Question

Ok

Answer 1

I would use elimination method. First write the system in matrix form,
\[\left[\begin{matrix}-2 &1 &-2 &10 \\3 & -2 &-2 & 6\\ 1 & 2 & 1 &-10\end{matrix}\right]\]Operate with rows,
\[\left[\begin{matrix} 1 & 2 & 1 &-10\\-2 &1 &-2 &10 \\3 & -2 &-2 & 6\\\end{matrix}\right]\]\[\left[\begin{matrix} 1 & 2 & 1 &-10\\0 &5 &0 &-10 \\0 & 8 &5 & -36\\\end{matrix}\right]\]\[\left[\begin{matrix} 1 & 2 & 1 &-10\\0 &5 &0 &-10 \\0 & 0 &-25 & 100\\\end{matrix}\right]\]
Then solve.

Answer 2

First operation: change row 1 for row 3
Second operation: 2*row1+row2 to substitute in row2
3*row1-row3 to substitute in row3
Third operation: 8*row2-5*row3 to substitue in row 3

Answer 3

Rewrite the system,
\[x+2y+z=-10\\
5y=-10\\
-25z=100\]
Now solve,
\[z=-100/25=-4\]\[z=-10/5=-2\]\[x-4-4=-10\Rightarrow x=-2\]