document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1010138>Question 1010138</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Morgan and Autumn start out 58.8 miles apart. They begin riding their bicycles toward each other with Morgan riding four mph faster than Autumn. If they meet in three hours, how fast was each person riding? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #626395 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=626395\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call Autumn's rate, r.  Then Morgan's rate is r+4.<BR>\n" );
document.write( "Now rate times time equals distance, so we have<BR>\n" );
document.write( "3r + 3(r+4) = 58.8 miles<BR>\n" );
document.write( "3r + 3r + 12 = 58.8<BR>\n" );
document.write( "6r + 12 = 58.8<BR>\n" );
document.write( "6r = 46.8<BR>\n" );
document.write( "r = 7.8 mph<BR>\n" );
document.write( "r+4 = 11.8 mph </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );