Line them up so that like letters line up vertically, you might
like to put 1 coefficients before letters not showing a coefficient:

eq.1    2x-1y+1z = 3
eq.2    3x-3y-1z = 4
eq.3    1x+3y+3z = 6



I.  Pick a pair of equations and a letter that both contain and 
    eliminate that letter from them.

I pick eq.1 and eq.2 and z to eliminate from them:

eq.1    2x-1y+1z = 3
eq.2    3x-3y-1z = 4

The coefficients are z in eq.1 and eq.2 are 1 and -1 which will 
cancel if we add corresponding terms in the two equations:

eq.1    2x-1y+1z = 3
eq.2    3x-3y-1z = 4 
--------------------
eq.4    5x-4y    = 7

II. Use the equation you haven't used yet with one of the other
    equations and eliminate that SAME letter from them.

We haven't used eq.3 yet, so we put eq.3 with, say eq.2, and we
eliminate the same variable that we eliminated in the first step: 

eq.2    3x-3y-1z = 4
eq.3    1x+3y+3z = 6

[Caution: Even though the y's would cancel if we added them just
as they are, DO NOT add them, because we must eliminate the SAME 
letter we eliminated before, which is z, not y.]

To eliminate z we must multiply both sides of eq.2 by 3 to make
the -1z term, become -3z, and then add them

        9x-9y-3z = 12
eq.3    1x+3y+3z =  6
        -------------
       10x-6y    = 18

That can be divided through by 2, so we will do that:

eq.5   5x-3y = 9

III.  Solve the resulting 2×2 system of equations:

So we put eq.5 with eq.4:

eq.4   5x-4y = 7
eq.5   5x-3y = 9

The 5x in eq.4 can be made to cancel the 5x in eq.5 by
multiplying eq.4 by -1

      -5x+4y = -7
eq.5   5x-3y =  9
      -----------
           y =  2
        
Now we substitute in either eq.4 or eq.5 to find x. Say,
pick eq.5 and substitute y = 2

eq.5    5x-3y = 9
      5x-3(2) = 9
         5x-6 = 9
           5x = 15
            x = 3


IV.  Substitute the values obtained into one of the original equations
     to find the remaining variable, which was the firet ofne eliminated.

So we substitute x = 4 and y = 2 in any of the original 3 equations 
to find z, which was the first letter we eliminated.  I'll use eq.1:

eq.1    2x-1y+1z = 3
    2(3)-1(2)+1z = 3
           6-2+z = 3
             4+z = 3
               z = -1

Solution (x,y,z) = (4,2,-1)

Edwin