a{ax + b) + b(cx — a) a~x + bcx c(ax + b) — a(cx — a) a2 + be

FIGURE 5

FIGURE 5

11. The condition is that y/(x + 2)2 + y2 = 2y/(x -4)2 + y2, or equivalently (.x-6)2 +r = 42.

13. (a) The two expressions give total price per unit (product price plus shipping costs) for the product delivered at (x. y) from A and B respectively, (b) The condition is that p + 10V*2 + y2 = p 4- 5yJ(x — 60)2 4- y2? which reduces to (x + 20)2 + y2 = 402.

15. x2 + y2 + Ax + By + C = 0 x2 + Ax + y2 + By + C = 0

x2 + Ax + (\A)2 + y2 + By + {{B)2 = \{A2 + B2-AC) (x + \A)2 + (y + \B)2 — j(A2 + B2 — 4C). The last is the equation of a circle centered at (—\A, -\B) with radius |VA2 + 52 -4C. If A2 + B2 — 4C, the graph consists only of the point (— ¡¡Ar — \B). For A2 + B2 < 4C, the solution set is empty.

1. (a) All x (b) x = 0 (c) All x (d) x = 0 (For * > 0r the equation y4 = * has two solutions.) (e) * = ±1 (f) All * # 3 (g) All * (h) All *

3. Suppose c is positive. Then f{x) + c is obtained by raising the graph of f(x) by c units. f(x + c) is obtained by shifting the graph of f{x) by c units to the left. —f(x) is obtained by reflecting the graph of f{x) in the *-axis. /(-*) is obtained by reflecting the graph of f(x) in the y-axis.

1. (a) Slope = (8 - 3)/(5 - 2) = 5/3 (b) -2/3 (c) 51/5

3. L) is >• = x + 2, with slope 1; L2 is y = —|x + 3, with slope —3/5; L3 is y = 1, with slope 0; L4 is y = 3x — 14, with slope 3; L5 is y = |x + 2, with slope 1/9.

7. (a) L\ is (y—3) = 2(x — 1) or y = 2x+1 (b)L2 is >—2 =

or y = x/5 + 12/5 (c) L5 is y = -x/2 (d) L4 is x/a + y/b = 1, or y = —bx/a + b. The graphs are shown in Figs. 9 and 10.

FIGURE 10

FIGURE 10

9. The point-point formula gives y—200 = ^^—^^(x —100) ory = fx+50.

11. (a) April 1960 corresponds to t = 9/4, when JV(9/4) = -17,400- (9/4) -151.000= 111.850. (b) -17,400r +151,000=0 implies t = 8.68, which corresponds roughly to September 1966.

13. For (a), shown in Fig. 11, the solution is x = 3, y = —2. For (b), shown in Fig. 12, the solution is x = 2, y = 0. For (c), shown in Fig. 13, there are no solutions, because the two lines are parallel.

15. See Fig. 14. C = 0.8824K-1.3941. The slope is an estimate of the marginal propensity to consume.

17. See Fig. 15. Each arrow shows the $ide of the line on which the relevant inequality is satisfied. The shaded triangle is the required set

FIGURE 11 FIGURE 12

X | ||

^ 6x - 8 v = 6 | ||

^ 3x + 4 v = 1 |

21 22 23 24 25 FIGURE 14

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