document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=62466>Question 62466</A>: <I><SPAN STYLE=\"word-break: break-all;\"> At 6:00 A.M. a Pony Express rider leaves San Antonio for Las Cruces.  His horse travels at 10 miles per hour.  A second Pony Express Rider leaves at 9:00 A.M. with the sole purpose of giving an additional package for the first Pony Express rider to deliver.  The second rider’s horse travels at 18 miles per hour.  At what time will the second Pony Express rider catch up to the first Pony Express rider.  (Assume that the horses maintain the same rate of speed and that no hindrances affect either horse or rider). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #43223 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=43223\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> DISTANCE=RATE*TIME SO<BR>\n" );
document.write( "D=10T FOR THE FIRST RIDER<BR>\n" );
document.write( "FOR THE SECOND RIDER WE HAVE <BR>\n" );
document.write( "D=18(T-3) SEEING AS THE DISTANCES ARE EQUAL WE HAVE <BR>\n" );
document.write( "10T=18(T-3)<BR>\n" );
document.write( "10T=18T-54<BR>\n" );
document.write( "10T-18T=-54<BR>\n" );
document.write( "-8T=-54<BR>\n" );
document.write( "T=-54/-8<BR>\n" );
document.write( "T=6.75 HOURS AFTER 6:00 AM WHICH MAKES IT 6+6.75 OR  12:45 P.M. </SPAN>\n" );
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