document.write( "Question 343497: I cannot figure out how to set up an equation to find the length and the width of the rectange. The word problem is:\r
\n" ); document.write( "\n" ); document.write( "The perimeter of a rectangle is 32 inches, and the area is 60 square inches. Find the length and width of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #245838 by ankor@dixie-net.com(22740)\"\" \"About 
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The perimeter of a rectangle is 32 inches, and the area is 60 square inches.
\n" ); document.write( " Find the length and width of the rectangle.
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\n" ); document.write( "Start with the perimeter equation
\n" ); document.write( "2L + 2W = 32
\n" ); document.write( "Simplify, divide by 2
\n" ); document.write( "L + W = 16
\n" ); document.write( "W = (16-L); use this form for substitution
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\n" ); document.write( "The area equation
\n" ); document.write( "L * W = 60
\n" ); document.write( "Replace W with (16-L)
\n" ); document.write( "L*(16-L) = 60
\n" ); document.write( "16L - L^2 = 60
\n" ); document.write( "Arrange as a quadratic equation
\n" ); document.write( "-L^2 + 16L - 60 = 0
\n" ); document.write( "Factors easier if we change the signs, mult by -1
\n" ); document.write( "L^2 - 16L + 60 = 0
\n" ); document.write( "Factors to
\n" ); document.write( "(L-6)(L-10) = 0
\n" ); document.write( "Two solutions
\n" ); document.write( "L = 10, then W=6
\n" ); document.write( "or
\n" ); document.write( "L = 6, then W=10
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\n" ); document.write( "You can confirm this by finding the perimeter and area when L=10, W=6 \n" ); document.write( "
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