linear systems in three variables 

2x + 3y - 6z =  4
3x - 2y - 9z = -7
2x + 5y - 6z =  8

Make this augmented matrix of coefficients:

[2  3 -6 |  4]
[3 -2 -9 | -7]
[2  5 -6 |  8]

The idea is to get 0's where the three red
numbers are.

[2  3 -6 |  4]
[3 -2 -9 | -7]
[2  5 -6 |  8]

[2  3 -6 |  4]
[3 -2 -9 | -7]
[2  5 -6 |  8]

Get a 0 where the red 3 is by multiplying row 1 by
-3 and adding it to 2 times row 2, but restore row 1:

-3×[2  3 -6 |  4]
 2×[3 -2 -9 | -7]
   [2  5 -6 |  8]

[2   3  -6 |   4]
[0 -13   0 | -26]
[2   5  -6 |   8] 

Get a 0 where the red 2 is by multiplying row 1 by
-1 and adding it to 1 times row 3, but restore row 1:

-1×[2   3  -6 |   4]
   [0 -13   0 | -26]
 1×[2   5  -6 |   8]

[2   3  -6 |   4]
[0 -13   0 | -26]
[0   2   0 |   4]

Get a 0 where the red 2 is by multiplying row 2 by
2 and adding it to 13 times row 3, but restore row 2:

   [2   3  -6 |   4]
 2×[0 -13   0 | -26]
13×[0   2   0 |   4] 
  
[2   3  -6 |   4]
[0 -13   0 | -26]
[0   0   0 |   0]

Get a 1 where the red 2 is by dividing row 1 thru by 2
Get a 1 where the red -13 is by dividing row 2 thru by -13
 
[1   -3 | 2]
[0   1  0 | 2]
[0   0  0 | 0]

Convert back into a system of equations:

1x +  y - 3z = 2
0x +   1y + 0z = 2
0x +   0y + 0z = 0

Erase the first 0 term of the 2nd equation,
Erase the first two 0 terms of the third equation:

 x + y - 3z = 2
        y + 0z = 2
            0z = 0

The last equation is true for any value of z
So we let z be arbitrary. Then z = a

Substitute in the 2nd equation:

     y + 0(a) = 2
            y = 2

Substitute z = a and y = 2 in the first equation:

 x + (2) - 3(a) = 2
        x + 3 - 3a = 2
                 x = 3a-1

So the set of solutions is

(x, y, z) = (3a-1, 2, a)

Edwin