x+2y+2z=1; 2x-3y-z=6; 3x+y+2z=5

Given the system of equations
$$2 z + \left(x + 2 y\right) = 1$$
$$- z + \left(2 x - 3 y\right) = 6$$
$$2 z + \left(3 x + y\right) = 5$$

We give the system of equations to the canonical form
$$x + 2 y + 2 z = 1$$
$$2 x - 3 y - z = 6$$
$$3 x + y + 2 z = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}1 & 2 & 2 & 1\\2 & -3 & -1 & 6\\3 & 1 & 2 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}1\\2\\3\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}1 & 2 & 2 & 1\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 2 + 2 & - 2 \cdot 2 - 3 & - 2 \cdot 2 - 1 & \left(-1\right) 2 + 6\end{matrix}\right] = \left[\begin{matrix}0 & -7 & -5 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 2 & 2 & 1\\0 & -7 & -5 & 4\\3 & 1 & 2 & 5\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\left(-1\right) 3 + 3 & 1 - 2 \cdot 3 & 2 - 2 \cdot 3 & \left(-1\right) 3 + 5\end{matrix}\right] = \left[\begin{matrix}0 & -5 & -4 & 2\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 2 & 2 & 1\\0 & -7 & -5 & 4\\0 & -5 & -4 & 2\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}2\\-7\\-5\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -7 & -5 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-2\right) 0}{7} & 2 - - -2 & 2 - - \frac{-10}{7} & 1 - \frac{\left(-2\right) 4}{7}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & \frac{4}{7} & \frac{15}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & \frac{4}{7} & \frac{15}{7}\\0 & -7 & -5 & 4\\0 & -5 & -4 & 2\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 5}{7} & -5 - - 5 & -4 - - \frac{25}{7} & 2 - \frac{4 \cdot 5}{7}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & - \frac{3}{7} & - \frac{6}{7}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & \frac{4}{7} & \frac{15}{7}\\0 & -7 & -5 & 4\\0 & 0 & - \frac{3}{7} & - \frac{6}{7}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{4}{7}\\-5\\- \frac{3}{7}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & - \frac{3}{7} & - \frac{6}{7}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}1 - \frac{\left(-4\right) 0}{3} & - \frac{\left(-4\right) 0}{3} & \frac{4}{7} - - \frac{-4}{7} & \frac{15}{7} - - \frac{-8}{7}\end{matrix}\right] = \left[\begin{matrix}1 & 0 & 0 & 1\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 1\\0 & -7 & -5 & 4\\0 & 0 & - \frac{3}{7} & - \frac{6}{7}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 35}{3} & -7 - \frac{0 \cdot 35}{3} & -5 - - 5 & 4 - - 10\end{matrix}\right] = \left[\begin{matrix}0 & -7 & 0 & 14\end{matrix}\right]$$
you get
$$\left[\begin{matrix}1 & 0 & 0 & 1\\0 & -7 & 0 & 14\\0 & 0 & - \frac{3}{7} & - \frac{6}{7}\end{matrix}\right]$$

It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$x_{1} - 1 = 0$$
$$- 7 x_{2} - 14 = 0$$
$$\frac{6}{7} - \frac{3 x_{3}}{7} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -2$$
$$x_{3} = 2$$