Question 377005: the length of a rectangle is 15 cm more than its width.A second rectangle whose perimeter is 72cm is 5cm wider but 2cm shorter than the first rectangle.What are dimensions of each rectangle? l=x+15cm Pis72cm=x+5cm-2cm??????????????????
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! : the length of a rectangle is 15 cm more than its width.
L - W = 15
:
A second rectangle whose perimeter is 72cm is 5cm wider but 2cm shorter than the first rectangle.
2(L-2) + 2(W+5) = 72
Simplify, divide by 2
L - 2 + W + 5 = 36
L + W + 3 = 36
L + W = 36 - 3
L + W = 33
:
Use elimination here, add the two equations
L - W = 15
L + W = 33
---------------adding eliminates W, find L
2L = 48
L = 24 cm, original length
then
W = 24 - 15
W = 9 cm original width
:
24 - 2 = 22 cm, 2nd rectangle length
9 + 5 = 14 cm, 2nd rectangle width
:
:
Check by finding the perimeter of the 2nd rectangle
2(22) + 2(14) =
44 + 28 = 72
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