SOLUTION: A rectangle is twice as long as it is wide. if its length is increased by 6in and its width is incresed by 2in, its area is increased by 32in squared. find the original dimensions.

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A rectangle is twice as long as it is wide. if its length is increased by 6in and its width is incresed by 2in, its area is increased by 32in squared. find the original dimensions.      Log On

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Question 58530: A rectangle is twice as long as it is wide. if its length is increased by 6in and its width is incresed by 2in, its area is increased by 32in squared. find the original dimensions.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W
(2W+6)(W+2)=2W*W+32
2W^2+10W+12=2W^2+32
2W^2-2W^2+10W=32-12
10W=20
W=2
L=2*2
L=4
PROOF
2*4=8 IN^2
(4+6)(2+2)=2*4+32
10*4=40
40=40