For which integer $(x, y)$ pairs does the following system of inequalities hold?

$$
\begin{gathered}
y+\frac{1}{2}>\left|x^{2}-2 x\right| \\
y \leq 2-|x-1|
\end{gathered}
$$
By comparing the two inequalities

$$
\left|x^{2}-2 x\right|-1 / 2<y \leq 2-|x-1|
$$

we obtain, from which

$$
\left|x^{2}-2 x\right|-1 / 2<2-|x-1|
$$

After rearrangement, we get

$$
\left|x^{2}-2 x\right|+|x-1|<5 / 2
$$

The left side is non-negative and an integer if $x$ is an integer, so its possible values are 0, 1, and 2. If $x$ is an integer, then $x^{2}-2 x$ and $x-1$ give different remainders when divided by 2, so the left side is odd. Therefore,

$$
\left|x^{2}-2 x\right|+|x-1|=1
$$

and the terms are non-negative integers, so one of them is 0 and the other is 1.

If $\left|x^{2}-2 x\right|=0$, then $x=0$ or $x=2$, and in both cases $|x-1|=1$. If $|x-1|=0$, then $x=1$ and in this case $\left|x^{2}-2 x\right|=0$, so the possible values of $x$ are 0, 1, and 2.

If $x=0$ or $x=2$, then from (1)

$$
-1 / 2<y \leq 1, \quad \text { or } \quad y=0 \quad \text { or } \quad y=1
$$

If $x=1$, then from (1)

$$
1 / 2<y \leq 2, \quad \text { or } \quad y=1 \quad \text { or } \quad y=2 \text {. }
$$

Thus, the system of inequalities has six integer solutions, which are

$$
(0,0) ; \quad(0,1) ; \quad(1,1) ; \quad(1,2) ; \quad(2,0) ; \quad(2,1)
$$

Remark. We received many graphical solutions from our solvers. Solutions that included a function discussion and verified the results read from the graph were considered correct, but solutions that only provided a drawing and (flawless) result were considered incomplete.