document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1149799>Question 1149799</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the speed of a boat in still water is 15 km/hr. it needs four<BR>\n" );
document.write( "more hours to travel 63 km against the current of a river<BR>\n" );
document.write( "than it needs to travel down the river. Determine the speed<BR>\n" );
document.write( " of the current of the river.\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 #771146 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=771146\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> d=v*t<BR>\n" );
document.write( "63=(15+c)*t where c is current speed<BR>\n" );
document.write( "t=63/(15+c)<BR>\n" );
document.write( "and <BR>\n" );
document.write( "63=(15-c)(t+4)<BR>\n" );
document.write( "and  t=[63/(15-c)]-4, which is (4c+3)/(15-c), from putting -4 over 15-c and getting -60+4c<BR>\n" );
document.write( "those t s are equal<BR>\n" );
document.write( "cross-multiply and get 2c^2+63c-450=0<BR>\n" );
document.write( "and this is (2c+75)(c-6)=0<BR>\n" );
document.write( "positive root is c=6 kph current\r<BR>\n" );
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document.write( "this would be 21 kph downstream and 3 hours to do 63 km and 9 kph upstream, which takes 7 hours. </SPAN>\n" );
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