15. Between two stations $A$ and $B$ on a double-track railway, there are two down trains departing at 6:00 and 6:45, and two up trains departing at 7:15 and 8:30. If these four trains pass each other simultaneously, find the equations formed by their speeds $x_{1}, x_{2}, x_{3}, x_{4}$ as follows:
$$
\frac{3 x_{2}}{x_{2}-x_{1}}=\frac{4 m+5 x_{3}}{x_{1}+x_{3}}=\frac{4 m+10 x_{4}}{x_{1}+x_{4}},
$$

where $m$ is the distance between $A$ and $B$.
Let $T_{1}, T_{2}$ be the trains going downhill, and $T_{3}, T_{4}$ be the trains going uphill. They meet at 6 o'clock in the morning. Then, the distances traveled by $T_{1}, T_{2}, T_{3}, T_{4}$ when they meet are $x_{1} y, x_{2}\left(y-\frac{3}{4}\right), x_{3}\left(y-\frac{5}{4}\right)$, $z_{1}\left(y-\frac{5}{2}\right)$ respectively.
$$
x_{1} y+x_{3}\left(y-\frac{5}{4}\right)=m=x_{1} y+x_{6}\left(y-\frac{5}{2}\right)
$$

Thus,
$$
\left(x_{1}+x_{3}\right) y=m+\frac{5}{4} x_{3}
$$

And
$$
\left(x_{1}+x_{4}\right) y=m+\frac{5}{2} x_{4} \text {. }
$$

From these equations, we get
$$
4 y=\frac{4 m+5 x_{3}}{x_{1}+x_{3}}=\frac{4 m+10 x_{4}}{x_{1}+x_{4}} .
$$

Considering $x_{1} y=x_{2}\left(y-\frac{3}{4}\right)$, we have
$$
4 y=\frac{3 x_{2}}{x_{2}-x_{1}} .
$$

From the above three equations, it is easy to solve for $y$ that satisfies the required result.