Решите систему уравнений x^2-y^2=3*(x+y); 1/(4x-3y)=1/7.

\[\left\{ \begin{matrix}
x^{2} - y^{2} = 3(x + y) \\
\frac{1}{4x - 3y} = \frac{1}{7}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\
\end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix}
(x + y)(x - y - 3) = 0 \\
4x - 3y = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix}
x + y = 0\ \ \ \ \ \\
4x - 3y = 7 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix}
x = - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
4 \cdot ( - y) - 3y = 7 \\
\end{matrix} \right.\ \]

\[- 4y - 3y = 7\]

\[- 7y = 7\]

\[y = - 1 \Longrightarrow x = 1.\]

\[2)\ \left\{ \begin{matrix}
x - y = 3\ \ \ \ \ \\
4x - 3y = 7 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix}
x = 3 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
4(3 + y) - 3y = 7 \\
\end{matrix} \right.\ \]

\[12 + 4y - 3y = 7\]

\[y = 7 - 12\]

\[y = - 5 \Longrightarrow x = 3 - 5 = - 2.\]

\[Ответ:(1;\ - 1);( - 2; - 5).\]