SOLUTION: The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle. I worked this problem and I ca

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Question 54812:
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.

I worked this problem and I came up with 9 in. Hope I am right!!
Found 2 solutions by stanbon, funmath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
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Let the width be "x"
Length = 2x+2
Perimeter = 2(length + width)
52 = 2(3x+2)
26=3x+2
3x=24
x=8
width = 8 inches
Cheers,
Stan H.

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.
You're almost right.
Let the width =x
then the length=2x+2
The perimeter=52 in.
The formula for the perimeter of a rectangle is: highlight%28P=2W%2B2L%29 where P=perimeter, W=width, and L=length.
P=52, W=x, and L=2x+2
52=2(x)+2(2x+2)
52=2x+4x+4
52=6x+4
52-4=6x+4-4
48=6x
48/6=6x/6
8=x
The width:x=8 in.
Happy Calculating!!!