document.write( "Question 590923: if it takes 3 hours to travel 15 miles downstream and 5 hours to travel upstream, how fast is the current if the speed of the boat remains constant? \n" ); document.write( "

Algebra.Com's Answer #375371 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+s+\"$ = the speed of the boat
\n" ); document.write( "Let $\"+c+\"$ = the speed of the current
\n" ); document.write( " $\"+s+%2B+c+\"$ = the speed going downstream
\n" ); document.write( " $\"+s+-+c+\"$ = the speed going upstream
\n" ); document.write( "----------
\n" ); document.write( "Going downstream:
\n" ); document.write( " $\"+s+%2B+c+=+15%2F3+\"$
\n" ); document.write( " $\"+s+%2B+c+=+5+\"$ mi/hr
\n" ); document.write( "Going upstream:
\n" ); document.write( " $\"+s+-+c+=+15%2F5+\"$
\n" ); document.write( " $\"+s+-+c+=+3+\"$ mi/hr
\n" ); document.write( "-----------
\n" ); document.write( "Add these equations:
\n" ); document.write( " $\"+s+%2B+c+=+5+\"$
\n" ); document.write( " $\"+s+-+c+=+3+\"$
\n" ); document.write( " $\"+2s+=+8+\"$
\n" ); document.write( " $\"+s+=+4+\"$
\n" ); document.write( "The speed of the boat is 4 mi/hr
\n" ); document.write( "check answer:
\n" ); document.write( " $\"+s+%2B+c+=+5+\"$
\n" ); document.write( " $\"+4+%2B+c+=+5+\"$
\n" ); document.write( " $\"+c+=+1+\"$ mi/hr
\n" ); document.write( "Going downstream:
\n" ); document.write( "r*t = d
\n" ); document.write( " $\"+%28+4+%2B+1%29%2A3+=+15+\"$
\n" ); document.write( "Going upstream:
\n" ); document.write( " $\"+%28+4+-+1%29%2A5+=+15+\"$
\n" ); document.write( "OK \n" ); document.write( "

\n" );