document.write( "Question 1179263: A rectangle is 3 inches longer than it is wide. Numerically, its area exceeds its perimeter by 14. Find the perimeter.
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Algebra.Com's Answer #808783 by MathLover1(20850)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"L=W%2B3\"........eq.1\r
\n" ); document.write( "\n" ); document.write( "area: \"A=LW\"\r
\n" ); document.write( "\n" ); document.write( "perimeter:\"P=2%28L%2BW%29\"\r
\n" ); document.write( "\n" ); document.write( "its area exceeds its perimeter by \"14\"\r
\n" ); document.write( "\n" ); document.write( "\"LW=2%28L%2BW%29%2B14\"..........substitute \"L\" from eq.1\r
\n" ); document.write( "\n" ); document.write( "\"%28W%2B3%29W=2%28W%2B3%2BW%29%2B14\"\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B3W=2%282W%2B3%29%2B14\"\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B3W=4W%2B6%2B14\"\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2%2B3W-4W-20=0\"\r
\n" ); document.write( "\n" ); document.write( "\"W%5E2-W-20=0\"\r
\n" ); document.write( "\n" ); document.write( "\"%28W+-+5%29+%28W+%2B+4%29+=+0\"\r
\n" ); document.write( "\n" ); document.write( "=>\"W=+5\" or \"W=+-4\"-> disregard negative solution\r
\n" ); document.write( "\n" ); document.write( "go to\r
\n" ); document.write( "\n" ); document.write( "\"L=W%2B3\"........eq.1, substitute \"W\"\r
\n" ); document.write( "\n" ); document.write( "\"L=5%2B3\"\r
\n" ); document.write( "\n" ); document.write( "\"L=8\"\r
\n" ); document.write( "\n" ); document.write( "then perimeter is:\r
\n" ); document.write( "\n" ); document.write( "\"P=2%28L%2BW%29\"
\n" ); document.write( "\"P=2%288%2B5%29\"
\n" ); document.write( "\"highlight%28P=26%29\" inches\r
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\n" ); document.write( "\n" ); document.write( "check if area exceeds its perimeter by \"14\": \r
\n" ); document.write( "\n" ); document.write( "\"A=LW\"
\n" ); document.write( "\"A=8%2A5\"
\n" ); document.write( "\"A=40\"\r
\n" ); document.write( "\n" ); document.write( "\"40-26=14\"\r
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