What is the crews rowing rate in still water and the rate of the current?

Let x be the crew rowing rate and y the rate of the current.
When the boat goes downstram the rates add up, then we have:
[tex]x+y[/tex]Now, we know that it took 2 hours to travel 12 miles, this means that the speed is 6 miles per hour, hence we have the equation:
[tex]x+y=6[/tex]Now, when the boats go downstream the rate of the current substract from the rowing rate, then we have:
[tex]x-y[/tex]Since it took them 3 hours to travel 12 miles the speed in this case is 4 miles per hour, then we have the equation:
[tex]x-y=4[/tex]Then we have the system of equations:
[tex]\begin{gathered} x+y=6 \\ x-y=4 \end{gathered}[/tex]To solve this system we add the equation, then we get:
[tex]\begin{gathered} 2x=10 \\ x=\frac{10}{2} \\ x=5 \end{gathered}[/tex]now that we know the value of x we plug it on the first equation and solve for y:
[tex]\begin{gathered} 5+y=6 \\ y=6-5 \\ y=1 \end{gathered}[/tex]Therefore the rowing rate is 5 miles per hour and the current rate is 1 mile per hur.