document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1013964>Question 1013964</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 5 in. more than 1/3 the width. The perimeter in 130 in. What is the area? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #630329 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=630329\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Perimeter P = 2L + 2W = 130.  We also know<BR>\n" );
document.write( "L = 5 + (1/3)W  or <BR>\n" );
document.write( "L - 5 = (1/3)W  or<BR>\n" );
document.write( "W = 3L - 15<BR>\n" );
document.write( "Now substitute this into the first equation and get<BR>\n" );
document.write( "2L + 2(3L - 15) = 130<BR>\n" );
document.write( "2L + 6L - 30 = 130<BR>\n" );
document.write( "8L - 30 = 130<BR>\n" );
document.write( "8L = 160<BR>\n" );
document.write( "L = 20 inches<BR>\n" );
document.write( "W = 3(20) - 15 = 45 inches </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );