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\n" ); document.write( "\n" ); document.write( "the perimeter of a rectangle is 500 yards. What are the dimensions of the rectangle if the length is 50 yards more than the width?
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Algebra.Com's Answer #420804 by checkley79(3341)\"\" \"About 
You can put this solution on YOUR website!
P=2L+2W
\n" ); document.write( "L=W+50
\n" ); document.write( "500=2(W+50)+2W
\n" ); document.write( "500=2W+100+2W
\n" ); document.write( "500=4W+100
\n" ); document.write( "4W=500-100
\n" ); document.write( "4W=400
\n" ); document.write( "W=400/4
\n" ); document.write( "W=100 ANS.
\n" ); document.write( "L=100+50=150 ANS.
\n" ); document.write( "PROOF:
\n" ); document.write( "500=2*150+2*100
\n" ); document.write( "500=300+200
\n" ); document.write( "500=500
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