Solution Manual for Algebra Foundations: Prealgebra, Introductory Algebra and Intermediate Algebra, 2nd Edition

Preview Extract
Chapter 2 Section 2.1 Practice Exercises 1. a. b. If 0 represents the surface of the earth, then 3805 below the surface of the earth is โˆ’3805. If zero degrees Fahrenheit is represented by 0ยฐF, then 85 degrees below zero, Fahrenheit is represented by โˆ’85ยฐF. 2. 3. a. 0 > โˆ’5 since 0 is to the right of โˆ’5 on a number line. b. โˆ’3 โˆ’12 since โˆ’7 is to the right of โˆ’12 on a number line. 4. a. |โˆ’6| = 6 because โˆ’6 is 6 units from 0. b. |4| = 4 because 4 is 4 units from 0. c. |โˆ’12| = 12 because โˆ’12 is 12 units from 0. 5. a. b. The opposite of 14 is โˆ’14. The opposite of โˆ’9 is โˆ’(โˆ’9) or 9. 6. a. โˆ’|โˆ’7| = โˆ’7 b. โˆ’|4| = โˆ’4 c. โˆ’(โˆ’12) = 12 7. โˆ’|x| = โˆ’|โˆ’6| = โˆ’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 …โˆ’3, โˆ’2, โˆ’1, 0, 1, 2, 3, … are called integers. 2. Positive numbers, negative numbers, and zero together are called signed numbers. 3. The symbols โ€œโ€ 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 โ€œโ€ means is greater than. 6. On a number line, the greater number is to the right of the lesser number. 7. A numberโ€™s distance from 0 on the number line is the numberโ€™s absolute value. 8. The numbers โˆ’5 and 5 are called opposites. 42 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations 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 also. 13. A negative sign can be translated into the phrase โ€œopposite of.โ€ 14. Eyre Exercise Set 2.1 2. If 0 represents the surface of the water, then 25 feet below the surface of the water is โˆ’25. 4. If 0 represents sea level, then 282 feet below sea level is โˆ’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 โˆ’14,040. 10. If 0 represents a loss of $0, then a loss of $241 million is โˆ’241 million. 12. If 0 represents 0ยฐ Celsius, then 10ยฐ below 0ยฐ Celsius is โˆ’10. Since 5ยฐ below 0ยฐ Celsius is โˆ’5 and โˆ’10 is less than โˆ’5, โˆ’10 (or 10ยฐ below 0ยฐ Celsius) is cooler. 14. From the map, the coldest temperature on record for the state of Arkansas is โˆ’29ยฐF. 16. 18. 20. 22. 24. โˆ’8 < 0 since โˆ’8 is to the left of 0 on a number line. 26. โˆ’12 โˆ’29 since โˆ’27 is to the right of โˆ’29 on a number line. 30. 13 > โˆ’13 since 13 is to the right of โˆ’13 on a number line. 32. |7| = 7 since 7 is 7 units from 0 on a number line. 34. |โˆ’19| = 19 since โˆ’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. |โˆ’10| = 10 since โˆ’10 is 10 units from 0 on a number line. Copyright ยฉ 2020 Pearson Education, Inc. 43 Chapter 2: Integers and Introduction to Solving Equations ISM: Algebra Foundations 80. โˆ’|โˆ’8| = โˆ’8 โˆ’|โˆ’4| = โˆ’4 Since โˆ’8 < โˆ’4, โˆ’|โˆ’8| < โˆ’|โˆ’4|. 40. The opposite of 8 is negative 8. โˆ’(8) = โˆ’8 42. The opposite of negative 6 is 6. โˆ’(โˆ’6) = 6 44. The opposite of 123 is negative 123. โˆ’(123) = โˆ’123 46. The opposite of negative 13 is 13. โˆ’(โˆ’13) = 13 48. |โˆ’11| = 11 82. โˆ’(โˆ’38) = 38 Since โˆ’22 < 38, โˆ’22 โˆ’17 since โˆ’4 is to the right of โˆ’17 on a number line. 72. โˆ’|17| = โˆ’17 โˆ’(โˆ’17) = 17 Since โˆ’17 < 17, โˆ’|17| < โˆ’(โˆ’17). โˆ’|โˆ’5|, โˆ’(โˆ’4), 23 , |10|. 76. โˆ’45 0, |โˆ’45| > |0|. 1 100. 70. |โˆ’8| = 8 |โˆ’4| = 4 Since 8 > 4, |โˆ’8| > |โˆ’4|. 20 + 15 35 104. 14 = 1, โˆ’(โˆ’3) = 3, โˆ’|7| = โˆ’7, and |โˆ’20| = 20, so the numbers in order from least to greatest are โˆ’|7|, 14 , โˆ’(โˆ’3), |โˆ’20|. 106. 33 = 27, โˆ’|โˆ’11| = โˆ’11, โˆ’(โˆ’10) = 10, โˆ’4 = โˆ’4, โˆ’|2| = โˆ’2, so the numbers in order from least to greatest are โˆ’|โˆ’11|, โˆ’4, โˆ’|2|, โˆ’(โˆ’10), and 33. Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations |0| = 0; since 0 4 is false. 9. โˆ’54 + 20 = โˆ’34 b. |โˆ’4| = 4; since 4 = 4, then |โˆ’4| > 4 is false. 10. 7 + (โˆ’2) = 5 c. |5| = 5; since 5 > 4, then |5| > 4 is true. 11. โˆ’3 + 0 = โˆ’3 d. |โˆ’100| = 100; since 100 > 4, then |โˆ’100| > 4 is true. 12. 18 + (โˆ’18) = 0 108. a. 13. โˆ’64 + 64 = 0 110. (โˆ’|โˆ’(โˆ’7)|) = (โˆ’|7|) = โˆ’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 + (โˆ’2) + (โˆ’15) = 4 + (โˆ’15) = โˆ’11 15. 5 + (โˆ’3) + 12 + (โˆ’14) = 2 + 12 + (โˆ’14) = 14 + (โˆ’14) =0 16. x + 3y = โˆ’6 + 3(2) = โˆ’6 + 6 = 0 17. x + y = โˆ’13 + (โˆ’9) = โˆ’22 18. Temperature at 8 a.m. = โˆ’7 + (+4) + (+7) = โˆ’3 + (+7) =4 The temperature was 4ยฐF at 8 a.m. 120. no; answers may vary Section 2.2 Practice Exercises 1. Calculator Explorations 1. โˆ’256 + 97 = โˆ’159 2. 811 + (โˆ’1058) = โˆ’247 2. 3. 6(15) + (โˆ’46) = 44 4. โˆ’129 + 10(48) = 351 5. โˆ’108,650 + (โˆ’786,205) = โˆ’894,855 3. 6. โˆ’196,662 + (โˆ’129,856) = โˆ’326,518 Vocabulary, Readiness & Video Check 2.2 4. |โˆ’3| + |โˆ’19| = 3 + 19 = 22 The common sign is negative, so (โˆ’3) + (โˆ’19) = โˆ’22. 5. โˆ’12 + (โˆ’30) = โˆ’42 1. If n is a number, then โˆ’n + n = 0. 2. Since x + n = n + x, we say that addition is commutative. 3. If a is a number, then โˆ’(โˆ’a) = a. 6. 9 + 4 = 13 7. |โˆ’1| = 1, |26| = 26, and 26 โˆ’ 1 = 25 26 > 1, so the answer is positive. โˆ’1 + 26 = 25 8. |2| = 2, |โˆ’18| = 18, and 18 โˆ’ 2 = 16 18 > 2, so the answer is negative. 2 + (โˆ’18) = โˆ’16 4. Since n + (x + a) = (n + x) + a, we say that addition is associative. 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, โˆ’6. Copyright ยฉ 2020 Pearson Education, Inc. 45 Chapter 2: Integers and Introduction to Solving Equations ISM: Algebra Foundations 6. Negative; the numbers have the same signโŽฏboth are negativeโŽฏand we keep this common sign in the sum. 7. The diverโ€™s current depth is 231 feet below the surface. Exercise Set 2.2 2. โˆ’6 + (โˆ’5) = โˆ’11 4. 10 + (โˆ’3) = 7 6. 9 + (โˆ’4) = 5 8. 15 + 42 = 57 10. |โˆ’5| + |โˆ’4| = 5 + 4 = 9 The common sign is negative, so โˆ’5 + (โˆ’4) = โˆ’9. 12. โˆ’62 + 62 = 0 14. |8| โˆ’ |โˆ’3| = 8 โˆ’ 3 = 5 8 > 3, so the answer is positive. 8 + (โˆ’3) = 5 16. โˆ’8 + 0 = โˆ’8 18. |โˆ’9| โˆ’ |5| = 9 โˆ’ 5 = 4 9 > 5, so the answer is negative. 5 + (โˆ’9) = โˆ’4 20. |โˆ’6| + |โˆ’1| = 6 + 1 = 7 The common sign is negative, so โˆ’6 + (โˆ’1) = โˆ’7. 22. |โˆ’23| + |โˆ’23| = 23 + 23 = 46 The common sign is negative, so โˆ’23 + (โˆ’23) = โˆ’46. 24. |โˆ’400| + |โˆ’256| = 400 + 256 = 656 The common sign is negative, so โˆ’400 + (โˆ’256) = โˆ’656. 26. |24| โˆ’ |โˆ’10| = 24 โˆ’ 10 = 14 24 > 10, so the answer is positive. 24 + (โˆ’10) = 14 46 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations 28. |โˆ’8| โˆ’ |4| = 8 โˆ’ 4 = 4 8 > 4, so the answer is negative. โˆ’8 + 4 = โˆ’4 30. |โˆ’89| โˆ’ |37| = 89 โˆ’ 37 = 52 89 > 37, so the answer is negative. โˆ’89 + 37 = โˆ’52 32. |62| โˆ’ |โˆ’32| = 62 โˆ’ 32 = 30 62 > 32, so the answer is positive. โˆ’32 + 62 = 30 34. |โˆ’375| โˆ’ |325| = 375 โˆ’ 325 = 50 375 > 325, so the answer is negative. 325 + (โˆ’375) = โˆ’50 36. |โˆ’56| + |โˆ’33| = 56 + 33 = 89 The common sign is negative, so โˆ’56 + (โˆ’33) = โˆ’89. 38. โˆ’1 + 5 + (โˆ’8) = 4 + (โˆ’8) = โˆ’4 40. โˆ’103 + (โˆ’32) + (โˆ’27) = โˆ’135 + (โˆ’27) = โˆ’162 42. 18 + (โˆ’9) + 5 + (โˆ’2) = 9 + 5 + (โˆ’2) = 14 + (โˆ’2) = 12 44. 34 + (โˆ’12) + (โˆ’11) + 213 = 22 + (โˆ’11) + 213 = 11 + 213 = 224 46. โˆ’12 + (โˆ’3) + (โˆ’5) = โˆ’15 + (โˆ’5) = โˆ’20 48. โˆ’35 + (โˆ’12) = โˆ’47 68. The sum of โˆ’49, โˆ’2, and 40 is โˆ’49 + (โˆ’2) + 40 = โˆ’51 + 40 = โˆ’11. 70. 0 + (โˆ’248) + 8 + (โˆ’16) + (โˆ’28) + 32 = โˆ’248 + 8 + (โˆ’16) + (โˆ’28) + 32 = โˆ’240 + (โˆ’16) + (โˆ’28) + 32 = โˆ’256 + (โˆ’28) + 32 = โˆ’284 + 32 = โˆ’252 The diverโ€™s final depth is 252 meters below the surface. 72. Since โˆ’3 (โˆ’1)50 since (โˆ’1)50 = 1. (โˆ’1)55 and (โˆ’7)23 are negative since there are an odd number of factors. Note that (โˆ’7)23 โˆ’10 since 0 is to the right of โˆ’10 on a number line. 28. โˆ’9(100) = โˆ’900 4. โˆ’4 < 4 since โˆ’4 is to the left of 4 on a number line. 5. โˆ’15 โˆ’7 since โˆ’2 is to the right of โˆ’7 on a number line. 7. |โˆ’3| = 3 because โˆ’3 is 3 units from 0. 8. |โˆ’9| = 9 because โˆ’9 is 9 units from 0. 29. โˆ’12 โˆ’ 6 โˆ’ (โˆ’6) = โˆ’12 + (โˆ’6) + 6 = โˆ’18 + 6 = โˆ’12 30. โˆ’4 + (โˆ’8) โˆ’ 16 โˆ’ (โˆ’9) = โˆ’4 + (โˆ’8) + (โˆ’16) + 9 = โˆ’12 + (โˆ’16) + 9 = โˆ’28 + 9 = โˆ’19 31. โˆ’105 is undefined. 0 32. 7(โˆ’16)(0)(โˆ’3) = 0 (since one factor is 0) 9. โˆ’|โˆ’4| = โˆ’4 33. Subtract โˆ’8 from โˆ’12 is โˆ’12 โˆ’ (โˆ’8) = โˆ’12 + 8 = โˆ’4. 10. โˆ’(โˆ’5) = 5 34. The sum of โˆ’17 and โˆ’27 is โˆ’17 + (โˆ’27) = โˆ’44. 11. The opposite of 11 is โˆ’11. 35. The product of โˆ’5 and โˆ’25 is โˆ’5(โˆ’25) = 125. 12. The opposite of โˆ’3 is โˆ’(โˆ’3) = 3. 13. The opposite of 64 is โˆ’64. 14. The opposite of 0 is โˆ’0 = 0. 36. The quotient of โˆ’100 and โˆ’5 is 37. Divide a number by โˆ’17 is 15. โˆ’3 + 15 = 12 16. โˆ’9 + (โˆ’11) = โˆ’20 17. โˆ’8(โˆ’6)(โˆ’1) = 48(โˆ’1) = โˆ’48 โˆ’100 = 20. โˆ’5 x or x รท (โˆ’17). โˆ’17 38. The sum of โˆ’3 and a number is โˆ’3 + x. 39. A number decreased by โˆ’18 is x โˆ’ (โˆ’18). Copyright ยฉ 2020 Pearson Education, Inc. 53 Chapter 2: Integers and Introduction to Solving Equations 40. The product of โˆ’7 and a number is โˆ’7 โ‹… x or โˆ’7x. 41. x + y = โˆ’3 + 12 = 9 42. x โˆ’ y = โˆ’3 โˆ’ 12 = โˆ’3 + (โˆ’12) = โˆ’15 43. 2y โˆ’ x = 2(12) โˆ’ (โˆ’3) = 24 โˆ’ (โˆ’3) = 24 + 3 = 27 ISM: Algebra Foundations 10. โˆ’4[โˆ’6 + 5(โˆ’3 + 5)] โˆ’ 7 = โˆ’4[โˆ’6 + 5(2)] โˆ’ 7 = โˆ’4[โˆ’6 + 10] โˆ’ 7 = โˆ’4(4) โˆ’ 7 = โˆ’16 โˆ’ 7 = โˆ’23 11. x 2 = (โˆ’15)2 = (โˆ’15)(โˆ’15) = 225 44. 3y + x = 3(12) + (โˆ’3) = 36 + (โˆ’3) = 33 45. 5x = 5(โˆ’3) = โˆ’15 46. โˆ’ x 2 = โˆ’(โˆ’15)2 = โˆ’(โˆ’15)(โˆ’15) = โˆ’225 12. 5 y 2 = 5(4) 2 = 5(16) = 80 y 12 = = โˆ’4 x โˆ’3 5 y 2 = 5(โˆ’4)2 = 5(16) = 80 13. x 2 + y = (โˆ’6)2 + (โˆ’3) = 36 + (โˆ’3) = 33 Section 2.5 Practice Exercises 1. (โˆ’2)4 = (โˆ’2)(โˆ’2)(โˆ’2)(โˆ’2) = 16 14. 4 โˆ’ x 2 = 4 โˆ’ (โˆ’8)2 = 4 โˆ’ 64 = โˆ’60 2. โˆ’24 = โˆ’(2)(2)(2)(2) = โˆ’16 15. average sum of numbers = number of numbers 17 + (โˆ’1) + (โˆ’11) + (โˆ’13) + (โˆ’16) + (โˆ’13) + 2 = 7 โˆ’35 = 7 = โˆ’5 The average of the temperatures is โˆ’5ยฐF. 3. 3 โ‹… 62 = 3 โ‹… (6 โ‹… 6) = 3 โ‹… 36 = 108 4. โˆ’25 โˆ’25 = =5 5(โˆ’1) โˆ’5 5. โˆ’18 + 6 โˆ’12 = =3 โˆ’3 โˆ’ 1 โˆ’4 Calculator Explorations 3 6. 30 + 50 + (โˆ’4) = 30 + 50 + (โˆ’64) = 80 + (โˆ’64) = 16 1. โˆ’120 โˆ’ 360 = 48 โˆ’10 7. โˆ’23 + (โˆ’4)2 + 15 = โˆ’8 + 16 + 1 = 8 + 1 = 9 2. 4750 = โˆ’250 โˆ’2 + (โˆ’17) 8. 2(2 โˆ’ 9) + (โˆ’12) โˆ’ 3 = 2(โˆ’7) + (โˆ’12) โˆ’ 3 = โˆ’14 + (โˆ’12) โˆ’ 3 = โˆ’26 โˆ’ 3 = โˆ’29 3. โˆ’316 + (โˆ’458) = โˆ’258 28 + (โˆ’25) 4. โˆ’234 + 86 = 74 โˆ’18 + 16 9. (โˆ’5) โ‹… โˆ’8 + (โˆ’3) + 23 = (โˆ’5) โ‹… 8 + (โˆ’3) + 23 = (โˆ’5) โ‹… 8 + (โˆ’3) + 8 = โˆ’40 + (โˆ’3) + 8 = โˆ’43 + 8 = โˆ’35 Vocabulary, Readiness & Video Check 2.5 1. To simplify โˆ’2 รท 2 โ‹… (3), which operation should be performed first? division 2. To simplify โˆ’9 โˆ’ 3 โ‹… 4, which operation should be performed first? multiplication 54 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations 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 โ‹… 6 โˆ’ 6 โ‹… 5 + (โˆ’10) = 42 โˆ’ 6 โ‹… 5 + (โˆ’10) = 42 โˆ’ 30 + (โˆ’10) = 12 + (โˆ’10) =2 4. To simplify 5[โˆ’9 + (โˆ’3)] รท 4, which operation should be performed first? addition 28. 7 โˆ’ (โˆ’5)2 = 7 โˆ’ 25 = โˆ’18 5. To simplify โˆ’2 + 3(10 โˆ’ 12) โ‹… (โˆ’8), which operation should be performed first? subtraction 30. 6. To evaluate x โˆ’ 3y for x = โˆ’7 and y = โˆ’1, replace x with โˆ’7 and y with โˆ’1 and evaluate โˆ’7 โˆ’ 3(โˆ’1). 32. 10 โ‹… 53 + 7 = 10 โ‹…125 + 7 = 1250 + 7 = 1257 7. A fraction bar means divided by and it is a grouping symbol. 8. To make sure that the entire value of โˆ’2, including the sign, is squared. 9. Finding the average is a good application of both order of operations and adding and dividing integers. โˆ’3 + 7 โ‹… 7 2 = 4 โ‹… 72 = 4 โ‹… 7 2 = 4 โ‹… 49 = 196 34. 82 โˆ’ (5 โˆ’ 2)4 = 82 โˆ’ 34 = 64 โˆ’ 81 = โˆ’17 36. |12 โˆ’ 19| รท 7 = |โˆ’7| รท 7 = 7 รท 7 = 1 38. โˆ’(โˆ’2)3 = โˆ’(โˆ’8) = 8 40. (2 โˆ’ 7) 2 รท (4 โˆ’ 3) 4 = (โˆ’5)2 รท 14 = 25 รท 1 = 25 2. โˆ’24 = โˆ’(2)(2)(2)(2) = โˆ’16 42. 3 โˆ’ 15 โ‹… (โˆ’4) รท (โˆ’16) = โˆ’12 โ‹… (โˆ’4) รท (โˆ’16) = 12 โ‹… (โˆ’4) รท (โˆ’16) = โˆ’48 รท (โˆ’16) =3 4. (โˆ’2)4 = (โˆ’2)(โˆ’2)(โˆ’2)(โˆ’2) = 16 44. (โˆ’20 โˆ’ 5) รท 5 โˆ’ 15 = (โˆ’25) รท 5 โˆ’ 15 = โˆ’5 โˆ’ 15 = โˆ’20 6. 5 โ‹… 23 = 5 โ‹… 8 = 40 46. 3 โ‹… (8 โˆ’ 3) + (โˆ’4) โˆ’ 10 = 3 โ‹… (5) + (โˆ’4) โˆ’ 10 = 15 + (โˆ’4) โˆ’ 10 = 11 โˆ’ 10 =1 Exercise Set 2.5 8. 10 โˆ’ 23 โˆ’ 12 = โˆ’13 โˆ’ 12 = โˆ’25 10. โˆ’8 + 4(3) = โˆ’8 + 12 = 4 12. 7(โˆ’6) + 3 = โˆ’42 + 3 = โˆ’39 14. โˆ’12 + 6 รท 3 = โˆ’12 + 2 = โˆ’10 16. 5 + 9 โ‹… 4 โˆ’ 20 = 5 + 36 โˆ’ 20 = 41 โˆ’ 20 = 21 18. 20 โˆ’ 15 5 = = โˆ’5 โˆ’1 โˆ’1 20. 88 88 = = โˆ’8 โˆ’8 โˆ’ 3 โˆ’11 22. 7(โˆ’4) โˆ’ (โˆ’6) = โˆ’28 + 6 = โˆ’22 24. [9 + (โˆ’2)]3 = [7]3 = 343 48. (4 โˆ’ 12) โ‹… (8 โˆ’ 17) = (โˆ’8) โ‹… (โˆ’9) = 72 50. (โˆ’4 รท 4) โˆ’ (8 รท 8) = (โˆ’1) โˆ’ (1) = โˆ’2 52. (11 โˆ’ 32 )3 = (11 โˆ’ 9)3 = 23 = 8 54. โˆ’3(4 โˆ’ 8)2 + 5(14 โˆ’ 16)3 = โˆ’3(โˆ’4)2 + 5(โˆ’2)3 = โˆ’3(16) + 5(โˆ’8) = โˆ’48 + (โˆ’40) = โˆ’88 56. 12 โˆ’ [7 โˆ’ (3 โˆ’ 6)] + (2 โˆ’ 3)3 = 12 โˆ’ [7 โˆ’ (โˆ’3)] + (2 โˆ’ 3)3 = 12 โˆ’ (7 + 3) + (โˆ’1)3 = 12 โˆ’ 10 + (โˆ’1) = 2 + (โˆ’1) =1 Copyright ยฉ 2020 Pearson Education, Inc. 55 Chapter 2: Integers and Introduction to Solving Equations 58. 10(โˆ’1) โˆ’ (โˆ’2)(โˆ’3) โˆ’10 โˆ’ 6 = 2[โˆ’8 รท (โˆ’2 โˆ’ 2)] 2[โˆ’8 รท (โˆ’4)] โˆ’16 = 2(2) โˆ’16 = 4 = โˆ’4 60. โˆ’2[6 + 4(2 โˆ’ 8)] โˆ’ 25 = โˆ’2[6 + 4(โˆ’6)] โˆ’ 25 = โˆ’2[6 + (โˆ’24)] โˆ’ 25 = โˆ’2(โˆ’18) โˆ’ 25 = 36 โˆ’ 25 = 11 62. x โˆ’ y โˆ’ z = โˆ’2 โˆ’ 4 โˆ’ (โˆ’1) = โˆ’2 โˆ’ 4 + 1 = โˆ’6 + 1 = โˆ’5 84. The two lowest scores are โˆ’14 and โˆ’10. โˆ’10 โˆ’ (โˆ’14) = โˆ’10 + 14 = 4 The difference between the two lowest scores is 4. โˆ’14 + (โˆ’10) + (โˆ’6) โˆ’30 = = โˆ’10 3 3 The average of the scores is โˆ’10. 86. average = 90. 90 รท 45 = 2 92. 45 + 90 = 135 94. 3 + 5 + 3 + 5 = 16 The perimeter is 16 centimeters. 66. x 2 + z = (โˆ’2)2 + (โˆ’1) = 4 + (โˆ’1) = 3 96. 17 + 23 + 32 = 72 The perimeter is 72 meters. 98. (7 โ‹… 3 โˆ’ 4) โ‹… 2 = (21 โˆ’ 4) โ‹… 2 = 17 โ‹… 2 = 34 4 x 4(โˆ’2) โˆ’8 = = = โˆ’2 68. 4 4 y 100. 2 โ‹… (8 รท 4 โˆ’ 20) = 2 โ‹… (2 โˆ’ 20) = 2 โ‹… (โˆ’18) = โˆ’36 102. answers may vary 70. z 2 = (โˆ’4)2 = 16 104. answers may vary 2 72. โˆ’ x = โˆ’(โˆ’3) = โˆ’9 106. (โˆ’17)6 = (โˆ’17)(โˆ’17)(โˆ’17)(โˆ’17)(โˆ’17)(โˆ’17) = 24,137,569 74. 3x 2 = 3(โˆ’3)2 = 3(9) = 27 76. 3 โˆ’ z 2 = 3 โˆ’ (โˆ’4)2 = 3 โˆ’ 16 = โˆ’13 78. 3z 2 โˆ’ x = 3(โˆ’4) 2 โˆ’ (โˆ’3) = 3(16) + 3 = 48 + 3 = 51 โˆ’18 + (โˆ’8) + (โˆ’1) + (โˆ’1) + 0 + 4 6 โˆ’24 = 6 = โˆ’4 80. average = 56 โˆ’40 + (โˆ’20) + (โˆ’10) + (โˆ’15) + (โˆ’5) 5 โˆ’90 = 5 = โˆ’18 82. average = 88. no; answers may vary 64. 5 x โˆ’ y + 4 z = 5(โˆ’2) โˆ’ 4 + 4(โˆ’1) = โˆ’10 โˆ’ 4 + (โˆ’4) = โˆ’14 + (โˆ’4) = โˆ’18 2 ISM: Algebra Foundations 108. 3x 2 + 2 x โˆ’ y = 3(โˆ’18)2 + 2(โˆ’18) โˆ’ 2868 = 3(324) + (โˆ’36) โˆ’ 2868 = 972 + (โˆ’36) โˆ’ 2868 = 936 โˆ’ 2868 = โˆ’1932 110. 5(ab + 3)b = 5(โˆ’2 โ‹… 3 + 3)3 Copyright ยฉ 2020 Pearson Education, Inc. = 5(โˆ’6 + 3)3 = 5(โˆ’3)3 = 5(โˆ’27) = โˆ’135 ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations 7. Section 2.6 Practice Exercises 1. โˆ’4 x โˆ’ 3 = 5 โˆ’4(โˆ’2) โˆ’ 3 โฑจ 5 8โˆ’3โฑจ 5 5 = 5 True Since 5 = 5 is true, โˆ’2 is a solution of the equation. 2. y โˆ’ 6 = โˆ’2 y โˆ’ 6 + 6 = โˆ’2 + 6 y=4 Check: y โˆ’ 6 = โˆ’2 4โˆ’6โฑจ โˆ’2 โˆ’2 = โˆ’2 True The solution is 4. 3. โˆ’2 = z + 8 โˆ’2 โˆ’ 8 = z + 8 โˆ’ 8 โˆ’10 = z Check: โˆ’2 = z + 8 โˆ’2 โฑจ โˆ’ 10 + 8 โˆ’2 = โˆ’2 True The solution is โˆ’10. 4. x = โˆ’2 + 90 + (โˆ’100) x = 88 + (โˆ’100) x = โˆ’12 The solution is โˆ’12. 5. 3 y = โˆ’18 3 y โˆ’18 = 3 3 3 โˆ’18 โ‹…y= 3 3 y = โˆ’6 Check: 3 y = โˆ’18 3(โˆ’6) โฑจ โˆ’ 18 โˆ’18 = โˆ’18 True The solution is โˆ’6. 6. โˆ’32 = 8 x โˆ’32 8 x = 8 8 โˆ’32 8 = โ‹…x 8 8 โˆ’4 = x Check: โˆ’32 = 8 x โˆ’32 โฑจ 8(โˆ’4) โˆ’32 = โˆ’32 True The solution is โˆ’4. 8. โˆ’3 y = โˆ’27 โˆ’3 y โˆ’27 = โˆ’3 โˆ’3 โˆ’3 โˆ’27 โ‹…y= โˆ’3 โˆ’3 y=9 Check: โˆ’3 y = โˆ’27 โˆ’3 โ‹… 9 โฑจ โˆ’ 27 โˆ’27 = โˆ’27 True The solution is 9. x =7 โˆ’4 x โˆ’4 โ‹… = โˆ’4 โ‹… 7 โˆ’4 โˆ’4 โ‹… x = โˆ’4 โ‹… 7 โˆ’4 x = โˆ’28 x =7 Check: โˆ’4 โˆ’28 โฑจ7 โˆ’4 7 = 7 True The solution is โˆ’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 โ€œexpression = expressionโ€ 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 ยฉ 2020 Pearson Education, Inc. 57 Chapter 2: Integers and Introduction to Solving Equations 9. an equal sign ISM: Algebra Foundations 12. s โˆ’ 7 = โˆ’15 s โˆ’ 7 + 7 = โˆ’15 + 7 s = โˆ’8 Check: s โˆ’ 7 = โˆ’15 โˆ’8 โˆ’ 7 โฑจ โˆ’ 15 โˆ’15 = โˆ’15 True The solution is โˆ’8. 14. 1= y+7 1โˆ’ 7 = y + 7 โˆ’ 7 โˆ’6 = y Check: 1 = y + 7 1โฑจ โˆ’6+ 7 1 = 1 True The solution is โˆ’6. 10. We can add the same number to both sides of an equation and weโ€™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. Exercise Set 2.6 2. y โˆ’ 16 = โˆ’7 9 โˆ’ 16 โฑจ โˆ’ 7 โˆ’7 = โˆ’7 True Since โˆ’7 = โˆ’7 is true, 9 is a solution of the equation. 4. a + 23 = โˆ’16 โˆ’7 + 23 โฑจ โˆ’ 16 16 = โˆ’16 False Since 16 = โˆ’16 is false, โˆ’7 is not a solution of the equation. 6. โˆ’3k = 12 โˆ’ k โˆ’3(โˆ’6) โฑจ 12 โˆ’ (โˆ’6) 18 โฑจ 12 + 6 18 = 18 True Since 18 = 18 is true, โˆ’6 is a solution of the equation. 8. 2(b โˆ’ 3) = 10 2(1 โˆ’ 3) โฑจ 10 2(โˆ’2) โฑจ 10 โˆ’4 = 10 False Since โˆ’4 = 10 is false, 1 is not a solution of the equation. 10. f + 4 = โˆ’6 f + 4 โˆ’ 4 = โˆ’6 โˆ’ 4 f = โˆ’10 f + 4 = โˆ’6 โˆ’10 + 4 โฑจ โˆ’ 6 โˆ’6 = โˆ’6 True The solution is โˆ’10. Check: 58 16. โˆ’50 + 40 โˆ’ 5 = z โˆ’10 โˆ’ 5 = z โˆ’15 = z Check: โˆ’50 + 40 โˆ’ 5 = z โˆ’50 + 40 โˆ’ 5 โฑจ โˆ’ 15 โˆ’10 โˆ’ 5 โฑจ โˆ’ 15 โˆ’15 = โˆ’15 True The solution is โˆ’15. 18. 6 y = 48 6 y 48 = 6 6 6 48 โ‹…y= 6 6 y =8 Check: 6 y = 48 6(8) โฑจ 48 48 = 48 True The solution is 8. 20. โˆ’2 x = 26 โˆ’2 x 26 = โˆ’2 โˆ’2 โˆ’2 26 โ‹…x = โˆ’2 โˆ’2 x = โˆ’13 Check: โˆ’2 x = 26 โˆ’2(โˆ’13) โฑจ 26 26 = 26 True The solution is โˆ’13. Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations 22. 24. 26. 28. n = โˆ’5 11 n 11 โ‹… = 11 โ‹… (โˆ’5) 11 11 โ‹… n = 11 โ‹… (โˆ’5) 11 n = โˆ’55 n = โˆ’5 Check: 11 โˆ’55 โฑจ โˆ’5 11 โˆ’5 = โˆ’5 True The solution is โˆ’55. 7 y = โˆ’21 7 y โˆ’21 = 7 7 7 โˆ’21 โ‹…y= 7 7 y = โˆ’3 Check: 7 y = โˆ’21 7 โ‹… (โˆ’3) โฑจ โˆ’ 21 โˆ’21 = โˆ’21 True The solution is โˆ’3. โˆ’9 x = 0 โˆ’9 x 0 = โˆ’9 โˆ’9 โˆ’9 0 โ‹…x = โˆ’9 โˆ’9 x=0 Check: โˆ’9 x = 0 โˆ’9 โ‹… 0 โฑจ 0 0 = 0 True The solution is 0. โˆ’31x = โˆ’31 โˆ’31x โˆ’31 = โˆ’31 โˆ’31 โˆ’31 โˆ’31 โ‹…x = โˆ’31 โˆ’31 x =1 Check: โˆ’31x = โˆ’31 โˆ’31 โ‹…1 โฑจ โˆ’ 31 โˆ’31 = โˆ’31 True The solution is 1. Chapter 2: Integers and Introduction to Solving Equations 30. 3 y = โˆ’27 3 y โˆ’27 = 3 3 3 โˆ’27 โ‹…y= 3 3 y = โˆ’9 The solution is โˆ’9. 32. n โˆ’ 4 = โˆ’48 n โˆ’ 4 + 4 = โˆ’48 + 4 n = โˆ’44 The solution is โˆ’44. 34. โˆ’36 = y + 12 โˆ’36 โˆ’ 12 = y + 12 โˆ’ 12 โˆ’48 = y The solution is โˆ’48. 36. x = โˆ’9 โˆ’9 x โˆ’9 โ‹… = โˆ’9 โ‹… (โˆ’9) โˆ’9 โˆ’9 โ‹… x = โˆ’9 โ‹… (โˆ’9) โˆ’9 x = 81 The solution is 81. 38. z = โˆ’28 + 36 z=8 The solution is 8. 40. 42. โˆ’11x = โˆ’121 โˆ’11x โˆ’121 = โˆ’11 โˆ’11 โˆ’11 โˆ’121 โ‹…x = โˆ’11 โˆ’11 x = 11 The solution is 11. n = โˆ’20 5 n 5 โ‹… = 5 โ‹… (โˆ’20) 5 5 โ‹… n = 5 โ‹… (โˆ’20) 5 n = โˆ’100 The solution is โˆ’100. Copyright ยฉ 2020 Pearson Education, Inc. 59 Chapter 2: Integers and Introduction to Solving Equations ISM: Algebra Foundations 44. โˆ’81 = 27 x โˆ’81 27 x = 27 27 โˆ’81 27 = โ‹…x 27 27 โˆ’3 = x The solution is โˆ’3. 46. A number increased by โˆ’5 is x + (โˆ’5). 48. The quotient of a number and โˆ’20 is x รท (โˆ’20) or x . โˆ’20 50. โˆ’32 multiplied by a number is โˆ’32 โ‹… x or โˆ’32x. 52. Subtract a number from โˆ’18 is โˆ’18 โˆ’ x. 54. 56. n + 961 = 120 n + 961 โˆ’ 961 = 120 โˆ’ 961 n = โˆ’841 The solution is โˆ’841. y = 1098 โˆ’18 y โˆ’18 โ‹… = โˆ’18 โ‹…1098 โˆ’18 โˆ’18 โ‹… y = โˆ’18 โ‹…1098 โˆ’18 y = โˆ’19, 764 The solution is โˆ’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 the number line but are on opposite sides of 0 are called opposites. 2. The absolute value of a number is that numberโ€™s distance from 0 on a number line. 3. The integers are …, โˆ’3, โˆ’2, โˆ’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 โ€œโ€ 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 ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations 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 โ€œexpression = expressionโ€ is called an equation. 11. The sign โ€œโ€ 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 ground level, then 1572 feet below the ground is โˆ’1572. 2. If 0 represents sea level, then an elevation of 11,239 feet is +11,239. 3. 4. 5. |โˆ’11| = 11 since โˆ’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. โˆ’|8| = โˆ’8 8. โˆ’(โˆ’9) = 9 9. โˆ’|โˆ’16| = โˆ’16 10. โˆ’(โˆ’2) = 2 11. โˆ’18 > โˆ’20 since โˆ’18 is to the right of โˆ’20 on a number line. 12. โˆ’5 โˆ’198, |โˆ’123| > โˆ’|โˆ’198|. 14. |โˆ’12| = 12 โˆ’|โˆ’16| = โˆ’16 Since 12 > โˆ’16, |โˆ’12| > โˆ’|โˆ’16|. 15. The opposite of โˆ’18 is 18. โˆ’(โˆ’18) = 18 16. The opposite of 42 is negative 42. โˆ’(42) = โˆ’42 17. False; consider a = 1 and b = 2, then 1 3, so the answer is positive. 5 + (โˆ’3) = 2 28. |18| โˆ’ |โˆ’4| = 18 โˆ’ 4 = 14 18 > 4, so the answer is positive. 18 + (โˆ’4) = 14 43. 12 โˆ’ 4 = 12 + (โˆ’4) = 8 44. โˆ’12 โˆ’ 4 = โˆ’12 + (โˆ’4) = โˆ’16 45. โˆ’7 โˆ’ 17 = โˆ’7 + (โˆ’17) = โˆ’24 29. |16| โˆ’ |โˆ’12| = 16 โˆ’ 12 = 4 16 > 12, so the answer is positive. โˆ’12 + 16 = 4 46. 7 โˆ’ 17 = 7 + (โˆ’17) = โˆ’10 30. |40| โˆ’ |โˆ’23| = 40 โˆ’ 23 = 17 40 > 23, so the answer is positive. โˆ’23 + 40 = 17 48. โˆ’6 โˆ’ (โˆ’14) = โˆ’6 + 14 = 8 47. 7 โˆ’ (โˆ’13) = 7 + 13 = 20 49. 16 โˆ’ 16 = 16 + (โˆ’16) = 0 31. |โˆ’8| + |โˆ’15| = 8 + 15 = 23 The common sign is negative, so โˆ’8 + (โˆ’15) = โˆ’23. 50. โˆ’16 โˆ’ 16 = โˆ’16 + (โˆ’16) = โˆ’32 32. |โˆ’5| + |โˆ’17| = 5 + 17 = 22 The common sign is negative, so โˆ’5 + (โˆ’17) = โˆ’22. 52. โˆ’5 โˆ’ (โˆ’12) = โˆ’5 + 12 = 7 33. |โˆ’24| โˆ’ |3| = 24 โˆ’ 3 = 21 24 > 3, so the answer is negative. โˆ’24 + 3 = โˆ’21 34. |โˆ’89| โˆ’ |19| = 89 โˆ’ 19 = 70 89 > 19, so the answer is negative. โˆ’89 + 19 = โˆ’70 35. 15 + (โˆ’15) = 0 51. โˆ’12 โˆ’ (โˆ’12) = โˆ’12 + 12 = 0 53. โˆ’(โˆ’5) โˆ’ 12 + (โˆ’3) = 5 + (โˆ’12) + (โˆ’3) = โˆ’7 + (โˆ’3) = โˆ’10 54. โˆ’8 + (โˆ’12) โˆ’ 10 โˆ’ (โˆ’3) = โˆ’8 + (โˆ’12) + (โˆ’10) + 3 = โˆ’20 + (โˆ’10) + 3 = โˆ’30 + 3 = โˆ’27 55. 600 โˆ’ (โˆ’92) = 600 + 92 = 692 The difference in elevations is 692 feet. 36. โˆ’24 + 24 = 0 62 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations 56. 142 โˆ’ 125 + 43 โˆ’ 85 = 142 + (โˆ’125) + 43 + (โˆ’85) = 17 + 43 + (โˆ’85) = 60 + (โˆ’85) = โˆ’25 The balance in his account is โˆ’25. 74. โˆ’72 = โˆ’9 8 75. โˆ’38 = 38 โˆ’1 57. 85 โˆ’ 99 = 85 + (โˆ’99) = โˆ’14 You are โˆ’14 feet or 14 feet below ground at the end of the drop. 76. 45 = โˆ’5 โˆ’9 58. 66 โˆ’ (โˆ’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 โˆ’5. (โˆ’5)(2) = โˆ’10 The total loss is 10 yards. 59. |โˆ’5| โˆ’ |โˆ’6| = 5 โˆ’ 6 = 5 + (โˆ’6) = โˆ’1 5 โˆ’ 6 = 5 + (โˆ’6) = โˆ’1 |โˆ’5| โˆ’ |โˆ’6| = 5 โˆ’ 6 is true. 78. A loss of $50 is represented by โˆ’50. (โˆ’50)(4) = โˆ’200 The total loss is $200. 60. |โˆ’5 โˆ’ (โˆ’6)| = |โˆ’5 + 6| = |1| = 1 5 + 6 = 11 Since 1 โ‰  11, the statement is false. 79. A debt of $1024 is represented by โˆ’1024. โˆ’1024 รท 4 = โˆ’256 Each payment is $256. 61. โˆ’3(โˆ’7) = 21 80. A drop of 45 degrees is represented by โˆ’45. โˆ’45 = โˆ’5 or โˆ’45 รท 9 = โˆ’5 9 The average drop each hour is 5ยฐF. 62. โˆ’6(3) = โˆ’18 63. โˆ’4(16) = โˆ’64 81. (โˆ’7)2 = (โˆ’7)(โˆ’7) = 49 64. โˆ’5(โˆ’12) = 60 82. โˆ’72 = โˆ’(7 โ‹… 7) = โˆ’49 65. (โˆ’5)2 = (โˆ’5)(โˆ’5) = 25 66. (โˆ’1)5 = (โˆ’1)(โˆ’1)(โˆ’1)(โˆ’1)(โˆ’1) = โˆ’1 84. โˆ’3 + 12 + (โˆ’7) โˆ’ 10 = 9 + (โˆ’7) โˆ’ 10 = 2 โˆ’ 10 = โˆ’8 67. 12(โˆ’3)(0) = 0 68. โˆ’1(6)(2)(โˆ’2) = โˆ’6(2)(โˆ’2) = โˆ’12(โˆ’2) = 24 69. โˆ’15 รท 3 = โˆ’5 70. โˆ’24 =3 โˆ’8 71. 0 =0 โˆ’3 72. โˆ’46 is undefined. 0 73. 100 = โˆ’20 โˆ’5 83. 5 โˆ’ 8 + 3 = โˆ’3 + 3 = 0 85. โˆ’10 + 3 โ‹… (โˆ’2) = โˆ’10 + (โˆ’6) = โˆ’16 86. 5 โˆ’ 10 โ‹… (โˆ’3) = 5 โˆ’ (โˆ’30) = 5 + 30 = 35 87. 16 รท (โˆ’2) โ‹… 4 = โˆ’8 โ‹… 4 = โˆ’32 88. โˆ’20 รท 5 โ‹… 2 = โˆ’4 โ‹… 2 = โˆ’8 89. 16 + (โˆ’3) โ‹…12 รท 4 = 16 + (โˆ’36) รท 4 = 16 + (โˆ’9) =7 90. โˆ’12 + 10 รท (โˆ’5) = โˆ’12 + (โˆ’2) = โˆ’14 91. 43 โˆ’ (8 โˆ’ 3)2 = 43 โˆ’ (5)2 = 64 โˆ’ 25 = 39 92. (โˆ’3)3 โˆ’ 90 = โˆ’27 โˆ’ 90 = โˆ’117 Copyright ยฉ 2020 Pearson Education, Inc. 63 Chapter 2: Integers and Introduction to Solving Equations 93. (โˆ’4)(โˆ’3) โˆ’ (โˆ’2)(โˆ’1) 12 โˆ’ 2 10 = = = โˆ’2 โˆ’10 + 5 โˆ’5 โˆ’5 94. 4(12 โˆ’ 18) 4(โˆ’6) โˆ’24 = = = โˆ’12 โˆ’10 รท (โˆ’2 โˆ’ 3) โˆ’10 รท (โˆ’5) 2 โˆ’18 + 25 + (โˆ’30) + 7 + 0 + (โˆ’2) 6 โˆ’18 = 6 = โˆ’3 95. average = โˆ’45 + (โˆ’40) + (โˆ’30) + (โˆ’25) 4 โˆ’140 = 4 = โˆ’35 ISM: Algebra Foundations 105. 10 x = โˆ’30 10 x โˆ’30 = 10 10 10 โˆ’30 โ‹…x = 10 10 x = โˆ’3 The solution is โˆ’3. 106. โˆ’8 x = 72 โˆ’8 x 72 = โˆ’8 โˆ’8 72 โˆ’8 โ‹…x = โˆ’8 โˆ’8 x = โˆ’9 The solution is โˆ’9. 96. average = 97. 2x โˆ’ y = 2(โˆ’2) โˆ’ 1 = โˆ’4 โˆ’ 1 = โˆ’5 108. 98. y 2 + x 2 = 12 + (โˆ’2)2 = 1 + 4 = 5 99. 3 x 3(โˆ’2) โˆ’6 = = = โˆ’1 6 6 6 5 y โˆ’ x 5(1) โˆ’ (โˆ’2) 5 + 2 7 = = = = โˆ’7 100. โˆ’y โˆ’1 โˆ’1 โˆ’1 101. 107. โˆ’20 + 7 = y โˆ’13 = y The solution is โˆ’13. 2n โˆ’ 6 = 16 2(โˆ’5) โˆ’ 6 โฑจ 16 โˆ’10 โˆ’ 6 โฑจ 16 โˆ’16 = 16 False Since โˆ’16 = 16 is false, โˆ’5 is not a solution of the equation. 109. 110. x โˆ’ 31 = โˆ’62 x โˆ’ 31 + 31 = โˆ’62 + 31 x = โˆ’31 The solution is โˆ’31. n = โˆ’11 โˆ’4 n โˆ’4 โ‹… = โˆ’4 โ‹… (โˆ’11) โˆ’4 โˆ’4 โ‹… n = โˆ’4 โ‹… (โˆ’11) โˆ’4 n = 44 The solution is 44. x = 13 โˆ’2 x โˆ’2 โ‹… = โˆ’2 โ‹…13 โˆ’2 โˆ’2 โ‹… x = โˆ’2 โ‹…13 โˆ’2 x = โˆ’26 The solution is โˆ’26. 102. 2(c โˆ’ 8) = โˆ’20 2(โˆ’2 โˆ’ 8) โฑจ โˆ’ 20 2(โˆ’10) โฑจ โˆ’ 20 โˆ’20 = โˆ’20 True Since โˆ’20 = โˆ’20 is true, โˆ’2 is a solution of the equation. 103. n โˆ’ 7 = โˆ’20 n โˆ’ 7 + 7 = โˆ’20 + 7 n = โˆ’13 The solution is โˆ’13. 111. n + 12 = โˆ’7 n + 12 โˆ’ 12 = โˆ’7 โˆ’ 12 n = โˆ’19 The solution is โˆ’19. 104. โˆ’5 = n + 15 โˆ’5 โˆ’ 15 = n + 15 โˆ’ 15 โˆ’20 = n The solution is โˆ’20. 112. n โˆ’ 40 = โˆ’2 n โˆ’ 40 + 40 = โˆ’2 + 40 n = 38 The solution is 38. 64 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations 113. โˆ’36 = โˆ’6 x โˆ’36 โˆ’6 x = โˆ’6 โˆ’6 โˆ’36 โˆ’6 = โ‹…x โˆ’6 โˆ’6 6= x The solution is 6. 129. 7 + 2(โˆ’3) = โˆ’14 โˆ’ 6 โˆ’20 = = โˆ’20 7 + (โˆ’6) 1 130. 5(7 โˆ’ 6)3 โˆ’ 4(2 โˆ’ 3)2 + 24 = 5(1)3 โˆ’ 4(โˆ’1)2 + 24 = 5(1) โˆ’ 4(1) + 16 = 5 โˆ’ 4 + 16 = 1 + 16 = 17 114. โˆ’40 = 8 y โˆ’40 8 y = 8 8 โˆ’40 8 = โ‹…y 8 8 โˆ’5 = y The solution is โˆ’5. 131. n โˆ’ 9 = โˆ’30 n โˆ’ 9 + 9 = โˆ’30 + 9 n = โˆ’21 The solution is โˆ’21. 132. n + 18 = 1 n + 18 โˆ’ 18 = 1 โˆ’ 18 n = โˆ’17 The solution is โˆ’17. 133. โˆ’4 x = โˆ’48 โˆ’4 x โˆ’48 = โˆ’4 โˆ’4 โˆ’4 โˆ’48 โ‹…x = โˆ’4 โˆ’4 x = 12 The solution is 12. 134. 9 x = โˆ’81 9 x โˆ’81 = 9 9 โˆ’81 9 โ‹…x = 9 9 x = โˆ’9 The solution is โˆ’9. 115. โˆ’6 + (โˆ’9) = โˆ’15 116. โˆ’16 โˆ’ 3 = โˆ’16 + (โˆ’3) = โˆ’19 117. โˆ’4(โˆ’12) = 48 118. โˆ’ โˆ’14 โˆ’ 6 84 = โˆ’21 โˆ’4 119. โˆ’76 โˆ’ (โˆ’97) = โˆ’76 + 97 = 21 120. โˆ’9 + 4 = โˆ’5 121. โˆ’18 โˆ’ 9 = โˆ’27 The temperature on Friday was โˆ’27ยฐC. 122. โˆ’11 + 17 = 6 The temperature at noon on Tuesday was 6ยฐC. 123. 12,923 โˆ’ (โˆ’195) = 12,923 + 195 = 13,118 The difference in elevations is 13,118 feet. 124. โˆ’32 + 23 = โˆ’9 His financial situation can be represented by โˆ’9. 125. (3 โˆ’ 7)2 รท (6 โˆ’ 4)3 = (โˆ’4)2 รท (2)3 = 16 รท 8 = 2 126. 3(4 + 2) + (โˆ’6) โˆ’ 32 = 3(6) + (โˆ’6) โˆ’ 32 = 3(6) + (โˆ’6) โˆ’ 9 = 18 + (โˆ’6) โˆ’ 9 = 12 โˆ’ 9 =3 135. n = 100 โˆ’2 n โˆ’2 โ‹… = โˆ’2 โ‹…100 โˆ’2 โˆ’2 โ‹… n = โˆ’2 โ‹…100 โˆ’2 n = โˆ’200 The solution is โˆ’200. 127. 2 โˆ’ 4 โ‹… 3 + 5 = 2 โˆ’ 12 + 5 = โˆ’10 + 5 = โˆ’5 128. 4 โˆ’ 6 โ‹… 5 + 1 = 4 โˆ’ 30 + 1 = โˆ’26 + 1 = โˆ’25 Copyright ยฉ 2020 Pearson Education, Inc. 65 Chapter 2: Integers and Introduction to Solving Equations 136. y = โˆ’3 โˆ’1 y โˆ’1 โ‹… = โˆ’1(โˆ’3) โˆ’1 โˆ’1 โ‹… y = โˆ’1 โ‹… (โˆ’3) โˆ’1 y=3 The solution is 3. ISM: Algebra Foundations 13. 14. Chapter 2 Getting Ready For the Test 1. The opposite of โˆ’2 is โˆ’(โˆ’2) = 2; A. 2. The absolute value of โˆ’2 is |โˆ’2| = 2 because โˆ’2 is 2 units from 0 on a number line; A. 3. The absolute value of 2 is |2| = 2 because 2 is 2 units from 0 on a number line; A. 5. For xy, the operation is multiplication; C. 6. For โˆ’12(+3), the operation is multiplication; C. 7. For โˆ’12 + 3, the operation is addition; A. โˆ’12 , the operation is division; D. 8. For +3 9. For 4 + 6 โ‹… 2, the operation of multiplication is performed first, since multiplication and division are performed before addition and subtraction; C. 10. For 4 + 6 รท 2, the operation of division is performed first, since multiplication and division are performed before addition and subtraction; D. 12. x โˆ’ 2 = โˆ’4 x โˆ’ 2 + 2 = โˆ’4 + 2 x = โˆ’2 Choice B is correct. x+2 = 4 x+2โˆ’2 = 4โˆ’2 x=2 Choice A is correct. Chapter 2 Test 1. โˆ’5 + 8 = 3 2. 18 โˆ’ 24 = 18 + (โˆ’24) = โˆ’6 3. 5 โ‹… (โˆ’20) = โˆ’100 4. The opposite of 2 is โˆ’2; B. 11. โˆ’3x = 6 โˆ’3 x 6 = โˆ’3 โˆ’3 x = โˆ’2 Choice B is correct. x =6 โˆ’3 โŽ› x โŽž โˆ’3 โŽœ โŽŸ = โˆ’3(6) โŽ โˆ’3 โŽ  x = โˆ’18 Choice D is correct. 4. โˆ’16 รท (โˆ’4) = 4 5. โˆ’18 + (โˆ’12) = โˆ’30 6. โˆ’7 โˆ’ (โˆ’19) = โˆ’7 + 19 = 12 7. โˆ’5 โ‹… (โˆ’13) = 65 8. โˆ’25 =5 โˆ’5 9. |โˆ’25| + (โˆ’13) = 25 + (โˆ’13) = 12 10. 14 โˆ’ |โˆ’20| = 14 โˆ’ 20 = 14 + (โˆ’20) = โˆ’6 11. |5| โ‹… |โˆ’10| = 5 โ‹… 10 = 50 12. โˆ’10 โˆ’ โˆ’5 = 10 = โˆ’2 โˆ’5 13. โˆ’8 + 9 รท (โˆ’3) = โˆ’8 + (โˆ’3) = โˆ’11 14. โˆ’7 + (โˆ’32) โˆ’ 12 + 5 = โˆ’7 + (โˆ’32) + (โˆ’12) + 5 = โˆ’39 + (โˆ’12) + 5 = โˆ’51 + 5 = โˆ’46 15. (โˆ’5)3 โˆ’ 24 รท (โˆ’3) = โˆ’125 โˆ’ 24 รท (โˆ’3) = โˆ’125 โˆ’ (โˆ’8) = โˆ’125 + 8 = โˆ’117 16. (5 โˆ’ 9)2 โ‹… (8 โˆ’ 2)3 = (โˆ’4) 2 โ‹… (6)3 = 16 โ‹… 216 = 3456 66 Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations Chapter 2: Integers and Introduction to Solving Equations 17. โˆ’(โˆ’7)2 รท 7 โ‹… (โˆ’4) = โˆ’49 รท 7 โ‹… (โˆ’4) = โˆ’7 โ‹… (โˆ’4) = 28 18. 3 โˆ’ (8 โˆ’ 2)3 = 3 โˆ’ 63 = 3 โˆ’ 216 = 3 + (โˆ’216) = โˆ’213 29. Subtract the depth of the lake from the elevation of the surface. 1495 โˆ’ 5315 = 1495 + (โˆ’5315) = โˆ’3820 The deepest point of the lake is 3820 feet below sea level. 30. average = 19. 4 82 4 64 โˆ’ = โˆ’ = 2 โˆ’ 4 = 2 + (โˆ’4) = โˆ’2 2 16 2 16 20. โˆ’3(โˆ’2) + 12 6 + 12 18 = = =2 โˆ’1(โˆ’4 โˆ’ 5) โˆ’1(โˆ’9) 9 31. a. b. 32. 21. 25 โˆ’ 30 2 2(โˆ’6) + 7 = โˆ’5 2 โˆ’12 + 7 = 2 (5) 25 = = โˆ’5 โˆ’5 โˆ’5 22. 5(โˆ’8) โˆ’ [6 โˆ’ (2 โˆ’ 4)] + (12 โˆ’ 16) 2 = 5(โˆ’8) โˆ’ [6 โˆ’ (โˆ’2)] + (12 โˆ’ 16)2 = 5(โˆ’8) โˆ’ (6 + 2) + (โˆ’4)2 33. = 5(โˆ’8) โˆ’ 8 + (โˆ’4)2 = 5(โˆ’8) โˆ’ 8 + 16 = โˆ’40 โˆ’ 8 + 16 = โˆ’48 + 16 = โˆ’32 23. 7 x + 3 y โˆ’ 4 z = 7(0) + 3(โˆ’3) โˆ’ 4(2) = 0 + (โˆ’9) โˆ’ 8 = โˆ’9 โˆ’ 8 = โˆ’17 24. 10 โˆ’ y 2 = 10 โˆ’ (โˆ’3)2 = 10 โˆ’ 9 = 1 25. 3z 3(2) 6 = = = โˆ’1 2 y 2(โˆ’3) โˆ’6 26. A descent of 22 feet is represented by โˆ’22. 4(โˆ’22) = โˆ’88 Mary is 88 feet below sea level. 27. 129 + (โˆ’79) + (โˆ’40) + 35 = 50 + (โˆ’40) + 35 = 10 + 35 = 45 His new balance can be represented by 45. 34. โˆ’12 + (โˆ’13) + 0 + 9 โˆ’16 = = โˆ’4 4 4 The product of a number and 17 is 17 โ‹… x or 17x. A number subtracted from 20 is 20 โˆ’ x. โˆ’9n = โˆ’45 โˆ’9n โˆ’45 = โˆ’9 โˆ’9 โˆ’9 โˆ’45 โ‹…n = โˆ’9 โˆ’9 n=5 The solution is 5. n =4 โˆ’7 n โˆ’7 โ‹… = โˆ’7 โ‹… 4 โˆ’7 โˆ’7 โ‹… n = โˆ’7 โ‹… 4 โˆ’7 n = โˆ’28 The solution is โˆ’28. x โˆ’ 16 = โˆ’36 x โˆ’ 16 + 16 = โˆ’36 + 16 x = โˆ’20 The solution is โˆ’20. 35. โˆ’20 + 8 + 8 = x โˆ’12 + 8 = x โˆ’4 = x The solution is โˆ’4. Cumulative Review Chapters 1โˆ’2 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. 28. Subtract the elevation of the Romanche Gap from the elevation of Mt. Washington. 6288 โˆ’ (โˆ’25,354) = 6288 + 25,354 = 31,642 The difference in elevations is 31,642 feet. 4. The place value of 3 is 693,298 is thousands. 5. The place value of 3 in 534,275,866 is tenmillions. Copyright ยฉ 2020 Pearson Education, Inc. 67 Chapter 2: Integers and Introduction to Solving Equations 6. The place value of 3 in 267,301,818 is hundredthousands. 7. a. โˆ’7 โˆ’4 since 0 is to the right of โˆ’4 on a number line. c. โˆ’9 > โˆ’11 since โˆ’9 is to the right of โˆ’11 on a number line. 8. a. 12 > โˆ’4 since 12 is to the right of โˆ’4 on a number line. b. โˆ’13 > โˆ’31 since โˆ’13 is to the right of โˆ’31 on a number line. c. โˆ’82 < 79 since โˆ’82 is to the left of 79 on a number line. 9. 13 + 2 + 7 + 8 + 9 = (13 + 7) + (2 + 8) + 9 = 20 + 10 + 9 = 39 ISM: Algebra Foundations 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 โˆ’ 2879 rounds to rounds to 4700 โˆ’ 2900 1800 18. 8394 โˆ’ 2913 rounds to rounds to 8000 โˆ’ 3000 5000 19. a. 10. 11 + 3 + 9 + 16 = (11 + 9) + (3 + 16) = 20 + 19 = 39 11. 12. 7826 โˆ’ 505 7321 Check: 3285 โˆ’ 272 3013 Check: b. 20(4 + 7) = 20 โ‹… 4 + 20 โ‹… 7 c. 2(7 + 9) = 2 โ‹… 7 + 2 โ‹… 9 20. a. 7321 + 505 7826 3013 + 272 3285 13. Subtract 7257 from the radius of Jupiter. 43, 441 โˆ’ 7 257 36,184 The radius of Saturn is 36,184 miles. 68 5(2 + 12) = 5 โ‹… 2 + 5 โ‹… 12 b. 9(3 + 6) = 9 โ‹… 3 + 9 โ‹… 6 c. 4(8 + 1) = 4 โ‹… 8 + 4 โ‹… 1 21. 631 ร— 125 3 155 12 620 63 100 78,875 22. 299 ร— 104 1 196 29 900 31, 096 23. a. 14. Subtract the cost of the camera from the amount in her account. 762 โˆ’ 237 525 She will have $525 left in her account after buying the camera. 5(6 + 5) = 5 โ‹… 6 + 5 โ‹… 5 b. c. 42 รท 7 = 6 because 6 โ‹… 7 = 42. 64 = 8 because 8 โ‹… 8 = 64. 8 7 3 21 because 7 โ‹… 3 = 21. Copyright ยฉ 2020 Pearson Education, Inc. ISM: Algebra Foundations 24. a. Chapter 2: Integers and Introduction to Solving Equations 35 = 7 because 7 โ‹… 5 = 35. 5 28. b. 64 รท 8 = 8 because 8 โ‹… 8 = 64. c. 12 4 48 because 12 โ‹… 4 = 48. 741 25. 5 3705 โˆ’35 20 โˆ’20 05 โˆ’5 0 Check: 741 ร— 5 3705 Cost of each total = รท number of tickets ticket cost = 324 รท 36 9 36 324 โˆ’324 0 Each ticket cost $9. 29. 92 = 9 โ‹… 9 = 81 30. 53 = 5 โ‹… 5 โ‹… 5 = 125 31. 61 = 6 32. 41 = 4 33. 5 โ‹… 62 = 5 โ‹… 6 โ‹… 6 = 180 34. 23 โ‹… 7 = 2 โ‹… 2 โ‹… 2 โ‹… 7 = 56 456 26. 8 3648 โˆ’32 44 โˆ’40 48 โˆ’48 0 35. 36. 7 โˆ’ 2 โ‹… 3 + 32 7 โˆ’ 2 โ‹… 3 + 9 7 โˆ’ 6 + 9 10 = = = =2 5(2 โˆ’ 1) 5(1) 5 5 6 2 + 4 โ‹… 4 + 23 2 37 โˆ’ 5 Check: 456 ร— 8 3648 36 + 4 โ‹… 4 + 8 37 โˆ’ 25 36 + 16 + 8 = 12 60 = 12 =5 = 37. x + 6 = 8 + 6 = 14 27. number of cards number of number of = รท for each person cards friends = 238 รท 19 12 R 10 19 238 โˆ’19 48 โˆ’38 10 Each friend will receive 12 cards. There will be 10 cards left over. 38. 5 + x = 5 + 9 = 14 39. a. |โˆ’9| = 9 because โˆ’9 is 9 units from 0. b. |8| = 8 because 8 is 8 units from 0. c. |0| = 0 because 0 is 0 units from 0. 40. a. |4| = 4 because 4 is 4 units from 0. b. |โˆ’7| = 7 because โˆ’7 is 7 units from 0. 41. โˆ’2 + 25 = 23 42. 8 + (โˆ’3) = 5 43. 2a โˆ’ b = 2(8) โˆ’ (โˆ’6) = 16 โˆ’ (โˆ’6) = 16 + 6 = 22 Copyright ยฉ 2020 Pearson Education, Inc. 69 Chapter 2: Integers and Introduction to Solving Equations 44. x โˆ’ y = โˆ’2 โˆ’ (โˆ’7) = โˆ’2 + 7 = 5 45. โˆ’7 โ‹… 3 = โˆ’21 46. 5(โˆ’2) = โˆ’10 47. 0 โ‹… (โˆ’4) = 0 48. โˆ’6 โ‹… 9 = โˆ’54 70 ISM: Algebra Foundations 49. 3(4 โˆ’ 7) + (โˆ’2) โˆ’ 5 = 3(โˆ’3) + (โˆ’2) โˆ’ 5 = โˆ’9 + (โˆ’2) โˆ’ 5 = โˆ’11 โˆ’ 5 = โˆ’16 50. 4 โˆ’ 8(7 โˆ’ 3) โˆ’ (โˆ’1) = 4 โˆ’ 8(4) โˆ’ (โˆ’1) = 4 โˆ’ 32 โˆ’ (โˆ’1) = 4 โˆ’ 32 + 1 = โˆ’28 + 1 = โˆ’27 Copyright ยฉ 2020 Pearson Education, Inc.

Document Preview (29 of 1033 Pages)

User generated content is uploaded by users for the purposes of learning and should be used following SchloarOn's honor code & terms of service.
You are viewing preview pages of the document. Purchase to get full access instantly.

Shop by Category See All


Shopping Cart (0)

Your bag is empty

Don't miss out on great deals! Start shopping or Sign in to view products added.

Shop What's New Sign in