7.2. There are nuts in three boxes. In the first box, there are six nuts fewer than in the other two boxes combined, and in the second box, there are ten fewer nuts than in the other two boxes combined. How many nuts are in the third box? Justify your answer.
Solution: Let there be $x$ nuts in the first box, $y$ and $z$ in the second and third boxes, respectively. Then the condition of the problem is defined by the equations $x+6=y+z$ and $x+z=y+10$. From the first equation, $x-y=z-6$, and from the second, $x-y=10-z$. Therefore, $z-6=10-z$, from which $z=8$.

Answer: 8 nuts.

| in progress | points |
| :--- | :--- |
| correct and fully justified answer | points |
| an incorrect answer due solely to arithmetic errors | 6 points |
| the system of equations describing the problem condition is correctly set up but not solved | 3 points |
| a specific example (examples) of the distribution of nuts in the boxes (and, therefore, the correct answer) is provided | 1 point |
| only the correct answer is given without justification (or with incorrect justification) | 0 points |