Example 5. Find two sets of smaller positive integer solutions for the system of equations
$$
\left\{\begin{array}{l}
2 x+3 y-4 z=8, \\
5 x-7 y+6 z=7
\end{array}\right.
$$
Eliminate $z$: (1) $\times 3+$ (2) $\times 2$, we get
$$
16 x-5 y=38 \text{. }
$$

Solve equation (3) for $y$, we get
$$
y=\frac{9 x-38}{5}=3 x-7+\frac{-3}{5}.
$$
Since $x, y$ in (4) are integers, we can let $\frac{x-3}{5}=k$ (an integer), i.e., $\square$
$$
x=5 k+3 \text{. }
$$

Substitute (5) into (4), we get
$$
y=16 k+2 \text{. }
$$

Substitute (4) and (6) into (1), we get
$$
z=\frac{1}{4}(58 k+4)=14 k+1+\frac{k}{2}.
$$

Since $z$ is an integer, $k$ must be even, so we can let $k=2 n$ (where $n$ is any integer).
Substitute (8) into (5), (6), and (7), we get
$$
x=10 n+3, y=32 n+2, z=29 n+1 \text{. }
$$

When $n=0,1$, we get two sets of smaller positive integer solutions for the system of equations: $(3,2,1)$ and $(13,34,30)$.