the dimensions of a rectangle are such that is lenght is 9 inches than its width. If the length were double and it's width was decreased by 4 inches, the area would be increased by 110 in^2. what are the length and the width of the rectangle?

You question doesn't say if the length is 9 inches less than the width or more than it.

I am assuming that the length is 9 inches less than the width.

L = W - 9

A = W*L = W^2 - 9W

New dimensions

L2 = 2*L = 2W18

W2 = W - 4

A2 = L2 * W2 = (2W - 18)(W - 4) = 2W^2 - 26W + 36

Difference in areas

A2 - A = 110

2W^2 - 26W + 36 - W^2 + 9W = 110

W^2 - 17W - 38 = 0

(W - 19)(W + 2) = 0

**Hence W = 19, L = 10**