SOLUTION: The length of a rectangle is 4 times as long as it is wide. If the length is decreased by 10 inches and the width is increased by 2 inches the perimeter will be 80 inches. Find the
Algebra ->
Rectangles
-> SOLUTION: The length of a rectangle is 4 times as long as it is wide. If the length is decreased by 10 inches and the width is increased by 2 inches the perimeter will be 80 inches. Find the
Log On
Question 261349: The length of a rectangle is 4 times as long as it is wide. If the length is decreased by 10 inches and the width is increased by 2 inches the perimeter will be 80 inches. Find the dimensions of the original field. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! L=4W
2L+2W=P
2(L-10)+2(W+2)=80
2(4W-10)+2W+4=80
8W-20+2W+4=80
10W=80+20-4
10W=96
W=96/10
W=9.6 ANS.
2(L-10)+2(9.6+2)=80
2L-20+2*11.6=80
2L-20+23.2=80
2L=80+20-23.2
2L=76.8
L=76.8/2
L=38.4 ANS.
2*38.4+2*9.6
76.8+19.2=96 ANS. FOR THE ORIGINAL RECTANGLE.
(