document.write( "Question 403593: Suppose that the width of a certain rectangle is 5 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle.
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Algebra.Com's Answer #285349 by mananth(16946)\"\" \"About 
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Suppose that the width of a certain rectangle is 5 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle.
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\n" ); document.write( "\n" ); document.write( "length = L
\n" ); document.write( "width = L-5
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\n" ); document.write( "Area = L(L-5)\r
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\n" ); document.write( "Perimeter = 2(L+L-5)
\n" ); document.write( "Perimeter = 2(2L-5)\r
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\n" ); document.write( "L(L-5)=2*2(2L-5)-16\r
\n" ); document.write( "\n" ); document.write( "L^2-5L=8L-20-16
\n" ); document.write( "L^2-5L-8L=-20-16
\n" ); document.write( "L^2-13L=-36
\n" ); document.write( "L^2-13L+36=0
\n" ); document.write( "L^2-9L-4L+36=0
\n" ); document.write( "L(L-9)-4(L-9)=0
\n" ); document.write( "(L-9)(L-4)=0\r
\n" ); document.write( "\n" ); document.write( "So L= 4 OR 9\r
\n" ); document.write( "\n" ); document.write( "The dimensions are 9 by 4 feet \n" ); document.write( "
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