Water can be let into a basin through four different taps. If the first and second taps are opened simultaneously, the basin fills up in 2 hours. If the second and third taps are opened at the same time, it takes 3 hours, and if the third and fourth taps are opened simultaneously, it takes 4 hours for the basin to fill up. How long will it take to fill the basin if water flows in through the fourth and first taps at the same time?
Solution. Let $m$ be the volume of the basin and $x, y, z$, and $u$ be the amount of water flowing out from the first, second, third, and fourth taps in one hour, respectively.

According to the problem, we can write the following equations:

$$
2(x+y)=m, \quad 3(y+z)=m, \quad 4(z+u)=m
$$

Thus:

$$
x+y=\frac{m}{2}, \quad y+z=\frac{m}{3}, \quad z+u=\frac{m}{4}
$$

To answer the question, we need to express $(x+u)$ in terms of $m$. Therefore, we derive $(x-z)$ from the difference of the first and second equations, and then add the third equation to the resulting equation:

$$
\begin{aligned}
& x-z=x+y-y-z=\frac{m}{2}-\frac{m}{3} \\
& x-z+z+u=\frac{m}{2}-\frac{m}{3}+\frac{m}{4}=\frac{(6-4+3) m}{12}=\frac{5}{12} m
\end{aligned}
$$

Thus, $\frac{12}{5}(x+u)=m$, which means that it takes $\frac{12}{5}$ hours, or 2 hours and 24 minutes, to fill the basin when the first and fourth taps are opened together.