SOLUTION: The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 64 in., find the width of the rectangle.

Algebra ->  Expressions-with-variables -> SOLUTION: The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 64 in., find the width of the rectangle.      Log On


   



Question 71190: The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 64 in., find the width of the rectangle.
Found 2 solutions by hailu, checkley75:
Answer by hailu(13) About Me  (Show Source):
You can put this solution on YOUR website!
L=8+2W
(L+W)2=64
W=?
substitute for (L)
(8+2W+W)2=64
(8+3W)2=64
16+6W=64
6W=64-16
6W=48
W=48/6
W=8
width is 8
therefore (L) is
L=8+2(8)
L=8+16
L=24

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
L=8+2W
64=2(8+2W)+2W
64=16+4W+2W
64-16=6W
48=6W
48/6=W
8=W ANSWER FOR THE WIDTH
PROOF
64=2(8+2*8)+2*8
64=2(8+16)+16
64=2*24+16
64=48+16
64=64