document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=630177>Question 630177</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Wilma drove at an average speed of 55 mi/h from her home in City A to visit her sister in City B.  She stayed in City B 20 hours, and on the trip back averaged 45 mi/h.  She returned home 41 hours after leaving.  How many miles is City A from City B? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #396759 by </B><A HREF=/tutors/aboutme.mpl?userid=graphmatics><B>graphmatics(170)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=graphmatics><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=graphmatics><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=396759\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the distance from City A to City B. The time from City A to City B is x/55. The time from City B to City A is x/45. So <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%2F55%2Bx%2F45=41-20+ ALIGN=MIDDLE  ALT=\"+x%2F55%2Bx%2F45=41-20+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+45%2Ax%2B55%2Ax=21%2A45%2A55+ ALIGN=MIDDLE  ALT=\"+45%2Ax%2B55%2Ax=21%2A45%2A55+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+100%2Ax=51975+ ALIGN=MIDDLE  ALT=\"+100%2Ax=51975+\"   DISABLED_style= \"max-width:90%; height:auto;\" > x=519.75 miles </SPAN>\n" );
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