Решите систему уравнений: y+4x=6; x^2+3xy-y^2=3.

\[\left\{ \begin{matrix} y + 4x = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3xy - y^{2} = 3 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y = 6 - 4x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 27x^{2} + 66x - 39 = 0\ \ \ |:( - 3) \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y = 6 - 4x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9x^{2} - 22x + 13 = 0 \\ \end{matrix} \right.\ \]

\[9x² - 22x + 13 = 0\]

\[D = 484 - 468 = 16\]

\[x_{1} = \frac{22 + 4}{18} = \frac{13}{9},\ \ \ \ \ \ \ \ \ \]

\[x_{2} = \frac{22 - 4}{18} = 1\]

\[\left\{ \begin{matrix} x = 1 \\ y = 2 \\ \end{matrix}\ \ \ \ \ \ \ или\ \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \frac{13}{9} \\ y = \frac{2}{9} \\ \end{matrix} \right.\ \]

\[Ответ:\ \ \ (1;2),\ \left( \frac{13}{9};\frac{2}{9} \right).\]