Section 8.1 Solutions
1. No, you can either have zero, one, or infinitely many. Examine graphs.
3. This means there is no realistic break-even point. By the time the company produces one unit they are already making profit.
5. You can solve by substitution (isolating x or y ), graphically, or by addition.
7. Yes
9. Yes
11. (−1,2)
13. (−3,1)
15. (−35,0)
17. No solutions exist.
19. (725,1325)
21. (6,−6)
23. (−12,110)
25. No solutions exist.
27. (−15,23)
29. (x,x+32)
31. (−4,4)
33. (12,18)
35. (16,0)
37. (x,2(7x−6))
39. (−56,43)
41. Consistent with one solution
43. Consistent with one solution
45. Dependent with infinitely many solutions
47. (−3.08,4.91)
49. (−1.52,2.29)
51. (A+B2,A−B2)
53. (−1A−B,AA−B)
55. (CE−BFBD−AE,AF−CDBD−AE)
57. They never turn a profit.
59. (1,250,100,000)
61. The numbers are 7.5 and 20.5.
63. 24,000
65. 790 sophomores, 805 freshman
67. 56 men, 74 women
69. 10 gallons of 10% solution, 15 gallons of 60% solution
71. Swan Peak: $750,000, Riverside: $350,000
73. $12,500 in the first account, $10,500 in the second account.
75. High-tops: 45, Low-tops: 15
77. Infinitely many solutions. We need more information.
Section 8.2 Solutions
1. A nonlinear system could be representative of two circles that overlap and intersect in two locations, hence two solutions. A nonlinear system could be representative of a parabola and a circle, where the vertex of the parabola meets the circle and the branches also intersect the circle, hence three solutions.
3. No. There does not need to be a feasible region. Consider a system that is bounded by two parallel lines. One inequality represents the region above the upper line; the other represents the region below the lower line. In this case, no points in the plane are located in both regions; hence there is no feasible region.
5. Choose any number between each solution and plug into C(x) and R(x). If [latex]C\left(x\right)
41.
43.
45.
47.
49. (−2√70383,−2√3529),(−2√70383,2√3529),(2√70383,−2√3529),(2√70383,2√3529)
51. No Solution Exists
53. x=0,y>0 and [latex]0
1. No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, 1x2+1 cannot be decomposed because the denominator cannot be factored.
3. Graph both sides and ensure they are equal.
5. If we choose x=−1, then the B-term disappears, letting us immediately know that A=3. We could alternatively plug in x=−53, giving us a B-value of −2.
7. 8x+3−5x−8
9. 1x+5+9x+2
11. 35x−2+44x−1
13. 52(x+3)+52(x−3)
15. 3x+2+3x−2
17. 95(x+2)+115(x−3)
19. 8x−3−5x−2
21. 1x−2+2(x−2)2
23. −64x+5+3(4x+5)2
25. −1x−7−2(x−7)2
27. 4x−32(x+1)+72(x+1)2
29. 4x+2x2−33x+2+72(3x+2)2
31. x+1x2+x+3+3x+2
33. 4−3xx2+3x+8+1x−1
35. 2x−1x2+6x+1+2x+3
37. 1x2+x+1+4x−1
39. 2x2−3x+9+3x+3
41. −14x2+6x+9+12x−3
43. 1x+1x+6−4xx2−6x+36
45. x+6x2+1+4x+3(x2+1)2
47. x+1x+2+2x+3(x+2)2
49. 1x2+3x+25−3x(x2+3x+25)2
51. 18x−x8(x2+4)+10−x2(x2+4)2
53. −16x−9x2+16x−1−7(x−1)2
55. 1x+1−2(x+1)2+5(x+1)3
57. 5x−2−310(x+2)+7x+8−710(x−8)
59. −54x−52(x+2)+112(x+4)+54(x+4)