document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=808189>Question 808189</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If one elevator is on the 139th floor and travels two seconds per floor, and the other elevator is on the 14th floor and travels three seconds per floor, on what floor will they meet. What formula would you use to solve this problem. thank you.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #486829 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=486829\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 139-14=125 floors difference<BR>\n" );
document.write( "u+d=125,<BR>\n" );
document.write( "3u=2d<BR>\n" );
document.write( "d=75<BR>\n" );
document.write( "u=50\r<BR>\n" );
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document.write( "50+14=64 50 floors going up<BR>\n" );
document.write( "139-75=64 75 floors going down\r<BR>\n" );
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document.write( "3*50=150 seconds going up<BR>\n" );
document.write( "2*75=150 seconds going down\r<BR>\n" );
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