document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1076701>Question 1076701</A>: <I><SPAN STYLE=\"word-break: break-all;\"> reggie made a business trip of 228 miles. he averaged 62 mph for the first part and 54 mph for the second part. If the trip took 4 hours, how long did he travel at each rate? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #691332 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=691332\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n be the amount of miles traveled at 62 mph. Then the amount of miles at the slower speed would be 228-n. So:<BR>\n" );
document.write( "n/62 + 228-n/54=4<BR>\n" );
document.write( "54n+62(228-n)=4*54*62<BR>\n" );
document.write( "14136-8n=13392<BR>\n" );
document.write( "8n=744<BR>\n" );
document.write( "n=93<BR>\n" );
document.write( "93/62=1.5 hours was driven at 62 mph for 93 miles<BR>\n" );
document.write( "4-1.5=2.5 hours was driven at 54 mph for 135 miles. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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