x +  y +  z = -1
 2x -  y + 2z = -5
 -x + 2y -  z =  4

Add the first equation term-by-term to the third 
equation and the x's and z's will both cancel:

  x +  y +  z = -1
 -x + 2y -  z =  4
------------------
      3y      =  3
       
Divide both sides by 3 and you get
            
             y = 1

Substitute y = 1 in the first equation:

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

Substitute y = 1 in the second equation:

   2x - y + 2z = -5
   2x - 1 + 2z = -5
       2x + 2z = -4

Divide through by 2

        x + z = -2

Wow!  We got the same equation:

Substitute y = 1 in the third equation:

  -x + 2y - z =  4
-x + 2(1) - z =  4
   -x + 2 - z =  4
       -x - z =  2

Divide through by -1

        x + z = -2

Wow!  We got the same equation a third time:  

This means the system is dependent and has infinitely
many solutions.

The way we handle such an equation is to choose an
arbitrary value k for the last letter z, that is z=k

We substitute k for z and have

       x + k = -2
           x = -2 - k

Then we have the general solution

          x = -2-k, y=1, z=k


 (x, y, z) = (-2-k, 1, k)  

 There are many many solutions.  Here are some sample solutions:

If k = 0, we have the solution (x, y, z) = (-2, 1, 0)
If k = 1, we have the solution (x, y, z) = (-3, 1, 1)
If k = -2, we have the solution (x, y, z) = (0, 1, -2)
If k = -7, we have the solution (x, y, z) = (5, 1, -7)

Edwin