Question 1179660
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A motorboat traveling with the current went 42 mi in 3.5 h. Traveling against the current, the boat went 18 mi in 3 h. 
Find the rate of the boat in calm water and the rate of the current.
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Let x be the rate of the motorboat in still water (in miles per hour)
and y be the rate of the current (in the same units).


Then the effective rate of the motorboat downstream is x + y
and  the effective rate of the motorboat   upstream is x - y.


From the problem, the effective rate of the motorboat downstream is the distance of 42 miles 
divided by the time of 3.5 hours  {{{42/3.5}}} = 12 mph.

                  The effective rate of the motorboat upstream is the distance of 18 miles 
divided by the time of 3 hours  {{{18/3}}} = 6 mph.


So, we have two equations to find 'k' and 'c'

    x + y = 12,    (1)

    x - y =  6.    (2)


To solve, add equations (1) and (2).  The terms 'y' and '-y' will cancel each other, and you will get

    2x = 12 + 6 = 18  --->   x = 18/2 = 9.

Now from equation (1)

     v = 12 - u = 12 - 9 = 3.


<U>ANSWER</U>.  The rate of the motorboat in still water is 9 mph.  The rate of the current is 3 mph km/h.
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Solved.