same d = rt
so the boat's still water speed is 20kph, if we... say the current of the river is "r", then, when the boat was going upstream, its speed was " 20 - r", and when it was going downstream, is " 20 + r", since the river's current is adding speed to it
again, if on the way up, it took "t" hours, on the way down, it took the slack of the 3hrs, or " 3 - t"
thus [tex]\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&36&20-r&t\\
downstream&22&20+r&3-t
\end{array}
\\\\\\
\begin{cases}
36=t(20-r)\implies \frac{36}{20-r}=\boxed{t}\\\\
22=(20+r)(3-t)\\
----------\\\\
22=(20+r)\left( 3- \boxed{\frac{36}{20-r}}\right)
\end{cases}[/tex]
solve for "r"
