Решите систему уравнений: y-x=2; y^2+4x=13.

Ответ:

\[\left\{ \begin{matrix} y - x = 2\ \ \ \ \ \ \ \\ y^{2} + 4x = 13 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = y - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 4(y - 2) = 13 \\ \end{matrix} \right.\ \]

\[y^{2} + 4y - 8 - 13 = 0\]

\[y^{2} + 4y - 21 = 0\]

\[D_{1} = 4 + 21 = 25\]

\[y_{1} = - 2 + 5 = 3;\]

\[y_{2} = - 2 - 5 = - 7;\]

\[x_{1} = 3 - 2 = 1;\]

\[x_{2} = - 7 - 2 = - 9.\]

\[Ответ:(1;3);( - 9; - 7).\]