document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=331364>Question 331364</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 14 more than the width. The perimeter of the rectangle is 264 meters. What are the dimensions of the rectangle?\r<BR>\n" );
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document.write( "Please help!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #237492 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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=237492\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Start with the formula for the perimeter of a rectangle:<BR>\n" );
document.write( "P = 2(L+W) Now substitute P = 264 and L = W+14<BR>\n" );
document.write( "264 = 2(W+14+W) Divide both sides by 2.<BR>\n" );
document.write( "132 = W+W+14 Simplify.<BR>\n" );
document.write( "132 = 2W+14 Subtract 14 from both sides.<BR>\n" );
document.write( "118 = 2W Finally, divide both sides by 2.<BR>\n" );
document.write( "59 = W and L = 59+14 = 73<BR>\n" );
document.write( "The dimensions are:<BR>\n" );
document.write( "Length = 73 meters.<BR>\n" );
document.write( "Width = 59 meters. </SPAN>\n" );
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