2B. In three gardens there is water. If half of the water in the first garden is poured into the second, then one third of the water in the second garden is poured into the third, and finally one quarter of the water in the third garden is poured into the first, then each garden will have 6 liters of water. How much water was there in each garden at the beginning?
Solution. Let's assume that at the beginning, the first tank had $x$ liters of water, the second tank had $y$ liters, and the third tank had $z$ liters.

After the first transfer, the first tank had $\frac{x}{2}$ liters left, the second tank had $y+\frac{x}{2}$ liters, and the third tank had $z$ liters. After the second transfer, the first tank had $\frac{x}{2}$ liters, the second tank had

$$
y+\frac{x}{2}-\frac{1}{3}\left(y+\frac{x}{2}\right)=\frac{1}{3}(2 y+x)
$$

liters, and the third tank had

$$
z+\frac{1}{3}\left(y+\frac{x}{2}\right)=\frac{1}{6}(6 z+2 y+x)
$$

liters. After the third transfer, the first tank had

$$
\frac{x}{2}+\frac{1}{4} \cdot \frac{1}{6}(6 z+2 y+x)=\frac{1}{24}(6 z+2 y+13 x)
$$

liters, the second tank had $\frac{1}{3}(2 y+x)$ liters, and the third tank had

$$
\frac{1}{6}(6 z+2 y+x)-\frac{1}{4} \cdot \frac{1}{6}(6 z+2 y+x)=\frac{1}{8}(6 z+2 y+x)
$$

liters. From the condition of the problem, we get the system

$$
\frac{1}{24}(6 z+2 y+13 x)=6, \frac{1}{3}(2 y+x)=6, \frac{1}{8}(6 z+2 y+x)=6 .
$$

The solution to the system is $x=8, y=5, z=5$.