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

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

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

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

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

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

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

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

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

\[Ответ:(3;2;1).\]