We form the augmented coefficient matrix

 [ 2   2   4 | 40]
 [ 0  -2   2 | 12]
 [ 2   3   3 | 34]

We must get three 0's in the lower 
left hand corner, where the numbers
are that I have colored red below:

 [ 2   2   4 | 40]
 [ 0  -2   2 | 12]
 [ 2   3   3 | 34]


Notice that one of them is already
0, so we get a zero where the red 2
is.

The top row will never change. 
The other two will. To get a 0 
where the red 2 is, We multiply -1 
times the first row and add it 
to 1 times the 3rd row. It's a 
good idea to put what you're
going to multiply a row by out
to the left of the row, even 
when it is just 1, like this:

-1[ 2   2   4 | 40]
  [ 0  -2   2 | 12]
 1[ 2   3   3 | 34]

The next matrix is then

 [ 2   2   4 | 40]
 [ 0  -2   2 | 12]
 [ 0   1  -1 | -6]

To get a 0 where the 1 is, 
we multiply 1 times the second 
row and add it to 2 times the 
3rd row.

 [ 2   2   4 | 40]
1[ 0  -2   2 | 12]
2[ 0   1  -1 | -6]

We get this:

 [ 2   2   4 | 40]
 [ 0  -2   2 | 12]
 [ 0   0   0 |  0]

Now that we have three 0's in the
lower left corner, the red places,
we go back to a system of equations:



or just:



Start at the bottom, and go back up.
This is called back-substitution.
Start with the bottom equation, solve for
z:



That's a strange sort of equation.  We see
that any arbitrarily chosen value of z will 
satisfy that equation.  This means that z 
is any arbitrarily chosen number.  Some books
use the letter "c" and others simply keep 
and write .

I will write as its solution

 because "a" stands for "arbitrary"

Substitute  for  in the
second equation:





Divide through by -2


Finally substitute both  for 
and  for  in the original
first equation:







Divide through by 2


So the solution is

(x,y,z)=(26-3a, -6+a, a)

Edwin