document.write( "Question 8539: The width of a rectangle is 2 feet less than 10 times its lenght. If the perimeter of the rectangle is 62 feet find its dimensions.\r
\n" ); document.write( "\n" ); document.write( "I began with 2l+2w=P\r
\n" ); document.write( "\n" ); document.write( "So.....2L+(2-10L)+(2-10L)=62\r
\n" ); document.write( "\n" ); document.write( "But that's not correct. How do I set this up to solve? \n" ); document.write( "
Algebra.Com's Answer #4760 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
\"2 feet less than 10 times its length\" means you must start with 10L and then take away 2 feet from this. It should be 10L-2, not 2-10L.\r
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\n" ); document.write( "\n" ); document.write( "Solve this:\r
\n" ); document.write( "\n" ); document.write( "2L + 10L-2 + 10L-2 = 62
\n" ); document.write( "22L - 4 = 62
\n" ); document.write( "22L = 66
\n" ); document.write( "L=3 = Length\r
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\n" ); document.write( "\n" ); document.write( "Then, 10L-2 = 10(3)-2= 28 = width.\r
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\n" ); document.write( "\n" ); document.write( "Interesting, the length and the width seem to be reversed. It could be an oversight by the person who made up the problem. I've done it myself.\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC\r
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