document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=324448>Question 324448</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The width of a rectangle is 3 less than the length. If the perimeter is 50 meters, find the length and width.. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #232216 by </B><A HREF=/tutors/aboutme.mpl?userid=graphmatics><B>graphmatics(170)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=graphmatics><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=graphmatics><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=232216\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If x is the length of the rectangle then the width of the rectangle y = x-3. The perimeter is twice the width plus twice the length so 2*x + 2*(x-3) = 50. We solve this equation for x.\r<BR>\n" );
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document.write( "2*x + 2*(x-3) - 50 <BR>\n" );
document.write( "2*x + 2*x - 2*3 = 50<BR>\n" );
document.write( "4*x -6 = 50<BR>\n" );
document.write( "4*x = 50 + 6<BR>\n" );
document.write( "4*x = 56<BR>\n" );
document.write( "x - 56/4<BR>\n" );
document.write( "x = 14\r<BR>\n" );
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document.write( "so y = 14 - 3<BR>\n" );
document.write( "   y = 11\r<BR>\n" );
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document.write( "So the length is 14 and the width is 11.  </SPAN>\n" );
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