On the board, an example of dividing two positive numbers was given. David noticed that if he increased the dividend by two and the divisor by seven, the quotient would remain unchanged.

By how much would the divisor need to increase so that increasing the dividend by three would again result in the same quotient?

(M. Petrová)
If we denote the original dividend by $n$ and the original divisor by $t$, then the condition from the problem statement is

$$
\frac{n}{t}=\frac{n+2}{t+7}
$$

By manipulations based on comparing fractions, we get:

$$
\begin{aligned}
n t+7 n & =n t+2 t, \\
\frac{n}{t} & =\frac{2}{7} .
\end{aligned}
$$

The question in the problem asks for which number $z$ the following holds:

$$
\frac{2}{7}=\frac{2+3}{7+z}
$$

By manipulations, we get the following answer:

$$
\begin{aligned}
14+2 z & =14+21, \\
2 z & =21, \\
z & =\frac{21}{2} .
\end{aligned}
$$

The divisor must be increased by $\frac{21}{2}=10.5$.

Evaluation. 2 points for auxiliary considerations and manipulations; 2 points for the result; 2 points for the quality of the commentary.

Notes. Even without the given manipulations, one can reason that equality (1) holds precisely for pairs of the form $n=2k$ and $t=7k$, where $k$ is any positive number. This arbitrariness does not play any role in the further steps (it cancels out).

Instead of equality (2), one can generally consider

$$
\frac{n}{t}=\frac{n+3}{t+z}
$$

and similarly derive that $\frac{n}{t}=\frac{3}{z}$. Together with the previous condition $\frac{n}{t}=\frac{2}{7}$, this gives $\frac{3}{z}=\frac{2}{7}$, from which it follows that $z=\frac{21}{2}$.