document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=43932>Question 43932</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the perimeter of a rectangle is 600. the lenght of the rectangle is 2 more than 3 times the width. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #28834 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=28834\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the perimeter be P.<BR>\n" );
document.write( "P = 2L + 2W = 600<BR>\n" );
document.write( "Then L = 2 + 3W<BR>\n" );
document.write( "Now substitute and solve for the width, W.<BR>\n" );
document.write( "2(2 + 3W) + 2W = 600<BR>\n" );
document.write( "4 + 6W + 2W = 600<BR>\n" );
document.write( "8W + 4 = 600<BR>\n" );
document.write( "8W = 596<BR>\n" );
document.write( "W = 74.5<BR>\n" );
document.write( "L = 2 + 3W = 2 + 3(74.5) =<BR>\n" );
document.write( "L = 225.5 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );