Решите систему уравнений: y=x^2-4x+1; 2x+y=4.

\[\left\{ \begin{matrix} y = x^{2} - 4x + 1 \\ 2x + y = 4\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2x + x^{2} - 4x + 1 = 4\]

\[x² - 2x + 1 - 4 = 0\]

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

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

\[= 4 + 12 = 16\]

\[x_{1} = \frac{2 + \sqrt{16}}{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3\]

\[x_{2} = \frac{2 - \sqrt{16}}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} =\]

\[= - 1\]

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

\[\Longrightarrow \text{\ \ \ \ \ \ \ \ \ \ y}_{1} = 3^{2} - 6 \cdot 3 + 7 =\]

\[= 9 - 18 + 7 = - 2.\]

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

\[\Longrightarrow y_{2} = ( - 1)^{2} - 6 \bullet ( - 1) + 7 =\]

\[= 1 + 6 + 7 = 14.\]

\[Ответ:(3;\ - 2),\ \ \ \ ( - 1;14).\]