SOLUTION: 20. 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. A) 7 in. B) 8 in. C) 9 in. D) 1

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: 20. 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. A) 7 in. B) 8 in. C) 9 in. D) 1      Log On

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Question 56777: 20. 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.
A) 7 in.
B) 8 in.
C) 9 in.
D) 10 in.
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Found 2 solutions by stanbon, tutorcecilia:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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.
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Let width be "w". Then length is "2w+8".
Perimeter = 2(length + width)
64 = 2(2w+8+w)
32=3w+8
3w=24
x=8 inches = width
2w+8=24 inches = length
Cheers,
Stan H.

Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
P=2(length+width) [Use the formula for the perimeter of a rectangle]
P = 64
Length = 2w+8
Width = w
64=2(2w+8+w) [Plug-in the values]
64=2(8+3w) [Simplify]
64=16+6w [Solve for w]
64-16=6w
48=6w
8=w
.
length = 2w+8=(2)(8)+8=24
width = 8
.
Check by plugging all of the values back into the formula for the perimeter of a rectangle:
64=2(24 + 8)
64=2(32)
64=64 [Checks out]