d = rt, distance = rate * time
so, we know the boat is going at a rate of "x", and the current has a rate of "y".
ok... from A to B, the boat is going downstream, the speed of the boat is not really "x", is " x + y ", because, is going with the stream and thus the current is adding "y" to its speed.
now, from B to A, the boat is going upstream, against the current, is not really going "x" fast either, is going " x - y ", because the current is eroding/subtracting from it's speed.
ok... since, since the towns A and B didn't really move, unless they're floating towns, but they're not, so, the distance from A to B is the same distance from B to A, say they're at a distance "d".
[tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
downstream&d&x+y&3\\
upstream&d&x-y&3.6
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=(x+y)3\\
d=(x-y)3.6\\
\boxed{3(x+y)}=3.6(x-y)
\end{cases}
\\\\\\
3(x+y)=3.6(x-y)\implies 3x+3y=3.6x-3.6y
\\\\\\
3y+3.6y=3.6x-3x\implies 6.6y=0.6x\implies y=\cfrac{0.6x}{6.6}
\\\\\\
y=\cfrac{1}{11}x\implies \textit{y is }\frac{1}{11}x\textit{ or about }9\%\textit{ the speed of \underline{x}}[/tex]
