document.write( "Question 4714: perimeter of a rectangle is 22 inches and its length is 4 inches less than twice its width. Set up a system of linear equations and solve to find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #2229 by rapaljer(4671)\"\" \"About 
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I think there is an error in the previously posted solution.\r
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\n" ); document.write( "\n" ); document.write( "Let W = Width of the rectangle
\n" ); document.write( "L = Length of the rectangle\r
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\n" ); document.write( "\n" ); document.write( "Perimeter: 2W + 2L = 22
\n" ); document.write( "2nd Equation: L = 2W - 4\r
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\n" ); document.write( "\n" ); document.write( "Solve by substituting L from the second equation into the first equation:
\n" ); document.write( "2W + 2(_____) = 22
\n" ); document.write( "2W + 2(2W -4) = 22
\n" ); document.write( "2W + 4W - 8 = 22
\n" ); document.write( "2W = 30
\n" ); document.write( "W= 5\r
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\n" ); document.write( "\n" ); document.write( "Substitute into second equation:
\n" ); document.write( "L = 2W - 4
\n" ); document.write( "L = 2(5) - 4
\n" ); document.write( "L = 10 - 4 = 6\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "P = 2W + 2L
\n" ); document.write( "P = 2(5) + 2(6)
\n" ); document.write( "P = 10 + 12 = 22\r
\n" ); document.write( "\n" ); document.write( "AND
\n" ); document.write( "L = 2W - 4
\n" ); document.write( "6 = 2(5) - 4
\n" ); document.write( "6 = 10 - 4\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC
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