document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=441532>Question 441532</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the excursion boat Holiday travels 35 km upstream and then back again in 4h 48min. if the speed of the Holiday in still water is 15km/h, what is the speed of the current?\r<BR>\n" );
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document.write( "how would i set up an equation to solve this?? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #304743 by </B><A HREF=/tutors/aboutme.mpl?userid=rwm><B>rwm(914)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rwm><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=304743\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the distance is the same up and down stream <BR>\n" );
document.write( "upstream means against the current s-c <BR>\n" );
document.write( "downstream is with the current s+c\r<BR>\n" );
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document.write( "trip takes 4 48/60 hrs<BR>\n" );
document.write( "r*t=d<BR>\n" );
document.write( "(s-c)*u=35<BR>\n" );
document.write( "(s+c)*d=35<BR>\n" );
document.write( "u+d=4 48/60<BR>\n" );
document.write( "s=15<BR>\n" );
document.write( "we can substitute 15 for s<BR>\n" );
document.write( "u=35-d<BR>\n" );
document.write( "35-d for u<BR>\n" );
document.write( "(15-c)*(35-d)=(15+c)*d<BR>\n" );
document.write( "c=5/2 or 2.5 current<BR>\n" );
document.write( "d=2<BR>\n" );
document.write( "s=15<BR>\n" );
document.write( "u=14/5= 2.8\r<BR>\n" );
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