posted by fernando .
the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were double and if the width were decreased by 1 inch, the area would be increased by 66inches ^2. What are the length and width of the rectangle?
Let w = width and w+3 = length.
2(w+3)(w-1) - w(w+3) = 66
Solve for w, then w+3.