document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168102>Question 1168102</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A boat sails downstream 36 miles and then returns upstream against the current. The downstream trip took 3/4 of an hour less than the upstream trip. If the speed of the current is 4 mph, calculate the speed of the boat in calm water and the time of each trip. (20 mph, 1.5h downstream, 2.25 upstream) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #792728 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=792728\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> trip downstream at s+4 mph so 36/(s+4) hours<BR>\n" );
document.write( "upstream at s-4 mph (s-4) so 36/(s-4)<BR>\n" );
document.write( "we know that the second is (3/4 hour) more than the first<BR>\n" );
document.write( "so 36/(s+4)+(3/4)=36/(s-4)<BR>\n" );
document.write( "clear fractions<BR>\n" );
document.write( "36(s-4)+(3/4)*(s^2-16)=36(s+4)<BR>\n" );
document.write( "36s-144+(3/4)s^2-12=36s+144<BR>\n" );
document.write( "(3/4)s^2-156=144<BR>\n" );
document.write( "(3/4)s^2-300=0<BR>\n" );
document.write( "s^2-400=0<BR>\n" );
document.write( "s=20 mph only positive root.<BR>\n" );
document.write( "24 mph downstream 1.5 hours to go 36 miles<BR>\n" );
document.write( "16 mph upstream and takes 2.25 hours<BR>\n" );
document.write( " </SPAN>\n" );
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