Solve the following system of equations:

$$
\begin{aligned}
(x+1)(3-4 y) & =(6 x+1)(3-2 y) \\
(4 x-1)(z+1) & =(x+1)(z-1) \\
(3-y)(z-2) & =(1-3 y)(z-6)
\end{aligned}
$$
Our system of equations can also be written as:

$$
\begin{array}{r}
15 x-8 x y+2 y=0 \\
5 x+3 x z-2 z=0 \\
8 y-y z-z=0
\end{array}
$$

In this form, it is evident that

$$
x_{1}=y_{1}=z_{1}=0
$$

is a trivial solution to our system of equations, and it is also evident that if any of the unknowns is 0, it necessarily follows that the other two are also 0. Therefore, apart from this trivial solution, only a triplet of roots where none of the roots is 0 can satisfy our system of equations.

Thus - excluding the above trivial values - we can divide our equations by \(xy\), \(xz\), and \(yz\) respectively. We obtain:

$$
\begin{aligned}
& \frac{15}{y}+\frac{2}{x}=8 \\
& \frac{5}{z}-\frac{2}{x}=-3 \\
& \frac{8}{z}-\frac{1}{y}=1
\end{aligned}
$$

Adding (1) and (2) and simplifying by 5, we get

$$
\frac{3}{y}+\frac{1}{z}=1
$$

Adding 3 times (3) to (4), we get

$$
\frac{25}{z}=4, \quad \text { from which } \quad z_{2}=\frac{25}{4}
$$

and thus

$$
y_{2}=\frac{25}{7}, \quad x_{2}=\frac{10}{19}
$$

Since we have performed only equivalent transformations, the roots we have found necessarily satisfy our system of equations.