document.write( "Question 163826: The width of a certain rectangle exceeds twice its length by three inches, and four times its lenght is twelve less than its perimeter. Find the dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "this is what i have so far, but am stuck.
\n" ); document.write( "P=2L+2W
\n" ); document.write( "W=2L+3
\n" ); document.write( "P=2L+2(2L+3)
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Algebra.Com's Answer #120686 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
You're getting there but you only used half of the information.
\n" ); document.write( "\"four times its lenght is twelve less than its perimeter\"
\n" ); document.write( "\"4L=P-12\"
\n" ); document.write( "\"4L%2B12=P\"
\n" ); document.write( "Now you have another equation for P.
\n" ); document.write( "Put that together with your other equation for P.
\n" ); document.write( "\"P=2L%2B2%282L%2B3%29\"
\n" ); document.write( "Equate them and solve for L.
\n" ); document.write( " \"4L%2B12=2L%2B2%282L%2B3%29\"
\n" ); document.write( "Re-post a question if you get stuck.
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