3. Solve the inequality $\frac{x+2}{x-3} \leq 1$ in the set of natural numbers.

(8)
Solution.

$$
\begin{aligned}
& \frac{x+2}{x-3} \leq 1 \quad \Longleftrightarrow \quad \frac{x+2-x+3}{x-3} \leq 0 \quad \Longleftrightarrow \quad \frac{5}{x-3} \leq 0 \\
& \Longleftrightarrow \quad x-3<0 \quad \Longleftrightarrow \quad x<3
\end{aligned}
$$

The solutions sought are $x=1$ and $x=2$.

(full solution 8 points) (including the solution $x=3 \text{ and/or } x=0, 4$ points)