2х-4у+9z=28; 7x+3y-6z=-1; 7x+9y-9z=5

Given the system of equations
$$9 z + \left(2 x - 4 y\right) = 28$$
$$- 6 z + \left(7 x + 3 y\right) = -1$$
$$- 9 z + \left(7 x + 9 y\right) = 5$$

We give the system of equations to the canonical form
$$2 x - 4 y + 9 z = 28$$
$$7 x + 3 y - 6 z = -1$$
$$7 x + 9 y - 9 z = 5$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}2 & -4 & 9 & 28\\7 & 3 & -6 & -1\\7 & 9 & -9 & 5\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}2\\7\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}2 & -4 & 9 & 28\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{2 \cdot 7}{2} & 3 - - 14 & - \frac{7 \cdot 9}{2} - 6 & - \frac{7 \cdot 28}{2} - 1\end{matrix}\right] = \left[\begin{matrix}0 & 17 & - \frac{75}{2} & -99\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -4 & 9 & 28\\0 & 17 & - \frac{75}{2} & -99\\7 & 9 & -9 & 5\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{2 \cdot 7}{2} & 9 - - 14 & - \frac{7 \cdot 9}{2} - 9 & 5 - \frac{7 \cdot 28}{2}\end{matrix}\right] = \left[\begin{matrix}0 & 23 & - \frac{81}{2} & -93\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & -4 & 9 & 28\\0 & 17 & - \frac{75}{2} & -99\\0 & 23 & - \frac{81}{2} & -93\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-4\\17\\23\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & 17 & - \frac{75}{2} & -99\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{\left(-4\right) 0}{17} & -4 - \frac{\left(-4\right) 17}{17} & 9 - - \frac{-150}{17} & 28 - - \frac{-396}{17}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & \frac{3}{17} & \frac{80}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & \frac{3}{17} & \frac{80}{17}\\0 & 17 & - \frac{75}{2} & -99\\0 & 23 & - \frac{81}{2} & -93\end{matrix}\right]$$
From 3 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{0 \cdot 23}{17} & 23 - \frac{17 \cdot 23}{17} & - \frac{81}{2} - - \frac{1725}{34} & -93 - - \frac{2277}{17}\end{matrix}\right] = \left[\begin{matrix}0 & 0 & \frac{174}{17} & \frac{696}{17}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & \frac{3}{17} & \frac{80}{17}\\0 & 17 & - \frac{75}{2} & -99\\0 & 0 & \frac{174}{17} & \frac{696}{17}\end{matrix}\right]$$
In 3 -th column
$$\left[\begin{matrix}\frac{3}{17}\\- \frac{75}{2}\\\frac{174}{17}\end{matrix}\right]$$
let’s convert all the elements, except
3 -th element into zero.
- To do this, let’s take 3 -th line
$$\left[\begin{matrix}0 & 0 & \frac{174}{17} & \frac{696}{17}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}2 - \frac{0}{58} & - \frac{0}{58} & \frac{3}{17} - \frac{174}{17 \cdot 58} & \frac{80}{17} - \frac{696}{17 \cdot 58}\end{matrix}\right] = \left[\begin{matrix}2 & 0 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 4\\0 & 17 & - \frac{75}{2} & -99\\0 & 0 & \frac{174}{17} & \frac{696}{17}\end{matrix}\right]$$
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{\left(-425\right) 0}{116} & 17 - \frac{\left(-425\right) 0}{116} & - \frac{75}{2} - \frac{\left(-425\right) 174}{17 \cdot 116} & -99 - \frac{\left(-425\right) 696}{17 \cdot 116}\end{matrix}\right] = \left[\begin{matrix}0 & 17 & 0 & 51\end{matrix}\right]$$
you get
$$\left[\begin{matrix}2 & 0 & 0 & 4\\0 & 17 & 0 & 51\\0 & 0 & \frac{174}{17} & \frac{696}{17}\end{matrix}\right]$$

It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$2 x_{1} - 4 = 0$$
$$17 x_{2} - 51 = 0$$
$$\frac{174 x_{3}}{17} - \frac{696}{17} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = 3$$
$$x_{3} = 4$$