2. Three teams scored 285 points in the Olympiad. If the team from School No. 24 had scored 8 points less, the team from School No. 46 had scored 12 points less, and the team from School No. 12 had scored 7 points less, then all of them would have scored equally. How many points did the teams from Schools No. 24 and No. 12 score together?
2. First method of solving. Let's make a brief record of the problem in the form of a drawing

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-080.jpg?height=308&width=1282&top_left_y=357&top_left_x=387)

1) $7+12+8=27$ (points) - this is how many fewer points all schools scored;
2) $285-27=258$ (points) - this would be their total sum;
3) $258: 3=86$ (points) - each school would have scored;
4) $86+7=93$ (points) — School No. 12 scored;
5) $93+(86+8)=187$ (points) — Schools No. 12 and No. 24 scored together.

Second method of solving. This problem can also be solved algebraically. Let $x$ be the number of points scored by School No. 12, then we can set up the equation:

$$
\begin{aligned}
& x+(x+1)+(x+5)=285 \\
& 3 x=279 \\
& x=93 \\
& 93 \times 2+1=187
\end{aligned}
$$

Answer: 187 points.