{x^2+xy+y^2=19 {x+xy+y=1

Решите задачу:

$\begin{cases}x^2+xy+y^2=19\\x+xy+y=1\end{cases}\Rightarrow\begin{cases}\left(\frac{1-y}{1+y}\right)^2+\left(\frac{1-y}{1+y}\right)y+y^2=19\\x=\frac{1-y}{1+y}\end{cases}\\\frac{1-2y+y^2+y+y^3+y^2+2y^3+y^4}{1+2y+y^2}=19\\y^4+y^3+2y^2-y+1=19y^2+38y+19\\y^4+y^3-17y^2-39y-18=0\\(y^2-4y-3)(y^2+5y+6)=0\\y_1=2-\sqrt7\\y_2=2+\sqrt7\\y_3=-2\\y_4=-3$
$\begin{cases}x_1=\frac{-1+\sqrt7}{3-\sqrt7}\\y_1=2-\sqrt7\end{cases}\quad\begin{cases}x_2=-\frac{1+\sqrt7}{3+\sqrt7}\\y_2=2+\sqrt7\end{cases}\quad\begin{cases}x_3=-3\\y_3=-2\end{cases}\quad\begin{cases}x_4=-2\\y_4=-3\end{cases}$