document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=698600>Question 698600</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangular field is 480 meters and the area is 10,800 meters squared. Find the dimensions\r<BR>\n" );
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document.write( "Can you please help me with this question? Thanks so much in advance :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #430979 by </B><A HREF=/tutors/aboutme.mpl?userid=sofiyac><B>sofiyac(983)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=sofiyac><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=sofiyac><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=430979\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's use 2 equations for this, L is Length and W is Width: <BR>\n" );
document.write( " 2L+2W=480<BR>\n" );
document.write( " L(W)=10800 rewritten as: L=10800/W<BR>\n" );
document.write( " Use substitution for this:<BR>\n" );
document.write( " 2(10800/W)+2W=480 or rewritten as: 21600/W+2W=480<BR>\n" );
document.write( " Multiply by W:<BR>\n" );
document.write( " 21600+2w^2=480W\r<BR>\n" );
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document.write( "Divide by 2 and subtract 240W:<BR>\n" );
document.write( " W^2 -240W +10800 =0<BR>\n" );
document.write( " Factor:<BR>\n" );
document.write( " (W-180)(W-60)<BR>\n" );
document.write( " Solve for W<BR>\n" );
document.write( " Width=60 meters<BR>\n" );
document.write( " Length=180 meters<BR>\n" );
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