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College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher

College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher

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College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher

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  1. Get Full Version With All Included At: https://easytestbanks.com/product/college-algebra- with-intermediate-algebra-1st-edition-solutions- manual-by-beecher/ Chapter 3 Systems of Equations 8. Exercise Set 3.1 C o n s i s t e n t ; i n d e p e n d e n t RC2. True; see page 152 in the text. RC4. True; see page 155 in the text. 2. y 5 x + y= 5 4 3 2 (4,1) 1 —5 —4 —3 —2 —1 10. x 2 3 4 5 x – y= 3 1 —1 —2 —3 —4 —5 Consistent; independent 4. y 5 4 3x + y = 5 3 2 1 x 12. —5 —4 —3 — 2 —1 2 3 4 5 (2, –1) 1 1 — — 2 — 3 — 4 x – 2y = 4 — 5 Consistent; independent 6. y 5 4 2 y = 6 – x 3 3 2) (3, – 2 1 14. x —5 —4 —3 —2 —1 1 2 3 4 5 1 — — — 2 3 Copyright· c 2017Pearson Education, Inc. 3x –2y = 6 4 — — 5

  2. y f ( x ) = x + 1 5 4 3 2 x 3 2 =– g ( x ) 1 x —5 —4 —3 —2 —1 (–3, –2) 1 2 3 4 5 — 1 — 2 — 3 — 4 — 5 Consistent; independent b 10 8 2b + a =11 6 4 (7,2) 2 2 a —10 —8 —6 —4 — 2 4 6 8 10 a–b = 5 2 — 4 — 6 — 8 — 0 — 1 Consistent; independent y 5 4 –1 x 1 3 f ( x ) = –4 + 2 1 (4,0) x —5 —4 —3 —2 —1 1 2 3 4 5 1 — — 2 — 3 — 4 g(x )= –1x — 5 2 – 2 Consistent; independent y 5 4 y–x = 5 3 2 1 x —5 —4 —3 —2 —1 1 2 3 4 5 — 1 — 2 — 3 2x – 2y = 10 — 4 — 5 No solution Inconsistent; independent Copyright· c 2017Pearson Education, Inc.

  3. 78 Chapter 3: Systems of Equations 16. 30. 5(10 −4x) = −3(7x −4) 50 −20x = −21x + 12 x = −38 32. (−0.26, 57.06) 34. y 5 4 y = 3 – x 3 2 2x + 2y = 6 1 —5 —4 —3 —2 —1 x 2 3 4 5 1 1 — y — 2 — 3 5 — 4 4 y =x 2 — 5 3 2 The graphs are the same. There are infinitely many solu- tions. Consistent; dependent 3 ( –3, 2) y= 2 2 1 (1 , 1) 1 —5 —4 —3 —2 —1 1 2 3 4 5 1 (0, 0) x — — 2 x – y= 0 — 3 18. y — 4 5 — 5 4 Exercise Set 3.2 —5 —4 —3 —2 —1 x =–3 x 2 3 4 5 1 1 — RC2. True; we can use the substitution method to solve any system of two equations in two variables. — 2 — 3 — 4 — 5 RC4. False; see Example 3 on page 161 in the text. Consistent; independent 2. x = 8 −4y, 3x + 5y = 3 (2) (1) 20. b 5 4 3(8 −4y) + 5y = 3 24 −12y + 5y = 3 24 −7y = 3 −7y = −21 b– a = 6 ( –5,1) Substituting for x in (2) 3 2 1 a —5 —4 —3 — 2 —1 2 3 4 5 a+ 2b = –3 1 — 1 — 2 y = 3 — 3 — 4 Substitute 3 for y in (1). x = 8 −4 · 3 = 8 −12 = −4 The solution is (−4, 3). 4. 9x −2y = 3, (1) 3x −6 = y — 5 Consistent; independent 22. Consistent; independent; the pair (1, 1) is a solution of both equations in system C and is not a solution of any other system shown, so system C corresponds to this graph. (2) 9x −2(3x −6) = 3 9x −6x + 12 = 3 3x + 12 = 3 Substituting for y in (1) 24. Consistent; independent; the system consists of a line with y-intercept (0, −1) and a horizontal line. Both lines pass through (4, −3). System E corresponds to this graph. 26. Consistent; independent; the pair ( 1, 3) is a solution of both equations in system A and is not a solution of any other system shown, so system A corresponds to this graph. 3 x + 2 = 2 x −5 20 =20 5 4 x 15x + 40 = 8x −100 7x = −140 x = −20 3x = −9 x = −3 − Substitute −3 for x in (2). 3(−3) −6 = y −9 −6 = y −15 = y The solution is (−3, −15). 28. .3+2 Σ .2 Σ 4 5 5 x − Copyright· c 2017Pearson Education, Inc.

  4. Exercise Set 3.2 79 Substitute 2p −8 for q in Equation (2) and solve for p. 5p + 7(2p −8) = 1 5p + 14p −56 = 1 19p −56 = 1 19p = 57 p = 3 Substitute 3 for p in Equation (3). q = 2 · 3 −8 = 6 −8 = −2 The solution is (3, −2). 14. 3x + y = 4, (1) 12 −3y = 9x (2) Solve the first equation for y. 3x + y = 4 y = −3x + 4 Substitute −3x + 4 for y in the second equation and solve for x. 12 −3y = 9x 12 −3(−3x + 4) = 9x 12 + 9x −12 = 9x 12 −12 = 0 0 = 0 We have a true equation. Any value of x will make this equation true. Thus the system of equations has infinitely many solutions. 6. m −2n = 3, (1) 4m + n = 1 We solve the first equation form. m −2n = 3 m = 2n + 3 (3) We substitute 2n + 3 for m in the second equation and solve for n. 4(2n + 3) + n = 1 8n + 12 + n = 1 9n + 12 = 1 9n = −11 11 n = −9 Now we substitute . Σ m = 2 −11 + 3 = −22 + 27 = 5 9 The solution is 9 9 (2) (1) 11 for 9 in Equation (3). − n 9 . 9 9 . 5 , −11 Σ 8. t = 4 −2s, (1) t + 2s = 6 (2) (4 −2s) + 2s = 6 Substituting for t in (2) 4 = 6 We get a false equation. The system has no solution. 10. 5x + 6y = 14, (1) −3y + x = 7 We solve the second equation for x. −3y + x = 7 x = 3y + 7 (3) Substitute 3y + 7 for x in the first equation and solve for y. 5(3y + 7) + 6y = 14 15y + 35 + 6y = 14 21y + 35 = 14 21y = −21 y = −1 Substitute −1 for y in Equation (3). x = 3(−1) + 7 = −3 + 7 = 4 The solution is (4, −1). 16. 5x + 3y = 4, (1) x −4y = 3 (2) We solve the second equation for x. x −4y = 3 x = 4y + 3 (3) Substitute 4y + 3 for x in the first equation and solve for y. 5x + 3y = 4 (1) 5(4y + 3) + 3y = 4 20y + 15 + 3y = 4 23y + 15 = 4 23y = −11 11 y = − 23 Substitute −23 for y in Equation (3). x=4 23 25 11 . 23 23 (2) (2) (2) 11 −11Σ . +3=−44+69=25 23 12. 4p −2q = 16, (1) 5p + 7q = 1 Solve the first equation for q. 4p −2q = 16 −2q = −4p + 16 q = 2p −8 23 23 (2) . Σ The solution is , − (1) 18. 4x + 13y = 5, (1) −6x + y = 13 (2) Solve the second equation for y. −6x + y = 13 y = 6x + 13 (3) (3) (2) Copyright· c 2017Pearson Education, Inc.

  5. 80 Chapter 3: Systems of Equations 5x + 2y = a Substitute 6x + 13 for y in Equation (1) and solve for x. 4x + 13(6x + 13) = 5 4x + 78x + 169 = 5 82x + 169 = 5 82x = −164 x = −2 −5x + 5y = −5b Multiplying by −5 7y = a −5b y = a −5b 7 We obtain 7 . a + 2b , a −5b Σ . This checks, so it is the so- 7 lution. Substitute −2 for x in Equation (3). y = 6(−2) + 13 = −12 + 13 = 1 The solution is (−2, 1). 20. Let l = the length and w = the width. Solve: 2l + 2w = 340, l = w + 50 The solution is (110, 60), so the length is 110 m and the width is 60 m. 34. Let b = the number of ounces of baking soda and v = the number of ounces of vinegar to be used. Solve: b = 4v, b + v = 16 The solution is (12.8, 3.2), so 12.8 oz of baking soda and 3.2 oz of vinegar should be used. Exercise Set 3.3 22. Let x and y represent the angles. Solve: x + y = 90, y = 5x + 6 The solution is (14, 76), so the measures of the angles are 14◦and 76◦. RC2. If a system of equations has no solution, then it is inconsistent. RC4. If the graphs of the equations in a system of two equa- tions in two variables are the same line, then the equa- tions are dependent. 24. Let x = the number of coach-class seats and y = the num- ber of first-class seats. Solve: x + y = 152, x = 5 + 6y The solution is (131, 21), so there are 131 coach-class seats and 21 first-class seats. RC6. If the graph of the equations in a system of two equa- tions in two variables intersect at one point, then the equations are independent. 2. x + y = 9, (1) 2x −y = −3 (2) = 6 x = 2 Substitute 2 for x in (1). 2 + y = 9 y = 7 The solution is (2, 7). 26. −9(y + 7) −6(y −4) = −9y −63 −6y + 24 = −15y −39 28. m = 3 −(−4) = 7 −2 −(−5) 30. −12(2x −3) = 16(4x −5) −24x + 36 = 64x −80 116 = 88x 29 = x 22 32. 5x + 2y = a, x −y = b We multiply by 2 on both sides of the second equation and then add. 5x + 2y = a 3x Adding 3 4. 2x −3y = 18, (1) 2x + 3y = −6 (2) 4x x = 3 Substitute 3 for x in (2). 2 · 3 + 3y = −6 6 + 3y = −6 3y = −12 y = −4 The solution is (3, −4). = 12 Adding 2x −2y = 2b 7x + 0 = a + 2b Adding 7x = a + 2b a + 2b x = Multiplying by 2 7 Next we multiply by 5 on both sides of the second equa- tion and then add. − Copyright· c 2017Pearson Education, Inc.

  6. Exercise Set 3.3 81 5 · 2 + 4y = 2 10 + 4y = 2 3 4y = −3 6. 2x + 3y = −9, (1) 5x −6y = −9 4x + 6y = −18 Multiplying (1) by 2 5x −6y = −9 9x = −27 x = −3 Substitute −3 for x in (1). 2(−3) + 3y = −9 −6 + 3y = −9 3y = −3 y = −1 The solution is (−3, −1). 8. 2a + 3b = 11, 4a −5b = −11 (2) 3 (2) 4 1 y = −3 . 2 , −1 Σ 3 3 The solution is . 14. 5x + 3y = 25, (1) 3x + 4y = 26 (2) 20x + 12y = 100 −9x −12y = −78 Multiplying (2) by −3 11x = 22 x = 2 Substitute 2 for x in (1). 5 · 2 + 3y = 25 10 + 3y = 25 3y = 15 y = 5 The solution is (2, 5). Multiplying (1) by 4 (1) −4a −6b = −22 Multiplying (1) by −2 4a −5b = −11 −11b = −33 b = 3 Substitute 3 for b in (1). 2a + 3 · 3 = 11 2a + 9 = 11 2a = 2 a = 1 The solution is (1, 3). 3x −5y = −2 −3x + 5y = 7 Rearranging 0 = 5 False equation There is no solution. 16. 18. 10x + y = 306, 10y + x = 90 y = −10x + 306 (3) 10(−10x + 306) + x = 90 Substituting for y (1) (2) 3x −2y = 1, −6x + 4y = −2 (2) 10. (1) Solving (1) for y 6x −4y = 2 Multiplying (1) by 2 −6x + 4y = −2 (2) 0 = 0 True for all (x, y) There are infinitely many solutions. in (2) −100x + 3060 + x = 90 −99x = −2970 12. 5x + 4y = 2, (1) 2x −8y = 4 10x + 8y = 4 2x −8y = 4 12x x = 30 Substitute 30 for x in (3). y = −10 · 30 + 306 = 6 The solution is (30, 6). 20. 2 x + 1 y = −11, 3 7 1 1 x −y = −10 7 3 14x + 3y = −231, 3x −7y = −210 (2) Multiplying (1) by 2 = 8 2 x = 3 Substitute 2 for x in (1). 3 (1) Multiplying by 21 (2) to clear fractions 98x + 21y = −1617 Multiplying (1) by 7 9x −21y = −630 Multiplying (2) by 3 107x = −2247 x = −21 Copyright· c 2017Pearson Education, Inc.

  7. 82 Chapter 3: Systems of Equations 49x −21y = 35 Multiplying (1) by 7 −12x + 21y = 39 Multiplying (2) by 3 37x = 74 x = 2 Substitute 2 for x in (2). −4 · 2 + 7y = 13 7y = 21 Substitute −21 for x in (1). 14(−21) + 3y = −231 3y = 63 y = 21 The solution is (−21, 21). 22. 2 x + 3 y = −17, 3 5 1 1 2 x −3 y = −1 We multiply the first equation by 15 and the second equa- tion by 6 to clear fractions. y = 3 The solution is (2, 3). 28. 1.3x −0.2y = 12, (1) 0.4x + 17y = 89 10x + 9y = −255, (1) 3x −2y = −6 20x + 18y = −510 Multiplying (1) by 2 27x −18y = −54 Multiplying (2) by 9 47x = −564 x = −12 Substitute −12 for x in (1). 10(−12) + 9y = −255 −120 + 9y = −255 9y = −135 y = −15 The solution is (−12, −15). 24. 3 2 5 1 x −y = −6 6 8 (2) (2) 13x −2y = 120, (3) 4x + 170y = 890 (4) Clearing decimals 1105x −170y = 10, 200 Multiplying (3) by 85 4x + 170y = 890 (4) 1109x = 11, 090 x = 10 Substitute 10 for x in (3). 13 · 10 −2y = 120 −2y = −10 y = 5 The solution is (10, 5). 30. Let x = the larger number and y = the smaller number. Solve: x + y = 2, 4 x + 3 y = 4, We multiply the first equation by 6 and the second equa- tion by 24 to clear fractions. x −y = 20 The solution is (11, −9), so the numbers are 11 and −9. 32. Let x = the larger number and y = the smaller number. Solve: x + y = 9, 2x + 3y = 2 The solution is (25, −16), so the numbers are 25 and −16. 34. Let x and y represent the measures of the angles. Solve: x + y = 180, x −y = 22 The solution is (101, 79), so the measures of the angles are 101◦and 79◦. 8x + 9y = 24, 20x −3y = −144 (2) 8x + 9y = 24 60x −9y = −432 Multiplying (2) by 3 68x = −408 x = −6 Substitute −6 for x in (1). 8(−6) + 9y = 24 −48 + 9y = 24 9y = 72 y = 8 The solution is (−6, 8). (1) 36. Let x = the number of two-point shots made and y = the number of foul shots made. Solve: x + y = 64, 2x + y = 100 The solution is (36, 28), so Chamberlain made 36 two-point shots and 28 foul shots. 0.7x −0.3y = 0.5, −0.4x + 0.7y = 1.3 7x −3y = 5, (1) Clearing decimals −4x + 7y = 13 (2) in both equations 26. 38. Let x = the number of redbud trees chosen and y = the number of dogwood trees chosen. Solve: Copyright· c 2017Pearson Education, Inc.

  8. Exercise Set 3.4 83 x + y = 18, 37x + 45y = 754 The solution is (7, 11), so 7 customers chose an Eastern Redbud and 11 chose a Kousa Dogwood. −4 · 2 + 3a = 7 3a = 15 a = 5 Thus, a = 5 and b = 2. 40. f (−1) = 3(−1)2 −(−1) + 1 = 3 + 1 + 1 = 5 Exercise Set 3.4 42. f (2a) = 3(2a)2 −2a + 1 = 12a2 −2a + 1 44.Domain: all realnumbers;range:{y|y≤5},or (−∞,5] 46. m = 10 −2 −2 −(−10) 8 y −y1 = m(x −x1) y −2 = 1(x −(−10)) y −2 = x + 10 y = x + 12 x + y −x −y = 1, 2 x −y + 2 6 After clearing fractions we have: 3x + 7y = 10, (1) RC2. 15 RC4. 0.2(10) = 2 = 8 = 1 2. Let x = the number of small stems sold and y = the num- ber of large stems sold. Solve: x + y = 118, 2.50x + 3.95y = 376.20 The solution is (62, 56), so Kevin sold 62 sunflowers with small stems and 56 with large stems. 48. 5 4. Let x = the number of ounces of lemon juice and y = the number of ounces of linseed oil to be used. Solve: y = 2x, x + y = 32 The solution is 102 211 , so 10 2 oz of lemon juice and 3 3 211 oz of linseed oil should be used. 3 6. Let x and y represent the number of truffles and cream mints, respectively, in each box. The 75 boxes are identi- cal, so each sold for $1872/75, or $24.96. Solve: x + y = 12, 2.95x + 1.79y = 24.96 The solution is (3, 9), so each box contained 3 truffles and 9 cream mints. x + y = 2 − . Σ 2x −y = −6 (2) 6x + 14y = 3 , 20 Multiplying (1) by 2 28x −14y = −84 Multiplying (2) by 14 34x = −64 32 x = −17 Substitute 17 . −32 Σ + 7y = 10 7y = 266 17 38 y = 17 The solution is 32 for in (1). − x 3 17 8. Let x = the number of pounds of Deep Thought Granola and y = the number of pounds of Oat Dream Granola to be used. Solve: x + y = 20, 0.25x + 0.1y = 0.19(20) The solution is (12, 8), so 12 lb of Deep Thought Granola and 8 lb of Oat Dream Granola should be used. 32 38 . 17 17 . Σ − , 50. There are many correct answers. Find one by expressing the sum of the two numbers: x + y = −1 52. Substitute 4 for x and solve for a and b. −4a −3b = −26, (1) −4b + 3a = 7 (2) 10. Let x = the number of pounds of soybean meal and y = the number of pounds of corn meal in the mixture. The amount of protein in the mixture is 0.12(350 lb), or 42 lb. Solve: x + y = 350, 0.16x + 0.09y = 42 The solution is (150, 200), so 150 lb of soybean meal and 200 lb of corn meal should be used. 3 for y in both equations and − − −12a − 12a −16b = 28 Multiplying (2) by 4 9b = −78 Multiplying (1) by 3 12. Let x and y represent the number of Acuminata bulbs and Cafe Brun bulbs, respectively, in the assortment. Solve: x + y = 12, 4.85x + 9.50y = 12(7.95) The solution is (4, 8), so the assortment contains 4 Acumi- nata bulbs and 8 Cafe Brun bulbs. −25b = −50 b = 2 Substitute 2 for b in (2). Copyright· c 2017Pearson Education, Inc.

  9. 84 Chapter 3: Systems of Equations 2900 = (r −w)5, . r −w Σ 29 Solve: 14. Let x = the number of liters of Arctic Antifreeze that should be used and let y = the number of liters in the x + 7.5 = y 0.18x + 0.1(7.5) = 0.15y The solution is (12.5, 20), so 12.5 L of Arctic Antifreeze should be used. 2900 = mixture. Solve: 2 6 The solution is (620,40), so the headwind is 40 mph and the plane’s air speed is 620 mph. plane 28. Distance Rate Time Slower 190 d t 16. Let x = the amount invested at 4% and y = the amount invested at 6%. Solve: x + y = 45, 000 0.04x + 0.06y = 2430 The solution is (13, 500, 31, 500), so $13,500 was invested at 4% and $31,500 was invested at 6%. plane Faster 200 t 780 −d Solve: 780 −d = 200t d = 190t, 18. Let x = the amount borrowed at 2.8% interest and y = the amount borrowed at 4.5%. Solve: x + y = 31, 000, 0.028x + 0.045y = 1024.40 The solution is (21, 800, 9200), so $21,800 was borrowed at 2.8% and $9200 was borrowed at 4.5%. We find that t = 2 hr. Distance Rate Time 30. Toward Los Angeles Toward Honolulu Solve: 370 d t 250 t 2553 −d d = 370t, 20. Let x and y represent the number of quarters and dollar coins, respectively, used to make change. Note that the change owed is $20 −$9.25 = $10.75. Solve: x + y = 19, 2553 −d = 250t We find that d ≈ 1524 mi. 32. {2, 4, 6, 7, 8, 9, 10} 34. |− 7x2| = |− 7|· |x2| = 7x2 . a4 . = |a4| = a4 c 38. . . 0.25x + y = 10.75 The solution is (11, 8), so there were 11 quarters and 8 dol- lar coins. 36. |c| |c| 22. Distance Rate Time Distance Rate Time Max 80 d t Jogging 8 d t t −1 Olivia 96 d 6 −d 1 −t Walking 4 Solve: d = 80t, d = 96t(t −1) Solve: d = 8t, 6 −d = 4(1 −t) We find that d = 480 km. We find that d = 4 km. 24. Let r = the speed of the boat in still water. Distance Down- stream Up- stream Solve: d = 3(r + 6), d = 5(r −6) We find that r = 24 mph. 40.Let b = the number of boys and g = the number of girls. Then Phyllis has b brothers and g −1 sisters, and Phil has b −1 brothers and g sisters. Solve: b = 2(g −1), b −1 = g The solution is (4, 3), so there are 4 boys and 3 girls in the family. Rate Time r + 6 3 d 5 d r −6 Chapter 3 Mid-Chapter Review 26. Let w = the speed of the headwind. Note that 4 hr 50 5 min = 4 6 Distance Stronger headwind Weaker headwind hr, or 29 6hr. 1.False; see pages 161 and 168 in the text. Rate Time 2.False; see page 155 in the text. 2900 5 r −w 3.True; see page 155 in the text. 29 6 w 4.True; a vertical line x = a and a horizontal line y = b intersect at exactly one point, (a, b). 2900 r −2 Copyright· c 2017Pearson Education, Inc.

  10. Chapter 3 Mid-Chapter Review 85 7.Graph the lines on the same set of axes. 5.Graph the lines on the same set of axes. y y 5 5 y = 2x – 3 4 4 y = 4– x 3 3 2 2 4x – 2y = 6 1 1 —5 —4 —3 —2 —1 —5 —4 —3 —2 —1 1 2 3 4 5 1 2 3 4 5 x x 1 — — 1 — — 2 2 (5,–1 ) — — 3 3 — — 4 4 y =x – 6 — — 5 5 The graphs are the same. There is an infinite number of solutions. Since the system of equations has a solution, it is consistent. Since there are infinitely many solutions, the equations are dependent. The solution appears to be (5, −1). Check: y = x −6 −1 ? 5 −6 −1 TRUE The solution is (5, −1). Since the system of equations has a solution, it is consis- tent. Since there is exactly one solution, the equations are independent. y = 4 −x −1 ? 4 −5 8.Graph the lines on the same set of axes. −1 TRUE . . y 5 4 3 2 y –2x= 6 2 1 6.Graph the lines on the same set of axes. x —5 —4 —3 —2 —1 1 2 3 4 5 — 1 x – y = 3 y — 2 — 3 5 x + y = 3 — 4 4 (0, 3) — 5 3 2 The lines are parallel. There is no solution. Since the sys- tem of equations has no solution, it is inconsistent. Since there is no solution, the equations are independent. 1 x —5 —4 —3 —2 —1 1 2 3 4 5 1 — — 2 9. x = y + 2, 2x −3y = −2 (2) Substitute y + 2 for x in the Equation (2) and solve for y. 2(y + 2) −3y = −2 2y + 4 −3y = −2 −y + 4 = −2 −y = −6 y = 6 Substitute 6 for y in Equation (1) and find x. x = 6 + 2 = 8 The solution is (8, 6). — (1) 3 — 4 3x+ y= 3 — 5 The solution appears to be (0, 3). Check: x + y = 3 0 + 3 ? 3 3x + y = 3 3 0 + 3 ? 3 · 0 + 3 3 TRUE . . 3 . TRUE . The solution is (0, 3). Since the system of equations has a solution, it is consis- tent. Since there is exactly one solution, the equations are independent. 10. y = x −5, x −2y = 8 (2) Substitute x −5 for y in the Equation (2) and solve for x. x −2(x −5) = 8 x −2x + 10 = 8 −x + 10 = 8 −x = −2 x = 2 (1) Copyright· c 2017Pearson Education, Inc.

  11. 86 Chapter 3: Systems of Equations Substitute 2 for x in Equation (1) and find y. y = 2 −5 = −3 The solution is (2, −3). 11. 4x + 3y = 3, (1) y = x + 8 (2) Substitute x + 8 for y in the Equation (1) and solve for x. 4 x + 3(x + 8) = 3 4x + 3x + 24 = 3 7x + 24 = 3 7x = −21 x = −3 Substitute −3 for x in Equation (2) and find y. y = −3 + 8 = 5 The solution is (−3, 5). 15. 3x −4y = 5, 5x −2y = −1 (2) First multiply by then add. (1) 2 on both sides of Equation (2) and − 3x −4y = 5 −10x + 4y = 2 −7x = 7 x = −1 Substitute −1 for x in Equation (1) and solve for y. 3(−1) −4y = 5 −3 −4y = 5 −4y = 8 y = −2 The solution is (−1, −2). 16. 3x + 2y = 11, (1) 2x + 3y = 9 (2) We use the multiplication principle on both equations and then add. 9x + 6y = 33 Multiplying (1) by 3 −4x −6y = −18 5x = 15 x = 3 Substitute 3 for x in Equation (1) and solve for y. 3 · 3 + 2y = 11 9 + 2y = 11 2y = 2 y = 1 The solution is (3, 1). 12. 3x −2y = 1, (1) x = y + 1 Substitute y + 1 for x in the Equation (1) and solve for y. 3(y + 1) −2y = 1 3y + 3 −2y = 1 y + 3 = 1 y = −2 Substitute −2 for y in Equation (2) and find x. x = −2 + 1 = −1 The solution is (−1, −2). 13. 2x + y = 2, (1) x −y = 4 (2) 3x = 6 Adding x = 2 Substitute 2 for x in Equation (1) and solve for y. 2 · 2 + y = 2 4 + y = 2 y = −2 The solution is (2, −2). x −2y = 13, (1) x + 2y = −3 (2) 2x = 10 Adding x = 5 Substitute 5 for x in Equation (2) and solve for y. 5 + 2y = −3 2y = −8 y = −4 The solution is (5, −4). (2) Multiplying (2) by −2 x −2y = 5, (1) 3x −6y = 10 (2) First we multiply by 3 on both sides of Equation (1) and then add. −3x + 6y = −15 3x −6y = 10 0 = −5 We get a false equation. There is no solution. 17. − 14. 4x −6y = 2, −2x + 3y = −1 (2) First we multiply by 2 on both sides of Equation (2) and then add. 4x −6y = 2 −4x + 6y = −2 0 = 0 We obtain an equation that is true for all values of x and y. Thus, if an ordered pair is a solution of one of the original equations, it is a solution of the other equation also. The system of equations has an infinite number of solutions. 18. (1) Copyright· c 2017Pearson Education, Inc.

  12. Chapter 3 Mid-Chapter Review 87 1 x + 1 y = 1, 2 3 1 3 x −y = 11 (2) 5 4 We multiply both sides of Equation 1 by 6 and both sides of Equation (2) by 20 to clear fractions. 3x + 2y = 6 (3) Solve. We substitute l −2 for w in Equation (1) and solve for l. 2l + 2(l −2) = 44 2l + 2l −4 = 44 19. (1) 4l −4 = 44 4l = 48 l = 12 4x −15y = 220 (4) We use the multiplication principle on both equations again and then add. 45x + 30y = 90 Multiplying (3) by 15 8x −30y = 440 Multiplying (4) by 2 53x = 530 x = 10 Substitute 10 for x in Equation (3) and solve for y. 3 · 10 + 2y = 6 30 + 2y = 6 2y = −24 y = −12 The solution is (10, −12). 0.1x −0.2y = −2.5 (2) Now substitute 12 for l in Equation (2) and find w. w = 12 −2 = 10 Check. The perimeter is 2 12 ft+2 10 ft = 24 ft+20 ft = 44 ft. Also, the width, 10 ft, is 2 ft less than the length, 12 ft. The answer checks. State. The length of the garden is 12 ft, and the width is 10 ft. · · 22. Familiarize. Let x and y represent the amounts invested at 2% and at 3%, respectively. Then the 2% investment earns 0.02x, and the 3% investment earns 0.03y. Translate. Total amount invested s ˛ ¸ x + y x was $5000. = 5000 20. 0.2x + 0.3y = 0.6, (1) Interest earned was $129. s ˛ 0.03y After clearing decimals, we have the following system of equations. x + y = 5000, (1) 2x + 3y = 12, 900 (2) Solve. Multiply both sides of Equation (1) by 2 and then add. −2x −2y = −10, 000 2x + 3y = 12, 900 y = 2900 Substitute 2900 for y in Equation (1) and solve for x. x + 2900 = 5000 x = 2100 Check. Total investment: $2100 + $2900 = $5000 Amount earned at 2%: 0.02($2100) = $42 Amount earned at 3%: 0.03($2900) = $87 Total earnings: $42 + $87 = $129 The answer checks. State. $2100 was invested at 2%, and $2900 was invested at 3%. ¸ x 29 We could begin by clearing the decimals. Instead, we will multiply by 2 on both sides of Equation (2) and then add. 0.2x + 0.3y = 0.6 −0.2x + 0.4y = 5 0.7y = 5.6 y = 8 Substitute 8 for y in Equation (2) and solve for x. 0.1x −0.2(8) = −2.5 0.1x −1.6 = −2.5 0.1x = −0.9 x = −9 The solution is (−9, 8). 21.Familiarize. Let l = the length of the garden, in feet, and let w = the width. Translate. Perimeter is 44 ft. s ˛ 2l + 2w = 44 Width is length less 2 ft. s˛ w = l − We have a system of equations. 2l + 2w = 44, (1) − 0.02x + = 1 − ¸x ¸ x 2 23.Familiarize. Let x = the number of liters of 20% solution and y = the number of liters of 50% solution to be used. The mixture contains 30% (84 L), or 0.3(84 L), or 25.2 L of acid. w = l −2 (2) Copyright· c 2017Pearson Education, Inc.

  13. 88 Chapter 3: Systems of Equations Translate. We organize the information in a table. 20% solution Number of liters Percent of acid Amount of acid Check. When r = 26, then r + 6 = 26 + 6 = 32, and the distance traveled in 5 hr is 32 5, or 160 mi. Also, r 6 = 26 6 = 20, and the distance traveled in 8 hr is 20 8, or 160 mi. Since the distances are the same, the answer checks. State. The speed of the boat in still water is 26 mph. · 50% solution Mixture − − · 84 x y 20% 50% 30% 3 2 25.Graphically: 1. Graph y = 4 x + 2 and y = 5 x −5 and find the point of intersection. The first coordinate of this point is the solution of the original equation. 2. Rewrite the equation as 7 x + 7 = 0. Then graph 7 y = 20 x + 7 and find the x-intercept. The first coordinate of this point is the solution of the original equation. Algebraically: 1. Use the addition and multiplication prin- ciples for equations. 2. Multiply by 20 to clear the fractions and then use the addition and multiplication principles for equations. 0.2x 0.5y 25.2 L We get one equation from the “Number of liters” row of the table. x + y = 84 The last row of the table yields a second equation. 0.2x + 0.5y = 25.2 After clearing decimals, we have the following system of equations. x + y = 84, (1) 2x + 5y = 252 (2) Solve. Multiply both sides of Equation (1) by 2 and then add. −2x −2y = −168 2x + 5y = 252 3y = 84 y = 28 Substitute 28 for y in Equation (1) and solve for x. x + 28 = 84 x = 56 Check. The total amount of the mixture is 56 L + 28 L, or 84 L. The amount of acid in the mixture is 0.2(56 L)+ 0.5(28 L) = 11.2 L + 14 L = 25.2 L. The answer checks. State. 56 L of the 20% solution and 28 L of the 50% solution should be used. 20 26.a) Answers may vary. − x + y = 1, x −y = 7 b) Answers may vary x + 2y = 5, 3x + 6y = 10 c) Answers may vary. x −2y = 3, 3x −6y = 9 27.Answers may vary. Form a linear expression in two vari- ables and set it equal to two different constants. See Ex- ercises 8 and 17 in this review for examples. 28.Answers may vary. Let any linear equation be one equation in the system. Multiply by a constant on both sides of that equation to get the second equation in the system. See Exercises 7 and 18 in this review for examples. 24.Familiarize. Let d = the distance traveled and r = the speed of the boat in still water, in mph. Then when the boat travels downstream its speed is r + 6, and its speed upstream is r −6. The distances are the same. We organize the information in a table. Exercise Set 3.5 Distance Rate Time RC2. (c) Downstream r + 6 5 d RC4. (a) r −6 Upstream 8 d Translate. Using d = rt in each row of the table, we get a system of equations. d = (r + 6)5, d = 5r + 30, (1) or d = (r −6)8 Solve. Substitute 8r 48 for d in Equation (1) and solve for r. 8r −48 = 5r + 30 3r −48 = 30 3r = 78 r = 26 2. 2x −y −4z = −12, (1) 2x + y + z = 1, x + 2y + 4z = 10 (2) (3) 2x −y −4z = −12 (1) 2x + y + z = d = 8r −48 (2) 1 (2) − −3z = −11 (4) 4x −2y −8z = −24 Multiplying (1) by 2 x + 2y + 4z = 10 (3) 4x Adding −4z = −14 (5) 5x Adding Copyright· c 2017Pearson Education, Inc.

  14. Exercise Set 3.5 89 16x −12z = −44 Multiplying (4) by 4 −15x + 12z = 42 Multiplying (5) by −3 x = −2 4(−2) −3z = −11 Substituting in (4) z = 1 8. x + y + z = 0, (1) 2x + 3y + 2z = −3, (2) −x + 2y −3z = −1 (3) −2x −2y −2z = 0 2x + 3y + 2z = −3 (2) y Multiplying (1) by −2 = −3 Substituting in (1) 2(−2) + y + 1 = 1 Substituting in (2) y −3 = 1 y = 4 The solution is (−2, 4, 1). x −y + z = 4, (1) 3x + 2y + 3z = 7, (2) 2x + 9y + 6z = 5 x + z = 3 −x −3z = 5 Substituting in (3) −2z = 8 z = −4 4. x −3 −4 = 0 Substituting in (1) x = 7 The solution is (7, −3, −4). 10. 2x + y −3z = −4, (1) 4x −2y + z = 9, 3x + 5y −2z = 5 (3) 2x −2y + 2z = 8 Multiplying (1) by 2 3x −2y + 3z = 7 (2) 5x + 5z = 15 (4) (2) (3) 9x −9y + 9z = 36 Multiplying (1) by 9 2x + 9y + 6z = 5 (3) 11x + 15z = 41 (5) 2x + y −3z = −4 (1) 12x −6y + 3z = 27 Multiplying (2) by 3 14x −5y = 23 (4) 8x −4y + 2z = 18 Multiplying (2) by 2 3x + 5y −2z = 5 (3) 11x + y = 23 (5) −15x −15z = −45 Multiplying (4) by −3 11x + 15z = 41 −4x = −4 x = 1 5 · 1 + 5z = 15 Substituting in (4) z = 2 14x −5y = 23 (4) 55x + 5y = 115 Multiplying (5) by 5 69x = 138 x = 2 1 −y + 2 = 4 Substituting in (1) y = −1 The solution is (1, −1, 2). 6. 6x −4y + 5z = 31, (1) 5x + 2y + 2z = 13, (2) x + y + z = 2 11 · 2 + y = 23 Substituting in (5) y = 1 4 · 2 −2 · 1 + z = 9 Substituting in (2) z = 3 The solution is (2, 1, 3). (3) 6x −4y + 5z = 31 (1) 4x + 4y + 4z = 8 Multiplying (3) by 4 10x + 9z = 39 (4) 12. 2x + y + 2z = 11, (1) 3x + 2y + 2z = 8, x + 4y + 3z = 0 (2) (3) 5x + 2y + 2z = 13 (2) −2x −2y −2z = −4 Multiplying (3) by −2 3x = 9 x = 3 2x + y + 2z = 11 (1) −2x −8y −6z = 0 Multiplying (3) by −2 −7y −4z = 11 (4) 3x + 2y + 2z = 8 (2) −3x −12y −9z = 0 Multiplying (3) by −3 −10y −7z = 8 (5) 10 · 3 + 9z = 39 Substituting in (4) z = 1 3 + y + 1 = 2 Substituting in (3) y = −2 The solution is (3, −2, 1). Copyright§ c 2017Pearson Education, Inc.

  15. 90 Chapter 3: Systems of Equations 2 · 4 + s + 3 = 6 8 + s + 3 = 6 s + 11 = 6 −49y −28z = 77 Multiplying (4) by 7 40y + 28z = −32 Multiplying (5) by −4 −9y = 45 y = −5 −7(−5) −4z = 11 Substituting in (4) z = 6 Substituting in (1) s = −5 The solution is (4, −5, 3). x + 4y −z = 5, 2x −y + 3z = −5, (2) 4x + 3y + z = 5 18. (1) x + 4(−5) + 3 · 6 = 0 Substituting in (3) x = 2 The solution is (2, −5, 6). 14. 2x + y + 2z = 3, (1) x + 6y + 3z = 4, (2) (3) 3x + 12y −3z = 15 Multiplying (1) by 3 2x − 5x + 11y = 10 (4) y + 3z = −5 (2) 3x −2y + z = 0 −12x −6y −12z = −18 Multiplying (1) by −6 x + 6y + 3z = x + 4y −z = 5 (1) 4x + 3y + z = 5 (3) 5x + 7y (3) = 10 (5) 4 (2) −11x 4x + 2y + 4z = 6 Multiplying (1) by 2 −9z = −14 (4) 5x + 11y = 10 −5x −7y = −10 Multiplying (5) by −1 4y = 0 y = 0 (4) 3x −2y + z = 0 (3) 7x + 5z = 6 (5) 5x + 11 · 0 = 10 Substituting in (4) 5x = 10 x = 2 −55x −45z = −70 Multiplying (4) by 5 63x + 45z = 54 Multiplying (5) by 9 = −16 x = −2 8x 4 · 2 + 3 · 0 + z = 5 8 + z = 5 Substituting in (3) 7(−2) + 5z = 6 Substituting in (5) z = 4 z = −3 The solution is (2, 0, −3). 20. 10x + 6y + z = 7, 2(−2) + y + 2 · 4 = 3 Substituting in (1) (1) y = −1 5x −9y −2z = 3, 15x −12y + 2z = −5 (3) 20x + 12y + 2z = 14 Multiplying (1) by 2 (2) The solution is (−2, −1, 4). 16. 2r + s + t = 6, 3r −2s −5t = 7, r + s −3t = −10 (3) 4r + 2s + 2t = 12 Multiplying (1) by 2 3r −2s −5t = 7 (2) 7r −3t = 19 (4) 2r + s + t = 6 (1) −r −s + 3t = 10 Multiplying (3) by −1 r + 4t = 16 (5) (1) (2) 5x −9y −2z = 3 (2) 25x + 3y = 17 (4) 5x −9y −2z = 3 (2) 15x −12y + 2z = −5 (3) 20x −21y 175x + 21y = 119 Multiplying (4) by 7 = −2 (5) 20x −21y = −2 (5) 195x = 117 7r −3t = 19 −7r −28t = −112 Multiplying (5) by −7 −31t = −93 t = 3 3 5 x = 25.3Σ +3y=17Substitutingin(4) 5 2 r + 4 · 3 = 16 Substituting in (5) r + 12 = 16 r = 4 y = 3 Copyright§ c 2017Pearson Education, Inc.

  16. Exercise Set 3.5 91 10.3Σ +6.2Σ 8.1Σ 2 Substituting1forain(2) 2 +6c=−1 +z=7 Substitutingin(1) 5 3 z = −3 4 + 6c = −1 6c = −5 The solution is . 3 2 −3 Σ . 5 5 , 3 , c = −6 22. 3p + 2r = 11, (1) q −7r = 4, p −6q The solution is . 1 2 −5 Σ . (2) 2 , 3 , 6 = 1 (3) a − 26. 5c = 17, b + 2c = −1, (2) 4a −b −3c = 12 (1) 6q −42r = 24 Multiplying (2) by 6 p −6q = 1 (3) p −42r = 25 (4) 3p + 2r = 11 (1) (3) b + 2c = −1 (2) 4a −b −3c = 12 (3) 4a −c = 11 (4) −3p + 126r = −75 Multiplying (4) by −3 128r = −64 1 r = −2 3p+2. =11Substitutingin(1) a −5c = 17 (1) −20a + 5c = −55 Multiplying (4) by −5 −19a = −38 a = 2 −1Σ 2 p = 4 Σ =4 q−7. 2 −5c = 17 Substituting in (1) c = −3 1 −2 Substitutingin(2) 1 q = 2 b + 2(−3) = −1 Substituting in (2) b = 5 Thesolutionis. 41−1Σ . , 2 , 2 The solution is (2, 5, −3). 28. F = 3ab F = a 3b F = 1 t(c −d) 2F = tc −td 2F −tc = −td 2F −tc = d, or −t tc −2F t 2F c = d t 32. Ax + By = c By = c −Ax 24. 4a + 9b = 8, (1) + 6c = −1, (2) 6b + 6c = −1 (3) Note that there is no c in Equation (1). We will use equa- tions (2) and (3) to obtain another equation with no c terms. 8a + 6c = −1 (2) −6b −6c = 1 Multiplying (3) by −1 8a −6b = 0 (4) Adding −8a −18b = −16 Multiplying (1) by −2 8a −6b = 0 (4) −24b = −16 2 b = 3 8a−6.2Σ =0 3 8a −4 = 0 8a = 4 1 a = 2 8a Dividing by 3b 30. 2 = d, or − Substituting2forbin(4) c −Ax B 5 y = 3 2 34. y = −3 x −4 The equation is in slope-intercept form, y = mx + b. The slopeis−2, and the y-interceptis. 0,−5Σ . 3 4 Copyright§ c 2017Pearson Education, Inc.

  17. 92 Chapter 3: Systems of Equations 4. Let x, y, and z represent the measures of angles A, B, and C, respectively. Solve: x + y + z = 180, 2x −5y = 10 −5y = −2x + 10 1 −5 (−5y) = −5 (−2x + 10) 2 y = 5 x −2 The equation is now in slope-intercept form, y = mx + b. 2 The slope is 5 , and the y-intercept is (0, −2). 38. w + x −y + z = 0, (1) w −2x −2y −z = −5, (2) w −3x −y + z = 4, (3) 2w −x −y + 3z = 7 (4) Pairing equations (1) and (2), (2) and (3), and (2) and (4) and eliminating z, we have 2w −x −3y = −5, (5) 2w −5x −3y = −1, (6) 5w −7x −7y = −8. (7) 36. 1 y = 2x, z = 80 + x The solution is (25, 50, 105), so the measures of angles A, B, and C are 25◦, 50◦, and 105◦, respectively. 6. We know the thousands digit must be 1. The United States did not exist as a union until the 1700’s, and the first U. S. transcontinental railroad was completed before 2000. 1 x y z We let x represent the hundreds digit, y the tens digit, and z the ones digit. Since z is a multiple of 3, we know that z must be 3, 6, or 9. We have the following possibilities: 1 x y 3 1 x y 6 1 x y 9 We also know that z is one more than x. When z = 3, x = 2. When z = 6, x = 5. When z = 9, x = 8. Now the possibilities are: 1 2 y 3 1 5 y 6 1 8 y 9 Only 1 8 y 9 is a reasonable possibility. 2w −x −3y = −5 (5) −2w + 5x + 3y = 1 Multiplying (6) by −1 4x = −4 x = −1 Substituting −1 for x in (5) and (7), we have 2w −3y = −6, (8) 5w −7y = −15. (9) Solving the system composed of (8) and (9), we have w = −3 and y = 0. −3 −1 −0 + z = 0 Substituting in (1) z = 4 The solution is (−3, −1, 0, 4). 1 + 8 + y + 9 = 24 y = 6 Thus, the year is 1869. 8. Let x, y, and z represent the number of tall, grande and venti coffees served, respectively. Solve: x + y + z = 50, 1 75 + 1 95 + 2 25 = 98 70 . x . y . z 12x + 16y + 20z = 10(80) The solution is (12, 26, 12), so Brandie served 12 tall cof- fees, 26 grande coffees, and 12 venti coffees. . , Exercise Set 3.6 10. Let x, y, and z represent the prices of a children’s book, a paperback, and a hardback, respectively. Solve: 22x + 10y + 5z = 63.50, 12y + 15z = 52.50, 8x + 6z = 29 The solution is (1.75, 1.25, 2.50), so a children’s book costs $1.75, a paperback costs $1.25, and a hardback costs $2.50. RC2. (d) RC4. (b) 2. Let x, y, and z represent the fat content of a Big Mac, a medium order of fries, and a 12-oz vanilla milkshake, respectively. Solve: x + y + z = 59, x = z + 13, y + z = x + 3 The solution is (28, 16, 15), so a Big Mac has 28 g of fat, the fries have 16 g of fat, and the milkshake has 15 g of fat. 12. Let x, y, and z represent the prices of upgrading the pro- cessor, the memory, and the graphics card, respectively. Solve: 480 + x + y = 745, 480 + y + z = 690, 480 + x + z = 805 The solution is (190, 75, 135), so the processor upgrade costs $190, the memory upgrade costs $75, and the up- grade to the graphics card costs $135. Copyright§ c 2017Pearson Education, Inc.

  18. Exercise Set 3.6 93 14. Let x, y, and z represent the number of calories in the sandwich, the soup, and the cookie respectively. Solve: x + y + z = 580, x = y + z −20, z = y + 120 The solution is (280,90,210), so the sandwich has 280 calo- ries, the soup has 90 calories, and the cookie has 210 calo- ries. 24. 16. Let x, y, and z represent the number of servings of roast beef, baked potato, and asparagus, respectively. Solve: 300x + 100y + 50z = 800 (Calories) 20x + 5y + 5z = 55 (protein) 20y + 44z = 220 (vitamin C) . 9 , 11 , 15 Σ 8 4 4 3 servings of roast beef, 24 baked potatoes, and 34 servings of asparagus. 26. It is possible for a vertical line to intersect the graph more than once, so this is not the graph of a function. 28. There is no vertical line that can intersect the graph more than once, so this is the graph of a function. 30. Let T , G, and H represent the number of tickets Tom, Gary, and Hal begin with, respectively. After Hal gives tickets to Tom and Gary, each has the following number of tickets: Tom: T + T, or 2T, Gary: G + G, or 2G, , so the meal should have 11 The solution is 8 3 18. Let x, y, and z represent the amount of the Perkins loan, the Stafford loan, and the bank loan, respectively. Solve: x + y + z = 32, 000, 0.05x + 0.04y + 0.07z = 1500, 0.05x = 0.07z + 220 The solution is (10, 000, 18, 000, 4000), so Terrance has a Perkins loan for $10,000, a Stafford loan for $18,000, and a bank loan for $4000. Hal: H −T −G. After Tom gives tickets to Gary and Hal, each has the following number of tickets: Gary: 2G + 2G, or 4G, (H −T −G) + (H −T −G), or 2(H −T −G), 2T −2G −(H −T −G), or 3T −H −G After Gary gives tickets to Hal and Tom, each has the following number of tickets: 2(H −T −G) + 2(H −T −G), or 4(H −T −G) Tom: (3T −H −G) + (3T −H −G), or 2(3T −H −G), Gary: 4G −2(H −T −G) −(3T −H −G), or 7G −H −T. Since Hal, Tom, and Gary each finish with 40 tickets, we write the following system of equations: 4(H −T −G) = 40, 2(3T −H −G) = 40, 7G −H −T = 40 Solving the system we find that T = 35, so Tom started with 35 tickets. Hal: 20. Let x, y, and z represent the number of 2-point field goals, 3-point field goals, and 1-point foul shots made, respec- tively. The total number of points scored from each of these types of goals is 2x, 3y, and z. Solve: 2x + 3y + z = 92, x + y + z = 50, x = 19 + z The solution is (32, 5, 13), so the Knicks made 32 two-point field goals, 5 three-point field goals, and 13 foul shots. Tom: Hal: 22. Let x, y, and z represent the number of orders that can be processed alone by Steven, Teri, and Isaiah, respectively. Solve: x + y + z = 740, x + y = 470, y + z = 520 The solution is (220, 250, 270), so Steven, Teri, and Isaiah can process 220, 250, and 270 orders, respectively, alone. Copyright§ c 2017Pearson Education, Inc.

  19. 94 Chapter 3: Systems of Equations 32. Let w, x, y, and z represent the ages of Tammy, Carmen, Dennis, and Mark respectively. Solve: w = x + y, x = 2 + y + z, y = 4z, w + x + y + z = 42 The solution is (20, 12, 8, 2), so Tammy is 20 years old. 18. y 4 2 —4 —2 x 2 4 —2 — 4 2x —5y < 10 20. Exercise Set 3.7 y 42y + x “ 4 2. Graph (c) is the graph of y < −2x. 4. Graph (a) is the graph of y ≥ x + 5. 6. Graph (d) is the graph of 3x + y < −6. 8. Graph (e) is the graph of 3x −9y < 9. 10. y 2 —4 —2 x 2 4 —2 — 4 22. y 4 2 2 —4 —2 x 2 4 2y < x —4 —2 x 2 4 —2 —2 — 4 2x —6y “8 + 2y 12. y 24. 4 y —x < 0 2 y —4 —2 x 2 4 4 —2 — 4 2 —4 —2 2 4 x —2 14. y —4 y > —3 26. y “x + 4 4 2 y x —4 —2 2 4 4 x “5 —2 2 —4 —4 —2 x 2 4 —2 — 4 y 16. y 28. 4 x —y “5 2 4 —4 —2 x 2 4 2 —2 — 4 —2 2 — 4 x 4 —2 — 4 —3 “x “3 Copyright§ c 2017Pearson Education, Inc.

  20. Exercise Set 3.7 95 30. 44. Graph: y ≥ x, y y ≤ 5 −x 4 2 y —4 —2 x 2 4 4 (5 , 5) 2 2 —2 — 4 2 x —4 —2 2 4 y “¦x + 2¦ —2 —4 32. Graph (c) is the correct graph. 34. Graph (d) is the correct graph. 46. Graph: y ≥ x, y ≤ 2 −x 36. Graph (e) is the correct graph. y 38. First we find the related equations. One line goes through (−1, 2) and (0, 0). We find its slope: = 0 −2 = 2 0 −(−1) Then the equation of the line is y = −2x +0, or y = −2x. The other line goes through (0, 3) and (3, 0). We find its slope: m = 0 −(−3) = 3 = 1 3 −0 3 Then the equation of this line is y = x −3. Observing the shading on the graph and the fact that the lines are solid, we can write the system of inequalities as y ≤ −2x, y ≥ x −3. Answers may vary. 4 2 − m (1, 1) 2 4 —4 —2 x —2 — 4 − 48. Graph: y ≤ −2, x ≥ 2 y 4 2 —4 —2 x 2 4 —2 40. The equation of the vertical line is x = 1 and the equa- tion of the horizontal line is y = 2. The lines are dashed and the shaded area is to the right of the vertical line and below the horizontal line, so the system of inequalities can be written x > −1, y < 2. − (2, —2) — 4 50. Graph: x ≥ −2, y ≤ 3 −2x y 8 42. One line goes through (−2, 0) and (0, 2). We find its slope: m = 2 −0 = 2 = 1 0 −(−2) 2 Then the equation of this line is y = x + 2. The other line is the vertical line x = 2. Observe that both lines are solid and that the shading lies below y = x + 2, to the left of x = 2, to the right of the y-axis, and above the x-axis. We can write this system of inequalities as y ≤ x + 2, x ≤2, x ≥0, y ≥0. (—2, 7) 6 4 2 —4 —2 x 2 4 —2 52. Graph: y + 3x ≥ 0, y + 3x ≤ 2 y 2 1 —2 —1 x 1 2 —1 —2 Copyright§ c 2017Pearson Education, Inc.

  21. 96 Chapter 3: Systems of Equations 62. We graph the system of inequalities and find the vertices: y ≤ 2x + 1, y ≥ −2x + 1, x −2 ≤ 0 54. Graph: y 5 4, 4) (— 5 (3, 4) 4 y (2, 5) 3 2 4 2 5) 4 (3, — (0, 1) 1 2 4 x —4 —2 —2 — 4 (2, — 3) 5 x 1 2 3 4 VertexQ = 28x −4y + 72 . 4 , 4 5 (3, 4) . 3,5Σ 4 Σ 782 ←− Minimum 5 140 56. Graph: x + 2y ≤12, 2x + y ≤12, x ≥ 0, y ≥ 0 151 ←− Maximum y 64. We graph the system of inequalities and find the vertices: 6 (0, 6) y (4, 4) 6 (0, 5.8) 4 (6, 0) 2 4 6 5 —2 x 4 —2 (2, 3) 3 58. Graph: 8x + 5y ≤ 40, 2 1 x + 2y ≤8, x ≥0, y ≥ 0 5 x 1 2 3 (4, 0) 4 (0, 0) VertexG = 16x + 14y (0, 0) (0, 5.8) (2, 3) 74 (4, 0) 64 y 0 ←− Minimum 81.2 ←− Maximum 8 6 4 40,— 24 (— 11 11 (0,4) 2 (0, 0) —2 — 2 4 6 8 x (5, 0) 66. Let x = the number of jumbo biscuits and y = the number of regular biscuits to be made per day. The income I is given by I = 0.45x + 0.30y subject to the constraints x + y ≤200, 2x + y ≤300, x ≥0, y ≥0. We graph the system of inequalities, determine the ver- tices, and find the value if I at each vertex. 60. Graph: y −x ≥1, y −x ≤3, 2 ≤ x ≤ 5 y (5, 8) 8 6 (5, 6) (2, 5) 4 (2, 3) 2 2 4 6 8 x Copyright§ c 2017Pearson Education, Inc.

  22. Exercise Set 3.7 97 50x + 20y ≥ 120, 8x + 6y ≥24, 5x + 8y ≥10, y 300 (0, 200) 200 x ≥0, y ≥0. (100, 100) 100 (150, 0) y 6 A 300 x 100 200 (0, 0) 5 Vertex (0, 0) (0, 200) (100, 100)0.45(100) + 0.30(100) = 75 (150, 0) 0.45(150) + 0.30(0) = 67.50 I = 0.45x + 0.30y 0.45(0) + 0.30(0) = 0 0.45(0) + 0.30(200) = 60 4 3 B 2 1 C 3 x 1 2 The company will have a maximum income of $75 when 100 of each type of biscuit are made. Vertex A(0, 6) B.12,12Σ 7 7 C(3, 0) The minimum cost of $513 is achieved by using 12 , or 15 7 sacks of soybean meal and 12 C = 20x + 10y 20 · 0 + 10 · 6 = 60 12 20 · 7 + 10 · 7 = 51 7 20 · 3 + 10 · 0 = 60 68. Let x = the number of units of lumber and y = the number of units of plywood produced per week. The profit P is given by P = 25x + 38y subject to the constraints x + y ≤ 400, 12 3 5 7 sacks of alfalfa. 7 7 , or 1 7 x ≥ 100, y ≥ 150. 72. Let x = the amount invested in City Bank and y = the amount invested in People’s Bank. Find the maximum value of I = 0.025x + 0.0175y subject to x + y ≤ 22, 000, 2000 ≤ x ≤ 14, 000 0 ≤ y ≤ 15, 000. We graph the system of inequalities, determine the vertices and find the value of P at each vertex. y 400 300 (100, 300) 200 (250, 150) y (100, 150) (2000, 15,000) 100 (7000, 15,000) (14,000, 8,000) 400 x 100 200 300 Vertex (100, 150) (100, 300) (250, 150) P = 25x + 38y 25 · 100 + 38 · 150 = 8200 25 · 100 + 38 · 300 = 13, 900 25 · 250 + 38 · 150 = 11, 950 x (2000, 0) (14,000, 0) The maximum profit of $13,900 is achieved by producing 100 units of lumber and 300 units of plywood. Vertex (2000, 0) (2000, 15, 000) 312.50 (7000, 15, 000) 437.50 (14, 000, 8, 000) 490 (14, 000, 0) The maximum interest income is $490 when $14,000 is invested in City Bank and $8000 is invested in People’s Bank. I = 0.025x + 0.0175y 50 70. Let x = the number of sacks of soybean meal to be used and y = the number of sacks of alfalfa. Find the minimum value of C = 20x + 10y subject to the constraints 350 Copyright§ c 2017Pearson Education, Inc.

  23. 98 Chapter 3: Systems of Equations 74. Let x = the number of P2 airplanes and y = the number of P3 airplanes to be used. Find the minimum value of C = 10x + 15y subject to 80x + 40y ≥2000, 30x + 40y ≥1500, 40x + 80y ≥2400, x ≥0, y ≥0. 78. Let x = the number of teachers and y = the number of teacher’s aides. Find the minimum value of C = 53, 000x + 23, 600y subject to x + y ≤50, x + y ≥20, y ≥12, x ≥2y. y y 60 50 (0, 50) ( 100 50 ) — , —— 3 3 40 (24, 12) (10, 30) 30 (38, 12) 20 (30, 15) 10 (60, 0) 10 20 30 40 50 60 70 80 x x Vertex (24, 12) . 100 , 50 Σ 3 (38, 12) The minimum cost of $1,555,200 is achieved when 24 teach- ers and 12 teacher’s aides are hired. C = 53, 000x + 23, 600y 1, 555, 200 VertexC = 10x + 15y (0, 50) (10, 30) (30, 15) (60, 0) 10 · 60 + 15 · 0 = 600 10 · 0 + 15 · 50 = 750 10 · 10 + 15 · 30 = 550 10 · 30 + 15 · 15 = 525 2, 160, 000 3 2, 297, 200 The minimum cost of $525 thousand is achieved using 30 P2’s and 15 P3’s. 80. Let x = the number of animal A and y = the number of animal B. Find the maximum value of T = x + y subject to x + 0.2y ≤ 1080, 0.5x + y ≤ 810, x ≥0, y ≥0. 76. Let x = the number of smaller gears and y = the number of larger gears produced each day. Find the maximum value of P = 45x + 30y subject to 4x + y ≤24, x + y ≤9, x ≥0, y ≥0. y 6000 y 5000 4000 3000 2000 30 20 (1020, 300) 1000 (0, 810) 10 (0, 9) (1080, 0) (5, 4) (6, 0) (0, 0) x 500 1000 1500 2000 (0, 0) 10 x 2 4 6 8 Vertex (0, 0) (0, 810) (1020, 300) 1020 + 300 = 1320 (1080, 0) 1080 + 0 = 1080 T = x + y 0 + 0 = 0 0 + 810 = 810 VertexP = 45x + 30y (0, 0) 0 (0, 9) 270 (5, 4) 345 (6, 0) 270 The maximum total number of 1320 is achieved when there are 1020 of A and 300 of B. The maximum profit of $345 per day is achieved when 5 smaller gears and 4 larger gears are produced. 82. |x −3|≥ 2 Copyright§ c 2017Pearson Education, Inc.

  24. Chapter 3 Summary and Review: Study Guide 99 x −3 ≤ −2 or x −3 ≥2 x ≤ 1 or Thesolutionsetis{x|x≤1orx≥5},or (−∞, 1] ∪[5, ∞). x −1 > 4 Rational inequality x + 2 x −1 4 x + 2 x −1 4 = 0 Related equation x + 2 The denominator of 90. y x ≥5 4 2 —4 —2 x 2 4 84. —2 — 4 0 −> ¦x —y¦ > 0 Chapter 3 Vocabulary Reinforcement − ( ) = x −1 f x 4 is 0 when = 2, x + 2 − − x so the function is not defined for x = 2. We solve the related equation f (x) = 0. x −1 4 = 0 x + 2 x −1 −4(x + 2) = 0 Multiplying by x + 2 x −1 −4x −8 = 0 −3x −9 = 0 −3x = 9 x = −3 Thus, the critical values are 3 and 2. They divide the x- axis into the intervals ( , 3), ( 3, 2), and ( 2, ). We test a value in each interval. (−∞, −3): f (−4) = −3 < 0 − 1.A solution of a system of two equations in two variables is an ordered pair that makes both equations true. − 2.A consistent system of equations has at least one solution. 3.A solution of a system of three equations in three variables is an ordered triple that makes all three equations true. 4.If, for a system of two equations in two variables, the graphs of the equations are different lines, then the equa- tions are independent. − − 5.The graph of an inequality like x> 2y is a half-plane. −∞− − − −∞ Chapter 3 Concept Reinforcement 2 (−3, −2): f (−2.5) = 3 > 0 (−2, ∞): f (0) = −9 < 0 Function values are positive on (−3, −2). This can also be determined from the graph of = x −1 set is {x|− 3 <x< −2}, or (−3, −2). 86. y < x + 1 y ≥ x2 1.False; see page 155 in the text. 2 2.True; see page 155 in the text. 3.The point (0, b) is the y-intercept of a linear equation. If both equations contain the point (0, b), then the equations have the same y-intercept. − 4. The solution y x + 2 4.The graph of x = 4 is a vertical line and the graph of y = 4 is a horizontal line, so the lines have a point of intersection. Thus the system of equations x = 4 and y = −4 is consistent. The given statement is false. − y 4 2 Chapter 3 Study Guide —4 —2 x 2 4 —2 — 4 1. Graph both lines on the same set of axes. y 88. y 5 x + y= 3 4 2 3 x + 3y= 1 1 2 1 —2 —1 x 1 2 x —5 —4 —3 —2 —1 1 2 3 4 5 (4,–1) —1 — 1 ¦x¦ + ¦y¦ “ 1 — 2 —2 — 3 — 4 — 5 The solution appears to be (4, −1). Copyright§ c 2017Pearson Education, Inc.

  25. 100 Chapter 3: Systems of Equations Check: 6% loan x 5% loan y Total x + 3y = 1 x + y = 3 Principal Interest rate Time Interest $23,000 4+ 3(−1) ? 1 4+ (−1) ? 3 4 −3 . 3 . TRUE 6% 5% 1 TRUE 1 yr 0.06x 1 yr 0.05y . $1237 is (4, −1). The solution Since the system of equations has a solution, it is consis- tent. Since there is exactly one solution, the equations are independent. The “Principal” row of the table gives us one equation: x + y = 23, 000. The “Interest” row of the table gives us a second equation: 0.06x + 0.05y = 1237. After clearing decinmals, we have the following system of equations. x + y = 23, 000 (1) 6x + 5y = 123, 700 (2) Solve. We multiply Equation (1) by −5 and then add. −5x −5y = −115, 000 6x + 5y = 123, 700 x = 8700 Substitute 8700 for x in Equation (1) and solve for y. 8700 + y = 23, 000 y = 14, 300 Check. $8700 + $14, 300 = $23, 000 and 0.06($8700) + 0.05($14, 300) = $522 + $715 = $1237, so the answer checks. State. $8700 was invested at 6%, and $14,300 was invested at 5%. 2. 2x + y = 2, (1) 3x + 2y = 5 We will use the substitution method. First solve Equation (1)for y. 2x + y = 2 y = −2x + 2 (3) Now substitute 2x + 2 for y in Equation (2) and solve for x. 3x + 2(−2x + 2) = 5 3x −4x +4 = 5 −x +4 = 5 −x = 1 x = −1 Substitute −1 for x in Equation (3) and find y. y = −2(−1)+2 = 2+2 = 4 The solution is (−1, 4). 3. 2x + 3y = 5, (1) 3x + 4y = 6 (2) We will use the elimination method. Use the multiplication principle on both equations and then add. 8x + 12y = 20 Multiplying (1) by 4 −9x −12y = −18 −x = 2 x = −2 Substitute 2 for x in one of the original equations and solve for y. 2(−2)+ 3y = 5 Using Equation (1) −4+ 3y = 5 3y = 9 y = 3 The solution is (−2, 3). 4. Familiarize. We let x and y represent the amounts of the investments. We will use the simple interest formula, I = Prt. Translate. We organize the information in a table. (2) − x −y + z = 9, 2x + y + 2z = 3, 4x + 2y −3z = −1 (3) We will start by eliminating y from two different pairs of equations. x −y + z = 9 (1) 2x + y + 2z = 3 (2) 3x + 3z = 12 (4) 5. (1) (2) Multiplying (2) by −3 − 2x −2y + 2z = 18 4x + 2y −3z = −1 6x −z = 17 Now solve the system of Equations (4) and (5). 3x + 3z = 12 (4) 18x −3z = 51 Multiplying (5) by 3 21x = 63 x = 3 Multiplying (1) by 2 (3) (5) 3 · 3+ 3z = 12 Substituting 3 for x in (4) 9+ 3z = 12 3z = 3 z = 1 Copyright§ c 2017Pearson Education, Inc.

  26. Chapter 3 Summary and Review: Review Exercises 101 2 3+ y +2 1 = 3 Substituting 3 for x and 1 for z in (2) · · Graph the system of inequalities and determine the ver- tices. 6+ y +2 = 3 8+ y = 3 y = −5 The solution is (3, −5, 1). 6. Graph: 3x −2y > 6 We first graph the line 3x 2y = 6. We draw the line dashed since the inequality symbol is >. To determine which half-plane to shade, test a point not on the line. We try (0, 0). 3x −2y > 6 3 · 0 −2 · 0 ? 6 − Vertex A: We solve the system x + 2y = 8 and y = 4. The coordinates of point A are (0, 4). Vertex B: We solve the system x coordinates of point B are (6, 4). Vertex C: We solve the system x y = 2 and x + 2y = 8. The coordinates of point C are (4, 2). Evaluate the objective function P at each vertex. 0 −0 . 0 FALSE . y = 2 and y = 4. The − Since 0 > 6 is fals contain (0, 0). e, we shade the half-plane that does not − y Vertex P = 4x −5y A(0, 4) B(6, 4) C(4, 2) 4 · 4 −5 · 2 = 6 4 · 0 −5 · 4 = −20 4 · 6 −5 · 4 = 4 x 6 x —2y > 3 The maximum value of P is 6 when x = 4 and y = 2. 7. Graph: x −2y ≤ 4, (1) x + y ≤ 4, (2) x −1 ≥ 0 Graph the lines x 2y = 4, x + y = 4, and x 1 = 0 (or x = 1) using solid lines. Indicate the region for each inequality by arrows and shade the region where they over- lap. Chapter 3 Review Exercises (3) − − 1.Graph both lines on the same set of axes. y 5 x – y= – 3 y 4 4x – y = – 9 3 2 (–2, 1) (1, 3) 1 x —5 —4 —3 —2 —1 1 2 3 4 5 — 1 3) 2 x — (1,—— 2 (4, 0) — 3 — 4 — 5 To find the vertices we solve three different systems of equations. From (1) and (2) we obtain the vertex (4, 0). From(1)and(3)weobtainthevertex. and (3) we obtain the vertex (1, 3). 8.Find the maximum value of P = 4x −5y subject to x −y ≤2, x + 2y ≥8, y ≤4. The solution (point of intersection) seems to be the point (−2, 1). Check: 4x −y = −9 4(−2) −1 ? −9 −8 −1 . 1,−3Σ . From(2) x −y = −3 −2 −1 ? −3 −3 . 2 TRUE −9 . TRUE The solution is (−2, 1). Copyright§ c 2017Pearson Education, Inc.

  27. 102 Chapter 3: Systems of Equations Substitute x + 2 for y in the second equation and solve for x. y −x = 8 (2) x +2 −x = 8 2 = 8 We get a false equation. There is no solution. Since the system of equations has a solution it is consis- tent. Since there is exactly one solution, the equations are independent. 2.Graph both lines on the same set of axes. y 5 6. 7x −4y = 6, (1) y −3x = −2 (2) First solve the second equation for y. y −3x = −2 y = 3x −2 (3) Now substitute 3x 2 for y in the first equation and solve for x. 7x −4y = 6 7x −4(3x −2) = 6 7x −12x +8 = 6 −5x +8 = 6 −5x = −2 2 x = 5 Next substitute 2 for x in Equation (3) and find y. 5 2 6 10 y = 3 · 5 −2 = 5 −5 = −5 Since checks, it is the solution. 5 5 x + 3y = −3, (1) 2x −3y = 21 (2) 3x = 18 Adding x = 6 Substitute 6 for x in Equation (1) and solve for y. 6+ 3y = −3 3y = −9 y = −3 The solution is (6, −3). 8. 3x −5y = −4, (1) 5x −3y = 4 (2) We multiply Equation (1) by 3 and Equation (2) by and then add. 9x −15y = −12 −25x + 15y = −20 −16x x = 2 Now substitute 2 for x in one of the equations and solve for y. 3x −5y = −4 (1) 3 · 2 −5y = −4 6 −5y = −4 −5y = −10 y = 2 4 3 2 1 (2) —5 —4 —3 —2 —1 x 1 2 3 4 5 1 — — 2 15x + 10y = –2 3x + 2y =–4 0 − — 3 — 4 — 5 (1) The graphs are the same. Any solution of one of the equa- tions is also a solution of the other. Each equation has an infinite number of solutions. Thus the system of equations has an infinite number of solutions. Since the system of equations has a solution, it is consis- tent. Since there are infinitely many solutions, the equa- tions are dependent. 3.Graph both lines on the same set of axes. 4 . 2 , −4 Σ y 5 4 y – 2x = 4 y –2x= 5 3 7. 2 1 x —5 —4 —3 —2 —1 1 2 3 4 5 1 — — 2 — 3 — 4 — 5 The lines are parallel. There is no solution. Since the system of equations has no solution, it is in- consistent. Since there is no solution, the equations are independent. 4. 2x −3y = 5, (1) x = 4y + 5 (2) Substitute 4y + 5 for x in Equation (1) and solve for y. 2(4y + 5) −3y = 5 8y + 10 −3y = 5 5y + 10 = 5 5y = −5 y = −1 Substitute −1 for y in Equation (2) and find x. x = 4(−1)+5 = −4 + 5 = 1 The solution is (1, −1). 5. y = x + 2, (1) y −x = 8 (2) 5 − = −32 Copyright§ c 2017Pearson Education, Inc.

  28. Chapter 3 Summary and Review: Review Exercises 103 Since (2, 2) checks, it is the solution. We have a system of equations: x + y = 45, 8.50x + 9.75y = 398.75 We can multiply the second equation on both sides by 100 to clear the decimals: x + y = 45, 850x + 975y = 39, 875 (2) Solve. We solve the system of equations using the elim- ination method. Begin by multiplying Equation (1) by −850. −850x −850y = −38, 250 Multiplying (1) 850x + 975y = 39, 875 9. 1 x + 2 y = 1, 3 3 2 x + 2 y = 6 We multiply by 9 on both sides of the first equation and by 2 on both sides of the second equation to clear the fractions. 9 1 (1) 3x + 2y = 9, (1) 3x + y = 12 (2) Now we multiply both sides of Equation (2) by 1 and then add. 3x + 2y = 9 −3x −y = −12 y = −3 Substitute −3 for y in Equation (2) and solve for x. 3x + (−3) = 12 3x = 15 x = 5 The solution is (5, −3). − 125y = y = 1625 13 Substitute 13 for y in (1) and solve for x. x + 13 = 45 x = 32 Check. The number of brushes sold is 32 + 13, or 45. The amount taken in was $8.50(32) + $9.75(13) = $272 + $126.75 = $398.75. The answer checks. State. 32 of the less expensive brushes were sold, and 13 of the more expensive brushes were sold. 10. 1.5x −3 = −2y, 3x + 4y = 6 First we rewrite the first equation in the form Ax + By = C. 1.5x + 2y = 3, (1) 3x + 4y = 6 (2) Now multiply Equation (1) by −2 and then add. −3x −4y = −6 3x + 4y = 6 12.Familiarize. Let x and y represent the number of liters of Orange Thirst and Quencho that should be used, re- spectively. The amount of orange juice in the mixture is 10%(10 L), or 0.1(10 L), or 1 L. Translate. We organize the information in a table. Orange Thirst Number of liters Percent of juice Liters of juice The first row of the table gives us one equation. x + y = 10 The last row yields a second equation. 0.15x + 0.05y = 1 After clearing decimals we have the following system of equations. x + y = 10, (1) 15x + 5y = 100 (2) Solve. We use the elimination method. First we multiply Equation (1) by −5 and then add. −5x −5y = −50 15x + 5y = 100 10x = 50 x = 5 Now substitute 5 for x in one of the equations and solve for y. Quencho Mixture x y 10 0 = 0 15% 5% 10% We get an equation that is true for all numbers x and y. The system has an infinite number of solutions. 0.15x 0.05y 1 11.Familiarize. Let x = the number of less expensive brushes sold and y = the number of more expensive brushes sold. Translate. We organize the information in a table. Less expen- sive More expen- sive Kind of brush Total Number sold Price Amount taken in x y 45 $8.50 $9.75 8.50x 9.75y 398.75 The “Numbersold” row of the table gives us one equation: x + y = 45 The “Amount taken in” row gives us a second equation: 8.50x + 9.75y = 398.75 Copyright§ c 2017Pearson Education, Inc.

  29. 104 Chapter 3: Systems of Equations x + y = 10 (1) 5+ y = 10 y = 5 Check. If 5 L of each type of juice are used, then the mixture contains 5 + 5, or 10 L. The amount of orange juice in the mixture is 0.15(5) + 0.05(5), or 0.75 + 0.25, or 1 L. The answer checks. State. 5 L of Orange Thirst and 5 L of Quencho should be used. 2x −y + z = 8 3x + y + 4z = 2 5x + 5z = 10 (5) Now solve the system of Equations (4) and (5). We mul- tiply Equation (5) by −1 and then add. 5x + 3z = 26 (4) (2) (3) −5x −5z = −10 −2z = 16 z = −8 13.Familiarize. We first make a drawing. 5x + 3(−8) = 26 Substituting −8 for z in (4) 5x −24 = 26 5x = 50 x = 10 Slow train d miles . Fast train d miles . (t + 1) hr✲ 44 mph t hr 52 mph ✲ 3 · 10 + y + 4(−8) = 2 Substituting 10 for x and 30 +y −32 = 2 y −2 = 2 y = 4 We obtain (10, 4, −8). This checks, so it is the solution. 15. 3x + 2y + z = 1, (1) −8 for z in (3) From the drawing we see that the distances are the same. We organize the information in a table. · Time d = r Rate t Distance d = 44(t+1) → Slow train Fast train d t +1 44 d t 52 → d =52t 2x −y −3z = 1, (2) −x + 3y + 2z = 6 (3) We start by eliminating x from two different pairs of equa- tions. 3x + 2y + z = 1 −3x + 9y + 6z = 18 Multiplying (3) by 3 11y + 7z = 19 (4) Translate. Using d = rt in each row of the table, we get a system of equations: d = 44(t + 1), d = 52t Solve. We solve the system of equations. 52t = 44(t + 1) Using substitution 52t = 44t + 44 8t = 44 11 1 t = , or 5 2 2 1 Check. At 52 mph, in 52 hr the fast train will travel 11 52 · 2 , or 286 mi. At 44 mph, in 52 + 1, or 62 hr, the slow train travels 44 13 , or 286 mi. Since the distances 2 are the same, the answer checks. State . The second train will travel 52 hr before it over- takes the first train. (1) 2x −y −3z = 1 −2x + 6y + 4z = 12 Multiplying (3) by 2 5y + z = 13 (5) (2) Now we solve the system of Equations (4) and (5). 11y + 7z = 19 (4) 1 1 −35y −7z = −91 Multiplying (5) by −7 −24y y = 3 5 · 3+ z = 13 Substituting 3 for y in (5) = −72 · 1 15 + z = 13 z = −2 −x +3 · 3+ 2(−2) = 6 Substituting 3 for y and 14. x + 2y + z = 10, (1) 2x −y + z = 8, (2) 3x + y + 4z = 2 (3) We start by eliminating y from two different pairs of equa- tions. x + 2y + z = 10 (1) −2 for z in (3) −x +9 −4 = 6 −x + 5 = 6 −x = 1 x = −1 4x −2y + 2z = 16 Multiplying (2) by 2 5x + 3z = 26 (4) We obtain (−1, 3, −2). This checks, so it is the solution. Copyright§ c 2017Pearson Education, Inc.

  30. Chapter 3 Summary and Review: Review Exercises 105 2 +y+2. −2Σ 3 Substituting2forxand 2x −5y −2z = −4, (1) 7x + 2y −5z = −6, (2) −2x + 3y + 2z = We start by eliminating x from two different pairs of equa- tions. 14x −35y −14z = −28 Multiplying (1) by 7 −14x −4y + 10z = 12 Multiplying (2) by −2 −39y −4z = −16 (4) =1 16. 2 −3 for z in (1) 4 (3) 2+ y −4 = 1 3 2 y + 3 = 1 1 3 y = . We obtain 2, 1 , −2 Σ 18.Familiarize. Let a, b, and c represent the measures of angles A, B, and C, respectively. Recall that the sum of the measures of the angles of a triangle is 180◦. Translate. T he sum of th e measures is 180◦. s ˛ ¸ b + c is 4 times · The measure of angle B 3 3 2x −5y −2z = −4 (1) −2x + 3y + 2z = 4 (3) −2y . This checks, so it is the solution. = 0 y = 0 Substitute 0 for y in Equation (4) and solve for z. −39 · 0 −4z = −16 −4z = −16 z = 4 Now substitute 0 for y and 4 for z in one of the original equations and solve for x. We use Equation (1). x 80 a + = 1 The measure of angle A the measure of angle C. s ¸ x s ¸ x ˛ ˛ 4 2x −5 · 0 −2 · 4 = −4 2x −8 = −4 = a c more than the measure of angle C. is 45◦ 2x = 4 x = 2 s ¸ x s ˛ ¸ x s ¸ x ˛ ˛ We obtain (2, 0, 4). This checks, so it is the solution. = 4 5 b c + 17. x + y + 2z = 1, (1) x −y + z = 1, (2) x + 2y + z = 2 (3) We start by eliminating y from two different pairs of equa- tions. x + y + 2z = 1 (1) We have a system of equations. a + b + c = 180, a = 4c, b = 45 + c Solve. Solvingthesystemweget. 90,671,221Σ . 2 2 2+ 22 1 ◦, Check. The sumofthemeasuresis90◦+671◦ x −y + z = 1 (2) 2x + 3z = 2 (4) 2 Σ or180◦. FourtimesthemeasureofangleCis4. 221◦ , or 90 , the measure of angle A; 45 more than the measure 2x −2y + 2z = 2 Multiplying (2) by 2 x + 2y + z = 2 3x + 3z = 4 (5) Now solve the system of Equations (4) and (5). 2x + 3z = 2 (4) 2 ◦ ◦ 1 1 ◦+ 45◦, or 672 ◦, the measure of angle B. of angle C is 222 The answer checks. 1 State. The measuresofanglesA,B,andCare90◦, 672 and 221 ◦, respectively. 2 19.Familiarize. Let x, y, and z represent the prices of 1 bag of caramel nut crunch popcorn, 1 bag of plain popcorn, and 1 bag of mocha choco latte popcorn, respectively. Translate. The total cost of 1 bag of each type of popcorn was $49, so we have x + y + z = 49. The price of the caramel nut crunch popcorn was six times the price of the plain popcorn and $16 more than the price of the mocha choco latte popcorn, so we have x = 6y and x = z + 16. ◦, −3x −3z = −4 Multiplying (5) by −1 x = 2 x = 2 − − 2 · 2+ 3z = 2 4+ 3z = 2 3z = −2 z = −3 Substituting 2 for x in (4) 2 Copyright§ c 2017Pearson Education, Inc.

  31. 106 Chapter 3: Systems of Equations x + y ≥ 1 0 + 0 ? 1 0 . We have a system of equations. x + y + z = 49, x = 6y, x = z + 16 Solve. Solving the equation we have (30, 5, 14). Check. The total cost was $30 + $5 + $14, or $90. $30 is six times $5 and is $16 more than $14. The answer checks. State. One bag of caramel nut crunch popcorn costs $30, one bag of plain popcorn costs $5, and one bag of mocha choco latte popcorn costs $14. FALSE Since 0 1 contain (0, 0) and obtain the graph. is false, we shade the half-plane that does not ≥ y x + y “1 x 20.Graph: 2x + 3y < 12 First graph the line 2x+3y = 12. Draw it dashed since the inequality symbol is <. Test the point (0, 0) to determine if it is a solution. 2x + 3y < 12 23.Graph: y ≥−3, x ≥ 2 2 · 0+3 · 0 ? 12 0 TRUE We graph the lines y = 3 and x = 2 using solid lines. We indicate the region for each inequality by arrows at the ends of the lines. Note where the regions overlap, and shade the region of solutions. − . Since 0 < 12 is t (0, 0) and draw the graph. rue, we shade the half-plane containing y y x x 2x + 3y < 12 (2, —3) To find the vertex we solve the system of related equations: y = −3, x = 2 We see that the vertex is (2, −3). 24.Graph: x + 3y ≥−1, x + 3y ≤ 4 We graph the lines x + 3y = 1 and x + 3y = 4 using solid lines. We indicate the region for each inequality by arrows at the ends of the lines. Note where the regions overlap, and shade the region of solutions. 21.Graph: y ≤0 First graph the line y = 0 (the x-axis). Draw it solid since the inequality symbol is . Test the point (1, 2) to determine if it is a solution. y ≤ 0 2 ? 0 FALSE Since 2 0 is false, we shade the half-plane that does not contain (1, 2) and obtain the graph. ≤ ≤ − y y x y “0 x 22. Graph: x + y ≥1 First graph the line x + y = 1. Draw it solid since the inequality symbol is . Test the point (0, 0) to determine if it is a solution. (1) (2) (3) ≥ 25.Graph: x −y ≤3, x + y ≥ −1, y ≤ 2 Copyright§ c 2017Pearson Education, Inc.

  32. Chapter 3 Summary and Review: Review Exercises 107 We have a system of equations. x + y = −2, (1) 2x + y = Solve. We multiply both sides of Equation (1) by 1 and then add. −x −y = 2 2x + y = 4 x = 6 Substitute 6 for x in Equation (1) and solve for y. 6+ y = −2 y = −8 Check. 6 + ( 8) = 2, and 2 6 + ( 8) = 12 The answer checks. We graph the lines x y = 3, x + y = 1, and y = 2 using solid lines. We indicate the region for each inequality by arrows at the ends of the lines. Note where the regions overlap, and shade the region of solutions. − − 4 (2) − y (—3, 2) (5, 2) x ( ) 1, —2 8 = 4. − − · − − To find the vertices we solve three different systems of equations. From (1) and (2) we obtain the vertex (1, 2). From (1) and (3) we obtain the vertex (5, 2). From (2) and (3) we obtain the vertex (−3, 2). 26.Let x = the number of questions answered from group A and y = the number of questions answered from group B. Find the maximum value of S = 7x + 12y subject to x + y ≥ 8, 8x + 10y ≤80, x ≥0, y ≥ 0 Graph the system of inequalities, determine the vertices, and find the value of T at each vertex. − State. The numbers are 6 and −8. Since one number is 6, C is the correct answer. 28.Familiarize. Let t = the time the cars travel, and let d = the distance the first car travels. Then the distance the second car travels is 275 d. We organize the information in a table. Distance Rate − Time d t First car 50 Second car 275 −d Translate. Using d = rt in each row of the table, we get a system of equations. d = 50t, 275 −d = 60t Solve. 275 −50t = 60t Using substitution 275 = 110t 2.5 = t Check. In 2.5 hr, the first car travels 50(2.5), or 125 mi, and the second car travels 60(2.5), or 150 mi. Then the cars are 125 mi + 150 mi, or 275 mi, apart after 2.5 hr. The answer checks. State. The cars will be 275 mi apart after 2.5 hr. A is the correct answer. 60 t y 11 10 9 8 (0, 8) 7 6 5 4 3 2 (8, 0) 1 (10, 0) 1 2 3 4 5 6 7 8 9 10 11 x Vertex S = 7x + 12y (0, 8) (8, 0) (10, 0) 7 · 10 + 12 · 0 = 70 The maximum score of 96 occurs when 0 questions from group A and 8 questions from group B are answered cor- rectly. 7 · 0+ 12 · 8 = 96 7 · 8+ 12 · 0 = 56 29.We graph the equations and find the points of intersection. y 5 y =x 2 + 2 y = x + 2 4 3 (1, 3) 27.Familiarize. Let x and y represent the numbers. Translate. The sum of the numbers s ˛ ↓ x + y Twice one number plus the other is 4. 2 (0, 2) 1 xis −2. ↓ = −2 x —5 —4 —3 —2 —1 1 2 3 4 5 ¸ — 1 ↓ — 2 — 3 — 4 — 5 s ¸ x + s ˛ y ¸ x ˛ The solutions are (0, 2) and (1,3). ↓ ↓ ↓ ↓ ↓↓ · x = 4 2 Copyright§ c 2017Pearson Education, Inc.

  33. 108 Chapter 3: Systems of Equations 30.Answers may vary. One day a florist sold a total of 23 hanging baskets and flats of petunias. Hanging baskets cost $10.95 each and flats of petunias cost $12.95 each. The sales totaled $269.85. How many of each were sold? Chapter 3 Test 1. Graph both lines on the same set of axes. 31.We know that machines A and B can polish 3400 lenses, so machine C can polish 5700 3400, or 2300 lenses. Machines B and C can polish 4200 lenses, so machine B can polish 4200 −2300, or 1900 lenses. Then machine A can polish 5700 −1900 −2300, or 1500 lenses. 32.Let x = the number of adults in the audience, y = the number of senior citizens, and z = the number of chil- dren. The total attendance is 100, so we have Equation (1), x + y + z = 100. The amount taken in was $100, so Equa- tion (2) is 10x + 3y + 0.5z = 100. There is no other in- formation that can be translated to an equation. Clearing decimals in Equation (2) and then eliminating z gives us Equation (3), 95 + 25y = 500. Dividing by 5 on both sides, we have Equation (4), 19x + 5y = 100. Since we have only two equations, it is not possible to eliminate z from another pair of equations. However, in Equation (4), note that 5 is a factor of both 5y and 100. Therefore, 5 must also be a factor of 19x, and hence of x, since 5 is not a factor of 19. Then for some positive integer n, x = 5n. (We require n to be positive, since the number of adults clearly cannot be negative and must also be nonzero since the exercise states that the audience consists of adults, senior citizens, and children.) We have: 19 · 5n + 5y = 100 19n + y = 20 Dividing by 5 Since n and y must both be positive, n = 1. (If n > 1, then 19n + y > 20.) Then x = 5 · 1, or 5. 19 · 5+ 5y = 100 Substituting in (4) y = 1 − y 5 4 3 2 (–2, 1) 1 x —5 —4 —3 —2 —1 y = 3x + 7 1 2 3 4 5 23x+ 2y =–4 3 — 1 — — — 4 — 5 The solution (point of intersection) seems to be the point (−2, 1). Check: y = 3x +7 1 ? 3(−2)+7 −6 + 7 TRUE The solution is (−2, 1). Since the system of equations has a solution, it is consis- tent. Since there is exactly one solution, the equations are independent. 3x + 2y = −4 3(−2)+2 · 1 ? −4 −6+2 . . .1 −4 . TRUE 2.Graph both lines on the same set of axes. y 5 4 4 3 y =3x– 2 y =3x + 5+1+ z = 100 Substituting in (1) z = 94 There were 5 adults, 1 senior citizen, and 94 children in the audience. 2 1 x —5 —4 —3 —2 —1 1 2 3 4 5 — 1 — 2 — 3 33. No; the symbol the half-plane above the line is shaded. For the inequality y 3, for example, the half-plane below the line y = 3 does not always yield a graph in which ≥ — 4 — 5 −≥ is shaded. − The lines are parallel. There is no solution. Since the system of equations has no solution, it is in- consistent. Since there is no solution, the equations are independent. Copyright§ c 2017Pearson Education, Inc.

  34. Chapter 3 Test 109 3.Graph both lines on the same set of axes. Now substitute 2y + 6 for x in the second equation and solve for y. 2(−2y + 6)+ 3y = 7 −4y + 12+ 3y = 7 −y + 12 = 7 −y = −5 y = 5 Next we substitute 5 for y in Equation 3 and find x. x = −2 · 5+6 = −10 + 6 = −4 The ordered pair ( 4, 5) checks in both equations. It is the solution. − y 5 y –3x= 6 6 x –2y= –12 4 3 2 1 —5 —4 —3 —2 —1 1 2 3 4 5 x 1 — — 2 — 3 − — 4 — 5 2x + 5y = 3, (1) −2x + 3y = 5 8y = 8 y = 1 Now substitute 1 for y in one of the equations and solve for x. 2x + 5y = 3 Equation (1) 2x +5 · 1 = 3 2x +5 = 3 2x = −2 x = −1 The ordered pair ( 1, 1) checks in both equations. It is the solution. 7. The graphs are the same. Any solution of one of the equa- tions is also a solution of the other. Each equation has an infinite number of solutions. Thus the system of equations has an infinite number of solutions. Since the system of equations has a solution, it is consis- tent. Since there are infinitely many solutions,the equa- tions are dependent. (2) Adding 4. 4x + 3y = −1, (1) y = 2x −7 Substitute 2x 7 for y in the first equation and solve for x. 4x + 3(2x −7) = −1 4x + 6x −21 = −1 10x −21 = −1 10x = 20 x = 2 Next we substitute 2 for x in Equation (2) to find y. y = 2x −7 = 2 · 2 −7 = 4 −7 = −3 The ordered pair (2, 3) checks in both equations. It is the solution. (2) − − x + y = −2, (1) 4x −6y = −3 Multiply Equation (1) by 6 and then add. 6x + 6y = −12 4x −6y = −3 10x = −15 3 x = −2 3 Now substitute −for x in one of the original equations 2 x + y = −2 Equatlon (1) 3 −2 + y = −2 1 y = −2 . −3 , −1 Σ is the solution. 8. (2) − 5. x = 3y + 2, 2x −6y = 4 (2) Substitute 3y + 2 for x in the second equation and solve for y. 2(3y + 2) −6y = 4 6y +4 −6y = 4 4 = 4 We have an equation that is true for all values of x and y. The system of equations has infinitely many solutions. (1) and solve for y. The ordered pair checks in both equations. It 2 2 6. x + 2y = 6, 2x + 3y = 7 (2) Solve the first equation for x. (1) 2 x −4 y = 1, (1) 2 −y = 2 5 We first multiply each equation by 15 to clear the fractions. 15.2x−4y =15·1→10x−12y=15(3) 15.1x−2y =15·2→5x−6y=30 9. 3 1 3 5 x + 2y = 6 x = −2y + 6 (3) (2) Σ 3 5 Σ (4) 3 5 Copyright§ c 2017Pearson Education, Inc.

  35. 110 Chapter 3: Systems of Equations r 20 = 120 20 = 100, and the distance traveled in 7 hr is 100 7, or 700 km. The distances are the same, so the answer checks. State. The speed of the plane in still air is 120 km/h. − − Now we multiply Equation (4) by −2 and then add. 10x −12y = 15 −10x + 12y = −60 0 = −45 We get a false equation. The system of equations has no solution. · 12.Familiarize. Let b = the number of buckets of wings and d = the number of chicken dinners sold. We organize the information in a table. Buckets Dinners Total Number sold Price $12 Amount collected Translate. The first and last rows of the table give us two equations. b + d = 28, (1) 12b + 7d = 281 (2) Solve. We use the elimination method. −7b −7d = −196 Multiplying (1) by −7 12b + 7d = 281 10.Familiarize. Let l = the length and w = the width, in feet. Translate. The width is 42 ft less than the length, so we have one equation: w = l −42. We use the formula P = 2l + 2w to get a second equation: 288 = 2l + 2w. We have a system of equations. w = l −42, (1) 288 = 2l + 2w (2) Solve. We use the substitution method. 288 = 2l + 2(l −42) 288 = 2l + 2l −84 288 = 4l −84 372 = 4l 93 = l Substitute 93 for l in Equation (1) and find w. w = 93 −42 = 51 Check. The length, 93 ft, is 42 ft more than the width, 51 ft. Also, 2 · 93+2 · 51 = 288 ft. The answer checks. State. The length is 93 ft, and the width is 51 ft. b d 28 $7 12b 7d 281 5b = 85 17 b = Substitute 17 for b in Equation (1) and solve for d. 17 + d = 28 d = 11 Check. The number of orders filled was 17 + 11, or 28. The amount taken in was $12 17+ $7 11, or $281. The answer checks. State. 17 buckets of wings and 11 chicken dinners were sold. · · 11.Familiarize. Let d = the distance traveled and r = the speed of the plane in still air, in km/h. Then with a 20- km/h tailwind the plane’s speed is r + 20. The speed of the plane traveling against the wind is r 20. We organize the information in a table. Distance 13.Familiarize. Let x = the number of liters of 20% solution and y = the number of liters of 45% solution to be used. We organize the information in a table. 20% solution Number of liters Percent of salt Amount of salt Translate. We get one equation from the first row of the table. x + y = 20 The last row of the table yields a second equation. 0.2x + 0.45y = 6 After clearing decimals, we have the following system of equations. x + y = 20, (1) 20x + 45y = 600 (2) − 45% solution Mixture Rate Time d r + 20 With wind 5 20 x y r −20 Against wind 7 d Translate. Using d = rt in each row of the table, we get a system of equations. d = (r + 20)5, d = (r −20)7, or 20% 45% 30% 0.3(20), or 6 0.2x 0.45y d = 5r + 100, d = 7r −140 Solve. We use the substitution method. 5r + 100 = 7r −140 100 = 2r −140 240 = 2r 120 = r Check. If r = 120, then r + 20 = 120 + 20 = 140, and the distance traveled in 5 hr is 140 · 5, or 700 km. Also, Copyright§ c 2017Pearson Education, Inc.

  36. Chapter 3 Test 111 Solve. We use the elimination method. −20x −20y = −400 Multiplying (1) by −20 20x + 45y = 600 y = 8 Substitute 8 for y in Equation (1) and solve for x. x = 12 Translate. Total time worked is 21.5 hr. s ˛¸ x s ˛¸ x x + y + z Totalamountearned s ˛¸ 21x + 19.50y + 24z = 469.50 21.5 = xis $469.50. 25y = 200 x +8 = 20 Plumber’s hours are 2 more than x = 2 carpenter’s hours. s ˛¸ y x ¸ s ¸ s x ˛ ˛ Check. 12 L + 8 L = 20 L. The amount of salt in the mixture is 0.2(12)+0.45(8) = 2.4+3.6, or 6 L. The answer checks. State. 12 L of 20% solution and 8 L of 45% solution should be used. z + We have a system of equations. x + y + z = 21.5, 21x + 19.50y + 24z = 469.50, z = 2 + y Solve. Solving the system, we get (3.5, 8, 10). Check. The total time worked was 3.5 +8 +10, or 21.5 hr. The amount earned was $21(3.5) + $19.50(8) + $24(10), or $469.50. The time worked by the plumber, 10 hr, is 2 more than the time worked by the carpenter, 8 hr. The answer checks. State. The electrician worked 3.5 hr. 6x + 2y −4z = 15, (1) −3x −4y + 2z = −6, (2) 4x −6y + 3z = We start by eliminating y from two different pairs of equa- tions. 12x + 4y −8z = 30 −3x −4y + 2z = −6 (2) 9x −6z = 24 18x + 6y −12z = 45 Multiplying (1) by 3 4x −6y + 3z = 8 (3) 22x −9z = 53 (5) Now solve the system of Equations (4) and (5). 9x −6z = 24 (4) 22x −9z = 53 (5) 27x −18z = 72 Multiplying (4) by 3 −44x + 18z = −106 Multiplying (5) by −2 17x = 34 x = 2 14. 8 (3) Multiplying (1) by 2 (4) 16.Graph: y ≥ x −2 We first graph the line y = x 2. We draw a solid line because the inequality symbol is . Test the point (0, 0) to determine if it is a solution. y ≥ x −2 0 ? 0 −2 −2 TRUE Since 0 2 is true, we shade the half-plane that contains (0, 0). − ≥ . ≥ − − − y 9 · 2 −6z = 24 Substituting 2 for x in (4) 18 −6z = 24 −6z = 6 z = −1 6 · 2+ 2y −4(−1) = 15 y “x —2 x Substituting 2 for x and −1 for z in (1) 12+ 2y + 4 = 15 16+ 2y = 15 17.Graph: x −6y < −6 We first graph the line x 6y = 6. We draw the line dashed since the inequality symbol is <. Test the point (0, 0) to determine if it is a solution. x −6y < −6 0 −6 · 0 ? −6 0 FALSE . − 2y = −1 y = −2 1 Weobtain. 15. Familiarize. Let x, y, and z represent the number of hours worked by the electrician, the carpenter, and the plumber, respectively. In x hours the carpenter earns 21x; in y hours the carpenter earns 19.50y; and in z hours the plumber earns 24z. Σ 2,−1,−1 2 . This checks,soitisthesolution. Since 0 < 6 not contain (0, 0). is false, we shade the half-plane that does − Copyright§ c 2017Pearson Education, Inc.

  37. 112 Chapter 3: Systems of Equations x + y ≤ 100, x ≥ 25, y ≥ 15 Graph the system of inequalities, determine the vertices, and find the value of P at each vertex. y 6y 6 x < x y 100 90 18. Graph: x + y ≥ 3, 80 (25, 75) 70 x −y ≥ 5 60 50 Graph the lines x + y = 3 and x y = 5 using solid lines. Indicate the region for each inequality by arrows and shade the region where they overlap. − 40 30 20 (85, 15) (25, 15) 10 y 10 20 30 40 50 60 70 80 90 100 x Vertex P = 6x + 8y (25, 15) (25, 75) (85, 15) 6 · 85 + 8 · 15 = 630 The maximum profit of $750 occurs when 25 pound cakes and 75 carrot cakes are prepared. 6 · 25 + 8 · 15 = 270 6 · 25 + 8 · 75 = 750 x (4, —1) To find the vertex we solve the system of related equations: x + y = 3, x −y = 5 Solving, we obtain the vertex (4, −1). 19. Graph: 2y −x ≥ −4, (1) 21. Familiarize. Let x, y, and z represent the amounts in- vested at 2%, 3%, and 5%, respectively. The amounts earned by the funds are 0.02x, 0.03y, and 0.05z. Translate. Amount invested was $30, 000. s x ↓ ˛¸ ↓ ↓ 000 x+y+z 30, 2y + 3x ≤ −6, (2) y ≤0, x ≤0 = (3) (4) were $990. Total earnings s ˛ ↓ 3y+0 Earnings at 5% s ˛ ↓ 5z ¸ x Shade the intersection of the graphs of the four inequalities above. ↓ ↓ 90 0.02 x+0. .05z= 0 9 were $280 more x ↓ 2 80 earnings at 2%. ˛ than ¸xs0. y ¸ ¸ x ↓ ↓ s˛ 0. 2x 0 = 0 ↓ + We have a system of equations. x + y + z = 30, 000, 0.02x + 0.03y + 0.05z = 990, 0.05z = 280+ 0.02x Solve. Solving the system, we get (11, 000, 9000, 10, 000). Check. The total amount invested was $11, 000 +$9000 + $10, 000, or $30,000. The amounts earned at 2%, 3%, and 5% are 0.02($11, 000), 0.03($9000), and 0.05($10, 000), or $220, $270, and $500, respectively. Thus the total earnings were $220 + $270 + $500, or $990. The amount earned at 5%, $500, is $280 more than $220, the amount earned at 2%. The answer checks. State. $10,000 was invested at 5%. (—2, 0) x 1,—— 2 9) 4 (—— To find the two vertices we solve two systems of equations, as follows: System of equations Vertex From (1) and (2) 2 . −1 , −9 Σ 4 (−2, 0) From (2) and (3) 20. Let x = the number of pound cakes prepared and y = the number of carrot cakes. Find the maximum value of P = 6x + 8y subject to Answer B is correct. Copyright§ c 2017Pearson Education, Inc.

  38. Chapter 3 Test 113 22. Substituting −1 for x and 3 for f (x), we have 3 = m(−1) + b, or 3 = −m + b. Substituting −2 for x and −4 for f (x), we have −4 = m(−2) + b, or −4 = −2m + b. Then we have a system of equations: 3 = −m + b, −4 = −2m + b We solve the system of equations. −3 = −4 = −2m + b (2) 7 = m 7 = m Substitute 7 for m in (1) and solve for b. 3 = −7+ b 10 = b Thus, we have m = 7 and b = 10. (1) (2) m −b Multiplying (1) by −1 − − Copyright§ c 2017Pearson Education, Inc.

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