document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=521970>Question 521970</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of a rectangle is 7 meters less than twice he width. If the perimeter is 40 meters, find the dimensions 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 #346634 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=346634\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Perimeter =	40	feet												<BR>\n" );
document.write( "Length  	1	(	2	x	-7	)							<BR>\n" );
document.write( "Width 	=		x	ft										<BR>\n" );
document.write( "Perimeter =2(L+W)														<BR>\n" );
document.write( "Perimeter =	2(((2x-7)+x)<BR>\n" );
document.write( "Perimeter =	2((2x-7	+x)			<BR>\n" );
document.write( "Perimeter =4x-14+2x				<BR>\n" );
document.write( "40	=6x-14								<BR>\n" );
document.write( "40	-=6x								<BR>\n" );
document.write( "40+14=6	x									<BR>\n" );
document.write( "54=6x											<BR>\n" );
document.write( "x=9	ft	Width 										<BR>\n" );
document.write( "Length  	=11ft											 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );