document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #95365 by checkley71(8403)\"\" \"About 
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W=L-12 OR 2(L-12)+2L=P OR 2L-24+2L=P OR 4L-24=P ORIGINAL PERIMETER.
\n" ); document.write( "2(L-12+30)+2(L+30)=2P OR 2L+36+2L+60=2P OR 4L+96=2P DOUBLED PERIMETER.
\n" ); document.write( "NOW MULTIPLY THE FIRST EQUATION BY 2 & SET THEM EQUAL.
\n" ); document.write( "2(4L-24)=4L+96
\n" ); document.write( "8L-48=4L+96
\n" ); document.write( "8L-4L=96+48
\n" ); document.write( "4L=144
\n" ); document.write( "L=144/4
\n" ); document.write( "L=36 IS THE ORIGINAL LENGTH.
\n" ); document.write( "W=36-12
\n" ); document.write( "W=24 FOR THE ORIGINAL WIDTH.
\n" ); document.write( "PROOF
\n" ); document.write( "2*24+2*36=48+72=120
\n" ); document.write( "2(24+30)+2(36+30)=2*54+2*66=108+132=240
\n" ); document.write( "240=2*120
\n" ); document.write( "240=240\r
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