SOLUTION: suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50. set up an equation for the perimeter involving only L, the

Algebra ->  Linear-equations -> SOLUTION: suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50. set up an equation for the perimeter involving only L, the      Log On


   

Question 103909: suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
set up an equation for the perimeter involving only L, the length of the rectangle.
sove this equation algebraically to find the length of the rectange. find the width as well.
thanks for the help-renee
Found 2 solutions by stanbon, checkley75:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
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Let length be L ;then width = L-5
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set up an equation for the perimeter involving only L, the length of the rectangle.
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Perimeter = 2L + 2(L-5) = 4L-10
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solve this equation algebraically to find the length of the rectangle. find the width as well.
50 = 4L-10
4L = 60
L = 15 inches (this is the length)
L-5 =10 inches (this is the width)
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Cheers,
Stan H.

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
W=L-5
PERIMETER=2L+2W
50=2L+2(L-5)
50=2L+2L-10
50=4L-10
4L=50+10
4L=60
L=60/4
L=15 ANSWER FOR THE LENGTH.
W=15-5
W=10 ANSWER FOR THE WIDTH.
PROOF
2*15+2*10=50
30+20=50
50=50