document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=882986>Question 882986</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A new youth center is being built in Roxbury. The perimeter of the rectangular fried us 340 yards. The length of the field us 6 yards less than triple the width. That are the dimensions of the playing field? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #533287 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=533287\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2L + 2W = P<BR>\n" );
document.write( "2L + 2W = 340<BR>\n" );
document.write( "L = 3W-6<BR>\n" );
document.write( "Substitute the known value of L from the bottom equation into the previous one.<BR>\n" );
document.write( "2(3W-6) + 2W = 340<BR>\n" );
document.write( "6W-12+2W = 340<BR>\n" );
document.write( "8W-12 = 340<BR>\n" );
document.write( "Add 12 to each side<BR>\n" );
document.write( "8W = 352<BR>\n" );
document.write( "Divide each side by 8<BR>\n" );
document.write( "W = 44<BR>\n" );
document.write( ".<BR>\n" );
document.write( "If the width is 44, the length is (3*44)-6, or 126<BR>\n" );
document.write( "Perimeter fencing = 126+126+44+44, or 340 </SPAN>\n" );
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