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

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

\[= - 5\]

\[3x^{2} - 2x - 6x + 4 - 4x + 5 = 0\]

\[3x^{2} - 12x + 9 = 0\ \ \ \ \ \ \ |\ :3\]

\[x^{2} - 4x + 3 = 0\]

\[D = ( - 4)^{2} - 4 \cdot 1 \cdot 3 =\]

\[= 16 - 12 = 4\]

\[x_{1} = \frac{4 + \sqrt{4}}{2} = \frac{4 + 2}{2} = \frac{6}{2} = 3\]

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

\[x_{1} = 3 \Longrightarrow \ \ \ \ y_{1} = 3 \cdot 3 - 2 =\]

\[= 9 - 2 = 7.\]

\[x_{2} = 1 \Longrightarrow \ \ \ \ y_{2} = 3 \cdot 1 - 2 =\]

\[= 3 - 2 = 1.\]

\[Ответ:(3;7),\ \ \ (1;1).\]