x+3y+3z=3
x+4y+3z=8
x+4y+4z=2
Write that as a matrix by dropping the letters
and putting a vertical line instead of equal signs:
The idea is to get three zeros in the three positions
in the lower left corner of the matrix, where the elements
I've colored red are:
To get a 0 where the red 1 on the left of the middle row is,
multiply R1 by -1 and add it to 1 times R2, and put it in place
of the present R2. That's written as
-R1+1R2->R2
To make it easy, write the multipliers to the left of the two
rows you're working with; that is, put a -1 by R1 and a 1 by R2
We are going to change only R2. Although R1 gets multiplied
by -1 we are going to just do that mentally and add it to R2, but
not really change R1.
-----
To get a 0 where the lower left red 1 is, multiply R1
by -1 and add it to 1 times R3. That's written as
-1R1+1R3->R3
Write the multipliers to the left of the two rows you're
working with; that is, put a -1 by R1 and a 1 by R3
We are going to change only R3.
---------------
To get a 0 where the red 1 on the bottom row is,
multiply R2 by -1 and add it to 1 times R3. That's
written as
-1R2+1R3->R3
Write the multipliers to the left of the two
rows you're working with; that is, put a -1 by R2 and a 1 by R3
We are going to change only R3.
Now that we have 0's in the three positions in the
lower left corner of the matrix, we change the matrix
back to equations:
or
The third equation is already solved for z, and
The second equation is already solved for y, so
Substitute -6 for z and 5 for y in the top equation:
So the solution is 
Edwin
