SOLUTION: can you help me with this problem? it is... the perimeter of a rectangle is 160ft. one fourth the length is the same as twice the width. find the dimension of the rectangle.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: can you help me with this problem? it is... the perimeter of a rectangle is 160ft. one fourth the length is the same as twice the width. find the dimension of the rectangle.       Log On


   

Question 140755: can you help me with this problem? it is... the perimeter of a rectangle is 160ft. one fourth the length is the same as twice the width. find the dimension of the rectangle.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
PERIMETER=2L+2W
L/4=2W
L=8W
160=2*8W+2W
160=16W+2W
160=18W
W=160/18
W=80/9 ANSWER FOR THE WIDTH.
L=8*80/9
L=640/9 ANSWER FOR THE LENGTH.
PROOF
160=2*80/9+2*640/9
160=160/9+1280/9
160=1440/9
160=160