Arectangular field is 4 times as long as it is wide. if the length is decreased by 10 feet and the width is increased by 2 feet, the perimeter will be 80 feet. find the dimensions of the original field. the original dimensions are feet long by feet wide.

L = 4W [A rectangular field is 4 times as long as it is wide.]
2(L-10) + 2(W+2) = 80 [If the length is decreased by 10 feet and the width is increased by 2 feet, the perimeter will be 80 feet. ]
Distributed: (2L - 20) + (2W + 4) = 80
Substitute the known value of L from the first equation into the second.
8W - 20 + 2W + 4 = 80
10 W -16 = 80
Add 16 to each side
10W = 96

Divide each side by 10
W = 9.6 <<--original dimension
L = 9.6 * 4 = 38.4 <<--original dimension

...

If the length (38.4) is decreased by 10 feet (which would make it 28.4) and the width (9.6) is increased by 2 feet (11.6), the perimeter will be 80 feet. 
28.4 + 28.4 + 11.6 + 11.6 = 80 feet.


Have a nice day :D

Neither is the answer