Petr and Pavel were picking apples and pears in the orchard. On Monday, Petr ate 2 more pears than Pavel and 2 fewer apples than Pavel. On Tuesday, Petr ate 4 fewer pears than on Monday. Pavel ate 3 more pears than Petr and 3 fewer apples than Petr on Tuesday. Pavel ate 12 apples over the two days and the same number of apples as pears on Tuesday. In the evening on Tuesday, both boys found out that the number of apples they had eaten together over the two days was the same as the number of pears they had eaten together. How many apples did Petr eat on Monday and how many pears did Pavel eat on Tuesday?
Let $x, y$ be the corresponding numbers of pears and apples that Pavel ate on Monday. According to the problem, we will patiently construct the following table:

|  | Monday | Tuesday |
| :---: | :---: | :---: |
| Pavel pears | $x$ | $x+1$ |
| Pavel apples | $y$ | $12-y$ |
| Petr pears | $x+2$ | $x-2$ |
| Petr apples | $y-2$ | $15-y$ |

However, while filling out the table, we have not yet used the following information:

a) On Tuesday, Pavel ate the same number of apples as pears,

b) the total number of apples eaten together is the same as the number of pears eaten.

According to information b), we set up an equation, and after its rearrangement, we obtain the value of $x$:

$$
\begin{aligned}
y+(12-y)+(y-2)+(15-y) & =x+(x+1)+(x+2)+(x-2), \\
25 & =4 x+1, \\
x & =6 .
\end{aligned}
$$

According to information a), we also set up an equation, rearrange it, and substitute:

$$
\begin{aligned}
x+1 & =12-y, \\
y & =11-x=5 .
\end{aligned}
$$

By substituting into the relevant fields of the table, we find that Petr ate 3 apples on Monday and Pavel ate 7 pears on Tuesday.

For verification, we provide the table with all substituted values:

|  | Monday | Tuesday |
| :---: | :---: | :---: |
| Pavel pears | 6 | $\mathbf{7}$ |
| Pavel apples | 5 | 7 |
| Petr pears | 8 | 4 |
| Petr apples | $\mathbf{3}$ | 10 |

Note. When compiling the data in the table, we can certainly proceed in different ways and the unused information may differ from that in the previous procedure. With the same notation of unknowns, we can thus obtain two different equations, which, when calculated correctly, lead to the same solution. Moreover, the unknowns can also be chosen differently, but at least two are always needed. The same number of unused information then determines the system of equations, which we subsequently solve.

Another solution. If we denote $x$ as the number of apples that Petr ate on Monday, and $y$ as the number of pears that Pavel ate on Tuesday, then the table might look like this:

|  | Monday | Tuesday |
| :---: | :---: | :---: |
| Pavel pears | $y-1$ | $y$ |
| Pavel apples | $12-y$ | $y$ |
| Petr pears | $y+1$ | $y-3$ |
| Petr apples | $x$ | $y+3$ |

In this case, we have not used the following information:

a) On Monday, Petr ate 2 apples fewer than Pavel,

b) the total number of apples eaten together is the same as the number of pears eaten.

The corresponding equations (after simple rearrangement) are:

$$
\begin{aligned}
x+y & =10, \\
-x+3 y & =18,
\end{aligned}
$$

and the only solution to this system is $x=3$ and $y=7$.