document.write( "Question 389070: A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the rate of the water current, in miles per hour? \n" ); document.write( "
Algebra.Com's Answer #275571 by CharlesG2(834)\"\" \"About 
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A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the rate of the water current, in miles per hour?\r
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\n" ); document.write( "\n" ); document.write( "distance = rate * time
\n" ); document.write( "30 miles = RA * 5 hours
\n" ); document.write( "30/5 = 6 mph = RA, rate against current
\n" ); document.write( "30 miles = RW * 3 hours
\n" ); document.write( "30/3 = 10 mph = RW, rate with current
\n" ); document.write( "BR = boat rate, CR = current rate
\n" ); document.write( "RA = BR - CR, RW = BR + CR
\n" ); document.write( "6 = BR - CR, 10 = BR + CR
\n" ); document.write( "replace 10 - CR for BR in 1st
\n" ); document.write( "6 = 10 - CR - CR = 10 - 2CR
\n" ); document.write( "-4 = -2CR
\n" ); document.write( "4 = 2CR
\n" ); document.write( "2 mph = CR = rate of the water current
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