Решите систему уравнений: y=x^2-6x+7; 2x+y=4.

\[\left\{ \begin{matrix} y = x^{2} - 6x + 7\ \ \\ 2x + y = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2x + x^{2} - 6x + 7 = 4\]

\[x^{2} - 4x + 7 - 4 = 0\]

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

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

\[= 16 - 12 = 4\]

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

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

\[x_{1} = 3 \Longrightarrow \ \ \ \ \ y_{1} =\]

\[= 3^{2} - 6 \cdot 3 + 7 = 9 - 18 + 7 =\]

\[= - 2.\]

\[x_{2} = 1 \Longrightarrow \ \ \ \ \ y_{2} =\]

\[= 1^{2} - 6 \cdot 1 + 7 = 1 - 6 + 7 = 2\]

\[Ответ:(3;\ - 2),\ \ \ (1;2).\]