SOLUTION: The length of a certain rectangle is 10 meters greater than twice its width. if its length were doubled and the width were halved, the perimeter would be increased by 80 meters. Fi

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Question 158252: The length of a certain rectangle is 10 meters greater than twice its width. if its length were doubled and the width were halved, the perimeter would be increased by 80 meters. Find the area of the original rectangle.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W+10
W=W
2(2W+10)+2W=2*2L+2W/2-80
4W+20+2W=4(2W+10)+W-80
6W+20=8W+40+W-80
6W-9W=-40-20
-3W=-60
W=-60/-3
W=20 FOR THE ORIGINAL WIDTH
L=2*20+10
L=40+10
L=50 FOR THE ORIGINAL LENGTH.
PROOF:
2(2*20+10)+2*20=4(2*20+10)+2*20/2-80
2(40+10)+40=4(40+10)+40/2-80
2*50+40=4*50+20-80
100+40=200+20-80
140=220-80
140=140