Use Gauss-Jordan elimination method to solve 

2x-5y+5z=17
x-2y+3z=9
-x+3y=-4

Line up the terms vertically:

2x - 5y + 5z = 17
 x - 2y + 3z =  9
-x + 3y      = -4

Fill up the "hole" in the bottom
equation with "+ 0z:

2x - 5y + 5z = 17
 x - 2y + 3z =  9
-x + 3y + 0z = -4 

Fill in all invisible 1's for
coefficients:

 2x - 5y + 5z = 17
 1x - 2y + 3z =  9
-1x + 3y + 0z = -4

Erase all the letters and
equal signs:

 2  - 5  + 5    17
 1  - 2  + 3     9
-1  + 3  + 0    -4 

Erase the plus signs
and move the minus
signs close to the
numbers as negative
signs:

 2   -5    5    17
 1   -2    3     9
-1    3    0    -4 

Draw a vertical line
where the equal signs
were and put brackets
around the whole thing.

[ 2   -5    5  | 17]
[ 1   -2    3  |  9]
[-1    3    0  | -4]

This is called the
augmented matrix

The idea is to get 0's
in the lower left three
positions, where the 
three red numbers are 
below:

[ 2   -5    5  | 17]
[ 1   -2    3  |  9]
[-1    3    0  | -4]

To get a 0 where the 1 is,
temporarily multiply the top 
row thru by 1 and the middle
row thru by -2.

 1[ 2   -5    5  | 17]
-2[ 1   -2    3  |  9]
  [-1    3    0  | -4]

  [ 2   -5    5  | 17]
  [-2    4   -6  |-18]
  [-1    3    0  | -4]

Add the top row to the middle
row and leave the top row as
it is:

  [ 2   -5    5  | 17]
  [ 0   -1   -1  | -1]
  [-1    3    0  | -4]

To get a 0 where the -1 at the
bottom left is, multiply the
top row by 1 and the bottom row
by 2:

 1[ 2   -5    5  | 17]
  [ 0   -1   -1  | -1]
 2[-1    3    0  | -4]

  [ 2   -5    5  | 17]
  [ 0   -1   -1  | -1]
  [-2    6    0  | -8]

Add the top row to the bottom
row, leaving the top row as it
is:

  [ 2   -5    5  | 17]
  [ 0   -1   -1  | -1]
  [ 0    1    5  |  9]

To get a 0 where the 1 is,
multiply the middle row by 1
and add to the bottom row:

  [ 2   -5    5  | 17]
 1[ 0   -1   -1  | -1]
 1[ 0    1    5  |  9]

  [ 2   -5    5  | 17]
  [ 0   -1   -1  | -1]
  [ 0    0    4  |  8]

Now that there are 0's
in those three positions,
we rewrite the augmented
matrix as a system of
equations, by putting the
variables and equal signs
back in:

  [ 2x  -5y   5z = 17]
  [ 0x  -1y  -1z = -1]
  [ 0x   0y   4z =  8]

Erase the brackets, the
terms with 0 coefficients,
the 1's, and move the negative signs
left as minus signs:

    2x - 5y + 5z = 17 
         -y -  z = -1 
              4z =  8 

Solve the bottom equation for z:

              4z = 8
               z = 2

Substitute z = 2 into the middle
equation:

         -y -  z = -1
         -y -  2 = -1
              -y =  1
               y = -1

Substitute y = -1 and z = 2 into
the top equation:

     2x - 5y + 5z = 17
2x - 5(-1) + 5(2) = 17
      2x + 5 + 10 = 17
          2x + 15 = 17
               2x =  2
                x =  1

So the solution is

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

Edwin