Solve the following system of inequalities over the set of integer pairs:

$$
11>2 a-b, \quad 25>2 b-a, \quad 42<3 b-a \quad \text { and } \quad 46<2 a+b .
$$
I. Solution. We know the following:

$$
\begin{aligned}
11 & >2 a-b \\
25 & >2 b-a \\
3 b-a & >42 \\
2 a+b & >46
\end{aligned}
$$

Rearranging inequalities (2) and (4) and writing them in sequence:

$$
\begin{aligned}
\frac{25+a}{2}>b & >46-2 a \\
25+a & >92-4 a \\
5 a & >67 \\
a & >13.4
\end{aligned}
$$

We know that $a$ is an integer, so $14 \leq a$.

Rearranging inequalities (1) and (2) and writing them in sequence:

$$
\begin{aligned}
\frac{25+a}{2}>b & >2 a-11 \\
25+a & >4 a-22 \\
47 & >3 a \\
\frac{47}{3}=15 \frac{2}{3} & >a
\end{aligned}
$$

Since $a$ is an integer, $a \leq 15$.

Summarizing: $14 \leq a \leq 15$, so $a=14$ or $a=15$.

Rearranging inequality (3):

$$
b>\frac{42+a}{3}
$$

Substituting $a=14$ into the rearranged form of inequalities (3) and (2):

$$
\begin{aligned}
& b>\frac{42+a}{3}=\frac{42+14}{3}=18 \frac{2}{3} \\
& b<\frac{25+a}{2}=\frac{25+14}{2}=19.5
\end{aligned}
$$

Since $b$ is an integer and equality is not allowed, $b=19$.

Substituting $a=15$ into the rearranged form of inequalities (3) and (2):

$$
\begin{aligned}
& b>\frac{42+a}{3}=\frac{42+15}{3}=19 \\
& b<\frac{25+a}{2}=\frac{25+15}{2}=20
\end{aligned}
$$

Since $b$ is an integer and equality is not allowed, there is no solution here.

Therefore, the system of inequalities has one solution: $a=14$ and $b=19$.

The obtained solution indeed satisfies the requirements of the problem.

II. Solution. Plotting the inequalities on a coordinate system, we determine the part of the plane that satisfies all conditions. The diagram shows that this is a quadrilateral within which only one integer pair (lattice point) is found. The problem does not allow solutions on the lines, as equality is not permitted.

![](https://cdn.mathpix.com/cropped/2024_05_02_4cbcab7acb3db0a95547g-2.jpg?height=607&width=612&top_left_y=125&top_left_x=751)

Thus, the solution to the problem is: $a=14$ and $b=19$.