Answer:
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.
Step-by-step explanation:
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
