document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1004356>Question 1004356</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 968 cm. The width is one-third as long as the length. Find the dimensions.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #620873 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=620873\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the length and width, L and W.<BR>\n" );
document.write( "We have W = (1/3)L or L = 3W.<BR>\n" );
document.write( "Perimeter P = 2L + 2W = 968<BR>\n" );
document.write( "Substituting the first equation into the second gives<BR>\n" );
document.write( "2(3W) + 2W = 968<BR>\n" );
document.write( "6W + 2W = 968<BR>\n" );
document.write( "8W = 968<BR>\n" );
document.write( "W = 121 cm<BR>\n" );
document.write( "and since this is one-third of the length,<BR>\n" );
document.write( "L = 363 cm<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );