document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133384>Question 1133384</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Becky and Delia, living 450 meters apart, leave their homes on a pedicab and travel toward each other. If Becky averages 3 meters per minute more than Delia and they meet after 30 minutes, then find their rates of cycling.\r<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 #750607 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=750607\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x m/min for slower<BR>\n" );
document.write( "x+3 m/minfor faster<BR>\n" );
document.write( "30 minutes for 450 meters is 15 meters/min<BR>\n" );
document.write( "combined speed is 2x+3 m/min<BR>\n" );
document.write( "2x+3=15<BR>\n" );
document.write( "2x=12<BR>\n" );
document.write( "x=6 m/min for Delia<BR>\n" );
document.write( "x+3=9 m/min for Becky </SPAN>\n" );
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