document.write( "Question 250284: a boat travels 210 miles downstream and back. the trip downstream took 10 hours. the trip back took 70 hours. what is the speed of the boat in still water? what is the speed of the current?
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Algebra.Com's Answer #185309 by checkley77(12844)\"\" \"About 
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D=RT
\n" ); document.write( "LET B=THE SPEED OF THE BOAT & C=THE SPEED OF THE CURRENT.
\n" ); document.write( "105=(B+C)10
\n" ); document.write( "105=(B-C)70
\n" ); document.write( "10(B+C)=70(B-C)
\n" ); document.write( "10B+10C=70B-70C
\n" ); document.write( "10B-70B=-70C-10C
\n" ); document.write( "-60B=-80C
\n" ); document.write( "B=-80C/-60
\n" ); document.write( "B=1.333C
\n" ); document.write( "105=(1.333C+C)10
\n" ); document.write( "105=2.333C*10
\n" ); document.write( "105=23.33C
\n" ); document.write( "C=105/23.33
\n" ); document.write( "C=4.5 MPH. FOR THE CURRENT.
\n" ); document.write( "1.333*4.5=6 MPH. FOR THE SPEED OF THE BOAT.
\n" ); document.write( "PROOF:
\n" ); document.write( "105=(1.333*-4.5-4.5)70
\n" ); document.write( "105=(6-4.5)*70
\n" ); document.write( "105=1.5*70
\n" ); document.write( "105=105\r
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