The solutions are x = 7 and x = 8. 33.
2 z b b 4ac 2a
(7) (7)2 4(3)(12)
2(3) 49 144 6 7 193 6 7
34.
The solutions are z 7 193 . 6
668
. 12
3z 2 7z 12 3z2 7z 12 0 a = 3, b = 7, c = 12
6z 2 7z 6 6z 7z 6 0 a = 6, b = 7, c = 6
7 193
0 x2 x 110 0 (x 11)(x 10) x 11 0 or x 10 0 x 11 x 10 The solutions are x = 11 and x = 10.
5 4x 75 4x 5 5 3 5 53 5 5 3 x 4 4 or
2
7
The solutions are z
28. (5 4x) 75
30.
7 (7)2 4(6)(6) 2(6)
49 144 12 7 193 12
2
z
z
or 7 5x 18
29.
b b2 4ac 2a
7 3 2 7 3 2 5 5
The solutions are x
z
3 2 5 x x20 4 2 2 3x 10x 8 0 (3x 2)(x 4) 0 3x 2 0 or x 4 0 3x 2 x4 2 x 3 2 The solutions are x and x = 4. 3 6 8 x2 x 0 5 5 5x2 6x 8 0 (5x 4)(x 2) 0 5x 4 0 or x 2 0 5x 4 x2 4 x 5 4 The solutions are x and x = 2. 5
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x2 0.6x 0.05 0 (x 0.5)(x 0.1) 0 x 0.5 0 or x 0.1 0 x 0.5 x 0.1 The solutions are x = 0.5 and x = 0.1.
35.
(8) (8)2 4(2)(3) 2(2) 8 64 24 4 8 40 4 8 2 10 4 4 10 2 4 10 The solutions are z . 2 z
x2 0.1x 0.06 0 (x 0.3)(x 0.2) 0 x 0.3 0 or x 0.2 0 x 0.3 x 0.2 The solutions are x = 0.3 and x = 0.2.
36.
37. 10x2 11x 2 0 a = 10, b = 11, c = 2 2 x b b 4ac 2a
x
(11) (11)2 4(10)(2) 2(10)
121 80 20 11 41 20
11
The solutions are x
11 41
.
b b2 4ac 2a
2 x (11) (11) 4(20)(1) 2(20) 11 121 80 40 11 41 40 11 41 The solutions are x . 40
1 3 39. z 2 2z 0 2 4 2z2 8z 3 0 a = 2, b = 8, c = 3
1 2 1 z z2 0 5 2 2 2z 5z 20 0 a = 2, b = 5, c = 20
(5) (5)2 4(2)(20) 2(2) 5 25 160 4 5 185 4 5 185 The solutions are z . 4 z
38. 20x2 11x 1 0 a = 20, b = 11, c = 1
40.
2 z b b 4ac 2a
20
x
b b2 4ac 2a
z
41. answers may vary Section 16.4 Practice Exercises 1. y 3x2 x
y
2
12
1
3
0
0
1
3
2
12
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669
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
3. y x2 2x 3 a = 1, b = 2, c = 3 b (2) 2 x 1 2a 2(1) 2 y 12 2(1) 3 1 2 3 4 The vertex is (1, 4). Let x = 0. y 02 2(0) 3 3 The y-intercept is (0, 3). Let y = 0. 2. y x2 9 Let x = 0. y 02 9 9 The y-intercept is (0, 9). Let y = 0. 0 x2 9 0 (x 3)(x 3) x 3 0 or x 3 0 x3 x 3 The x-intercepts are (3, 0) and (3, 0).
0 x2 2x 3 0 (x 3)(x 1) x 3 0 or x 1 0 x3 x 1 The x-intercepts are (3, 0) and (1, 0). x
y
1
0
0
3
1
4
x
y
2
3
3
0
3
0
2
5
1
8
0
9
1
8
2
5
3
0
y
(–1, 0)
6 5 4 3 2 1
–6 –5 –4 –3 –2 –1–1
(0, –3)
–2 –3 –4 –5 –6
(3, 0) 1 2 3 4 5 6
(1, –4)
4. y x2 4x 1 a = 1, b = 4, c = 1 b (4) 4 x 2 2a 2(1) 2 y 22 4(2) 1 4 8 1 3 vertex = (2, 3) 2 x b b 4ac 2a
670
Copyright © 2025 Pearson Education, Inc.
x
ISM: Prealgebra and Introductory Algebra
x
Chapter 16: Quadratic Equations
Vocabulary, Readiness & Video Check 16.4
(4) (4)2 4(1)(1)
1. If a parabola opens upward, the lowest point is called the vertex; if a parabola opens downward, the highest point is called the vertex. If a graph can be folded along a line such that the two sides coincide or form mirror images of each other, we say that the graph is symmetric about that line and that line is the line of symmetry.
2(1) 4 16 4 2 4 12 2 42 3 2 2 3 3.7 or 0.3
2. The vertex; it is a very useful point to plot since it is the highest or lowest point on your graph.
x
y
2
3
2 3 3.7
0
2 3 0.3
0
0
1
4
1
3. For example, if the vertex is in quadrant III or IV and the parabola opens downward, then there won’t be any x-intercepts and there’s no need to let y = 0 and solve the equation for x. Exercise Set 16.4 2.
x
y 3x2
2
3(2)2 3(4) 12
1
3(1)2 3(1) 3
0
3(0)2 3(0) 0
1
3(1)2 3(1) 3
2
3(2)2 3(4) 12
Calculator Explorations 1. x2 7 x 3 0
x 0.41, 7.41 x 0.09, 5.59
2.
2x2 11x 1 0
3.
1.7x2 5.6x 3.7 0
4.
5.8x2 2.3x 3.9 0 No real solutions
5. 5.8x2 2.6x 1.9 0 6.
7.5x2 3.7 x 1.1 0
x 0.91, 2.38
x 0.39, 0.84 x 0.21, 0.70
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671
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
4.
8. y x2 9
x
y 4x2
2
4(2)2 4(4) 16
1
4(1)2 4(1) 4
0
4(0)2 4(0) 0
x = 0: y 02 9 9 y-intercept: (0, 9)
1
4(1)2 4(1) 4
2
4(2)2 4(4) 16
y x2 0x 9 a = 1, b = 0, c = 9 b 0 0 0 2a 2(1) 2
0 x2 9
y = 0:
9 x2 9 x x-intercepts: none
x = 0: y 02 9 9 The vertex is (0, 9).
6. y x2 16 y = 0:
0 x2 16
16 x2 16 x 4 x x-intercepts: (4, 0), (4, 0) x = 0: y 02 16 16 y-intercepts: (0, 16) y x2 0x 16 a = 1, b = 0, c = 16 b 0 0 0 2a 2(1) 2 x = 0: y 02 16 16 The vertex is (0, 16).
10. y x2 2x 1 y = 0: 0 x2 2x 1 0 (x2 2x 1) 0 (x 1)2 0 (x 1)2 0 x 1 1 x x-intercept: (1, 0) x = 0: y 02 2(0) 1 1 y-intercept: (0, 1) y x2 2x 1 a = 1, b = 2, c = 1 b (2) 2 1 2a 2(1) 2 x = 1: y (1)2 2(1) 1 1 2 1 0 The vertex is (1, 0).
672
Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x
x
b b2 4ac 2a (6) (6)2 4(1)(10)
2(1) 6 36 40 2 6 4 2 There are no x-intercepts, since real number.
12. y x2 7x 10 y = 0: 0 x2 7x 10 0 (x 5)(x 2) 0 x 5 or 0 x2 5 x 2 x x-intercepts: (5, 0) and (2, 0) x = 0: y 02 7(0) 10 10 y-intercept: (0, 10) y x2 7x 10 a = 1, b = 7, c = 10 b 7 7 2a 2(1) 2 2 7 7 7 10 x : y 7 2 2 2 49 49 10 4 2 49 98 40 4 4 4 9 9 1 47 31, The vertex is 2 , 4 or 2 2 4 .
4 is not a
x = 0: y 02 6(0) 10 10 y-intercept: (0, 10) y x2 6x 10 a = 1, b = 6, c = 10 b (6) 6 3 2a 2(1) 2 x = 3: y 32 6(3) 10 9 18 10 1 vertex: (3, 1)
16. y 3 x2 2 y = 0: 0 3 x
x2 3 x 3 x-intercepts:
3, 0 and 3, 0
x = 0: y 3 02 3 y-intercept: (0, 3) y 3 x2 a = 1, b = 0, c = 3 b 0 0 0 2a 2(1) 2 14. y x2 6x 10 y = 0: 0 x 6x 10 a = 1, b = 6, c = 10 2
x = 0: y 3 02 3 vertex: (0, 3)
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x4 0 x4 x-intercepts: (0, 0) and (4, 0) x = 0 or
x = 0: y 02 4(0) 0 y-intercept: (0, 0) y x2 4x 0 a = 1, b = 4, c = 0 b (4) 4 2 2a 2(1) 2 18. y
x = 2: y (2)2 4(2) 4 8 4
1 2 x 2
vertex = (2, 4)
1 y = 0: 0 x2 3 0 x2 0x x-intercept: (0, 0) The y-intercept is also (0, 0). 1 y x2 0x 0 2 1 a , b = 0, c = 0 2 b 0 0 0 2a 2 12 1 1 2 1 x = 0: y (0) (0) 0
22. y x2 2x 3 y = 0: 0 x2 2x 3 0 (x 3)(x 1)
2 2 vertex: (0, 0) For additional points use x = 2. 1 1 2 x = 2: y (2) (4) 2 2 2 1 2 1 x = 2: y (2) (4) 2 2 2 (2, 2) and (2, 2) are also on the graph.
x 3 0 or x 1 0 x3 x 1 x-intercepts: (3, 0) and (1, 0) x = 0: y 02 2(0) 3 3 y-intercept: (0, 3) y x2 2x 3 a = 1, b = 2, c = 3 b (2) 2 1 2a 2(1) 2 x = 1: y 12 2(1) 3 1 2 3 4 vertex: (1, 4)
20. y x2 4x y = 0: 0 x2 4x 0 x(x 4)
674
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
1 2 24. y 3 x y = 0: 0
2 1 3 : y 2 1 4 4 4 1 1 y 3 8 4 1 2 24 y 8 8 8 25 y 8 25 1 vertex: , 4 8
x
1 2 x 3
0 x2 0x x-intercept: (0, 0) The y-intercept is also (0, 0). 1 y x2 0x 0 3
1 a , b = 0, c = 0 3 0 b 0 0 2 1 2a 2 3
1
3
1 2 1 x = 0: y (0) (0) 0 3 3 vertex: (0, 0)
28. y x2 6x 8 y = 0: 0 x2 6x 8 0 x2 6x 8 0 (x 4)(x 2) x 4 0 or x 2 0 x4 x2 x-intercepts: (4, 0) and (2, 0) x = 0: y 02 6(0) 8 8
26. y 2x2 x 3 y = 0: 0 2x2 x 3 0 (2x 3)(x 1) 2x 3 0 or x 1 0 2x 3 x 1 3 x 2 x-intercepts: 3, 0 and (1, 0) 2 x = 0: y 2(0)2 0 3 3
y-intercept: (0, 8) y x2 6x 8 a = 1, b = 6, c = 8 b 6 6 3 2a 2(1) 2 x = 3: y (3)2 6(3) 8 9 18 8 1 vertex: (3, 1)
y-intercept: (0, 3) y 2x2 x 3 a = 2, b = 1, c = 3 b 1 1 2a 2(2) 4
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
30.
3 3 1 3 7 21 8 1 8 7 8 1 8 7
x x x x2 32. 5 x 2 2 5 x 5 2 10
x
34.
36.
x
x(x)
x 1x
x x 1x
2a2
2a2
a
x2
x2 1 2a2 3 a
a3 a a 3 3 a a 3 3a
40. y x2 2x 5 2a
a
y = 0: 0 x2 2x 5 a = 1, b = 2, c = 5
y = 0: 0 x2 4x 3 a = 1, b = 4, c = 3
2 x b b 4ac 2a
x
(4) (4)2 4(1)(3) 2(1) 4 16 12 2 4 28
(2) (2)2 4(1)(5) x 2(1) 2 4 20 2 2 24 2 22 6 2 1 6 x-intercepts: 1 6, 0
2 4 27 2 2 7
x-intercepts:
2 x b b 4ac 2a
38. y x2 4x 3
2 7, 0 (0.6, 0),
2 7, 0 (4.6, 0)
1 6, 0 (3.4, 0)
x = 0: y 02 2(0) 5 5 y-intercept: (0, 5) b
(2)
2
1
x = 0: y 02 4(0) 3 3
2a
y-intercept: (0, 3) b (4) 4 2 2a 2(1) 2
x = 1: y 12 2(1) 5 1 2 5 6
2(1)
vertex: (1, 6)
x = 2: y 22 4(2) 3 4 8 3 7 vertex: (2, 7)
676
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ISM: and Introductory Algebra 42. Prealgebra answers may vary
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Chapter 16: Quadratic Equations
677
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
44. With a < 0, the parabola opens downward. A downward opening parabola that crosses the x-axis once is graph E.
3m2 5m 2 0 (3m 1)(m 2) 0 3m 1 0 or m 2 0 3m 1 m2 1 m 3 1 The solutions are m and m = 2. 3
46. With a > 0, the parabola opens upward. An upward-opening parabola that does not touch or cross the x-axis (no x-intercept) is graph C. 48. With a < 0, the parabola opens downward. A downward-opening parabola that crosses the x-axis twice is graph B. Chapter 16 Vocabulary Check
2. The graph of y x2 is called a parabola. b
, where y ax2 bx c, is 2a called the vertex formula.
4. The process of solving a quadratic equation by writing it in the form (x a)2 c is called completing the square. 2 5. The formula x b b 4ac is called the 2a quadratic formula.
6. The lowest point on a parabola that opens upward is called the vertex. 7. The zero-factor property states that if the product of two numbers is zero, then at least one of the two numbers is zero. Chapter 16 Review 1. x2 121 0 2
x 121 x 121 x 11 The solutions are x = 11. 2. y2 100 0 y2 100 y 100 y 10 The solutions are y = 10. 678
7m2 2m 5
4.
1. If x2 a, then x a or x a. This property is called the square root property.
3. The formula
3m2 5m 2
3.
7m 2m 5 0 (7m 5)(m 1) 0 7m 5 0 or m 1 0 7m 5 m 1 5 m 7 5 The solutions are m and m = 1. 7 2
5. x2 36 x 36 x 6 The solutions are x = 6. 6. x2 81 x 81 x 9 The solutions are x = 9. 7. k 2 50 k 50 k 5 2 The solutions are k 5 2. 8. k 2 45 k 45 k 3 5 The solutions are k 3 5. 9. (x 11)2 49 x 11 49 x 11 7 x 11 7 or x 11 7 x 18 x4 The solutions are x = 18 and x = 4.
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ISM: Prealgebra and Introductory Algebra
10. (x 3)2 100 x 3 100 310 10or x 3 10 x x3 x7 x 13
Chapter 16: Quadratic Equations
15.
x2 9x 8 2 2 9 9 2 x 9x 8 29 2 81 2 8 x 4 2 2 9 49 x 2 4 9 49 x 2 4 9 7 x 2 2 9 7 16 9 7 2 x 8 or x 1 2 2 2 2 2 2 The solutions are x = 8 and x = 1.
16.
x2 8x 20 2 2 8 8 2 x 8x 20 2 2 (x 4)2 36
The solutions are x = 7 and x = 13. 11. (4 p 5)2 41 4 p 5 41 4 p 5 41 5 41 p 4 The solutions are p
5 41 . 4
12. (3 p 7)2 37 3 p 7 37 3 p 7 37 7 37 p 3
x 4 36 x 4 6 x = 4 + 6 = 2 or x = 4 6 = 10 The solutions are x = 7 and x = 10.
The solutions are p 7 37 . 3 13.
h 16t 2 100 16t 2
17.
2
6.25 t2 2.5 t
5280 ft
2
4 x 2 4x 4 1 2 2 (x 2)2 5
Reject 2.5 because time is not negative. It will take Kara 2.5 seconds to hit the water. 14. 5 miles
x2 4x 1
x2 5 x 2 5
26, 400 feet
The solutions are x 2 5.
1 mile
y 16t2 18.
2 26, 400 16t 1650 t2 40.6 t
Reject 40.6 because time is not negative. A 5-mile freefall will take 40.6 seconds.
x2 8x 3 2 2 8 8 2 x 8x 3 2 2 2 (x 4) 19 x 4 19 x 4 19 The solutions are x 4 19.
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x2 6x 7 0
19.
x2 6x 7 2 2 6 6 2 x 6x 7 2 (x 23)2 2
4 y2 33y 11 y2 y 2 3 342 14 3 y2 y 4 8 4 8 2 3 25 y 8 64 3 5 y 8 64 3 5 y 3 5 28 1 8 3 5 8 y or y 1 8 8 8 4 8 8 8 1 The solutions are y and y = 1. 4
x 3 2 x 3 2 The solutions are 3 2. x2 6x 7 0
20.
x2 6x2 7 2 6 6 2 x 6x 7 2 2 (x 3)2 2 x3 2 x 3 2
2 y2 y 1 0
2
y
2 y2 y 1 1 1 y2 y 22 2 2 1 1 1 1
x
b b2 4ac 2a
x
30 302 4(9)(25) 2(9)
y 2 2 41 29 4 y 4 16 1 9 y 4 16 1 3 y 4 4 1 3 2 1 y or 4 4 4 2 1 3 4 y 1 4 4 4 1 The solutions are y and y = 1. 2
680
23. 9x2 30x 25 0 a = 9, b = 30, c = 25
The solutions are 3 2. 21.
4 y2 3y 1 0
22.
900 900 18 30 0 18 30 18 5 3 5 The solution is x . 3
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ISM: Prealgebra and Introductory Algebra
24. 16x2 72x 81 0 a = 16, b = 72, c = 81 x
x
2
27. x2 10x 7 0 a = 1, b = 10, c = 7
b b2 4ac 2a
x
(72) (72)2 4(16)(81)
2(16) 72 5184 5184 32 72 0 32 72 32 9 4 9 The solution is x . 4 25.
Chapter 16: Quadratic Equations
7x2 35
x
2 28. x 4x 7 0 a = 1, b = 4, c = 7
x
b b 4ac 2a
0 02 4(7)(35) 2(7) 980 14 14 5 14 5
x
26.
11x2 33 11x2 33 0 a = 11, b = 0, c = 33 x
b b2 4ac 2a
0 02 4(11)(33) 2(11) 1452 22 22 3 22 3
x
b b2 4ac
2a 4 42 4(1)(7) x 2(1) 4 16 28 2 4 44 2 4 2 11 2 2 11
2
The solutions are x 5.
(10) (10)2 4(1)(7)
2(1) 10 100 28 2 10 72 2 10 6 2 2 53 2 The solutions are x 5 3 2.
7x 35 0 a = 7, b = 0, c = 35 x
b b2 4ac 2a
The solutions are x 2 11.
29. 3x2 x 1 0 a = 3, b = 1, c = 1 x
b b2 4ac 2a 2
1 (1) 4(3)(1) x 2(3) 1 1 12 6 1 13 6
The solutions are x
1 13 . 6
The solutions are x 3. Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
30. x2 3x 1 0 a = 1, b = 3, c = 1 x
2300 2x2 1.2x 1787 0 2x2 1.2x 513
b b2 4ac 2a 3 32 4(1)(1)
a = 2, b = 1.2, c = 513
x
2(1) 3 9 4 2 3 13 2
The solutions are x
3 13
.
1 12 4(2)(5) 2(2)
1 1 40 4 1 39 4 The equation has no solution, since the square root of 39 is not a real number. 2 32. 7 x 3x 1 0
b b2 4ac
3 9 28 14 3 19 14 The equation has no solution since the square root of 19 is not a real number.
2 682
0.3 or x
25, 300 191x2 1581x 28, 461 0 191x2 1581x 3161 a = 191, b = 1581, c = 3161
x
2a (3) (3)2 4(7)(1) x 2(7)
1 13 0.4 or x 6
y 191x2 1581x 28, 461
36.
2 x b b 4ac 2a (1581) (1581)2 4(191)(3161)
a = 7, b = 3, c = 1
3 13
1.2 1.22 4(2)(513) 2(2)
1.2 4105.44 16 or 4 1.2 4105.44 x 16 4 Since time will not be negative in this context, discard the solution x 16. The solution is x 16, so the model predicts that there will be 2300 Target stores in the United States in 2029 (2013 + 16).
34. x
x
x
2 x b b 4ac 2a
33. x
b b2 4ac 2a
1.2 1.44 4104 4 1.2 4105.44 4
2 31. 2x x 5 0 a = 2, b = 1, c = 5
x
x
2
x
y 2x2 1.2x 1.2x 1787
35.
1 13 0.8 6 3 13
2(191) 1581 2, 499, 561 2, 415, 004 382 1581 84, 557 382 1581 84, 557 x 5 or 382 1581 84, 557 3 382 The solutions are x 3 and x 5, so the model predicts that the global production of silver was 25,300 metric tons in 2019 (2016 + 3) and 2021 (2016 + 5). x
3.3
2 Copyright © 2025 Pearson Education, Inc.
ISM: Prealgebra and Introductory Algebra
37. y 5x2
Chapter 16: Quadratic Equations
x
y 1 x2
2
1 (2)2 1 (4) 2
1
1 (1)2 1 (1) 1
y = 0: 0 5x2 0 x2 0 x x-intercept: (0, 0) The y-intercept is also (0, 0).
2 2 2
5(2)2 5(4) 20
1
5(1) 5(1) 5
0
5(0)2 5(0) 0
1
5(1)2 5(1) 5
2
5(2)2 5(4) 20
2
2
0
1 (0) 1 (0) 0
1
1 (1)2 1 (1) 1
2
1 (2)2 1 (4) 2
y 5x
2
2 2
2
2
x
2
2 2
2 2
2
2
2 39. y x 25
y = 0:
38. y 1 x 2 2 y = 0: 0
1
x2
2 2
0x 0x x-intercept: (0, 0) The y-intercept is also (0, 0).
0 x2 25
25 x2 25 x 5 x x-intercepts: (5, 0) and (5, 0) x = 0: y 02 25 25 y-intercept: (0, 25) y x2 0x 25 a = 1, b = 0, c = 25 b 0 0 0 2a 2(1) 2 x = 0: y 02 25 25 vertex: (0, 25)
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
40. y x2 36 y = 0:
0 x2 36
36 x2 36 x 6 x x-intercepts: (6, 0) and (6, 0) x = 0: y 02 36 36 y-intercept: (0, 36) y x2 0x 36 a = 1, b = 0, c = 36 b 0 0 0 2a 2(1) 2 x = 0: y 02 36 36 vertex: (0, 36)
42. y x2 8 0 x2 8 y = 0: 8 x2 8 x x-intercepts: none x = 0: y 02 8 8 y-intercept: (0, 8) y x2 0x 8 a = 1, b = 0, c = 8 b 0 0 0 2a 2(1) 2 x = 0: y 02 8 8 vertex: (0, 8)
41. y x2 3 y = 0:
0 x2 3
3 x2 3 x x-intercepts: none x = 0: y 02 3 3 y- intercept: (0, 3) y x2 0x 3 a = 1, b = 0, c = 3
43. y 4x2 8 y = 0:
2
0 4x 8 0 4(x2 2)
b 0 0 0 2a 2(1) 2
0 x2 2 2 x2 2x
x = 0: y 02 3 3 vertex: (0, 3)
x-intercepts:
2, 0 and 2, 0
x = 0: y 4(0)2 8 8 y-intercept: (0, 8) y 4x2 0x 8 a = 4, b = 0, c = 8 684
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ISM: Prealgebra and Introductory Algebra b 0 0 0 2a 2(4) 8
Chapter 16: Quadratic Equations
45. y x2 3x 10 y = 0: 0 x2 3x 10 0 (x 5)(x 2) 0 x 5 or 0 x 2 5 x 2x x-intercepts: (5, 0) and (2, 0)
x = 0: y 4(0) 8 8 vertex: (0, 8) 2
x = 0: y 02 3(0) 10 10 y-intercept: (0, 10) y x2 3x 10
44. y 3x2 9 y = 0:
0 3x2 9 0 3(x2 3) 0 x2 3
3 x2 3x x-intercepts: 3, 0 and 3, 0
x = 0: y 3(0)2 9 9 y-intercept: (0, 9)
a = 1, b = 3, c = 10 b 3 3 2a 2(1) 2 2 3 3 3 10 x : y 3 2 2 2 9 9 10 4 2 9 18 40 4 4 4 49 4 49 3 vertex: , 2 4
y 3x2 0x 9 a = 3, b = 0, c = 8 b 0 0 0 2a 2(3) 6 x = 0: y 3(0)2 9 9 vertex: (0, 9) 46. y x2 3x 4 y = 0: 0 x2 3x 4 0 (x 4)(x 1) 0 x 4 or 0 x 1 4 x 1 x x-intercepts: (4, 0) and (1, 0) x = 0: y 02 3(0) 4 4 y-intercept: (0, 4) y x2 3x 4 a = 1, b = 3, c = 4 b 3 3 2a 2(1) 2
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685
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
2 3 3 4 : y 3 2 2 2 9 9 4 4 2 9 18 16 4 4 4 25 4 25 3 vertex: , 2 4
x
3
2 48. y 3x x 2 y = 0: 0 3x2 x 2 0 (3x 2)(x 1) 0 3x 2 or 0 x 1 2 3x 1 x 2 x 3 x-intercepts: 2, 0 and (1, 0) 3 2 x = 0: y 3(0) 0 2 2
47. y x2 5x 6 y = 0: 0 x2 5x 6
y-intercept: (0, 2)
0 x 5x 6 0 (x 3)(x 2) 0 x 3 or 0 x 2 3 x 2 x x-intercepts: (3, 0) and (2, 0)
y 3x2 x 2
2
x = 0: y 02 5(0) 6 6 y-intercept: (0, 6) y x2 5x 6 a = 1, b = 5, c = 6 5 b (5) 5 2a 2(1) 2 2 2 5 5 5 6 x : y 5 2 2 2
y
25
25
a = 3, b = 1, c = 2 b (1) 1 2a 2(3) 6 2 1 1 1 x : y 3 2 6 6 6 3 1 y 2 36 6 3 6 72 y 36 36 36 75 25 y 36 25 12 1 vertex: , 6 12
6 4 2 25 50 24 y 4 4 4 1 y 5 14 vertex: , 2 4
686
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
49. y 2x2 11x 6
4 42 4(1)(8) 2(1) 4 16 32 2 4 48 2 4 4 3 2 2 2 3
x
y = 0: 0 2x2 11x 6 0 (2x 1)(x 6) 0 2x 1 or 0 x 6 1 2x 6x 1 x 2 x-intercepts: 1, 0 and (6, 0) 2 x = 0: y 2(0)2 11(0) 6 6
x-intercepts:
y-intercept: (0, 6)
x = 0: y 02 4(0) 8 8
y 2x 11x 6
y-intercept: (0, 8)
2 2 3, 0 and 2 2 3, 0
2
a = 2, b = 11, c = 6 b (11) 11 2a 2(2) 4 2 11 11 11 x 4 : y 2 11 4 6 4 242 121 y 6 16 4 242 484 96 y 16 16 16 338 169 y 16 8 11 169 vertex: , 4 8
y x2 4x 8 a = 1, b = 4, c = 8 b 4 4 2 2a 2(1) 2 2 x = 2: y (2) 4(2) 8 4 8 8 12 vertex: (2, 12)
51. y 2x2 matches graph A. 52. y x2 matches graph D. 53. y x2 4x 4 matches graph B. 50. y x2 4x 8
54. y x2 5x 4 matches graph C.
y = 0: 0 x2 4x 8 a = 1, b = 4, c = 8
55. The graph crosses the x-axis once so there is one real solution.
2 x b b 4ac 2a
56. The graph crosses the x-axis twice so there are two real solutions. 57. The graph doesn’t cross the x-axis so there are no real solutions. 58. The graph crosses the x-axis twice so there are two real solutions.
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687
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
59. x2 49 x 49 x 7 The solutions are x = 7.
64.
2 60. y 75
2
x x 4 81 2 2 8 33
y 75 y 5 3 The solutions are y 5 3. 61. (x 7)2 64 x 7 64 x 7 8 x 7 8 or x 7 8 x 15 x 1 The solutions are x = 15 and x = 1. 62.
x2 4x 6 2
2 4 2 x 4x 6 4 2 2 (x 2)2 10 x 2 10 x 2 10
2
63.
688
x 8 64 1 33 x 8 64 1 33 x 8 8 1 33 x 8 8 1 33 x 8 1 33 The solutions are x . 8 65. 4x2 3x 2 0 a = 4, b = 3, c = 2
The solutions are x 2 10.
4x2 x 2 0 1 1 x2 x 0 4 2 1 1 x2 x 42 2 2 1 1 1 1
3x x 2 1 2 x2 x 32 3 2 2 1 1 2 1 x x 3 61 2 325 6 x 6 36 1 25 x 6 36 1 5 x 6 6 1 5 x 6 6 1 5 4 2 1 5 6 x or x 1 6 6 6 3 6 6 6 2 The solutions are x and x = 1. 3
2 x b b 4ac 2a (3) (3)2 4(4)(2) x 2(4) 3 9 32 8
3 41 8 3 41 The solutions are x . 8
66. 5x2 x 2 0 a = 5, b = 1, c = 2 2 x b b 4ac 2a
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ISM: Prealgebra and Introductory Algebra
x
1 12 4(5)(2)
x-intercepts: (2, 0) and (2, 0) x = 0: y 4 02 4 y-intercept: (0, 4)
2(5)
1 1 40 10 1 41 10
The solutions are x
Chapter 16: Quadratic Equations
y 4 0x x2 a = 1, b = 0, c = 4 b 0 0 0 2a 2(1) 2
1 41 .
x = 0: y 4 02 4 vertex: (0, 4)
10 67. 4x2 12x 9 0 a = 4, b = 12, c = 9 x x
b b2 4ac 2a 12 122 4(4)(9)
2(4) 12 144 144 8 12 0 8 12 8 3 2
70. y x2 4 y = 0:
4 x2 4 x x-intercepts: none
3 The solution is x . 2
x = 0: y 02 4 4 y- intercept: (0, 4) y x2 0x 4 a = 1, b = 0, c = 4 b 0 0 0 2a 2(1) 2
68. 2x2 x 4 0 a = 2, b = 1, c = 4 x x
0 x2 4
b b2 4ac 2a 1 12 4(2)(4)
x = 0: y 02 4 4
2(2) 1 1 32 4 1 31 4 The equation has no solution, since the square root of 31 is not a real number.
vertex: (0, 4)
2 69. y 4 x
y = 0: 0 4 x2 x2 4 x 4 x 2 Copyright © 2025 Pearson Education, Inc.
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
b (2) 2 1 2a 2(1) 2
71. y x2 6x 8 y = 0: 0 x2 6x 8 0 (x 4)(x 2) x40 or x 2 0 x 4 x 2 x-intercepts: (4, 0) and (2, 0)
x = 1: y 12 2(1) 4 1 2 4 5 vertex: (1, 5)
x = 0: y 02 6(0) 8 8 y- intercept: (0, 8) y x2 6x 8 a = 1, b = 6, c = 8 b 6 6 3 2a 2(1) 2 x = 3: y (3)2 6(3) 8 9 18 8 1 Chapter 16 Getting Ready for the Test
vertex: (3, 1)
1. To solve x2 6x 1 by completing the square, the correct next step is to add 2 6 2 (3) 9 to both sides of the equation 2 to obtain x2 6x 9 1 9; B. 2. To solve x2 5x 4 by completing the square, 2 5 25 the correct next step is to add to both 4 2 sides of the equation to obtain 25 25 x2 5x 4 ; D. 4 4
72. y x2 2x 4 y = 0: 0 x2 2x 4 a = 1, b = 2, c = 4 x b b2 4ac 2a (2) (2)2 4(1)(4) x 2(1) 2 4 16 2 2 20 2 22 5 2 1 5
x-intercepts:
12 37 3 4 3 7 3 4 7 4 7 3 3 3 9 33
3.
Choice C is correct. 4.
5 102 5 1 5 2 2 5 1 5 5 1 2 2 5 1 1 2 2 1 1 2 2 Choice A is correct.
1 5, 0 and 1 5, 0
x = 0: y 0 2(0) 4 4
2
x-intercept: (0, 4) y x2 2x 4
5.
7 x2 3 x in standard form is 7 x2 x 3 0.
The value of c is 3; B.
a = 1, b = 2, c = 4 690
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
6. 7 x2 3 x in standard form is 7 x2 x 3 0. The value of b is 1; C.
Chapter 16 Test 1.
7.
y x2 3 x2 0x 3 a = 1, b = 0, c = 3 b 0 0 x 0 2a 2(1) 2 x = 0: y 02 3 0 3 3
2.
y x2 2x x2 2x 0 a = 1, b = 2, c = 0 b (2) 2 x 1 2a 2(1) 2 x = 1: y (1) 2(1) 1 2 1 2
The vertex of the graph of y x2 2x is at
3. 5k 2 80
(1, 1); D.
k 2 16 k 16 k 4 The solutions are k = 4.
9. y x2 2x 4 a = 1, b = 2, c = 4 b 2 2 x 1 2a 2(1) 2
2
x = 1: y (1)2 2(1) 4 1 2 4 3 The vertex of the graph of y x2 2x 4 is at (1, 3); D. 10. x = 0: y 0 0 1 1 The y-intercept is (0, 1). 2
4. (3m 5) 8 3m 5 8 3m 5 2 2 3m 5 2 2 52 2 m 3 The solutions are m
y = 0: 0 x2 x 1 a = 1, b = 1, c = 1 5.
2
1 1 4(1)(1) 2(1)
5 2 2
.
3
b b2 4ac x 2a x
2x2 11x 21 2x2 11x 21 0 (2x 3)(x 7) 0 2x 3 0 or x 7 0 2x 3 x7 3 x 2 3 The solutions are x and x = 7. 2
The vertex of the graph of y x2 3 is at (0, 3); B. 8.
x2 400 0 (x 20)(x 20) 0 x 20 0 or x 20 0 x 20 x 20 The solutions are x = 20.
1 1 4 2
1
3 2
Since 3 is not a real number, there are no real-number solutions, so the graph has no x-intercepts. The only intercept of the graph of y x2 x 1 is (0, 1); B.
x2 26x 160 0 x2 26x 160 2 2 26 26 2 x 26x 160 2 2 x2 26x (13)2 160 (13)2 x2 26x 169 160 169 (x 13)2 9 x 13 9 or x 13 9 x 13 3 x 13 3 x 16 x 10 The solutions are x = 10 and x = 16.
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691
ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
6.
3x2 12x 4 0 4 x2 4x 0 3 4 x2 4x 3 4 2 4 x2 4x 4 2 2 3 2 4 2 2 2 2 x 4x (2) 3 16 (x 2)2 3 16 x2 3 16 3 x2 3 3 4 3 x2 3 4 3 x 2 3 4 3 The solutions are x 2 . 3 2
8. p2
p
1
0 3 5 1 a = 1, b , c 3 3 3
2 p b b 4ac 2a
53 4(1) 13 p 2
53
2(1) 5 25 4 9 3
3
2
5 25 12 3 9 9
2
5 37 9 3
2
5 37
3
2 x b b 4ac 2a 2 ( 3) (3) 4(1)(10) x 2(1) 3 9 40 2 3 49 2 37 2 3 7 10 3 7 4 x 5 or x 2 2 2 2 2 The solutions are x = 5 and x = 2.
3
2
5 37 6
The solutions are p 5 37 . 6
7. x 3x 10 0 a = 1, b = 3, c = 10
692
5
(3x 5)(x 2) 6
9.
3x 6x 5x 10 6 3x2 x 10 6 2
3x2 x 4 0 a = 3, b = 1, c = 4
b b2 4ac x 2a x
1 12 4(3)(4) 2(3) 1 1 48 6 1 49
6 1 7 6 1 7 1 7 4 x 1 or x 6 6 3 4 The solutions are x = 1 and x . 3
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ISM: Prealgebra and Introductory Algebra
10. (3x 1)2 16 3x 1 16 3x 1 4 3x 1 4 1 4 x 3 1 4 5 1 4 3 x 1 or x 3 3 3 3 5 The solutions are x and x = 1. 3
x
x
b b 4ac 2a
x (7) (7) 4(3)(2) 2(3) 7 49 24 6 7 25 6 75 6 2
and x = 2.
15. Let x = the length of the base. 4x = height 1 A bh 2 1 18 (x)(4x) 2 36 4x2
.
x2 4x 5 0 (x 5)(x 1) 0 x 5 0 or x 1 0 x5 x 1 The solutions are x = 5 and x = 1.
2
3
2 x (6) (6) 4(2)(1) 2(2) 6 36 8 4 6 28 4 62 7 4 3 7 2 3 7 The solutions are x . 2
6
13. 3x2 7 x 2 0 a = 3, b = 7, c = 2
1
b b2 4ac x 2a
12.
6 1
14. 2x2 6x 1 0 a = 2, b = 6, c = 1
2 (7) (7) 4(3)(2) 2(3)
7 73
75
3
2 x b b 4ac 2a
The solutions are x
2 or x
The solutions are x
a = 3, b = 7, c = 2
7 49 24 6 7 73 6
75 6
11. 3x2 7 x 2 0
x
Chapter 16: Quadratic Equations
9 x2
9x 3 x 4x = 4(3) = 12 The base is 3 feet. The height is 12 feet. 16. y 5x2 y = 0: 0 5x2 0 x2 0x x-intercept: (0, 0) The y-intercept is also (0, 0). y 5x2 0x 0 a = 5, b = 0, c = 0 b 0 0 0 2a 2(5) 10 x = 0: y 5(0)2 0
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
vertex: (0, 0)
17. y x2 4 y = 0: 0 x2 4 0 (x 2)(x 2) 0 x2 or 0 x 2 2 x 2x x-intercepts: (2, 0), (2, 0)
b (7) 7 2a 2(1) 2 2 7 7 7 x 2 : y 7 10 2 2 49 49 10 4 2 49 98 40 4 4 4 9 4 9 7 vertex: , 2 4
x = 0: y 02 4 4 y-intercept: (0, 4) y x2 0x 4 a = 1, b = 0, c = 4 b 0 0 0 2a 2(1) 2
19. y 2x 2 4x 1 y = 0: 0 2x2 4x 1 a = 2, b = 4, c = 1
x = 0: y 02 4 4 vertex: (0, 4)
x
x
18. y x2 7x 10 y = 0: 0 x2 7x 10 0 (x 5)(x 2) x 5 0 or x 2 0 x5 x2 x-intercepts: (5, 0) and (2, 0) x = 0: y 02 7(0) 10 10 y-intercept: (0, 10) 2 y x 7x 10 a = 1, b = 7, c = 10
694
b b2 4ac 2a 4 42 4(2)(1)
2(2) 4 16 8 4 4 24 4 4 2 6 4 2 6 2 2 x-intercepts: 2 6 6 and , 0 , 0 2 2 x = 0: y 2(0)2 4(0) 1 1 y-intercept: (0, 1) y 2x2 4x 1 a = 2, b = 4, c = 1
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ISM: Prealgebra and Introductory Algebra b 4 4 1 2a 2(2) 4
Chapter 16: Quadratic Equations
4.
5.03
0.0503
100
x = 1: y 2(1)2 4(1) 1 2 4 1 3 vertex: (1, 3)
5. Add the percentages for Mexico and Canada: 47% + 11% = 58% 58% of the visitors to the U.S. were from Mexico or Canada. 6. Add the percentages for India and the United Kingdom: 2% + 2% = 4% 4% of the visitors to the U.S. were from India or the United Kingdom. 7. Perimeter = 11 + 3 + 11 + 3 = 28 The perimeter is 28 inches. 8. Perimeter = 6 + 8 + 11 = 25 The perimeter is 25 feet.
2 20. d n 3n 22 n 3n 9 2 18 n2 3n
9. Area = bh = (3.4 mi)(1.5 mi) = 5.1 sq mi The area is 5.1 square miles. 1 1 10. Area bh (17 in.)(8 in.) 68 sq in. 2 2 The area is 68 square inches.
0 n 3n 18 0 (n 6)(n 3) n 6 0 or n 3 0 n6 n 3 A 6-sided polygon has 9 diagonals. 2
11. 3210 ml
4321 L 43.21 L 12. 4321 cl 4321 cl 1 L 1 100 cl 100
h 16t 2
21.
193.83 16t 2 193.83 t2 16 193.83 193.83 t or t 16 16 3.5 t 3.5 t Since the time of the dive is not a negative number, discard the solution 3.5. The dive took approximately 3.5 seconds. Cumulative Review Chapters 116 1.
786.1 0.7861 1000 818
2.
0.818
1000 3.
0.12 10
0.012
3210 ml 1L 3210 1000 ml 1000 L 3.21 L 1
13. 8(2 t) 5t 16 8t 5t 16 3t 16 t 3 14.
5
1 x 1 x 4 52 1 4 x 1 2 4 x 4 10x 4 4x 1 6x 4 1 6x 5 5 x 6
15. 30 1 0 0 16. 2 (2 ) 1
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
x2 8x 7
17. r 2 r 42 (r 7)(r 6) 27. 18. y 3y 70 ( y 7)( y 10)
(x 7)(x 1)
(x 5)(x 1)
x2 4x 5
x7 x 5
2
2
2 28. x 6x 5 (x 1)(x 5) x 5 (x 1)(x 7) x 7 x2 6x 7
2
19. 10x 13xy 3y (2x 3y)(5x y) 20. 72x2 35xy 3y2 (9x y)(8x 3y)
29.
21. 8x2 14x 5 8x2 4x 10x 5 4x(2x 1) 5(2x 1) (2x 1)(4x 5) 22. 15x2 4x 4 15x2 10x 6x 4 5x(3x 2) 2(3x 2) (3x 2)(5x 2) 23. a.
30.
4x3 49x x(4x2 49) x(2x 7)(2x 7)
b.
162x4 2 2(81x4 1) 2(9x2 1)(3x 1)(3x 1)
b.
9x3 x x(9x2 1) x(3x 1)(3x 1)
4n 5 7 6 2 4n 5 7 6 6 2 6 4n 5 21 4n 16 n4 The number is 4.
2(9x2 1)(9x2 1)
24. a.
n 5 n 6 3 2 n 5 n 6 6 6 3 2 n 10 3n 10 2n 5 n The number is 5.
x
y = 3x
a.
1
3
5(x2 1)(x2 1)
b.
0
0
5(x2 1)(x 1)(x 1)
c.
3
9
31.
5x4 5 5(x4 1)
25. (5x 1)(2x2 15x 18) 0 (5x 1)(2x 3)(x 6) 0 5x 1 0 or 2x 3 0 or x 6 0 5x 1 2x 3 x 6 1 3 x x 5 2 1 3 The solutions are x , x , and x = 6. 5 2
10x 1 0 10x 1 1 x 10 3 1 The solutions are x = 4, x , and x . 4 10
696
or
y = 2x + 7
a.
0
7
b.
7 2
0
c.
5
3
33. a. 26. (x 4)(40x2 34x 3) 0 (x 4)(4x 3)(10x 1) 0 x40 or 4x 3 0 x 4 4x 3 3 x 4
y7 2
x
32.
y
1
x 1 5 2x 10 y 3 10 y 2x 3 2 3 y x 10 10 1 3 y x 5 10 These two equations have the same slope, therefore they are parallel.
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
b.
x y 3 y x 3 x y 4 y x 4 These two equations have slopes whose product is 1, therefore they are perpendicular.
c.
3x + y = 5 y = 3x + 5 2x + 3y = 6 3y = 2x + 6 2 y x2 3 These two equations have different slopes (and their product is 1), so they are neither parallel nor perpendicular.
34. a.
y = 2x + 3 y = 2x + 5 These two equations have the same slope, therefore they are parallel.
b. 2x + y = 3 y = 2x + 3 x + 2y = 9 2y = x + 9 y
1
x
9
2 2 These two equations have slopes whose product is 1, therefore they are perpendicular. c.
35. a.
3x 2y = 8 2y = 3x 8 3 y x4 2 3x + 2y = 1 2y = 3x + 1 3 1 y x 2 2 These two lines have different slopes (and their product is 1), so they are neither parallel nor perpendicular.
b.
8x 3y 4 38. y 7x 3 Substitute 7x 3 for y in the first equation. 8x 3(7x 3) 4 8x 21x 9 4 13x 9 4 13x 13 x 1 Solve for y. y = 7x 3 = 7(1) 3 = 7 3 = 4 The solution for this system is (1, 4). 39.
36 6 because 62 36.
40.
81 9 because 9 81.
41.
9 3 3 9 . because 100 10 10 100
42.
4 16 16 4 because . 25 5 5 25
43.
2 1 3 2 2 3 2 2 3 1 3 1 3 2 1 3 1 3
{(2, 3), (5, 7), (9, 3), (0, 0)} This is a function because each x-value is assigned to only one y-value. 1 (9, 1), 0, , (2, 1), (9, 0) 2 This is not a function because the x-value 9 is paired with two different y-values.
2
2
2
{(1, 1), (2, 3), (7, 3), (8, 6)} This is a function because each x-value is assigned to only one y-value.
b. {(0, 2), (1, 5), (0, 3), (7, 7)} This is not a function because the x-value 0 is paired with two different y-values. 36. a.
2x y 10 37. x y 2 Substitute y + 2 for x in the first equation. 2( y 2) y 10 2 y 4 y 10 3y 6 y2 Solve for x. x=y+2=2+2=4 The solution for this system is (4, 2).
44.
7 5 2
5 2 7 5 2 5 4 52 7 5 2 1 7
5 2
45. (x 3)2 16 x 3 16 x 3 4 x 34 x = 3 + 4 = 7 or x = 3 4 = 1 The solutions are x = 7 and x = 1.
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ISM: Prealgebra and Introductory Algebra
Chapter 16: Quadratic Equations
46. (x 4)2 9 x4 9 x 4 3 x 4 3 x = 4 + 3 = 1 or x = 4 3 = 7 The solutions are x = 1 and x = 7. 47.
1 2 x x2 2 1 2 x x2 0 2 1 a , b = 1, c = 2 2 x
x
b b2 4ac 2a
(1) (1)2 4 12 (2)
2 12
2x2
48.
x
7
2 2 7 5 2(2x ) 2 2 x 2 4x2 5x 7 2
4x2 5x 7 0 a = 4, b = 5, c = 7 2 x b b 4ac 2a (5) (5)2 4(4)(7) x 2(4) 5 25 112 8 5 137 8 5 137 The solutions are x . 8
1 1 4 1 1 5
The solutions are x 1 5.
698
5
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Appendix Appendix B Exercise Set
18. s3 216 s3 63 (s 6)[s2 (s)(6) 62 ]
2. b3 8 b3 23
(s 6)(s2 6s 36)
(b 2)[b2 (b)(2) 22 ] (b 2)(b2 2b 4) 3
3
20. 8t3 s3 (2t)3 s3 (2t s)[(2t)2 (2t)(s) s2 ] (2t s)(4t 2 2ts s2 )
3
4. 64x 1 (4x) 1
(4x 1)[(4x)2 (4x)(1) 12 ] (4x 1)(16x2 4x 1)
Appendix C Practice Exercises 1. Let x represent the liters of 20% solution.
6. 6r3 162 6(r3 27)
Number of Liters
Dye Strength
Amount
20% solution
x
20%
0.2x
50% solution
6x
50%
0.5(6 x)
40% solution
6
40%
0.4(6)
6(r3 33 ) 2
2
6(r 3)[r (r)(3) 3 ] 6(r 3)(r2 3r 9) 8. 8x3 y3 (2x)3 y3 2
2
(2x y)[(2x) (2x)( y) y ] (2x y)(4x2 2xy y2 ) 10. a3 216 a3 63 2
2
(a 6)[a (a)(6) 6 ] (a 6)(a2 6a 36) 12. 375 y6 24 y3 3y3(125 y3 8) 3y3[(5 y)3 23 ] 3y3 (5 y 2)[(5 y)2 (5 y)(2) 22 ] 3
0.2x 0.5(6 x) 0.4(6) 0.2x 3 0.5x 2.4 0.3x 3 2.4 0.3x 0.6 x2 6x=62=4 2 liters of the 20% solution should be mixed with 4 liters of the 50% solution. 2. Let x represent the number of $5 bills.
2
3y (5 y 2)(25 y 10 y 4)
Number of Bills
Value of Bills (in dollars)
$5 bills
x
5x
$20 bills
x 47
20(x 47)
Denomination 14. 125 r3 53 r3 (5 r)[52 (5)(r) r 2 ] (5 r)(25 5r r 2 ) 16. 54r3 2 2(27r3 1) 2[(3r)3 13] 2(3r 1)[(3r)2 (3r)(1) 12 ] 2(3r 1)(9r 2 3r 1)
698
The total value was $1710. 5x 20(x 47) 1710 5x 20x 940 1710 25x 940 1710 25x 2650 x 106 x 47 = 106 47 = 59 There were 106 $5 bills and 59 $20 bills.
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ISM: Prealgebra and Introductory Algebra
Appendix 0.25x 0.60(10) 0.30(x 10) 0.25x 6 0.3x 3 6 0.05x 3 3 0.05x 60 x 60 cc of the 25% antibiotic solution should be used.
3. Let x represent the time it took to hike up the mountain. Rate Time = Distance Up
1.5
x
1.5x
Down
4
x1
4(x 1)
The distance up is the same as the distance down. 1.5x 4(x 1) 1.5x 4x 4 2.5x 4 x 1.6 x 1 = 1.6 1 = 0.6 The total time is 1.6 + 0.6 = 2.2 hours. 4. Let x represent the speed of the eastbound train.
r
t
=
4. Let x represent the amount of cashews. Number of Cost per = Value Pounds Pound
$3 per lb Peanuts
20
3
3(20)
$5 per lb Cashews
x
5
5x
$3.50 per lb mixture
20 + x
3.50
3.50(20 + x)
d
East Train
x
1.5
1.5x
West Train
x 10
1.5
1.5(x 10)
The total distance after 1.5 hours is 171 miles. 1.5x 1.5(x 10) 171 1.5x 1.5x 15 171 3x 15 171 3x 186 x 62 x 10 = 62 10 = 52 The eastbound train is going 62 miles per hour and the westbound train is going 52 miles per hour. Appendix C Exercise Set 2. Let x represent the amount of the 25% antibiotic solution. Number of Antibiotic = Amount of Strength Cubic cm Antibiotic
3(20) 5x 3.50(20 x) 60 5x 70 3.5x 60 1.5x 70 1.5x 10 10 2 x 6 1.5 3 2 6 pounds of cashews should be used. 3 Number of Coins or Bills
Value of Coins or Bills (in dollars)
6.
quarters
z
0.25z
8.
halfdollars
(20 z)
0.50(20 z)
10.
$100 bills
97z
100(97z) or 9700z
12.
$10 bills
(15 y)
10(15 y)
14. Let x represent the number of $50 bills.
25% solution
x
60% solution
10
30% solution
x + 10
25% 60% 30%
0.25x Number of Bills
Value of Bills
$20 bills
6x
20(6x)
$50 bills
x
50x
0.60(10) 0.30(x + 10)
Total
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3910
699
ISM: Prealgebra and Introductory Algebra
Appendix 20(6x) 50x 3910 120x 50x 3910 170x 3910 x 23 6x = 6(23) = 138 There were 23 $50 bills and 138 $20 bills.
28. Let x represent the time the bus is driving. 60x 40(x 1.5) 60x 40x 60 20x 60 x3 The bus will overtake the car after 3 hours.
16. Let x represent the amount of water (which is 0% antifreeze). 0x 0.70(30) 0.60(x 30) 21 0.6x 18 3 0.6x 5 x 5 gallons of water should be added.
30. Let x represent the amount of time they were going 50 miles per hour. 50x 40(7.2 x) 50x 288 40x 90x 288 x 3.2 50x = 50(3.2) = 160 The distance to Disneyland was 160 miles.
18. Let x represent the amount of chocolate-covered peanuts. 5x 2(10) 3(x 10) 5x 20 3x 30 2x 20 30 2x 10 x5 5 pounds of chocolate-covered peanuts should be used. 20. Let x represent the time that the two cars drive. 65x 41x 530 106x 530 x5 The cars are 530 miles apart after 5 hours. 22. Let x be the speed of the slower train. 1.5x 1.5(x 8) 162 1.5x 1.5x 12 162 3x 12 162 3x 150 x 50 x + 8 = 50 + 8 = 58 The slower train is traveling 50 miles per hour and the faster train is traveling 58 miles per hour. 24. Let x represent the time both cars travel. 65x 45x 330 110x 330 x3 The cars will be 330 miles apart after 3 hours.
32. Let x represent the amount of time they walk. 4x 5x 20 9x 20 20 2 x 2 9 9 2 They can talk for 2 hours. 9 34. Let x represent the time they both walk. 3x (3 1)x 12 3x 4x 12 7x 12 12 5 x 1 7 7 5 They will meet in 1 hours. 7 36. Let x represent the time he spent driving 70 miles per hour. 70x 60(4 x) 255 70x 240 60x 255 10x 240 255 10x 15 x 1.5 1 He was speeding for 1 hours. 2
26. Let x represent the time both cars travel. 50x 40x 20 10x 20 x2 The cars will be 20 miles apart after 2 hours.
700
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ISM: Prealgebra and Introductory Algebra
Appendix
38. Since there are x quarters, there are (x + 136) dimes, 8x nickels, and (16x + 32) pennies. 0.01(16x 32) 0.05(8x) 0.10(x 136) 0.25x 44.86 0.16x 0.32 0.4x 0.1x 13.6 0.25x 44.86 0.91x 13.92 44.86 0.91x 30.94 x 34 x + 136 = 34 + 136 = 170 8x = 8(34) = 272 16x + 32 = 16(34) + 32 = 544 + 32 = 576 The jar contained 34 quarters, 170 dimes, 272 nickels, and 576 pennies. Appendix D Exercise Set y x 3 2. y 1 x
y 2x 1 4. y x 2
y 2x 4 6. y x 5
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701
Appendix
ISM: Prealgebra and Introductory Algebra
y x 5 8. y 3x 3
x 3 16. y 2
2x y x 2 y 3
18. y 2 x 1
10.
12. y 2x 0 5x 3y 2
20. 3x 2 y 6 x2
14. 4x y 2 2x 3y 8
22. 2x 5 y 10 y 1
702
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ISM: Prealgebra and Introductory Algebra
Appendix
3 y x3 2 24. y 3 x 6 2 x 3
26. The graph of the system
y 5
is graph A.
y 5 28. The graph of the system is graph B. x 3
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703
Practice Final Exam Chapters 18
12. 16 3 16 12 3 12 3 3 16 12 3 3 16 3 4 3 3 64 1 or 21 3 3
1. 23 52 2 2 2 5 5 200 2. 16 9 3 4 7 16 3 4 7 16 12 7 28 7 21 3. 18 24 = 18 + (24) = 6
13.
4. 5 (20) = 100 49 7 because 72 49.
5.
18
19 23 11
2 16
6. (5) 24 (3) 125 24 (3) 125 (8) 125 8 117
0.23 1.63 14. 15.
8. 62 0 is undefined. 9.
8
2
15 y 15 y
11 8 11
3
7. 0 49 = 0
11 11 3
8 2 10 25 2 15 y 15 y 3 5 y 3y
0.3 10.2 4.01 102 40 800 40.902
1.86 0.3
6.2
1 decimal place 2 decimal places
1 2 3 decimal places
16. 0.6% = 0.6(0.01) = 0.006 10.
11 3 5 11 2 3 3 5 12 8 24 12 2 8 3 24 22 9 5 24 24 24 22 9 5 24 18 24 3 6 46 3 4
17. 6.1 = 6.1(100%) = 610% 18.
3 100 300 % % 37.5% 8 8 1 8
19. 0.345
20.
345 5 69 69 1000 5 200 200
13 113 1 1 5 5 0.5 26 2 13 2 25 10
21. 34.8923 rounded to the nearest tenth is 34.9.
11. 3a 16 3a 16 3 a 8 2 1 1 8 6a3 8 6a3 8 2 3 a a a a a a2
3
22. Let x be 2. 5(x3 2) 5(23 2) 5(8 2) 5(6) 30 23. 10 y2 10 (3)2 10 9 1
704
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ISM: Prealgebra and Introductory Algebra
24. x y
1
3
Practice Final Exam
7 30.
2 8 1 31 2 8
1 8 2 31 18 2 31 1 2 4 2 31 4 31
25. (3z 2) 5z 18 1(3z 2) 5z 18 1 3z (1) 2 5z 18 3z 2 5z 18 3z 5z 2 18 8z 20
31.
26. perimeter = 3(5x + 5) = 3 5x + 3 5 = 15x + 15 The perimeter is (15x + 15) inches. 27.
28.
n 7 4 n 7 7 4 77 n 7 4 7 n 28 The solution is 28. 4x 7 15 4x 7 7 15 7 4x 8 4x 8 4 4 x 2
29. 4(x 11) 34 10 12 4x 44 34 10 12 4x 10 2 4x 10 10 2 10 4x 12 4x 12 4 4 x3
32.
x 24 x 5 5 x 24 5 5 x 5 5 x 5 5 x 24 5 x 5x 24 6x 24 6x 24 6 6 x 4
2(x 5.7) 6x 3.4 2x 11.4 6x 3.4 2x 11.4 11.4 6x 3.4 11.4 2x 6x 14.8 2x 6x 6x 6x 14.8 4x 14.8 4x 14.8 4 4 x 3.7 5 4 y 1 y 2 5( y 2) 4( y 1) 5 y 10 4 y 4 y 10 4 y 6
33. Perimeter (20 10 20 10) yards 60 yards Area (length)(width) (20 yards)(10 yards) 200 square yards 34. average 12 (13) 0 9 16 4 4 4 35. The difference of three times a number and five times the same number is 4 translates to 3x 5x 4 2x 4 2x 4 2 2 x 2 The number is 2. 36. 258 10 3 258 43 258 4 43 6 4 24 4 1 4 1 43 1 43 Expect to travel 24 miles on 1 gallon of gas.
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ISM: Prealgebra and Introductory Algebra
Practice Final Exam
37. Let x be the number of children participating in the event. Since the number of adults participating in the event is 112 more than the number of children, the number of adults is x + 112. Since the total number of participants in the event is 600, the sum of x and x + 112 is 600. x x 112 600 2x 112 600 2x 112 112 600 112 2x 488 2x 488 2 2 x 244 244 children participated in the event. 38. Let x be the number of defective bulbs out of 510 bulbs. x defective defective 3 bulbs 85 510 bulbs 3 510 85 x 1530 85x 1530 85x 85 85 18 x There should be 18 defective bulbs in 510. 39. Amount of discount 15% $120 0.15 $120 $18 Sale price = $120 $18 = $102 The amount of the discount is $18; the sale price is $102. 40. hypotenuse (leg)2 (other leg)2 2
2
4 4 16 16 32 5.66 The hypotenuse is 5.66 centimeters. 41. The complement of an angle that measures 78 is an angle that measures 90 78 = 12. 42. x and the angle marked 73 are vertical angles, so mx 73. x and y are alternate interior angles, so my mx 73. x and z are corresponding angles, so mz mx 73.
706
43. The unmarked vertical side has length 11 in. 7 in. = 4 in. The unmarked horizontal side has length 23 in. 6 in. = 17 in. P = (6 + 4 + 17 + 7 + 23 + 11) in. = 68 in. Extending the unmarked vertical side downward divides the region into two rectangles. The region’s area is the sum of the areas of these: A 11 in. 6 in. 7 in.17 in. 66 sq in. 119 sq in. 185 sq in. 44. Circumference: C 2 r 2 9 in. 18 in. 56.52 in. Area: A r2 (9 in.)2 81 sq in. 254.34 sq in. 2 1 gal 4 qt 1 2 2 4 qt 10 qt 45. 2 gal 2 1 1 gal 2 1
46. 2.4 kg
2.4 kg 1000 g 2.4 1000 g 2400 g 1 1 kg
Chapters 916 47. 34 (3 3 3 3) 81 3 48. 4
1 43
1 64
49. 7 2(5 y 3) 7 10 y 6 7 6 10 y 1 10 y 10 y 1 50.
5x3 x2 5x 2 (8x3 4x2 x 7)
5x3 x2 5x 2 8x3 4x2 x 7 3x3 5x2 4x 5
51. (4x 2)2 (4x)2 2(4x)(2) 22 16x2 16x 4
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ISM: Prealgebra and Introductory Algebra
52. (3x 7)(x2 5x 2) 3x(x2 5x 2) 7(x2 5x 2) 3x(x2 ) 3x(5x) 3x(2) 7(x2 ) 7(5x) 7(2) 3x3 15x2 6x 7x2 35x 14 3x3 22x2 41x 14
Practice Final Exam
60.
53. 6t 2 t 5 6t2 5t 6t 5 t(6t 5) 1(6t 5) (6t 5)(t 1) 54. 180 5x2 5(36 x2 ) 5(62 x2 ) 5(6 x)(6 x) 2
55. 3a 3ab 7a 7b 3a(a b) 7(a b) (a b)(3a 7)
x2 2 61. x 5 x 7x 10 x2 5x 2x 10 2x 10
3 2 2 56. 3x 21x 30x 3x(x 7x 10) 3x(x 5)(x 2)
57.
4x2 y3
2
0 x 7x 10 x 2 x5 2
42 x22 y32
x3 y4 32 42 x y 16x4 y6 x6 y8 16x46 y6(8)
62.
16x2 y14 16 y14
x2 y 5 1 2 5 1 2 y 5 y 1 y 2 y 2 58. 1 2 2 1 2 2 y y y y 2 y 2
2
x 9 59.
x2 3x
2
x 4x 1 x2 9 2x 10 2 2x 10 x 3x x2 4x 1 (x 3)(x 3) 2(x 5) 2 x(x 3) x 4x 1 2(x 3)(x 5) x(x2 4x 1)
5a 2 a2 a 6 a 3 5a 2 (a 3)(a 2) a 3 5a 2(a 2) (a 3)(a 2) (a 3)(a 2) 5a 2(a 2) (a 3)(a 2) 5a 2a 4 (a 3)(a 2) 3a 4 (a 3)(a 2)
63.
4(n 5) (4 2n) 4n 20 4 2n 4n 20 2n 4 2n 2n 2n 20 4 2n 20 20 4 20 2n 16 2n 16 2 2 n 8 2
(3x 5)(x 2) 6
3x 6x2 5x 10 6 3x x 10 6 3x2 x 4 0 a = 3, b = 1, c = 4 x
b b2 4ac 2a
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707
ISM: Prealgebra and Introductory Algebra
Practice Final Exam
1 12 4(3)(4) 2(3) 1 1 48 6
x
1 49
6 1 7 6 1 7 1 7 4 x 1 or x 6 6 3 4 The solutions are x = 1 and x . 3 64.
5(x 1) 6 3(x 4) 1 5x 5 6 3x 12 1 5x 11 3x 11 5x 11 3x 3x 11 3x 2x 11 11 2x 1111 1111 2x 22 2x 22 2 2 x 11 {x|x 11}
65. 2x2 6x 1 0 a = 2, b = 6, c = 1 x
b b 4ac 2a 2
2
(6) (6) 4(2)(1) 2(2) 6 36 8 4 6 28 4 62 7 4 3 7 2 3 7 The solutions are x . 2 x
708
66. The LCD is 3 5 y = 15y. 4 5 1 y 3 5 4 5 1 15 y 15 y 5 y 3 60 25 y 3y 60 22 y 60 y 22 30 y 11 30 The solution is . 11 67. The LCD is 2(a 3). a 3 3 a3 a3 2 3 a 3 2(a 3) 2(a 3) a 3 a 3 2 2a 6 3(a 3) 2a 6 3a 9 5a 15 a3 Since a = 3 causes the denominator a 3 to be 0, the equation has no solution. 68.
2x 2 x 5
2x 2 (x 5)2 2
2x 2 x2 10x 25 0 x2 12x 27 0 (x 3)(x 9) x 3 0 or x 9 0 x3 x9 x = 3 does not check, so the solution is x = 9. 69. 5x 7y = 10 y = 0: 5x 7(0) 10 5x 0 10 5x 10 x2 x-intercept: (2, 0) x = 0: 5(0) 7 y 10 0 7 y 10 7 y 10 10 y 7
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ISM: Prealgebra and Introductory Algebra
10 y-intercept: 0, 7
Practice Final Exam 74. m y2 y1 3 (5) 3 5 8 x x 1 2 1 2
5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
70. The graph of y = 1 is a horizontal line with a y-intercept of (0, 1). y 5 4 3 2 1 –5 –4 –3 –2 –1–1
1 2 3 4 5
x
–2 –3 –4 –5
71. Graph the boundary line, y = 4x, with a solid line. Test (1, 1): y 4x 1 4(1) 1 4 True Shade the half-plane containing (1, 1).
y2 y1
2 (5) 7 1 72. m 1 6 7 x2 x1 73.
1
y y1 m(x x1) y (5) 8(x 2) y 5 8x 16 y 8x 11 8x y 11
y
y = mx + b 3x y 5 y 3x 5 m=3
1 75. m ; b = 12 8 1 y x 12 8 8 y x 8(12) 8 y x 96 x 8 y 96 76. 3x 2 y 14 y x 5 Substitute x + 5 for y in the first equation and solve for x. 3x 2 y 14 3x 2(x 5) 14 3x 2x 10 14 x 10 14 x 4 y = x + 5 = 4 + 5 = 1 The solution of the system is (4, 1). 77. 4x 6 y 7 2x 3y 0 Multiply the second equation by 2. 4x 6 y 7 4x 6 y 7 2(2x 4x 6 y 0 3y) 2(0) Add the equations to eliminate x. 4x 6 y 7 4x 6 y 0 07 Since the statement 0 = 7 is false, the system has no solution. 78.
f (x) x3 x a.
f (1) (1)3 (1) 1 1 0
b.
f (0) 03 0 0 0 0
c.
f (4) 43 4 64 4 60
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709
ISM: Prealgebra and Introductory Algebra
Practice Final Exam 79. The x-value 3 is assigned to two y-values, 3 and 2, so the relation is not a function. Note that the x-value 0 is also assigned to two y-values, 5 and 0. 80.
54 9 6 9 6 3 6
81.
24x8 4x8 6 4x8 6 2x4 6
82.
18 75 7 3 8 9 2 25 3 7 3 4 2 9 2 25 3 7 3 4 2 3 2 5 3 7 3 2 2 (3 2) 2 (5 7) 3 2 2 3
83.
40x4 2x
40x4 2x
20x3 4x2 5x 4x2 5x 2x 5x
84.
6 5 2 6 2 5
2
87. y x2 7x 10 y = 0: 0 x2 7x 10 0 (x 5)(x 2) x 5 0 or x 2 0 x5 x2 x-intercepts: (5, 0) and (2, 0) x = 0: y 02 7(0) 10 10 y-intercept: (0, 10) y x2 7x 10 a = 1, b = 7, c = 10 b (7) 7 2a 2(1) 2 2 7 7 7 x 2 : y 7 10 2 2 49 49 10 4 2 49 98 40 4 4 4 9 4 9 7 vertex: , 2 4
12 10 4 3 10 2 3 10
85.
86.
8 5y
8
8 62
5y
5y 5y
6 2 2 6 22 8 2 6 8
64
6 2 2 4 6 2
8
4 6 8
710
8 5y 5y
8 6 2 6 2 6 2
88. Let n be the number. 1 n5 6 n 5 n 6 n 5 n n n(6) n n2 5 6n n2 6n 5 0 (n 5)(n 1) 0 n 5 0 or n 1 0 n5 n 1 The number is 1 or 5.
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ISM: Prealgebra and Introductory Algebra
89. Let x be the number. 2 x x 35 3 3 2 x x 35 3 3 5 x 35 3 3 5 3 x 35 5 3 5 x 21 The number is 21.
Practice Final Exam
91. Let x be the speed of the boat in still water. Let x + 2 be the speed of the boat going downstream. Let x 2 be the speed of the boat going upstream. distance
90. Let t be the number of farms in Texas (in thousands) and let m be the number of farms in Missouri (in thousands). t m 342 t m 152 Substitute m + 152 for t in the first equation and solve for m. t m 342 m 152 m 342 2m 152 342 2m 190 m 95 t = m + 152 = 95 + 152 = 247 There were 247 thousand farms in Texas and 95 thousand in Missouri.
=
rate
time
Upstream
14
x2
14 x2
Downstream
16
x+2
16 x2
14 16 x2 x2 14(x 2) 16(x 2) 14x 28 16x 32 60 2x 30 x The speed of the boat in still water is 30 mph.
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