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

\[(2y - 5)y - 3y - 4 \bullet (2y - 5) =\]

\[= - 10\]

\[2y^{2} - 5y - 3y - 8y + 20 + 10 =\]

\[= 0\]

\[2y^{2} - 16y + 30 = 0\ \ \ \ \ \ \ |\ :2\]

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

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

\[= 64 - 60 = 4\]

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

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

\[y_{1} = 5 \Longrightarrow \ \ \ \ \ \ x_{1} = 2 \cdot 5 - 5 =\]

\[= 10 - 5 = 5.\]

\[y_{2} = 3 \Longrightarrow \ \ \ \ \ x_{2} = 2 \cdot 3 - 5 =\]

\[= 6 - 5 = 1.\]

\[Ответ:(5;5),\ \ \ (1;3).\]