4. Miro, Aleš, and Lovro played poker three times. The first time, Miro lost and had to pay Aleš and Lovro, each as much money as they had at the beginning of the game. The second time, Aleš lost and had to pay Miro and Lovro, each as much money as they currently had. The third time, Lovro lost and had to pay Miro and Aleš, each as much money as they currently had. After the three games, each of them had 24 euros. Who among them lost the most and how much?
4. Let $x$ be the initial amount of money that Aleš has, $y$ be the initial amount of money that Lovro has, and $z$ be the initial amount of money that Miro has. We can tabulate the changes in the amount of money each person has over time:

| Game | Miro | Aleš | Lovro |
| :---: | :---: | :---: | :---: |
| before the game | $x$ | $y$ | ![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-07.jpg?height=55&width=438&top_left_y=1165&top_left_x=1005) |
| after the 1st game | $x-y-z$ | $2 y$ | $2 z$ |
| after the 2nd game | $2 x-2 y-2 z$ | $2 y-(x-y-z)-2 z=$  $=-x+3 y-z$ | ![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-07.jpg?height=110&width=438&top_left_y=1265&top_left_x=1005) |
| after the 3rd game | $4 x-4 y-4 z$ | $-2 x+6 y-2 z$ | $4 x-(2 x-2 y-2 z)-(-x+3 y-z)=$  $=-x-y+7 z$ |

We write the system of equations:

$$
\begin{gathered}
4 x-4 y-4 z=24 \\
-2 x+6 y-2 z=24 \\
-x-y+7 z=24
\end{gathered}
$$

which we solve. The solution is $x=39, y=21$ and $z=12$. We determine that Miro lost the most, specifically 15 euros.

Written system of equations ...................................................................................................

Solving the system ......................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-07.jpg?height=60&width=1639&top_left_y=1917&top_left_x=206)

Written answer .................................................................................................

NOTE: The contestant earns 1 point if they write two equations of the system and 0 points if they write only one equation of the system.

The contestant earns 1 point for two correct solutions and 0 points if only one solution is correct.