Solve the following system of equations:

$$
\begin{gathered}
\frac{x y}{x+y}=\frac{1}{a}, \quad \frac{y z}{y+z}=\frac{1}{b}, \quad \frac{z u}{z+u}-\frac{1}{c}=0 \\
\frac{x y z u}{x+y+z+u}=\frac{1}{d}
\end{gathered}
$$

where $a=1, b=2, c=-1, d=1$ (general competition),

and $\quad a=1, b=3, c=-2, d=1$ (mathematics class competition)
The last equation can only have a solution where none of the unknowns is 0, so the reciprocals of the unknowns exist. In the third equation, moving the constant to the right side, and taking the reciprocal of the first three equations, we get a system of linear equations for the reciprocals of the unknowns:

$$
\frac{1}{x}+\frac{1}{y}=a, \quad \frac{1}{y}+\frac{1}{z}=b, \quad \frac{1}{z}+\frac{1}{u}=c
$$

From here, any three unknowns can be expressed in terms of the fourth, and substituting into the fourth equation, we get a single-variable equation. However, it will be useful to first take the reciprocal of the last equation and use the first and third equations to transform the equation, keeping $x$ and $y$, as well as $z$ and $u$, together:

$$
\frac{x+y+z+u}{x y z u}=\frac{x+y}{x y} \cdot \frac{1}{z u}+\frac{z+u}{z u} \cdot \frac{1}{x y}=a \frac{1}{z u}+c \frac{1}{x y}=d
$$

Now we express the reciprocals of the other unknowns in terms of $\frac{1}{y}$ and substitute into the last equation:

$$
\begin{aligned}
\frac{1}{x}= & a-\frac{1}{y}, \quad \frac{1}{z}=b-\frac{1}{y}, \quad \frac{1}{u}=c-\frac{1}{z}=c-b+\frac{1}{y} \\
& a\left(b-\frac{1}{y}\right)\left(c-b+\frac{1}{y}\right)+c\left(a-\frac{1}{y}\right) \frac{1}{y}=d
\end{aligned}
$$

From this, after rearrangement, we get the following equation for $\frac{1}{y}$:

$$
(a+c)\left(\frac{1}{y}\right)^{2}-2 a b\left(\frac{1}{y}\right)+a b^{2}-a b c+d=0
$$

With the first constant values, we get a linear equation:

$$
\begin{gathered}
-\frac{4}{y}+7=0, \quad \frac{1}{y}=\frac{7}{4}, \quad \text { so } \\
\frac{1}{x}=-\frac{3}{4}, \quad \frac{1}{z}=\frac{1}{4}, \quad \frac{1}{u}=-\frac{5}{4}
\end{gathered}
$$

and the solution to the system of equations is

$$
x=-\frac{4}{3}, \quad y=\frac{4}{7}, \quad z=4, \quad u=-\frac{4}{5}
$$

This is indeed a solution, because $x+y, \quad y+z, \quad z+u \quad$ and $x+y+z+u$ are all non-zero, and thus every transformation is reversible.

In the case of the mathematical class problem (1), it is as follows:

$$
\left(\frac{1}{y}\right)^{2}+6 \cdot \frac{1}{y}-16=0
$$

Adding 25 to both sides

$$
\left(\frac{1}{y}+3\right)^{2}=25
$$

from which the following two possibilities arise: $\frac{1}{y}=2$, and $\frac{1}{y}=-8$. Accordingly, we get two values for the reciprocals of the other unknowns:

$$
\begin{aligned}
& \frac{1}{y}=2 \text { case } \\
& \frac{1}{x}=-1, \quad \frac{1}{z}=1 \quad \text { and } \\
& \frac{1}{u}=-3 \\
& \frac{1}{y}=8 \text { case } \\
& \frac{1}{x}=9, \quad \frac{1}{z}=11 \\
& \frac{1}{u}=-13
\end{aligned}
$$

From this, the following two sets of roots arise:

$$
\begin{array}{lllll}
x=-1, & y=\frac{1}{2}, & z=1 & \text { and } & u=-\frac{1}{3} \\
x=\frac{1}{9}, & y=-\frac{1}{8}, & z=\frac{1}{11} & \text { and } & u=-\frac{1}{13} .
\end{array}
$$

Again, both sets of numbers are solutions to the system of equations.