document.write( "Question 443324: I can't figure this out: The dimensions of a rectangle are the length is 3 in more than its width, if the length were doubled and the width decreased by 1 in., the area would be increased by 126^2, what are the length and width of the rectangle? Please HELP! \n" ); document.write( "
Algebra.Com's Answer #305793 by swincher4391(1107)\"\" \"About 
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You say the area would be increased by 126^2, do you mean 126in^2?
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\n" ); document.write( "You will have a system of equations:
\n" ); document.write( "A = lw
\n" ); document.write( "[l = w+3]
\n" ); document.write( "A = (w+3)*w
\n" ); document.write( "===============
\n" ); document.write( "2l * w-1 = A + 126
\n" ); document.write( "2(w+3) * w-1 = A + 126
\n" ); document.write( "(2w+6)*(w-1) = A +126
\n" ); document.write( "2w^2 +6w -2w -6 = A +126
\n" ); document.write( "========================
\n" ); document.write( "2w^2 +4w -132 = A
\n" ); document.write( "=====================
\n" ); document.write( "2w^2 + 4w -132 = w(w+3)
\n" ); document.write( "2w^2 + 4w + -132= w^2 +3w
\n" ); document.write( "w^2 + w + -132 = 0
\n" ); document.write( "(w-11)(w+12)
\n" ); document.write( "================
\n" ); document.write( "w = 11 or w =-12\r
\n" ); document.write( "\n" ); document.write( "-12 inches doesn't make any sense, so 11 inches is the only one that makes sense.\r
\n" ); document.write( "\n" ); document.write( "So the width is 11 inches.
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\n" ); document.write( "Plug this back in for
\n" ); document.write( "l = w+3
\n" ); document.write( "l = 11 +3
\n" ); document.write( "l = 14
\n" ); document.write( "========================
\n" ); document.write( "The dimensions are 14 x 11 or \"highlight%28l=14%29\" \"highlight%28w=11%29\".
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