document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=83130>Question 83130</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A passanger train can travel 325 mi in the same time a freight train takes to travek 200 mi.  If the speed of the passanger train is 25 mi.h faster then the speed of the freight train, find the speed of each. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #59719 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=59719\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> TIME=DISTANCE/RATE. SEEING THAT THE TIMES ARE EQUAL THEN WE HAVE THE FOLLOWING EQUATION:<BR>\n" );
document.write( "325/(R+25)=200/R NOW CROSS MULTIPLY:<BR>\n" );
document.write( "325R=200(R+25)<BR>\n" );
document.write( "325R=200R+5000<BR>\n" );
document.write( "325R-200R=5000<BR>\n" );
document.write( "125R=5000<BR>\n" );
document.write( "R=5000/125<BR>\n" );
document.write( "X=40 MPH FOR THE FREIGHT TRAIN.<BR>\n" );
document.write( "PROOF<BR>\n" );
document.write( "325(40+25)=200/40<BR>\n" );
document.write( "325/65=200/40<BR>\n" );
document.write( "5=5\r<BR>\n" );
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document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );