SOLUTION: A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 32. Find the original dimensions.

Question 151995: A rectangle is twice as long as it is wide. If its length is decreased by 4 and its width is decreased by 2, its area is decreased by 32. Find the original dimensions.
Answer by orca(409) (Show Source): You can put this solution on YOUR website!
Suppose its width is x, then its length is 2x.
So its area is 2x^2.
The new rectangle's width is x + 2, length is 2x + 4, so its area is (x+2)(2x+4).
As the area of new rectangle is 32 more than the original one, we have
(x+2)(2x+4) = 2x^2 + 32
To Solve the equation for x, first we simplify it by dividing both sides by 2.
(x+2)(x+2) = x^2 + 16
Or written as
(x+2)^2 = x^2 + 16
Expanding the left side, we have
x^2 + 4x + 4 = x^2 + 16
4x = 12
x = 3
So the dimensions of the original rectangle is 3 and 6.

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