SOLUTION: the width of a rectangle is 5 inches shorter than the length and that the perimeter or the rectangle is 50 inches. set up and equation for the perimeter involving the length. solve

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Question 303652: the width of a rectangle is 5 inches shorter than the length and that the perimeter or the rectangle is 50 inches. set up and equation for the perimeter involving the length. solve this equation algebraically to find the length and width.
Thank-you
Shirley Sims
shirleysims174@aol.com
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
the width of a rectangle is 5 inches shorter than the length and that the
W = L - 5
:
perimeter or the rectangle is 50 inches.
2L + 2W = 50
:
set up an equation for the perimeter involving the length.
Replace W with (L-5)
2L + 2(L-5) = 50
solve this equation algebraically to find the length and width.
2L + 2L - 10 = 50
4L = 50 + 10
4L = 60
L = 60%2F4
L = 15 in is the lenght
then
W = L - 5
W = 15 - 5
W = 10 in is the width
:
:
Check solution by finding the perimeter with these values for L & W
2(15) + 2(10) = 50