document.write( "Question 450773: The perimeter of a rectangle is 110 inches. The length exceeds the width by 37 inches. Find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #314701 by simranb(63)\"\" \"About 
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Let the width be x
\n" ); document.write( "then,
\n" ); document.write( "Perimeter=2(l+b)=110 -eq 1
\n" ); document.write( "perimeter=110
\n" ); document.write( "width=x
\n" ); document.write( "length=x+37
\n" ); document.write( "substitute these values to eq 1
\n" ); document.write( "2(x+x+37)=110
\n" ); document.write( "2(2x+37)=110
\n" ); document.write( "4x+74=110
\n" ); document.write( "4x=110-74
\n" ); document.write( "4x=46
\n" ); document.write( "x=46/4
\n" ); document.write( "x=11.5 inches
\n" ); document.write( "so,width=11.5 inches
\n" ); document.write( "Length=11.5+37=48.5 inches
\n" ); document.write( "Hope this helps!
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