document.write( "Question 387782: A speedboat takes 4 hours longer to go 80 miles up a river than to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current? \n" ); document.write( "
Algebra.Com's Answer #275599 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Rate*Time=Distance
\n" ); document.write( "\"R%2At=D\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "Let the current speed be C.
\n" ); document.write( "With the current,
\n" ); document.write( "\"%2815%2BC%29%2At=80\"
\n" ); document.write( "\"t=80%2F%2815%2BC%29\"
\n" ); document.write( "Against the current,
\n" ); document.write( "\"%2815-C%29%2A%28t%2B4%29=80\"
\n" ); document.write( "\"t%2B4=80%2F%2815-C%29\"
\n" ); document.write( "Subsituting,
\n" ); document.write( "\"80%2F%2815%2BC%29%2B4=80%2F%2815-C%29\"
\n" ); document.write( "Use the common denominator, \"%2815-C%29%2815%2BC%29\"
\n" ); document.write( "
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"%2880%2815-C%29%2B4%2815-C%29%2815%2BC%29-80%2815%2BC%29%29%2F%28%2815-C%29%2815%2BC%29%29=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"%281200-80C%2B900-4C%5E2-1200-80C%29%2F%28%2815-C%29%2815%2BC%29%29=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"%28900-160C-4C%5E2%29%2F%28%2815-C%29%2815%2BC%29%29=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"%28225-40C-C%5E2%29%2F%28%2815-C%29%2815%2BC%29%29=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"%28%285-C%29%2845%2BC%29%29%2F%28%2815-C%29%2815%2BC%29%29=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "Only one solution since \"C%3E=0\"
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "\"highlight%28C=5%29\"mph\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );