2. The boat crossed the river in 11 hours, without stopping it traveled $168 \mathrm{~km}$ downstream, and $48 \mathrm{~km}$ upstream against the current of the river. Another time, in the same amount of time, it traveled $144 \mathrm{~km}$ downstream, and $60 \mathrm{~km}$ upstream against the current of the river. What is the speed of the boat, and what is the speed of the river?
Solution. Let $x$ be the speed of the boat and $y$ the speed of the river. From the conditions of the problem, we obtain the following system of two equations with two unknowns:

$$
\left\{\begin{array}{l}
\frac{168}{x+y}+\frac{48}{x-y}=11 \\
\frac{144}{x+y}+\frac{60}{x-y}=11
\end{array}\right.
$$

This system is transformed into the system

$$
\left\{\begin{array}{l}
168(x-y)+48(x+y)=11\left(x^{2}-y^{2}\right) \\
144(x-y)+60(x+y)=11\left(x^{2}-y^{2}\right)
\end{array}\right.
$$

from which, by subtracting the second equation from the first, we get

$$
24(x-y)-12(x+y)=0, \text { i.e. } x=3 y
$$

By substituting into one of the equations, we get $\frac{168}{4 y}+\frac{48}{2 y}=11$, from which we have $y=6$. Therefore, the speed of the boat is $18 \mathrm{~km} / \mathrm{h}$, and the speed of the river is $6 \mathrm{~km} / \mathrm{h}$.