document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=972803>Question 972803</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #595040 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=595040\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going? <BR>\n" );
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document.write( "d = r*t<BR>\n" );
document.write( "d = 50t = 30(t+1)<BR>\n" );
document.write( "50t = 30t + 30<BR>\n" );
document.write( "t = 1.5 hours<BR>\n" );
document.write( "d = 50*1.5 = 75 miles </SPAN>\n" );
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