document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=624650>Question 624650</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 46 feet. The width is 3ft less than twice the length. Find the length and width of the rectangle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #392958 by </B><A HREF=/tutors/aboutme.mpl?userid=oscargut><B>oscargut(2103)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oscargut><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=oscargut><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=392958\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solution:\r<BR>\n" );
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document.write( "Let L= length and W =width\r<BR>\n" );
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document.write( "W = 2L-3\r<BR>\n" );
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document.write( "Perimeter = 2(L+W)=2(L+2L-3) = 2(3L-3)=6L-6 = 46\r<BR>\n" );
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document.write( "6L = 46 + 6 \r<BR>\n" );
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document.write( "6L = 52\r<BR>\n" );
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document.write( "L = 52/6\r<BR>\n" );
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document.write( "L = 26/3\r<BR>\n" );
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document.write( "W = 2(26/3)-3 = 52/3-9/3 = 43/3\r<BR>\n" );
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document.write( "Answer: Width = 43/3 and Length = 26/3\r<BR>\n" );
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document.write( "You can send me more problems to my e-mail:  mthman@gmail.com </SPAN>\n" );
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