document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=913064>Question 913064</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Your friend is driving from Los Angeles to San Fransisco at an average rate of speed 55mph. After she has aleady driven 60 miles, you decide that you want to go to San Fransisco too. You are averaging 70mph.  How many hours will it take for you to catch up to her? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #554275 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=554275\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Gaining speed 70 -55= 15 mph<BR>\n" );
document.write( "Headstart 55*1.09090909 =60.0 miles<BR>\n" );
document.write( "Time to catch up 60.0/15=4.0 hours<BR>\n" );
document.write( "Algebraic solution<BR>\n" );
document.write( "55(t+1.09090909)=70t<BR>\n" );
document.write( "55t+60.0=70t<BR>\n" );
document.write( "60.0=70t-55t<BR>\n" );
document.write( "60.0=15t<BR>\n" );
document.write( "t=4.0  hours<BR>\n" );
document.write( "Distance to meeting 4.0*70=280.0 miles<BR>\n" );
document.write( "code drtch1<BR>\n" );
document.write( "no more info<BR>\n" );
document.write( " </SPAN>\n" );
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