## Chapter 39Solutions for twelfth week’s assignments

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### 39.1 Exercises for 7.1

Problem 1
A and C are linear.
Problem 2
For B, there is an exponent of two ({x}^{2}). And for D, there is an exponent of negative one ({1\over x}).
Problem 3
Substituting, 3(6 + 4) = 3 ⋅ 10 = 30 on the left, and 5 ⋅ 6 = 30 on the right. Hence 6 is a solution.
Problem 4
Without evaluating the expressions, we can see that substituting - 2 on the left yields an even number and that substituting - 2 on the right yields an odd number. Thus \mathbf{-2} cannot be a solution.

But if we want to evaluate the expressions anyways, we have 5(-2 + 4) - 3(-2 + 6) = 5 ⋅ 2 - 3 ⋅ 3 = 10 - 12 = -2 on the left, and 9(-2 + 1) = 9 ⋅-1 = -9. Because - 2\mathrel{≠} - 9, -2 is not a solution.

Problem 9
7k + 8 = 1 ⇒ 7k = -7 ⇒ k = -1.
Problem 10
5m - 4 = 21 ⇒ 5m = 25 ⇒ m = 5.
Problem 17
2(x + 3) = -4(x + 1) ⇒ 2x + 6 = -4x - 4 ⇒ 6x = -10 → x = {-10\over 6} = {-5\over 3} .
Problem 18
4(x - 9) = 8(x + 3) ⇒ 4x - 36 = 8x + 24 ⇒-60 = 4x ⇒ x = -15.
Problem 25
-[2z-(5z+2)] = 2+(2z+7) ⇒-2z+5z+2 = 9+2z ⇒ 3z+2 = 9+2z ⇒ z = 7.
Problem 26
- [6x - (4x + 8)] = 9 + (6x + 3) ⇒ -6x + 4x + 8 = 12 + 6x ⇒ -2x + 8 = 12 + 6x ⇒-8x = 4 ⇒ x = {-1\over 2} .
Problem 36
{3x\over 4} + {5x\over 2} = 13 ⇒ {3x+10x\over 4} = 13 ⇒ {13\over 4} x = 13 ⇒ x = 4.
Problem 37
{8x\over 3} -{2x\over 4} = -13 ⇒ {32x-6x\over 12} = -13 ⇒ {26\over 12}x = -13 ⇒ x = {-13⋅12\over 13⋅2} = -6.
Problem 61
t = {d\over r}
Problem 62
r = {I\over pt}
Problem 68
r = {C\over 2π}
Problem 69
h = {S-2π{r}^{2}\over 2πr} = {S\over 2πr} - r (either is fine, the latter is better for calculators)
Problem 76
• In part a, x = 93. Then y = .1 ⋅ 93 - 8.5 = 9.3 - 8.5 = .8, or 800 000 tickets.
• For part b, solve .75 = .1x - 8.5 for x. Then x = (.75 + 8.5)∕.1 = 9.25∕.1 = 92.5. So the model predicts that the season spanning the latter half of 1992 through the first half of 1993 sold 750 000 tickets. But reading into a model like this is tricky. I expect the authors intend the answer to be the 1992–1993 season.

### 39.2 Exercises for 7.2

Problem 21
1. Let x be the number of big-store shoppers, poor people.
2. Then x - 70 = y, where y is the number of happy small-store shoppers1 .
3. x + y = 442, or x + (x - 70) = 442.
4. From the above, 2x - 70 = 442 ⇒ 2x = 512 ⇒ x = 256.
5. So there are x = 256 big-store shoppers and y = 256 - 70 = 186 small-store shoppers.
6. The number of big-store shoppers was 70 more than the number of small-store shoppers, and the total number of these two bookstore types was \mathbf{256 + 186 = 442}.
Problem 24
Let W be the number of wins and L be the number of losses. Then W = 3L - 2 and W + L = 82. So 3L - 2 + L = 82 ⇒ 4L = 84 ⇒\mathbf{L = 21} is the number of losses, and \mathbf{W = 82 - 21 = 61} is the number of wins. This solution makes sense; both numbers are non-negative integers.
Problem 26
Let D be the number of votes for G.W. Bush, and S be the number of votes for A. Gore. Then D + S = 537 and D = S + 5. Then 2S + 5 = 537, so \mathbf{S = 266} and \mathbf{D = 271}. Again, this makes sense because the numbers are non-negative integers.
Problem 43
 Percent Investment Interest 0.03 x 0.03 ⋅ x 0.04 12\kern 1.66702pt 000 - x 0.04 ⋅ (12\kern 1.66702pt 000 - x) 12 000 440

So 0.03 ⋅ x + 0.04 ⋅ (12\kern 1.66702pt 000 - x) = 440. Solving for x, \mathbf{x = 4000}.

He invested $4 000 at 3% interest and$8 000 at 4% interest. Checking, this totals to 0.03 ⋅ 4000 + 0.04 ⋅ 8000 = 120 + 320 = 440.