I corrected the error which the other tutor observed 
and changed the x to z.

Line them up like this:

(1)      x + 2y +  z = -3 
(2)     2x - 2y +  z =  4 
(3)      x -  y + 2z =  5 

We pick a letter to eliminate and then pick two of the
equations to eliminate it from.

We notice that if we add (1) and (2) the y's cancel out,
so we pick y to eliminate and we pick (1) and (2) to
eliminate it from.  So we add (1) and (2)

(1)      x + 2y +  z = -3 
(2)     2x - 2y +  z =  4  
-------------------------
(4)     3x      + 2z =  1

Now we pick a different pair to eliminate the same letter
y from.

If we multiply (3) by 2 and add it to (1) the y's will
cancel

(1)      x + 2y +  z = -3 
        2x - 2y + 4z = 10
-------------------------
(5)     3x      + 5z =  7

So we have this system of two equations in two unknowns:

(4)     3x + 2z =  1
(5)     3x + 5z =  7

If we multiply (4) by -1 and add (5) to it the x's will 
cancel:

       -3x - 2z = -1
(5)     3x + 5z =  7
--------------------
             3z =  6
              z =  2

Substitute z = 2 in (4)

(4)   3x + 2(2) =  1
         3x + 4 =  1
             3x = -3
              x = -1

Substitute z = 2 and x = -1 in (1)

(1)      x + 2y +  z = -3
        -1 + 2y +  2 = -3
              2y + 1 = -3
                  2y = -4
                   y = -2


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

Edwin