Dan boated downstream with the current for 3 miles, then truned around and cam back to his starting point. If the total time was 4 hours and the current was 0.4 mile per hour, what is the speed dan can boat in still water?

Ok, first, let the speed of the boat in still water be x m/s.

The speed of the boat downstream is (x + 0.4) m/s
The speed of the boat upstream is (x - 0.4) m/s

Now, using the distance, speed time formula;

Time = \frac{Distance}{Speed}

We get Time downstream = 3/(x+0.4) Let this be Td
Time upstream = 3/(x - 0.4) Ket this be Tu

Total time = 4 hours. Let this be Tt

So, T_t = T_d + T_u

Hence;

4 = \frac{3}{(x+0.4)} + \frac{3}{(x - 0.4)}

Can you solve it?