Решите систему уравнений: (x-2)/4+(y-2)/4=2; (x-2)/3-(y-2)/9=4/3.

Ответ:

\[\left\{ \begin{matrix} \frac{x - 2}{4} + \frac{y - 2}{4} = 2\ \ \ | \cdot 4 \\ \frac{x - 2}{3} - \frac{y - 2}{9} = \frac{4}{3}\ \ | \cdot 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x - 2 + y - 2 = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 \cdot (x - 2) - (y - 2) = 3 \cdot 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x + y - 4 = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x - 6 - y + 2 = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = 12\ \ \ \ (1) \\ 3x - y = 16\ \ (2) \\ \end{matrix} \right.\ \]

\[4x = 28\ \]

\[x = 7\]

\[\left\{ \begin{matrix} x = 7\ \ \ \ \ \ \ \ \ \ \\ x + y = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 7\ \ \ \ \ \ \ \ \ \ \\ y = 12 - x \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 7\ \ \ \ \ \ \ \ \ \ \\ y = 12 - 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 7 \\ y = 5 \\ \end{matrix} \right.\ \]

\[Ответ:(7;5).\]