document.write( "Question 457513: The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.\r
\n" ); document.write( "\n" ); document.write( "Ok. I think I start out by
\n" ); document.write( "2l + 2w = 122
\n" ); document.write( "w+25=1\r
\n" ); document.write( "\n" ); document.write( "This is confusing, can someone help?
\n" ); document.write( "

Algebra.Com's Answer #313913 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
The perimeter of a rectangle is 122 inches. The length exceeds the withd by 25 inches. find the length and width.
\n" ); document.write( "Ok. I think I start out by
\n" ); document.write( "2l + 2w = 122
\n" ); document.write( "w+25=1
\n" ); document.write( "-----------------------
\n" ); document.write( "Let the width be \"x\"
\n" ); document.write( "Then length = \"x+25\"
\n" ); document.write( "----
\n" ); document.write( "Perimeter = 2(width + length)
\n" ); document.write( "122 = 2(x + x+25)
\n" ); document.write( "61 = 2x+25
\n" ); document.write( "2x = 36
\n" ); document.write( "x = 18 inches (width)
\n" ); document.write( "x+25 = 43 inches (length)
\n" ); document.write( "=============================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
\n" ); document.write( "
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