Solve the following system of equations, where $x, y, z, t$ are non-negative integers:

$$
\begin{aligned}
x+y+z+t & =5 \\
x+2 y+5 z+10 t & =17
\end{aligned}
$$
Subtracting the first equation from the second, we obtain the equivalent system of equations:

$$
\begin{aligned}
& x+y+z+t=5 \\
& y+4 z+9 t=12
\end{aligned}
$$

The solutions are non-negative, so according to (3), $t \leq 1$. We distinguish two cases based on whether $t=1$ or $t=0$.

a) If $t=1$, then our system of equations becomes:

$$
\begin{gathered}
x+y+z=4 \\
y+4 z=3
\end{gathered}
$$

According to (3a), $z<1$, so $z=0$, and $y=3$. Finally, according to (1a), $x=1$.

b) If $t=0$, then our system of equations is:

$$
\begin{gathered}
x+y+z=5 \\
y+4 z=12
\end{gathered}
$$

According to (3b), $y=12-4 z$, so $y$ is divisible by 4. Since according to (1b), $y \leq 5$, then $y=4$ or $y=0$. If $y=4$, then $z=2$, but according to (1b), $y+z \leq 5$, so this does not yield a solution.

If $y=0$, then $z=3$, and according to (1b), $x=2$.

We have examined all possibilities, so the system of equations has two solutions. These are:

$$
\begin{array}{llll}
x_{1}=1 ; & y_{1}=3 ; & z_{1}=0 ; & t_{1}=1 \\
x_{2}=2 ; & y_{2}=0 ; & z_{2}=3 ; & t_{2}=0
\end{array}
$$

Remark. Our problem can be reformulated as the following "change-making" problem: in how many ways can 17 forints be paid with 5 coins, if the denominations available are 1, 2, 5, and 10 forints.