SOLUTION: Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.

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Question 167583: Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!
Remember: P%5BR%5D=2%28L%2BW%29, WORKING EQN
But, W=L-2in, EQN 1
Subst, EQN 1 in our WORKING EQN:
P%5BR%5D=2%28L%2BL-2%29
P%5BR%5D=2%282L-2%29
80=4L-4
80%2B4=4L -------> cross%284%29L%2Fcross%284%29=cross%2884%2921%2Fcross%284%29
highlight%28L=21in%29
Via EQN 1,
W=21-2=highlight%2819in=W%29
check Via WORKING EQN:
80in=2%2821%2B19%29
80in=2%2840%29
80in=80in
Thank you,
Jojo