document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=558818>Question 558818</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Tracey drives to work at 50 mph and is 1 minute late.  If she leaves at the same time and drives 55 mph, she’ll be 1 minute early.  How far does Tracey live from work?<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 #363202 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=363202\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Suppose Tracey takes x minutes to get to work (on time). Then it takes x+1 minutes going at 50 mph and x-1 minutes going at 55 mph. Since distance(D) = rate*time, we have\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE D = 50(x+1) = 55(x-1)\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 50x + 50 = 55x - 55 \Rightarrow x = 21\">\r<BR>\n" );
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document.write( "Therefore Tracey takes 22 minutes going at 50 mph and 20 minutes going at 55 mph. It can be checked that the total distance is equal in both cases, 55/3 or 18 1/3 miles. </SPAN>\n" );
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