document.write( "Question 134610: suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80. \r
\n" ); document.write( "\n" ); document.write( "a)set up an equation for the perimeter involving only L, the length of the rectangle. \r
\n" ); document.write( "\n" ); document.write( "b)solve this equation algebraically to find the lenghth of the rectangle. Find the width as well.
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Algebra.Com's Answer #98474 by solver91311(24713)\"\" \"About 
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Formula for perimeter: \"P=2L+%2B+2W\"\r
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\n" ); document.write( "\n" ); document.write( "Width is 2 inches shorter than length, so: \"W=L-2\"\r
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\n" ); document.write( "\n" ); document.write( "Substituting:\r
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\n" ); document.write( "\n" ); document.write( "\"P=2L+%2B+2%28L-2%29\"
\n" ); document.write( "\"P=4L-4\"\r
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\n" ); document.write( "\n" ); document.write( "Given P = 80,\r
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\n" ); document.write( "\n" ); document.write( "\"4L-4=80\"
\n" ); document.write( "\"4L=84\"
\n" ); document.write( "\"L=21\", hence \"W=19\"\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "\"P=2%2821%29+%2B+2%2819%29=42%2B38=80\" Answer checks.
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