[Algebra]
Problem: In a soccer league with 5 teams, each team plays 20 games(that is, 5 games with each of the other 4 teams). For each team, every game ends in a win (W), a loss (L), or a tie (T). The numbers of wins, losses and ties for each team at the end of the season are shown in the table. Determine the values of $x, y$ and $z$.

| Team | W | L | T |
| :---: | ---: | ---: | ---: |
| A | 2 | 15 | 3 |
| B | 7 | 9 | 4 |
| C | 6 | 12 | 2 |
| D | 10 | 8 | 2 |
| E | $x$ | $y$ | $z$ |

Solution: In total, there are $\frac{1}{2} \times 5 \times 20=50$ games played, since each of 5 teams plays 20 games (we divide by 2 since each game is double-counted).

In each game, there is either a loss or a tie.

The number of games with a loss is $44+y$ from the second column, and the number of games with a tie is $\frac{1}{2}(11+z)$ (since any game ending in a tie has 2 ties).



So

$$
\begin{aligned}
50 & =44+y+\frac{1}{2}(11+z) \\
100 & =88+2 y+11+z \\
1 & =2 y+z
\end{aligned}
$$

Since $y$ and $z$ are non-negative integers, $z=1$ and $y=0$. So $x=19$ since Team E plays 20 games.
In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.

$$
\begin{aligned}
25+x & =44+y \\
x-y & =19
\end{aligned}
$$

Also, since team E plays 20 games, then

$$
x+y+z=20
$$

So from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.

Lastly, we know that the total of all of the teams numbers of ties must be even, ie. $11+z$ is even, ie. $z$ is odd.

Since $x$ is at least 19, then $z$ can be at most 1 by (2).

Therefore, $z=1$. Thus, $x=19$ and $y=0$.
In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.

$$
\begin{aligned}
25+x & =44+y \\
x-y & =19
\end{aligned}
\tag{1}
$$

Also, since team E plays 20 games, then

$$
x+y+z=20
\tag{2}
$$

So from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.

Consider the possibility that $x=20$. From (2), then $y=z=0$, which does not agree with (1).

Thus, the only possibility is $x=19$. From (1), $y=0$, and so $z=1$ from (2). (These three values agree with both equations (1) and (2).)

Final answer: 19,0,1