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| 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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