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Solution Manual For Intermediate Algebra, 13th Edition

Solution Manual For Intermediate Algebra, 13th Edition is your ultimate textbook solutions guide, providing answers to the most difficult questions.

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     S OLUTIONS M ANUAL I NTERMEDIATE A LGEBRA T HIRTEENTH E DITION Marvin L. Bittinger Indiana University Purdue University Indianapolis Judith A. Beecher Barbara L. Johnson Ivy Tech Community College of Indiana

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    Contents Just-In-Time Review . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . 113 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 183 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 211 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 259 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 295

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    Just-In-Time Review 1. The Set of Real Numbers 1. We specify conditions by which we know whether a number is in the set. { x | x is a whole number less than or equal to 5 } , or { x | x is a whole number less than 6 } 2. We specify conditions by which we know whether a number is in the set. { x | x is an integer greater than 3 and less than 11 } 3. Consider the following numbers:20, − 10, − 5 . 34, 18.999, 0 . 583, 11 45 , √ 7, − √ 2, √ 16, 0, − 2 3 , 9 . 34334333433334 .... a) The natural numbers are numbers used for counting. The natural numbers in the list above are 20 and √ 16 ( √ 16 = 4). b) The whole numbers are the set of natural numbers with 0 also included. The whole numbers in the list above are 20, √ 16, and 0. c) The integers are the set of whole numbers and their opposites. The integers in the list above are 20, − 10, √ 16, and 0. d) The irrational numbers are the numbers that can- not be represented as a quotient of two inte- gers. Their decimal notation neither terminates nor has a repeating block of digits. The irra- tional numbers in the list above are √ 7, − √ 2, and 9.34334333433334.... e) Rational numbers can be named as quotients of integers with nonzero denominators. Decimal no- tation for rational numbers either terminates or has a block of repeating digits. The rational num- bers in the list above are 20, − 10, − 5 . 34, 18.999, 0 . 583, 11 45 , √ 16, 0, and − 2 3 . f) The set of real numbers is composed of the ratio- nal numbers and the irrational numbers. All of the numbers in the list are real numbers. 4. Consider the following numbers: − 6, 0, 1, − 1 2 , − 4, 7 9 , 12, − 6 5 , 3.45, 5 1 2 , √ 3, √ 25, − 12 3 , 0 . 131331333133331 .... a) The natural numbers are numbers used for counting. The natural numbers in the list above are 1, 12, and √ 25 ( √ 25 = 5). b) The whole numbers are the set of natural numbers with 0 also included. The whole numbers in the list above are 0, 1, 12, and √ 25. c) The integers are the set of whole numbers and their opposites. The integers in the list above are − 6, 0, 1, − 4, 12, √ 25, and − 12 3 ( − 12 3 = − 4 ) . d) The irrational numbers are the numbers that cannot be represented as a quotient of two integers. Their decimal notation neither terminates nor has a re- peating block of digits. The irrational numbers in the list above are √ 3, and 0.131331333133331.... e) Rational numbers can be named as quotients of in- tegers with nonzero denominators. Decimal nota- tion for rational numbers either terminates or has a block of repeating digits. The rational numbers in the list above are − 6, 0, 1, − 1 2 , − 4, 7 9 , 12, − 6 5 , 3.45, 5 1 2 ( 5 1 2 = 11 2 ) , √ 25, and − 12 3 . f) The set of real numbers is composed of the ratio- nal numbers and the irrational numbers. All of the numbers in the list are real numbers. 2. Order for the Real Numbers 1. Since − 5 is to the left of − 4 on the number line, we have − 5 < − 4. 2. Since − 1 4 is to the right of − 1 2 on the number line, we have − 1 4 > − 1 2 . 3. Since 87 is to the right of 67, we have 87 > 67. 4. Since − 9 . 8 is to the left of − 4 2 3 , we have − 9 . 8 < − 4 2 3 . 5. Since 6.78 is to the right of − 6 . 77, we have 6 . 78 > − 6 . 77. 6. − 4 5 = − 0 . 8; since − 0 . 8 is to the right of − 0 . 86, we have − 0 . 8 > − 0 . 86, or − 4 5 > − 0 . 86. 7. We convert to decimal notation: 14 29 = 0 . 48275 . . . and 17 32 = 0 . 53125. We see that 14 29 is to the left of 17 32 , so we have 14 29 < 17 32 . 8. We convert to decimal notation: − 12 13 = − 0 . 923 . . . and − 14 15 = − 0 . 93. We see that − 12 13 is to the right of − 14 15 , so we have − 12 13 > − 14 15 . 9. Since 1.8 is to the right of 1.08, we have 1 . 8 > 1 . 08. 10. Since 0 is to the right of − 4, we have 0 > − 4. 11. x > 6 The inequality 6 < x has the same meaning. 12. − 4 < 7 The inequality 7 > − 4 has the same meaning.

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    2 Just-in-Time Review 13. 6 ≥ − 9 . 4 is true because 6 > − 9 . 4 is true. 14. − 18 ≤ − 18 is true because − 18 = − 18 is true. 15. − 7 . 6 ≤ − 10 4 5 is false because neither − 7 . 6 < − 10 4 5 nor − 7 . 6 = − 10 4 5 is true. 16. − 24 27 = − 0 . 8 and − 25 28 = − 0 . 892 . . . so − 24 27 > − 25 28 . Thus, − 24 27 ≥ − 25 28 is true. 3. Graphing Inqualities on the Number Line 1. x > − 1 We shade all the numbers to the right of − 1 and use a left parenthesis at − 1 to indicate that it is not a solution. 2. x ≤ 5 We shade all the numbers to the left of 5 and use a right bracket at 5 to indicate that it is also a solution. 3. x > 0 We shade all the numbers to the right of 0 and use a left parenthesis at 0 to indicate that it is not a solution. 4. x ≤ − 5 2 We shade all the numbers to the left of − 5 2 and use a right bracket at − 5 2 to indicate that it is also a solution. 4. Absolute Value 1. The distance of − 4 from 0 is 4, so | − 4 | = 4. 2. The distance of 0 from 0 is 0, so | 0 | = 0. 3. The distance of 252.7 from 0 is 252.7, so | 252 . 7 | = 252 . 7. 4. The distance of 31 from 0 is 31, so | 31 | = 31. 5. The distance of − 1 2 from 0 is 1 2 , so ∣ ∣ ∣ ∣ − 1 2 ∣ ∣ ∣ ∣ = 1 2 . 6. The distance of − 0 . 03 from 0 is 0.03, so | − 0 . 03 | = 0 . 03. 5. Add Real Numbers 1. 8 + ( − 3) We find the difference of the absolute values, 8 − 3 = 5. Since the positive number has the larger absolute value, the answer is positive. 8 + ( − 3) = 5 2. − 8 + 3 We find the difference of the absolute values, 8 − 3 = 5. Since the negative number has the larger absolute value, the answer is negative. − 8 + 3 = − 5 3. − 8 + ( − 3) The sum of two negative numbers is negative. We add the absolute values, 8 + 3 = 11, and make the answer negative. − 8 + ( − 3) = − 11 4. 8 + 3 We add the numbers. The result is positive. 8 + 3 = 11 5. 7 + ( − 2) We find the difference of the absolute values, 7 − 2 = 5. Since the positive number has the larger absolute value, the answer is positive. 7 + ( − 2) = 5 6. − 8 + ( − 8) The sum of two negative numbers is negative. We add the absolute values, 8 + 8 = 16, and make the answer negative. − 8 + ( − 8) = − 16 7. − 13 + 13 We have a positive number and a negative number with the same absolute value, so the sum is 0. − 13 + 13 = 0 8. − 26 + 0 One number is 0. The sum is the other number. − 26 + 0 = − 26 9. − 6 + ( − 15) The sum of two negative numbers is negative. We add the absolute values, 6 + 15 = 21, and make the answer negative. − 6 + ( − 15) = − 21 10. 9 + ( − 2) We find the difference of the absolute values, 9 − 2 = 7. Since the positive number has the larger absolute value, the answer is positive. 9 + ( − 2) = 7

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    Just-in-Time Review 3 11. − 8 . 4 + 9 . 6 We find the difference of the absolute values, 9 . 6 − 8 . 4 = 1 . 2. Since the positive number has the larger absolute value, the answer is positive. − 8 . 4 + 9 . 6 = 1 . 2 12. − 2 . 62 + ( − 6 . 24) The sum of two negative numbers is negative. We add the absolute values, 2 . 62 + 6 . 24 = 8 . 86, and make the answer negative. − 2 . 62 + ( − 6 . 24) = − 8 . 86 13. − 2 5 + 3 4 We find the difference of the absolute values. 3 4 − 2 5 = 3 4 · 5 5 − 2 5 · 4 4 = 15 20 − 8 20 = 7 20 Since the positive number has the larger absolute value, the answer is positive. − 2 5 + 3 4 = 7 20 14. − 5 6 + ( − 7 8 ) The sum of two negative numbers is negative. We add the absolute values. 5 6 + 7 8 = 5 6 · 4 4 + 7 8 · 3 3 · 20 24 + 21 24 = 41 24 The answer negative. − 5 6 + ( − 7 8 ) = − 41 24 6. Opposites, or Additive Inverses 1. The opposite, or additive inverse, of 10 is − 10, because 10 + ( − 10) = 0. 2. The opposite, or additive inverse, of − 2 3 is 2 3 , because − 2 3 + 2 3 = 0. 3. The opposite, or additive inverse, of − 0 . 007 is 0.007, be- cause − 0 . 007 + 0 . 007 = 0. 4. The opposite, or additive inverse, of 0 is 0, because 0 + 0 = 0. 5. When x = 9, then − x = − 9. When x = 9, then − ( − x ) = − ( − 9) = 9. 6. When x = − 23, then − x = − ( − 23) = 23. When x = − 23, then − ( − x ) = − ( − ( − 23)) = − (23) = − 23. 7. Subtract Real Numbers 1. 5 − 11 = 5 + ( − 11) = − 6 2. 5 − ( − 11) = 5 + 11 = 16 3. − 5 − 11 = − 5 + ( − 11) = − 16 4. − 5 − ( − 11) = − 5 + 11 = 6 5. 17 − ( − 17) = 17 + 17 = 34 6. − 6 − 14 = − 6 + ( − 14) = − 20 7. − 9 − ( − 9) = − 9 + 9 = 0 8. 8 − 13 = 8 + ( − 13) = − 5 9. 31 − ( − 16) = 31 + 16 = 47 10. 15 . 8 − 27 . 4 = 15 . 8 + ( − 27 . 4) = − 11 . 6 11. − 18 . 01 − 11 . 24 = − 18 . 01 + ( − 11 . 24) = − 29 . 25 12. 5 6 − ( − 1 12 ) = 5 6 + 1 12 = 10 12 + 1 12 = 11 12 13. − 1 3 − ( − 1 12 ) = − 1 3 + 1 12 = − 4 12 + 1 12 = − 3 12 = − 1 4 14. 1 3 − 4 5 = 1 3 + ( − 4 5 ) = 5 15 + ( − 12 15 ) = − 7 15 8. Multiply Real Numbers 1. 3( − 7) The product of a positive number and a negative number is negative. We multiply the absolute values, 3 · 7 = 21, and make the answer negative. 3( − 7) = − 21 2. ( − 4 . 2)( − 6 . 3) The product of two negative numbers is positive. We mul- tiply the absolute values, 4 . 2(6 . 3) = 26 . 46, and make the answer positive. ( − 4 . 2)( − 6 . 3) = 26 . 46 3. − 5 · 9 The product of a negative number and a positive number is negative. We multiply the absolute values, 5 · 9 = 45, and make the answer negative. − 5 · 9 = − 45 4. − 13 7 · ( − 5 2 ) The product of two negative numbers is positive. We mul- tiply the absolute values, 13 7 · 5 2 = 65 14 , and make the an- swer positive. − 13 7 · ( − 5 2 ) = 65 14 5. 0 · ( − 11) = 0

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    4 Just-in-Time Review 6. − 3 5 · 4 7 The product of a negative number and a positive number is negative. We multiply the absolute values, 3 5 · 4 7 = 12 35 , and make the answer negative. − 3 5 · 4 7 = − 12 35 7. − 4( − 13) The product of two negative numbers is positive. We mul- tiply the absolute values, 4 · 13 = 52, and make the answer positive. − 4( − 13) = 52 8. − 3 ( 2 3 ) The product of a negative number and a positive number is negative. We multiply the absolute values, 3 · 2 3 = 6 3 = 2, and make the answer negative. − 3 ( 2 3 ) = − 2 9. − 9 11 · ( − 11 9 ) The product of two negative numbers is positive. We mul- tiply the absolute values, 9 11 · 11 9 = 99 99 = 1, and make the answer positive. − 9 11 · ( − 11 9 ) = 1 10. − 3( − 4)(5) = 12(5) The product of two negative numbers is positive. = 60 The product of two positive numbers is positive. 9. Divide Real Numbers 1. When a negative number is divided by a positive number, the answer is negative. − 8 4 = − 2 2. When a negative number is divided by a negative number, the answer is positive. − 77 − 11 = 7 3. 5 0 Not defined:Division by 0. 4. 0 32 = 0 because 0 · 32 = 0. 5. The reciprocal of 3 4 is 4 3 , because 3 4 · 4 3 = 1. 6. The reciprocal of − 7 8 is − 8 7 , because − 7 8 · ( − 8 7 ) = 1. 7. The reciprocal of 25 is 1 25 , because 25 · 1 25 = 1. 8. The reciprocal of 0.2 is 1 0 . 2 . This can also be expressed as follows: 1 0 . 2 = 1 0 . 2 · 10 10 = 10 2 = 5. 9. 2 7 ÷ ( − 11 3 ) = 2 7 · ( − 3 11 ) = − 6 77 10. − 10 3 ÷ − 2 15 = − 10 3 · ( − 15 2 ) = 150 6 , or 25 11. − 48 ÷ 0 . 4 = − 48 0 . 4 = − 48 0 . 4 = − 120 12. 8 19 ÷ ( − 2) = 8 19 · ( − 1 2 ) = − 8 38 = − 2 · 4 2 · 19 = − 4 19 13. The opposite of − 3 8 is 3 8 , because − 3 8 + 3 8 = 0. The reciprocal of − 3 8 is − 8 3 , because − 3 8 ( − 8 3 ) = 1. 14. The opposite of 6 is − 6, because 6 + ( − 6) = 0. The reciprocal of 6 is 1 6 , because 6 · 1 6 = 1. 10. Exponential Notation (Part 1) 1. y · y · y · y · y · y ︸ ︷︷ ︸ = y 6 6 factors 2. 8 · 8 · 8 · 8 ︸ ︷︷ ︸ = 8 4 4 factors 3. (3 . 8)(3 . 8)(3 . 8)(3 . 8)(3 . 8) ︸ ︷︷ ︸ = (3 . 8) 5 5 factors 4. ( − 4 5 )( − 4 5 )( − 4 5 ) ︸ ︷︷ ︸ = ( − 4 5 ) 3 3 factors 5. ( 3 4 ) 1 = 3 4 (For any number a , a 1 = a .) 6. 17 0 = 1 (For any nonzero number a , a 0 = 1.) 7. ( − 2) 6 = ( − 2) · ( − 2) · ( − 2) · ( − 2) · ( − 2) · ( − 2) ︸ ︷︷ ︸ = 64 6 factors 8. ( − 7) 3 = ( − 7) · ( − 7) · ( − 7) ︸ ︷︷ ︸ = − 343 3 factors 9. ( 1 3 ) 3 = 1 3 · 1 3 · 1 3 ︸ ︷︷ ︸ = 1 27 3 factors 10. − (3 4 ) = − (3 · 3 · 3 · 3) = − (81) = − 81

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    Just-in-Time Review 5 11. (0 . 1) 6 = (0 . 1)(0 . 1)(0 . 1)(0 . 1)(0 . 1)(0 . 1) = 0 . 000001 12. ( − 5 7 ) 2 = ( − 5 7 )( − 5 7 ) = 25 49 13. y − 5 = 1 y 5 14. ( − 4) − 3 = 1 ( − 4) 3 = 1 − 64 = − 1 64 15. ( 3 4 ) − 2 = 1 ( 3 4 ) 2 = 1 9 16 = 1 · 16 9 = 16 9 16. 1 a − 2 = a 2 17. ( 1 4 ) − 2 = 1 ( 1 4 ) 2 = 1 1 16 = 1 · 16 1 = 16 18. 1 3 4 = 3 − 4 19. 1 b 3 = b − 3 20. 1 ( − 16) 2 = ( − 16) − 2 11. Order of Operations 1. 9[8 − 7(5 − 2)] = 9[8 − 7 · 3] = 9[8 − 21] = 9[ − 13] = − 117 2. (9 − 12) 2 = ( − 3) 2 = 9 3. 4 ÷ (8 − 10) 2 + 1 = 4 ÷ ( − 2) 2 + 1 = 4 ÷ 4 + 1 = 1 + 1 = 2 4. [64 ÷ ( − 4)] ÷ ( − 2) = − 16 ÷ ( − 2) = 8 5. 9 2 − 12 2 = 81 − 144 = − 63 6. [24 ÷ ( − 3)] ÷ ( − 1 2 ) = − 8 ÷ ( − 1 2 ) = − 8 · ( − 2) = 16 7. 9 ÷ ( − 3) + 16 ÷ 8 = − 3 + 2 = − 1 8. 20 + 4 3 ÷ ( − 8) = 20 + 64 ÷ ( − 8) = 20 + ( − 8) = 12 9. 256 ÷ ( − 32) ÷ ( − 4) = − 8 ÷ ( − 4) = 2 10. 9[(8 − 11) − 13] = 9[ − 3 − 13] = 9[ − 16] = − 144 11. (18 − 2 · 3) − 9 = (18 − 6) − 9 = 12 − 9 = 3 12. (13 · 2 − 8 · 4) 2 = (26 − 32) 2 = ( − 6) 2 = 36 13. 12 − 4(5 − 1) = 12 − 4(4) = 12 − 16 = − 4 14. [2 · (5 − 3)] 2 = [2 · 2] 2 = 4 2 = 16 15. [5(8 − 6) + 12] − [24 − (8 − 4)] = [5 · 2 + 12] − [24 − 4] = [10 + 12] − [24 − 4] = 22 − 20 = 2 16. − 32 − 18 ÷ [4 − ( − 2)] = − 32 − 18 ÷ [4 + 2] = − 32 − 18 ÷ 6 = − 32 − 3 = − 35 17. 2 5 − [4 · 3 − 2(1 + 4 · 2)] = 2 5 − [4 · 3 − 2(1 + 8)] = 2 5 − [4 · 3 − 2 · 9] = 2 5 − [12 − 18] = 2 5 − [ − 6] = 2 5 + 6 = 32 + 6 = 38 18. 10 − 3 { 8 ÷ [4 − (11 − 6)] } = 10 − 3 { 8 ÷ [4 − 5] } = 10 − 3 { 8 ÷ [ − 1] } = 10 − 3 {− 8 } = 10 + 24 = 34 19. 4 | 6 − 7 | − 5 · 4 6 · 7 − 8 | 4 − 1 | = 4 | − 1 | − 5 · 4 6 · 7 − 8 | 3 | = 4 · 1 − 5 · 4 6 · 7 − 8 · 3 = 4 − 20 42 − 24 = − 16 18 = − 8 9

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    6 Just-in-Time Review 20. (8 − 3) 2 + (7 − 10) 2 3 2 − 2 3 = 5 2 + ( − 3) 2 3 2 − 2 3 = 25 + 9 9 − 8 = 34 1 = 34 12. Translate to an Algebraic Expression 1. x divided by 6 We have x ÷ 6, or x 6 , or x/ 6. 2. Seven more than four times a number Let y represent the number. We have 7 + 4 y , or 4 y + 7. 3. 25 less than t We have t − 25. 4. One third of a number Let b represent the number. We have 1 3 b , or b 3 . 5. The sum of w and twice q We have w + 2 q , or 2 q + w . 6. − 18 multiplied by m We have − 18 · m , or − 18 m , or m ( − 18). 7. m subtracted from s We have s − m . 8. Nineteen percent of some number Let x represent the number. We have 19% x , or 0 . 19 x , or 19 100 x . 13. Evaluate Algebraic Expressions 1. Substitute − 8 for y and carry out the calculation. 57 y = 57( − 8) = − 456 2. Substitute 30 for x and − 6 for y and carry out the calcu- lation. x y = 30 − 6 = − 5 3. Substitute 20 for p and 30 for q and carry out the calcula- tions. 5 p + q = 5 20 + 30 = 5 50 = 1 10 4. Substitute 7 for m and 18 for n and carry out the calcula- tions. 18 m n = 18 · 7 18 = 18 18 · 7 1 = 7 5. Substitute 3 for x and − 2 for y and carry out the calcula- tions. 4 x − y = 4 · 3 − ( − 2) = 12 − ( − 2) = 12 + 2 = 14 6. Substitute 4 for b and 6 for c and carry out the calculations. 2 c ÷ 3 b = 2 · 6 ÷ 3 · 4 = 12 ÷ 3 · 4 = 4 · 4 = 16 7. Substitute 3 for r and 27 for s and carry out the calcula- tions. 25 − r 2 + s ÷ r 2 = 25 − 3 2 + 27 ÷ 3 2 = 25 − 9 + 27 ÷ 9 = 25 − 9 + 3 = 16 + 3 = 19 8. Substitute 15 for m and 3 for n and carry out the calcula- tions. m + n (5 + n 2 ) = 15 + 3(5 + 3 2 ) = 15 + 3(5 + 9) = 15 + 3 · 14 = 15 + 42 = 57 14. Equivalent Fraction Expressions 1. Since 8 x = 8 · x , we multiply by 1 using x/x as a name for 1. 7 8 = 7 8 · 1 = 7 8 · x x = 7 x 8 x 2. Since 50 y = 10 · 5 y , we multiply by 1 using 5 y 5 y as a name for 1. 3 10 = 3 10 · 5 y 5 y = 15 y 50 y 3. Since 51 b = 17 · 3 b we multiply by 1 using 3 b 3 b as a name for 1. 1 17 = 1 17 · 3 b 3 b = 3 b 51 b 4. Since 98 w = 49 · 2 w , we multiply by 1 using 2 w 2 w as a name for 1. 300 49 = 300 49 · 2 w 2 w = 600 w 98 w 5. 25 x 15 x = 5 · 5 x 3 · 5 x We look for the largest common factor of the numerator and de- nominator and factor each. = 5 3 · 5 x 5 x Factoring the expression = 5 3 · 1 ( 5 x 5 x = 1 ) = 5 3

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    Just-in-Time Review 7 6. 36 y 18 y = 18 y · 2 18 y · 1 We look for the largest common factor of the numerator and de- nominator and factor each. = 18 y 18 y · 2 1 Factoring the expression = 1 · 2 1 ( 18 y 18 y = 1 ) = 2 7. − 100 a 25 a = − 25 a · 4 25 a · 1 Factoring numerator and denominator = − 25 a 25 a · 4 1 Factoring the expression = − 1 · 4 1 ( 25 a 25 a = 1 ) = − 4 1 = − 4 8. − 625 t 15 t = − 125 · 5 t 3 · 5 t Factoring numerator and denominator = − 125 3 · 5 t 5 t Factoring the expression = − 125 3 · 1 ( 5 t 5 t = 1 ) = − 125 3 , or − 125 3 15. The Commutative Laws and the Associative Laws 1. w + 3 = 3 + w Commutative law of addition 2. rt = tr Commutative law of multiplication 3. pq + 14 = 14 + pq Commutative law of addition pq + 14 = qp + 14 Commutative law of multiplication 4. m + ( n + 2) = ( m + n ) + 2 Associative law of addition 5. (7 · x ) · y = 7 · ( x · y ) Associative law of multiplication 6. ( a + b ) + 8 = a + ( b + 8) Associative law = a + (8 + b ) Commutative law ( a + b ) + 8 = a + ( b + 8) Associative law = a + (8 + b ) Commutative law = ( a + 8) + b Associative law ( a + b ) + 8 = ( b + a ) + 8 Commutative law = b + ( a + 8) Associative law Other answers are possible. 7. 9 · ( x · y ) = (9 · x ) · y Associative law = ( x · 9) · y Commutative law = x · (9 · y ) Associative law 9 · ( x · y ) = (9 · x ) · y Associative law = y · (9 · x ) Commutative law = y · ( x · 9) Commutative law 9 · ( x · y ) = ( x · y ) · 9 Commutative law = ( y · x ) · 9 Commutative law 16. The Distributive Laws 1. 3( c + 1) = 3 · c + 3 · 1 = 3 c + 3 2. 3( x − y ) = 3 · x − 3 · y = 3 x − 3 y 3. − 2(3 c + 5 d ) = − 2 · 3 c − 2 · 5 d = − 6 c − 10 d 4. 5 x ( y − z + w ) = 5 x · y − 5 x · z + 5 x · w = 5 xy − 5 xz + 5 xw 5. 1 2 h ( a + b ) = 1 2 h · a + 1 2 h · b = 1 2 ha + 1 2 hb 6. 2 πr ( h + 1) = 2 πr · h + 2 πr · 1 = 2 πrh + 2 πr 7. 4 a − 5 b + 6 = 4 a + ( − 5 b ) + 6 The terms are 4 a , − 5 b , and 6. 8. 2 x − y − 1 = 2 x + ( − y ) + ( − 1) The terms are 2 x , − y , and − 1. 9. 9 a + 9 b = 9 · a + 9 · b = 9( a + b ) 10. 7 x − 21 = 7 · x − 7 · 3 = 7( x − 3) 11. ab + a = a · b + a · 1 = a ( b + 1) 12. 18 a − 24 b − 48 = 6 · 3 a − 6 · 4 b − 6 · 8 = 6(3 a − 4 b − 8) 13. 8 m + 4 n − 24 = 4 · 2 m + 4 · n − 4 · 6 = 4(2 m + n − 6) 14. xy − xz + xw = x · y − x · z + x · w = x ( y − z + w )

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    8 Just-in-Time Review 17. Collecting Like Terms 1. 7 x + 5 x = (7 + 5) x = 12 x 2. 8 b − 11 b = (8 − 11) b = − 3 b 3. 14 y + y = 14 y + 1 y = (14 + 1) y = 15 y 4. 12 a − a = 12 a − 1 a = (12 − 1) a = 11 a 5. t − 9 t = 1 t − 9 t = (1 − 9) t = − 8 t 6. 5 x − 3 x + 8 x = (5 − 3 + 8) x = 10 x 7. 3 c + 8 d − 7 c + 4 d = (3 − 7) c + (8 + 4) d = − 4 c + 12 d 8. 4 x − 7 + 18 x + 25 = (4 + 18) x + ( − 7 + 25) = 22 x + 18 9. 1 . 3 x + 1 . 4 y − 0 . 11 x − 0 . 47 y = (1 . 3 − 0 . 11) x + (1 . 4 − 0 . 47) y = 1 . 19 x + 0 . 93 y 10. 2 3 a + 5 6 b − 27 − 4 5 a − 7 6 b = ( 2 3 − 4 5 ) a + ( 5 6 − 7 6 ) b − 27 = ( 10 15 − 12 15 ) a + ( − 2 6 ) b − 27 = − 2 15 a − 1 3 b − 27 18. Removing Parentheses and Collecting Like Terms 1. − ( − 2 c ) = − 1( − 2 c ) = [ − 1( − 2)] c = 2 c 2. − ( b + 4) = − 1( b + 4) = − 1 · b + ( − 1)4 = − b − 4 3. − ( x − 8) = − 1( x − 8) = − 1 · x + ( − 1)( − 8) = − x + 8 , or 8 − x 4. − ( t − y ) = − 1( t − y ) = − 1 · t − ( − 1) · y = − t + [ − ( − 1) y ] = − t + y, or y − t 5. − ( r + s + t ) = − 1( r + s + t ) = − 1 · r + ( − 1) · s + ( − 1) · t = − r − s − t 6. − (8 x − 6 y + 13) = − 8 x + 6 y − 13 Changing the sign of every term inside parentheses 7. 4 m − (3 m − 1) = 4 m − 3 m + 1 = m + 1 8. 7 a − [9 − 3(5 a − 2)] = 7 a − [9 − 15 a + 6] = 7 a − [15 − 15 a ] = 7 a − 15 + 15 a = 22 a − 15 9. − 2( x + 3) − 5( x − 4) = − 2( x + 3) + [ − 5( x − 4)] = − 2 x − 6 + [ − 5 x + 20] = − 2 x − 6 − 5 x + 20 = − 7 x + 14 10. 8 x − ( − 3 y + 7) + (9 x − 11) = 8 x + 3 y − 7 + 9 x − 11 = 17 x + 3 y − 18 11. [7( x + 5) − 19] − [4( x − 6) + 10] = [7 x + 35 − 19] − [4 x − 24 + 10] = [7 x + 16] − [4 x − 14] = 7 x + 16 − 4 x + 14 = 3 x + 30 12. 5 {− 2 + 3[4 − 2(3 + 5)] } = 5 {− 2 + 3[4 − 2(8)] } = 5 {− 2 + 3[4 − 16] } = 5 {− 2 + 3[ − 12] } = 5 {− 2 − 36 } = 5 {− 38 } = − 190 19. Exponential Notation 1. 8 − 2 · 8 − 4 = 8 − 2+( − 4) = 8 − 6 = 1 8 6 2. a 3 a − 2 = a 3 − ( − 2) = a 3+2 = a 5 3. 10 − 3 10 6 = 10 − 3 − 6 = 10 − 3+( − 6) = 10 − 9 = 1 10 9

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    Just-in-Time Review 9 4. b 2 · b − 5 = b 2+( − 5) = b − 3 = 1 b 3 5. − 24 x 6 y 7 18 x − 3 y 9 = − 24 18 x 6 − ( − 3) y 7 − 9 = − 24 18 x 6+3 y − 2 = − 4 3 x 9 y − 2 = − 4 x 9 3 y 2 6. (14 m 2 n 3 )( − 2 m 3 n 2 ) = 14 · ( − 2) · m 2 · m 3 · n 3 · n 2 = − 28 m 2+3 n 3+2 = − 28 m 5 n 5 7. ( − 3 t − 4 a )( − 5 t − a ) = − 3( − 5) · t − 4 a · t − a = 15 · t − 4 a +( − a ) = 15 t − 5 a , or 15 t 5 a 8. − 18 x − 2 y 3 − 12 x − 5 y 5 = − 18 − 12 x − 2 − ( − 5) y 3 − 5 = 18 12 x − 2+5 y − 2 = 3 2 x 3 y − 2 = 3 x 3 2 y 2 9. ( r 9 ) 5 = r 9 · 5 = r 45 10. ( 4 − 3 3 4 ) 3 = (4 − 3 ) 3 (3 4 ) 3 = 4 − 3 · 3 3 4 · 3 = 4 − 9 3 12 = 1 4 9 · 3 12 11. (5 a 2 b 2 ) 3 = 5 3 ( a 2 ) 3 ( b 2 ) 3 = 125 a 2 · 3 b 2 · 3 = 125 a 6 b 6 12. ( a 4 c 7 ) 2 = ( a 4 ) 2 ( c 7 ) 2 = a 8 c 14 13. (4 3 ) 2 = 4 3 · 2 = 4 6 14. ( 2 x 3 y − 2 3 y − 3 ) 3 = (2 x 3 y − 2 ) 3 (3 y − 3 ) 3 = 2 3 ( x 3 ) 3 ( y − 2 ) 3 3 3 ( y − 3 ) 3 = 8 x 9 y − 6 27 y − 9 = 8 x 9 y − 6 − ( − 9) 27 = 8 x 9 y 3 27 15. (8 x ) 4 y = 8 x · 4 y = 8 4 xy 16. ( − 4 x 4 y − 2 5 x − 1 y 4 ) − 4 = ( − 4 x 4 y − 2 ) − 4 (5 x − 1 y 4 ) − 4 = ( − 4) − 4 ( x 4 ) − 4 ( y − 2 ) − 4 5 − 4 ( x − 1 ) − 4 ( y 4 ) − 4 = ( − 1) − 4 · 4 − 4 x − 16 y 8 5 − 4 x 4 y − 16 = 4 − 4 x − 16 − 4 y 8 − ( − 16) 5 − 4 = 4 − 4 x − 20 y 24 5 − 4 = 5 4 y 24 4 4 x 20 20. Scientific Notation 1. 4 . 7,000,000,000. ↑ 10 places Large number, so the exponent is positive. 47 , 000 , 000 , 000 = 4 . 7 × 10 10 2. 0.00000001.6 ↑ 8 places Small number, so the exponent is negative. 0 . 000000016 = 1 . 6 × 10 − 8 3. 0.0000002.63 ↑ 7 places Small number, so the exponent is negative. 0 . 000000263 = 2 . 63 × 10 − 7 4. 2 . 600,000,000,000. ↑ 12 places Large number, so the exponent is positive. 2 , 600 , 000 , 000 , 000 = 2 . 6 × 10 12 5. 200 billionths = 200 × 0 . 000000001 = 0 . 0000002 0.0000002. ↑ 7 places Small number, so the exponent is negative. 0 . 0000002 = 2 × 10 − 7 6. 6.73000000. ↑ 8 places Positive exponent so the number is large. 6 . 73 × 10 8 = 673 , 000 , 000

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    10 Just-in-Time Review 7. We move the decimal point 28 places to the left. 9 . 11 × 10 − 28 g = 0 . 000000000000000000000000000911 g 8. We move the decimal point 9 places to the right. 1 . 86 × 10 9 = 1 , 860 , 000 , 000 active Facebook users 9. (6 . 5 × 10 3 )(5 . 2 × 10 − 8 ) = (6 . 5 × 5 . 2)(10 3 × 10 − 8 ) = 33 . 8 × 10 − 5 = (3 . 38 × 10) × 10 − 5 = 3 . 38 × 10 − 4 10. 3 . 6 × 10 − 3 7 . 2 × 10 − 12 = 3 . 6 7 . 2 × 10 − 3 10 − 12 = 0 . 5 × 10 9 = (5 × 10 − 1 ) × 10 9 = 5 × 10 8

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    Chapter 1 Solving Linear Equations and Inequalities Exercise Set 1.1 RC2. (f) RC4. (c) CC2. 4 − 2 3 x = 5 2 a) The LCM of the denominators is 6, so we multiplyby6. b) 6 ( 4 − 2 3 x ) = 6 · 5 2 24 − 4 x = 15 CC4. 0 . 06 − 7 x = 1 . 2 x a) The greatest number of decimal places is 2, so we multiplyby100. b) 100(0 . 06 − 7 x ) = 100(1 . 2 x ) 6 − 700 x = 120 x 2. 47 − x = 23 47 − 24 ? 23 23 ∣ ∣ ∣ TRUE 24 is a solution of the equation. 4. 3 x + 14 = − 27 3( − 10) + 14 ? − 27 − 30 + 14 ∣ ∣ ∣ − 16 ∣ ∣ ∣ FALSE − 10 is not a solution of the equation. 6. − x 8 = − 3 − 32 8 ? − 3 − 4 ∣ ∣ ∣ FALSE 32 is not a solution of the equation. 8. 4 − 5 x = 59 4 − 5( − 11) ? 59 4 + 55 ∣ ∣ ∣ 59 ∣ ∣ ∣ TRUE − 11 is a solution of the equation. 10. 9 y + 5 = 86 9 · 9 + 5 ? 86 81 + 5 ∣ ∣ ∣ 86 ∣ ∣ ∣ TRUE 9 is a solution of the equation. 12. x + 5 = 5 + x − 13 + 5 ? 5 + ( − 13) − 8 ∣ ∣ ∣ − 8 TRUE − 13 is a solution of the equation. 14. x + 7 = 14 x + 7 − 7 = 14 − 7 x + 0 = 7 x = 7 16. − 27 = y − 17 − 27 + 17 = y − 17 + 17 − 10 = y + 0 − 10 = y 18. − 8 + r = 17 8 − 8 + r = 8 + 17 0 + r = 25 r = 25 20. − 37 + x = − 89 37 − 37 + x = 37 − 89 0 + x = − 52 x = − 52 22. z − 14 . 9 = − 5 . 73 z − 14 . 9 + 14 . 9 = − 5 . 73 + 14 . 9 z + 0 = 9 . 17 z = 9 . 17 24. x + 1 12 = − 5 6 x + 1 12 − 1 12 = − 5 6 − 1 12 x + 0 = − 10 12 − 1 12 x = − 11 12

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    12 Chapter 1: Solving Linear Equations and Inequalities 26. 5 x = 30 5 x 5 = 30 5 1 · x = 30 5 x = 6 28. − 4 x = 124 − 4 x − 4 = 124 − 4 1 · x = 124 − 4 x = − 31 30. − x 3 = − 25 − 1 3 x = − 25 − 3 ( − 1 3 ) x = − 3( − 25) x = 75 32. − 120 = − 8 y − 120 − 8 = − 8 y − 8 − 120 − 8 = 1 · y 15 = y 34. 0 . 39 t = − 2 . 73 0 . 39 t 0 . 39 = − 2 . 73 0 . 39 1 · t = − 2 . 73 0 . 39 t = − 7 36. − 7 6 y = − 7 8 − 6 7 ( − 7 6 ) ( y ) = − 6 7 ( − 7 8 ) 1 · y = 42 56 y = 3 4 38. 4 x − 7 = 81 4 x = 88 x = 22 40. 6 z − 7 = 11 6 z = 18 z = 3 42. 5 x + 7 = − 108 5 x = − 115 x = − 23 44. − 9 2 y + 4 = − 91 2 − 9 y + 8 = − 91 − 9 y = − 99 y = 11 46. 2 x + x = − 1 3 x = − 1 x = − 1 3 48. − 2 x + 6 x − 2 = 11 4 x = 13 x = 13 4 50. 9 5 y + 4 10 y = 66 10 18 y + 4 y = 66 Multiplying by 10 22 y = 66 y = 3 52. 0 . 8 t − 0 . 3 t = 6 . 5 0 . 5 t = 6 . 5 t = 13 54. 15 x + 40 = 8 x − 9 15 x = 8 x − 49 7 x = − 49 x = − 7 56. 3 x − 15 = 15 + 3 x − 15 = 15 False equation No solution 58. 9 t − 4 = 14 + 15 t 9 t − 18 = 15 t − 18 = 6 t − 3 = t 60. 6 − 7 x = x − 14 20 − 7 x = x 20 = 8 x 20 8 = x 5 2 = x 62. 5 x − 8 = − 8 + 3 x − x 5 x − 8 = − 8 + 2 x 3 x = 0 x = 0 64. 6 y + 20 = 10 + 3 y + y 6 y + 20 = 10 + 4 y 2 y = − 10 y = − 5

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    Exercise Set 1.1 13 66. − 3 t + 4 = 5 − 3 t 4 = 5 False equation No solution 68. 5 − 2 y = − 2 y + 5 5 = 5 True equation All real numbers are solutions. 70. 3 4 − 5 8 m = 1 2 m − 3 8 6 − 5 m = 4 m − 3 Multiplying by 8 9 = 9 m 1 = m 72. 0 . 2 t + 1 . 7 = 5 . 8 + 0 . 3 t 2 t + 17 = 58 + 3 t Multiplying by 10 − 41 = t 74. 2 − 0 . 01 x = 0 . 4 x + 2 . 5 200 − x = 40 x + 250 Multiplying by 100 − 50 = 41 x − 50 41 = x 76. 2 3 x + 1 − 1 6 = 1 2 x − 5 6 x 4 x + 6 − 1 = 3 x − 5 x 4 x + 5 = − 2 x 5 = − 6 x − 5 6 = x 78. 3( y + 6) = 9 y 3 y + 18 = 9 y 18 = 6 y 3 = y 80. 27 = 9(5 y − 2) 27 = 45 y − 18 45 = 45 y 1 = y 82. 210( x − 3) = 840 210 x − 630 = 840 210 x = 1470 x = 7 84. 8 x − (3 x − 5) = 40 8 x − 3 x + 5 = 40 5 x = 35 x = 7 86. 3(4 − 2 x ) = 4 − (6 x − 8) 12 − 6 x = 4 − 6 x + 8 12 − 6 x = 12 − 6 x 12 = 12 True equation All real numbers are solutions. 88. − 40 x + 45 = 3[7 − 2(7 x − 4)] − 40 x + 45 = 3[7 − 14 x + 8] − 40 x + 45 = 3[ − 14 x + 15] − 40 x + 45 = − 42 x + 45 2 x = 0 x = 0 90. 1 6 (12 t + 48) − 20 = − 1 8 (24 t − 144) 2 t + 8 − 20 = − 3 t + 18 5 t = 30 t = 6 92. 6[4(8 − y ) − 5(9 + 3 y )] − 21 = − 7[3(7 + 4 y ) − 4] 6[32 − 4 y − 45 − 15 y ] − 21 = − 7[21 + 12 y − 4] 6[ − 13 − 19 y ] − 21 = − 7[17 + 12 y ] − 78 − 114 y − 21 = − 119 − 84 y 20 = 30 y 2 3 = y 94. 3 4 ( 3 x − 1 2 ) + 2 3 = 1 3 54 x − 9 + 16 = 8 Multiplying by 24 54 x = 1 x = 1 54 96. 9(4 x + 7) − 3(5 x − 8) = 6 ( 2 3 − x ) − 5 ( 3 5 + 2 x ) 36 x + 63 − 15 x + 24 = 4 − 6 x − 3 − 10 x 21 x + 87 = − 16 x + 1 37 x = − 86 x = − 86 37 98. a − 9 a 23 = a − 32 = 1 a 32 100. − 2 x 8 y 3 102. − 5 + 6 x 104. − 10 x + 35 y − 20 106. 4( − x − 6 y ), or − 4( x + 6 y ) 108. 5( − 2 x + 7 y − 4), or − 5(2 x − 7 y + 4) 110. {− 8 , − 7 , − 6 , − 5 , − 4 , − 3 , − 2 , − 1 } ; { x | x is a negative integer greater than − 9 } 112. − 0 . 00458 y + 1 . 7787 = 13 . 002 y − 1 . 005 − 13 . 00658 y = − 2 . 7837 y ≈ 0 . 214

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