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

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Question 57559: The length of a rectangle is 8 inches more than twice its width. If the perimeter of the rectangle is 64 inches, what is the width of the rectangle
Found 2 solutions by checkley71, funmath:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W+8
2L+2W=64
2(2W+8)+2W=64
4W+16+2W=64
6W=64-16
6W=48
X=48/6
X=8
PROOF
L=2*8+8=16+8=24
2*24+2*8=64
48+16=64
64=64

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 8 inches more than twice its width. If the perimeter of the rectangle is 64 inches, what is the width of the rectangle?
:
The formula for the perimeter of a rectangle is: highlight%28P=2L%2B2W%29, where P=perimeter, L=lenght, and W=width.
:
Given:
P=64 in
Let width, W=x
Then length is 8 more than twice the width:L=2x+8
:Substitute those values into your formula and solve for x:
64=2%282x%2B8%29%2B2%28x%29
64=4x%2B16%2B2x
64=6x%2B16
64-16=6x%2B16-16
48=6x
48%2F6=6x%2F6
8=x
The width, x=8 in.
Happy Calculating!!!