Get a 0 where the 3 is by multiplying row 1 by -3,
and row 2 by 5, adding them and replacing row 2 by
the result.  That's written -3R1+5R2->R2

        -15   6    9  |  0
         15  -5  -20  |  0
      ---------------------
          0   1  -11  |  0

  

Get a 0 where the 4 is by multiplying row 1 by -4,
and row 3 by 5, adding them and replacing row 3 by
the result.  That's written -4R1+5R3->R3

        -20   8   12  |  0
         20  -5  -20  |  0
      ---------------------
          0   1  -11  |  0



Get a 0 where the 1 on the bottom row is by 
multiplying row 2 by -1, and row 3 by 1, 
adding them and replacing row 3 by
the result.  That's written -R2+R3->R3

          0   1  -11  |  0
          0  -1   11  |  0
      ---------------------
          0   0    0  |  0



Convert this back to a system of equations:



Which when simplified is:



z can be assigned any value, some people set z equal
to some other letter, and other just leave it as z.
I'll just leave it as z.

Solve the second equation for y

-y - 11z = 0
      -y = -11z
       y = 11z

Substitute 11z for y in the first equation:

    5x - 2y - 3z = 0
5x - 2(11z) - 3z = 0
   5x - 22z - 3z = 0
        5x - 25z = 0
              5x = 25z
               x = 5z

We write the solution as

(x, y, z) = (5z, 11z, z)

Some books and teachers require students to use a different letter
for z, such as

(x, y, z) = (5a, 11a, a) or (5k, 11k, k), etc.

This system is dependent and has infinitely many solutions. Substitute
different numbers for z (or for a or k) and get different solutions.  

Edwin