SOLUTION: The width of a rectangle is 15 less than twice its length. If the perimeter of the rectangle is 30 feet, what are the dimensions?

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Question 1194662: The width of a rectangle is 15 less than twice its length. If the perimeter of the rectangle is 30 feet, what are the dimensions?
Found 3 solutions by MathLover1, josgarithmetic, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
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The width of a rectangle W is 15 less than twice its length L.
W=2L-15ft
If the perimeter of the rectangle is 30 feet, we have
2%28L%2BW%29=30ft...substitute W
2%28L%2B2L-15ft%29=30ft
L%2B2L-15ft=15ft
3L=15ft%2B15ft
3L=30ft
L=10ft
no find the width

W=2%2A10ft-15ft
W=20ft-15ft
W=5ft

so, the dimensions are:
length is 10ft
width is 5ft


Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
These will all work the same no matter what the given values, if the values are meaningful.

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The width of a rectangle is k less than r times its length, L. If the perimeter of the rectangle is p feet, what are the dimensions?
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Dimensions,
L, length
rL-k, width
-
p, perimeter
UNKNOWN VARIABLE, L

highlight_green%28%28rL-k%29%2BL=p%2F2%29
rL-k%2BL=p%2F2
%28r%2B1%29L-k=p%2F2
%28r%2B1%29L=p%2F2%2Bk
highlight_green%28L=%28p%2F2%2Bk%29%2F%28r%2B1%29%29
and you can plug in the given values and compute L.

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L=%2830%2F2%2B15%29%2F%282%2B1%29
L=%2815%2B15%29%2F3
L=10

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The width of a rectangle is 15 less than twice its length. If the perimeter of the rectangle is 30 feet, what are the dimensions?
With L being length, width = 2L - 15



     Length or highlight_green%28matrix%281%2C6%2C+L%2C+%22=%22%2C+30%2F3%2C+%22=%22%2C+10%2C+feet%29%29 

You should now be able to find width, and you'll then have the dimensions (length & width).