Найдите решение системы: x+y+z=6; x+y-z=4; x-y-z=0.

\[\left\{ \begin{matrix} x + y + z = 6 \\ x + y - z = 4 \\ x - y - z = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = 6 - z \\ 6 - z - z = 4\ \\ x - y = z\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 2z = - 2\ \ \ \ \ \ \ \\ x + y = 6 - z \\ x - y = z\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} z = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + y = 6 - 1 \\ x - y = 1\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} z = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 1 + y\ \ \ \ \ \ \ \\ 1 + y + y = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} z = 1\ \ \ \ \ \ \ \ \ \\ y = 2\ \ \ \ \ \ \ \ \\ x = 1 + 2 \\ \end{matrix}\ \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} z = 1 \\ y = 2 \\ x = 3 \\ \end{matrix} \right.\ \]

\[Ответ:(3;2;1).\]