SOLUTION: The width of a rectangle is 12 units less than its length. If you add 30 units to both the length and width, you double the perimeter. Find the length and width of the original rec

Question 130570: The width of a rectangle is 12 units less than its length. If you add 30 units to both the length and width, you double the perimeter. Find the length and width of the original rectangle.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
W=L-12 OR 2(L-12)+2L=P OR 2L-24+2L=P OR 4L-24=P ORIGINAL PERIMETER.
2(L-12+30)+2(L+30)=2P OR 2L+36+2L+60=2P OR 4L+96=2P DOUBLED PERIMETER.
NOW MULTIPLY THE FIRST EQUATION BY 2 & SET THEM EQUAL.
2(4L-24)=4L+96
8L-48=4L+96
8L-4L=96+48
4L=144
L=144/4
L=36 IS THE ORIGINAL LENGTH.
W=36-12
W=24 FOR THE ORIGINAL WIDTH.
PROOF
2*24+2*36=48+72=120
2(24+30)+2(36+30)=2*54+2*66=108+132=240
240=2*120
240=240

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