281. The system of equations:

$$
\begin{gathered}
x^{4}+y^{4}=272 \\
x+y=6
\end{gathered}
$$

has 4 pairs of roots. Why? If it is known that $x_{1}=4, y_{1}=2$; $x_{2}=3+\sqrt{-55}$, then how to write the value of $y_{2}$ and the other pairs of roots $x_{3}, y_{3} ; x_{4}, y_{4}$ without calculations?
281. The coefficients of the equations are rational numbers, and therefore there must exist an irrational root conjugate to the value $x_{2}$, namely $y_{2}=3-\sqrt{55}$. The left sides of the equations are symmetric with respect to $x$ and $y$. Therefore, we can write:

$$
x_{3}=2, y_{3}=4 ; x_{4}=2-\sqrt{55}, y_{4}=3+\sqrt{55}
$$

There are no more solutions, because our system reduces to a fourth-degree equation with one unknown, which has four roots.