document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=911226>Question 911226</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a lacrosse field is 15 yards less than twice its width, and the perimeter is 330 yards. The defensive area of the field is 5/21 of the total field area. Find the defensive area of the lacrosse field. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #552928 by </B><A HREF=/tutors/aboutme.mpl?userid=ichigo449><B>ichigo449(30)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ichigo449><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=552928\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the length of the field and y the width. The first condition in the question says that x= 2y-15. Then the perimeter gives us a linear equation: <BR>\n" );
document.write( "2y + 2(2y-15) = 330<BR>\n" );
document.write( "This can be simplified to 6y-30 = 330 or 6y = 300 so y = 60. Then,  by x = 2y-15, x = 105. Now from area of a rectangle, x*y, we get the total field area to be: 6300. Since the defensive area is 5/21 the total area we conclude that it must be 1500 yards square.  <BR>\n" );
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