9-2. The school volleyball team played several matches. After they won another match, the share of victories increased by $1 / 6$. To increase the share of victories by another 1/6, the volleyball players had to win two more consecutive matches. What is the minimum number of victories the team needs to achieve to increase the share of wins by another 1/6?
Answer: 6.

Solution: Let the team initially play $n$ matches, of which $k$ were won. Then, after the next win, the share of victories increased by $\frac{k+1}{n+1}-\frac{k}{n}=\frac{1}{6}$. Similarly, after two more wins, the increase was $\frac{k+3}{n+3}-\frac{k+1}{n+1}=\frac{1}{6}$. Simplifying each equation, we get the system

$$
\left\{\begin{array}{c}
n-k=\frac{n(n+1)}{6} \\
2(n-k)=\frac{(n+1)(n+3)}{6}
\end{array}\right.
$$

Dividing the second equation by the first, we get that $\frac{n+3}{n}=2$, from which $n=3$. Substituting this value into the first equation, we find that $k=3-3 \cdot 4 / 6=1$. Thus, at the beginning, the share of victories was $1 / 3$, after the next win - 2/4, and after two more wins, the team had $1+1+2=4$ victories in $3+1+2=6$ games. If the team wins $m$ more games, their share of victories will be $\frac{4+m}{6+m}$, which should match $\frac{4}{6}+\frac{1}{6}=\frac{5}{6}$. Solving the corresponding equation, we get that $m=6$.

Criteria: Only the answer - 0 points. The system of equations is set up - 5 points. Complete solution - 7 points.