document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=465176>Question 465176</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of the rectangle is 3 meters more than its width.If the perimeter is 54 meters,what are its dimensions?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #318743 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=318743\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the length of the rectangle is 3 meters more than its width.If the perimeter is 54 meters,what are its dimensions?\r<BR>\n" );
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document.write( "Let L =  W+3<BR>\n" );
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document.write( "Perimeter = 2(L+W)<BR>\n" );
document.write( "Substitute and solve for \"W\":<BR>\n" );
document.write( "54 = 2(W+3 + W)<BR>\n" );
document.write( "27 = 2W+3<BR>\n" );
document.write( "2W = 24<BR>\n" );
document.write( "Width = 12 meters<BR>\n" );
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document.write( "Solve for \"L\":<BR>\n" );
document.write( "Length = W + 3<BR>\n" );
document.write( "Length = 15 meters<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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