SOLUTION: A rectangle is twice as long as it is wide. If the length is decreased by 4 inches and its width is decreased by 3 inches, the area is decreased by 88 square inches. Find the origi

Algebra ->  Rectangles -> SOLUTION: A rectangle is twice as long as it is wide. If the length is decreased by 4 inches and its width is decreased by 3 inches, the area is decreased by 88 square inches. Find the origi      Log On


   



Question 577839: A rectangle is twice as long as it is wide. If the length is decreased by 4 inches and its width is decreased by 3 inches, the area is decreased by 88 square inches. Find the original dimenions
Answer by dfrazzetto(283) About Me  (Show Source):
You can put this solution on YOUR website!
From original problem:
Original rectangle: L = 2W
A(original) = length x width = (2W)(W) = 2W^2
When L-4 and W-3, A(original) is less 88 in^2
L-4 = 2W-4 so:
(2W-4)(W-3) = 2W^2 - 88
2W^2 - 4W - 6W + 12 = 2W^2 - 88 (2W^2 cancels out
12 - 10W = -88
10W = 100
W = 10
L = 2W = 20
Check:
A(original) = L x W = 20 x 10 = 200 square inches
A(new) = (L-4)(W-3) = 16 x 7 = 112 square inches
200 - 112 = 88 square inches √