document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=190440>Question 190440</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dave averaged 30 miles per hour on his bike trip to a friend's house.  His return trip took an hour longer, as he averaged 20 miles per hour.  What was his one-way distance to his friends?\r<BR>\n" );
document.write( "\n" );
document.write( "d = s * t </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #142926 by </B><A HREF=/tutors/aboutme.mpl?userid=msaxby><B>msaxby(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=msaxby><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=142926\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let t = time to the friends house.\r<BR>\n" );
document.write( "\n" );
document.write( "Since d = s *t,  30*t = distance to the friend's house.<BR>\n" );
document.write( "and   20*(t+1) = the distance back from the friend's house.\r<BR>\n" );
document.write( "\n" );
document.write( "SO:   30*t = 20*(t+1)<BR>\n" );
document.write( "      30*t = 20*t + 20 <BR>\n" );
document.write( "      10*t = 20<BR>\n" );
document.write( "         t = 2 (hours).<BR>\n" );
document.write( "Since Dave averaged 30 miles per hour, it is 30*2 miles to his friend's house or 60 miles. \r<BR>\n" );
document.write( "\n" );
document.write( "(Check:  Averaging 20 mph would take 60 / 20 or 3 hours - that's 1 hour longer.) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );