4. Find the positive integer solutions for
$$\left\{\begin{array}{l}
5 x+7 y+2 z=24 \\
3 x-y-4 z=4
\end{array}\right.$$
4. Solution: This is a system of three linear equations. We can use the elimination method to eliminate one of the unknowns, thus obtaining a system of two linear equations. By multiplying the first equation by 2 and adding it to the second equation, we can eliminate \( z \) and get
$$13 x+13 y=52$$

which simplifies to
$$x+y=4$$

It is easy to see that \( x=4, y=0 \) is one solution to the above equation. The complete set of solutions is
$$\left\{\begin{array}{l}
x=4-t \\
y=t
\end{array} \quad t=0, \pm 1, \pm 2, \cdots\right.$$

Substituting this solution into the second equation of the original system, we get
$$3(4-t)-t-4 z=4$$

which simplifies to
$$z=2-t, \quad t=0, \pm 1, \cdots$$

Now we need to find all positive integer solutions. Let \( x>0, y>0, z>0 \), which gives us
$$4-t>0, \quad t>0, \quad 2-t>0$$

This simplifies to \( 0<t<2 \). Therefore, \( t \) can only be 1. Substituting \( t=1 \) into the solution, we get the unique positive integer solution to the original system of equations: \( x=3, y=1, z=1 \).