document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=27692>Question 27692</A>: <I><SPAN STYLE=\"word-break: break-all;\"> On Monday, Roger drove to work in 42 minutes. On Tuesday, he averaged 5 miles per hour more, and it took him 3 minutes less to get to work. How far (in miles) does he travel to 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 #15091 by </B><A HREF=/tutors/aboutme.mpl?userid=wuwei96815><B>wuwei96815(245)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=wuwei96815><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=15091\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Monday  distance=speed x time = s x 42minutes (equal to 42/60minutes or 0.7 hours) \r<BR>\n" );
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document.write( "Tuesday  distance=speed x time = (s+5mph) x (42minutes-3minutes) = (s+5) x 39 (equal to 39/60 minutes or 0.65 hours)\r<BR>\n" );
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document.write( "We know that the two distances are equal even though the speeds and times are different so we can write the following equation:<BR>\n" );
document.write( "s x 0.7 hours = (s+5mph) x 0.65 hours<BR>\n" );
document.write( "0.7s = 0.65s + 3.25<BR>\n" );
document.write( "0.7s -0.65s = 3.25<BR>\n" );
document.write( "0.05s = 3.25<BR>\n" );
document.write( "s = 65 mph\r<BR>\n" );
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document.write( "Going back to the original equation:<BR>\n" );
document.write( "d = s x t<BR>\n" );
document.write( "d = 65mph x 0.7 hours<BR>\n" );
document.write( "d = 45.5 miles </SPAN>\n" );
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