The dimensions of a rectangle are such that its length is 3 inches more than its width. If the length were doubled and if the width were decreased by 1 inch, the area would be decreased by 66 inches. What are the length and width of the rectangle.

Question

The dimensions of a rectangle are such that its length is 3 inches more than its width.  If the length were doubled and if the width were decreased by 1 inch, the area would be decreased by 66 inches.  What are the length and width of the rectangle.

anonymous · Answer

3 inches longer or shorter than the width?

anonymous · Answer

3inches more than width

anonymous · Answer

Sorry for the time it's taking but I think I got it now

anonymous · Answer

No, problem.  Im n and out of the house doing chores.

anonymous · Answer

L = Length
W = Width
A = Area

L*W = A
(W + 3)(W) = A
W^2 + 3W = A

2(W + 3))*(W - 1) = 2(W^2 + 2W - 3) 
                          =  2W^2 + 4W - 6

The change in area is equal to the new Area minus the original Area

(2W^2 + 4W - 6) - (W^2 + 3W) = 66
W^2 + W - 6 = 66
W^2 + W - 72 = 0
(W - 8)(W + 9) = 0

I.e width  => W = 8 or W = -9 
As Width cannot be negative W = 8

and L = 8 + 3
L = 11

I hope that works out right!

anonymous · Answer

Well, if not I really appreciate the time and effort that you put in
But, I think that it is and I will let you know