SOLUTION: The length of a rectangle is 4 inches greater than twice its width. If the length and width are both increased by 2 in., its area is increased by 48in.^2 What is the width of the o

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 4 inches greater than twice its width. If the length and width are both increased by 2 in., its area is increased by 48in.^2 What is the width of the o      Log On


   



Question 663452: The length of a rectangle is 4 inches greater than twice its width. If the length and width are both increased by 2 in., its area is increased by 48in.^2 What is the width of the original rectangle?
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
l = 2w + 4
(l+2)(w+2)=area+48
area = (l)(w)
w(2w+4) = 2w^2 + 4w
Using (l+2)(w+2)=area + 48
(2w+4+2)(w+2) = 2w^2 + 4w + 48
(2w+6)(w+2) = 2w^2 + 4w + 48
2w^2 + 4w + 6w + 12 = 2w^2 + 4w + 48
2w^2 + 10w + 12 = 2w^2 + 4w + 48
2w^2 - 2w^2 + 10w - 4w =48 - 12
6w = 36
width = 6 in.
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