Решите систему уравнений x^2+12xy+36y^2=16; x-6y=-8.

Ответ:

\[\left\{ \begin{matrix} x^{2} + 12xy + 36y^{2} = 16 \\ x - 6y = - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 6y - 8\ \ \ \ \ \ \ \ \\ (x + 6y)^{2} = 16 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 6y - 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (6y - 8 + 6y)^{2} = 16 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 6y - 8\ \ \ \ \ \ \ \ \ \ \\ (12y - 8)^{2} = 16 \\ \end{matrix} \right.\ \]

\[(12y - 8)^{2} = 16\]

\[12y - 8 = 4\ \ \ \ \ \ \ \ 12y - 8 = - 4\]

\[12y = 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ 12y = 4\]

\[y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = \frac{1}{3}\]

\[\left\{ \begin{matrix} y = 1\ \ \ \\ x = - 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \frac{1}{3}\text{\ \ \ } \\ x = - 6 \\ \end{matrix} \right.\ \]

\[Ответ:( - 2;1);\ \ \left( - 6;\frac{1}{3} \right).\]