document.write( "Question 1090190: A given rectangle has a length 2 feet longer than it's width. If the width is shortened by 2 feet and the length is doubled,a new rectangle is formed. The perimeter of the new rectangle is 12 more than the perimeter of the original rectangle. What are the dimensions of the original rectangle? \n" ); document.write( "
Algebra.Com's Answer #704596 by josgarithmetic(39617)\"\" \"About 
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Original rectangle dimensions, w+2 and w;
\n" ); document.write( "Perimeter \"p=2%28w%2B2%29%2B2w\".\r
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\n" ); document.write( "\n" ); document.write( "Second rectangle dimensions, 2(w+2) and w-2 for length and width;
\n" ); document.write( "Perimeter \"2%2A2%28w%2B2%29%2B2%28w-2%29=p%2B12\".\r
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\n" ); document.write( "\n" ); document.write( "STEPS\r
\n" ); document.write( "\n" ); document.write( "\"p=2w%2B4%2B2w\"
\n" ); document.write( "\"p=4w%2B4\"
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\n" ); document.write( "\"4w%2B8%2B2w-4=p%2B12\"
\n" ); document.write( "\"6w%2B4=p%2B12\"
\n" ); document.write( "\"6w-p=8\"
\n" ); document.write( "\"6w-%284w%2B4%29=8\"
\n" ); document.write( "\"6w-4w-4=8\"
\n" ); document.write( "\"2w-4=8\"
\n" ); document.write( "\"w-2=4\"
\n" ); document.write( "\"highlight%28w=6%29\"-------width, original rectangle
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\n" ); document.write( "\"w%2B2=highlight%288%29\"------length, original rectangle
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