document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=429237>Question 429237</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A doctor drove 200 miles to attend a conference. Bad weather on the return trip caused her average speed to be 10 mph less than it was going to the convention. If the return trip took 1 hour longer, how fast did she drive in both directions?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #298206 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=298206\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let X represent the doctor's original speed. Then his time taken to attend the conference is:200/x. Likewise, his return trip took 200/x-10 hours.  So, our equation looks like:\r<BR>\n" );
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document.write( "200/X + 1=200/X-10\r<BR>\n" );
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document.write( "Multiplying through by (X)(X-10), we get:<BR>\n" );
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document.write( "200X-2000+X^2-10X=200X<BR>\n" );
document.write( "X^2-10X-2000=0<BR>\n" );
document.write( "(X-50)(X+40)=0\r<BR>\n" );
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document.write( "Solving for X, and tossing out the negative result (-40), we get his speed going there to be 50 mph. His speed coming back was 50-10 mph, or 40 mph. </SPAN>\n" );
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