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

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

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

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

\[x^{2} + 2x = 0\]

\[x(x + 2) = 0\]

\[x_{1} = 0;\ \ \ x_{2} = - 2.\]

\[x_{1} = \ 0 \Longrightarrow \ \ \ \ \ \ y_{1} =\]

\[= 0^{2} + 4 \cdot 0 + 1 = 0 + 0 + 1 = 1\]

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

\[= ( - 2)^{2} + 4 \bullet ( - 2) + 1 =\]

\[= 4 - 8 + 1 = - 3.\]

\[Ответ:\ \ (0;1),\ \ \ ( - 2;\ - 3).\]