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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? \n" );
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Algebra.Com's Answer #420804 by checkley79(3341)![]() ![]() ![]() 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 \n" ); document.write( " |