Решите систему уравнений: xy-2y-4x=-5; x-3y=-10.

\[3y^{2} - 24y + 45 = 0\ \ \ \ \ \ \ \ \ |\ :3\]

\[y^{2} - 8y + 15 = 0\]

\[D = ( - 8)^{2} - 4 \cdot 1 \cdot 15 =\]

\[= 64 - 60 = 4\]

\[y_{1} = \frac{8 + \sqrt{4}}{2} = \frac{8 + 2}{2} = \frac{10}{2} = 5\]

\[y_{2} = \frac{8 - \sqrt{4}}{2} = \frac{8 - 2}{2} = \frac{6}{2} = 3\]

\[y_{1} = 5 \Longrightarrow \ \ \ \ \ \ \ x_{1} = 3 \cdot 5 - 10 =\]

\[= 15 - 10 = 5.\]

\[y_{2} = 3 \Longrightarrow \ \ \ \ \ \ \ x_{2} = 3 \cdot 3 - 10 =\]

\[= 9 - 10 = - 1\]

\[Ответ:(5;5),\ \ \ ( - 1;3).\]