document.write( "Question 630585: A tugboat goes 180 miles upstream in 15hours. The return trip downstream takes 10 hours. Find the speed of the tugboat without a current and the speed of the current \n" ); document.write( "

Algebra.Com's Answer #397022 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+s+\"$ = the speed of the boat without a current
\n" ); document.write( "Let $\"+c+\"$ = the speed of the current
\n" ); document.write( "Going upstream:
\n" ); document.write( "(1) $\"+180+=+%28+s+-+c+%29%2A15+\"$
\n" ); document.write( "Going downstream:
\n" ); document.write( "(2) $\"+180+=+%28+s+%2B+c+%29%2A10+\"$
\n" ); document.write( "--------------------
\n" ); document.write( "(1) $\"+180+=+15s+-+15c+\"$
\n" ); document.write( "(1) $\"+15s+=+180+%2B+15c+\"$
\n" ); document.write( "(1) $\"+s+=+12+%2B+c+\"$
\n" ); document.write( "Substitute this into (2)
\n" ); document.write( "(2) $\"+180+=+%28+12+%2B+c+%2B+c+%29%2A10+\"$
\n" ); document.write( "(2) $\"+180+=+120+%2B+20c+\"$
\n" ); document.write( "(2) $\"+20c+=+180+-+120+\"$
\n" ); document.write( "(2) $\"+20c+=+++60+\"$
\n" ); document.write( "(2) $\"+c+=+3+\"$
\n" ); document.write( "and, since
\n" ); document.write( "(1) $\"+180+=+%28+s+-+c+%29%2A15+\"$
\n" ); document.write( "(1) $\"+180+=+%28+s+-+3+%29%2A15+\"$
\n" ); document.write( "(1) $\"+180+=+15s+-+45+\"$
\n" ); document.write( "(1) $\"+15s+=+180+%2B+45+\"$
\n" ); document.write( "(1) $\"+15s+=+225+\"$
\n" ); document.write( "(1) $\"+s+=+15+\"$
\n" ); document.write( "The speed of the boat without a current is 15 mi/hr
\n" ); document.write( "The speed of the current is 3 mi/hr
\n" ); document.write( "check answers:
\n" ); document.write( "(2) $\"+180+=+%28+s+%2B+c+%29%2A10+\"$
\n" ); document.write( "(2) $\"+180+=+%28+15+%2B+3+%29%2A10+\"$
\n" ); document.write( "(2) $\"+180+=+18%2A10+\"$
\n" ); document.write( "(2) $\"+180+=+180+\"$
\n" ); document.write( "OK \n" ); document.write( "

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