Решите систему уравнений: y^2-xy=12; 3y-x=10.

\[y^{2} - 3y^{2} + 10y - 12 = 0\]

\[- 2y^{2} + 10y - 12 = 0\ \ \ \ \ \ \ \ |\ :( - 2)\]

\[y^{2} - 5y + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2;\ \ \ \ x_{2} = 3.\]

\[\left\{ \begin{matrix} x = 2\ \ \ \\ y = - 4 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = 3\ \ \ \\ y = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(2; - 4);\ \ (3; - 1).\]