document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=99078>Question 99078</A>: <I><SPAN STYLE=\"word-break: break-all;\"> on a trip, palmer s. friedrich drove at a rate of 55 mph. he desided to return by a route that was 5 miles longer. his rate returning was 45 mph. if it took him 1 hour longer returning, how long was the second route?\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 #72089 by </B><A HREF=/tutors/aboutme.mpl?userid=edjones><B>edjones(8007)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=edjones><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=edjones><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=72089\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> s=speed t=time d=distance<BR>\n" );
document.write( "s*t=d<BR>\n" );
document.write( "Going: d=d s=55 t=h<BR>\n" );
document.write( "Returning: d=d+5 s=45 t=h+1<BR>\n" );
document.write( "Distance going = distance returning -5. Why -5 because it wouldn't be an equation if we didn't make both sides equal. We could have added 5 to the left instead of subtracting 5 from the right, the answer would have been the same.<BR>\n" );
document.write( "55h=45(h+1)-5<BR>\n" );
document.write( "55h=45h+45-5<BR>\n" );
document.write( "55h=45h+40<BR>\n" );
document.write( "subtract 45h from each side: 10h=40<BR>\n" );
document.write( "divide 10 into each side: h=4 hours going and 5 hrs returning.<BR>\n" );
document.write( "45*5=225 mi (distance of the 2nd route) Ans.<BR>\n" );
document.write( "Check:<BR>\n" );
document.write( "55*4=220 mi (distance of the 1st route)<BR>\n" );
document.write( "225-220=5 mi longer for the second route.<BR>\n" );
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