The dimensions of a rectangle are such that its length is 5 inches more than its width. If the length were doubled and if the width were decreased by 2 inches, the area would be increased by 136 in^2. what are the length and width of the rectangle?

Question

anonymous · Answer

l = w+5

(2l)(w-2) = lw + 136
Substitute for l. Solve quadratic in w. Then go from there.

anonymous · Answer

L = 5 + W
2L(W-2) = LW + 136
2(W + 5)(W-2) = (W + 5)W + 136
2w^2 + 6w - 20 = w^2 + 5w + 136
w^2 + w -156 = 0
Factors into
(w - 12)(w + 13)
Solutions are 12 and -13. You can't have a negative length so that means the width is 12.
L = 5 + w
L = 5 + 12
L = 17, W = 12