The dimensions of a rectangle are such that its length is 5 in. More than its width. If the length were doubled and if the width were decreased by 2 in., the area would be increased by 162 in. What are the length and width of the rectangle?

Let the width of the original rectangle be W, then the length = (W + 5)
The area of this rectangle = W(W + 5) = W² +5W

The modified rectangle has a length 2(W + 5) and a width (W - 2)
The area of the modified rectangle = 2(W + 5)(W - 2) = 2W² + 6W - 20
As this area is 162 sq in greater than the original rectangle then we can write :-
2W² + 6W - 20 = W² + 5W + 162
W² + W - 182 = 0
The can be factored
(W +14)(W - 13) = 0
We are only concerned with the positive root that occurs when (W - 13) = 0, thus W = 13
The dimensions of the original rectangle are Width 13 in, Length 18 in (W + 5)
The dimensions of the modified rectangle are Width 11 in (W - 2) and Length 36 in [(2(W + 5)]