document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=41086>Question 41086</A>: <I><SPAN STYLE=\"word-break: break-all;\"> two cars leave the same point traveling in opposite direction. the second car travels 10 mph fasterthan the first and after 3 hours they are 330 miles apart. how fast is each car traveling?thanks very much </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #26546 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=26546\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the slower car travel at a rate of r.  The faster travels at a rate of r+10.<BR>\n" );
document.write( "We always use rate*time = distance in these travel problems.<BR>\n" );
document.write( "Here their combined distance after 3 hours is 330 miles...so<BR>\n" );
document.write( "3r + 3(r+10) = 330<BR>\n" );
document.write( "3r + 3r + 30 = 330<BR>\n" );
document.write( "6r + 30 = 330<BR>\n" );
document.write( "6r = 300<BR>\n" );
document.write( "r = 50<BR>\n" );
document.write( "r+10 = 60 </SPAN>\n" );
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