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

\[x(2x - 2) - 3 \bullet (2x - 2) - 4x =\]

\[= - 10\]

\[2x^{2} - 2x - 6x + 6 - 4x + 10 =\]

\[= 0\]

\[2x^{2} - 12x + 16 = 0\ \ \ \ \ \ \ \ \ |\ :2\]

\[x^{2} - 6x + 8 = 0\]

\[D = ( - 6)^{2} - 4 \cdot 1 \cdot 8 =\]

\[= 36 - 32 = 4\]

\[x_{1} = \frac{6 + \sqrt{4}}{2} = \frac{6 + 2}{2} = \frac{8}{2} = 4\]

\[x_{2} = \frac{6 - \sqrt{4}}{2} = \frac{6 - 2}{2} = \frac{4}{2} = 2\]

\[x_{1} = 4 \Longrightarrow \ \ \ \ \ \ \ y_{1} = 2 \cdot 4 - 2 =\]

\[= 8 - 2 = 6.\]

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

\[= 4 - 2 = 2.\]

\[Ответ:(4;6),\ \ \ (2;2).\]