document.write( "Question 62022: The perimeter of a rectangle is 82 yards. The width is 15 yardsless than the length. Find the length and width of the rectangle.\r
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Algebra.Com's Answer #42755 by DJ_Sammie(24)\"\" \"About 
You can put this solution on YOUR website!
l=length
\n" ); document.write( "l-15=width
\n" ); document.write( " 2l+2w=P
\n" ); document.write( "p=perimeter
\n" ); document.write( "2l+2(l-15)=82
\n" ); document.write( "2l+2l-30=82
\n" ); document.write( "4l-30=82
\n" ); document.write( " +30 +30
\n" ); document.write( " 4l=112
\n" ); document.write( " 4l/4=112/4
\n" ); document.write( " l=28
\n" ); document.write( "length=28yards
\n" ); document.write( "width=l-15
\n" ); document.write( "width=28-15
\n" ); document.write( "width=13yards
\n" ); document.write( "The width is 13 yards and the length is 28 yards.
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