3. The boat traveled 165 km upstream against the current and then returned. On the return trip, it took 4 hours less than the trip there. Find the boat's own speed if the river's current speed is 2 km/h.
# Solution:

$$
\begin{aligned}
& v_{\text {down}}=v_{\mathrm{c}}+v_{\text {current}} \\
& v_{\text {up}}=v_{\mathrm{c}}-v_{\text {current}}
\end{aligned}=>v_{\text {down}}-v_{\text {up}}=2 v_{\text {current}}=2 \cdot 2=4
$$

Then $\quad v_{\text {up}}=x$ km/h, $v_{\text {down}}=x+4$ km/h

Preliminary (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics

Let the time for the boats $t_{\text {up}}=y$ h, $t_{\text {down}}=y-4$ h

We get $x y=165$ and $(x+4)(y-4)=165=>x y=(x+4)(y-4)=>$ $=>4 y-4 x=16=>y-x=4 \Rightarrow y=x+4$

Then $x(x+4)=165$

$$
\begin{aligned}
& x^{2}+4 x+4-4-165=0 \\
& (x+2)^{2}-169=0 \\
& {\left[\begin{array}{l}
x+2-13=0 \\
x+2+13=0
\end{array} ; \begin{array}{c}
x=11 \\
x=-15
\end{array}\right.}
\end{aligned}
$$

-15 is an extraneous root

$$
v_{\mathrm{c}}=v_{\text {up}}+v_{\mathrm{current}}=11+2=13(\mathrm{km} / \mathrm{h})
$$

Answer: 13 km/h.