He did it my Cramer's rule (determinants). I didn't 
know if you'd had that or just the elimination method.
So I thought I'd do it by the first method that's
usually taught.


Line them up so that like letters, signs, 
and equal signs, and numbers line up
vertically.

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

The idea is to reduce that to a linear system with
only 2 variables.

Notice that one of the variables has already been 
eliminated in all three.  Pick one of the three
to be one of the equations in the system of 2
equations with 2 variables.  It doesn't matter 
which of the three you pick.  Say we pick the 
third one, from which y has already been 
eliminated.

 3x + 2z = 3

Now even though a variable has been eliminated from
the other two, you still must eliminate the same
variable from the other two that is already eliminated 
in the one we picked, 3x + 2z = 3.  So we eliminate
y from the other two equations:


4x - 3y      = 1
     2y - 3z = 2
 
To eliminate y from them, we multiply the first
equation by 2 and the second one by 3, and then
add them:

8x - 6y      = 2
     6y - 9z = 6
----------------
8x      - 9z = 8 

So the linear system in 2 variables is

3x + 2z = 3
8x - 9z = 8

We can eliminate z by multiplying the first one just
above by 9 and the second one just above by 2
and adding:

27x + 18z = 27
16x - 18z = 16
--------------
43x       = 43
        x = 1

Substitute x = 1 in

  3x + 2z = 3
3(1) + 2z = 3
   3 + 2z = 3
       2z = 0
        z = 0

Substitute x = 1 in the first original 
equation:

  4x - 3y = 1
4(1) - 3y = 1
   4 - 3y = 1
      -3y = -3
        y = 1 

Answer (x,y,z) = (1,1,0)

Edwin