document.write( "Question 266830: the perimetere of a rectangle is 20 in and the area is 16sq in. find the diminsions \n" ); document.write( "
Algebra.Com's Answer #196112 by ankor@dixie-net.com(22740)\"\" \"About 
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the perimeter of a rectangle is 20 in
\n" ); document.write( "2L + 2W = 20
\n" ); document.write( "Simplify, divide by 2
\n" ); document.write( "L + W = 10
\n" ); document.write( "W = (10-L); this form for substitution
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\n" ); document.write( "and the area is 16sq in.
\n" ); document.write( "L * W = 16
\n" ); document.write( "Replace W with (10-L)
\n" ); document.write( "L * (10-L) = 16
\n" ); document.write( "-L^2 + 10L - 16 = 0; our old friend, the quadratic equation
\n" ); document.write( "Multiply by -1, change the signs, makes it easier to factor
\n" ); document.write( "L^2 - 10L + 16 = 0
\n" ); document.write( "Factor
\n" ); document.write( "(L-8)(L-2) = 0
\n" ); document.write( "Two solutions
\n" ); document.write( "L = 8; then W = 2
\n" ); document.write( "L = 2: then W = 8
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\n" ); document.write( "Check solution in perimeter equation
\n" ); document.write( "2(8) + 2(2) = 20
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\n" ); document.write( "\n" ); document.write( "find the dimensions: 8 in by 2 in \n" ); document.write( "
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