document.write( "Question 231238: Translate the problem into a pair of linear equations in two vaiables
\n" ); document.write( "The perimeter of a rectangle is 72 m. If the width were doubled and the length were increased by 9 m, then the perimeter would be 114 m. What are the dimensions?
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Algebra.Com's Answer #171230 by checkley77(12844)\"\" \"About 
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2L+2W=72
\n" ); document.write( "2(L+9)+2(2W)=114
\n" ); document.write( "2L+18+4W=114
\n" ); document.write( "2L+4W=114-18
\n" ); document.write( "2L+4W=96
\n" ); document.write( "2L+2W=72 SUBTRACT THIS EQUATION
\n" ); document.write( "--------------
\n" ); document.write( "2W=24
\n" ); document.write( "W=24/2
\n" ); document.write( "W=12 ANS. FOR THE ORIGINAL WIDTH.
\n" ); document.write( "2L+2*12=72
\n" ); document.write( "2L+24=72
\n" ); document.write( "2L=72-24
\n" ); document.write( "2L=48
\n" ); document.write( "L=48/2
\n" ); document.write( "L=24 ANS. FOR THE ORIGINAL LENGTH.
\n" ); document.write( "PROOF:
\n" ); document.write( "2*24+2*12=72
\n" ); document.write( "48+24=72
\n" ); document.write( "72=72
\n" ); document.write( "2(24+9)+2(2*12)=114
\n" ); document.write( "2*33+2*24=114
\n" ); document.write( "66+48=114
\n" ); document.write( "114=114\r
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