I'm assuming that you meant '2', not 'z' for the right side of your second equation:

3x - 2y + z = 1

-x + y - z = 2

5x + 2y + 10z = 39

The variable z seems to be the easiest to remove:

Using the first and second equations:

                     3x - 2y + z = 1

                     -x +  y  - z = 2

Adding down:  2x  - y      = 3

Using the second and third equations:

-x + y - z = 2              --->   multiply by 10   --->     -10x + 10y - 10z  =  20

5x + 2y + 10z = 39                                      --->        5x  + 2y + 10z  = 39 

Adding down:                                                          -5x  + 12y          = 59

Combining these equations:

 2x   -    y  =   3           --->   multiply by 12   --->   24x - 12y  =  36

-5x + 12y  =  41                                          --->    -5x + 12y  =  59

 Adding down:                                                       19x           =   95     --->   x  =  5

Replacing x = 5 into 2x - y = 3   --->   10 - y  =  3     --->     y  =  7

Replacing x = 5 and y = 7 into 3x - 2y + z  =  1   --->   15 - 14 + z  =  1    --->     z  =  0

Nice answer Gino :)

I'd like to see this one answered using matrices.  

I'd like to do it myself but there arer too many questions and i do not have time      

this is going to be a major effort has I have to revise matrices from the beginning LOL

Talk about making life difficult fo myself.

I viewed this you tube clip before I started.

http://www.youtube.com/watch?v=pKZyszzmyeQ

$$\begin{pmatrix} 3&2&1 \\ -1&1&-2 \\ 5&2&10 \\ \end{pmatrix} & \begin{pmatrix} x\\ y\\ z \end{pmatrix}&=& \begin{pmatrix} 1\\ 0\\ 39 \end{pmatrix}$$

Now  det=3(14)-2(0)+1(-7)=42-7=35

$$\dfrac{1}{35}& \begin{pmatrix} 14&-18&-5 \\ 0& 25&5\\ -7&4&5 \end{pmatrix} & \begin{pmatrix} 3&2&1 \\ -1&1&-2 \\ 5&2&10 \\ \end{pmatrix} & \begin{pmatrix} x& \\ y& \\ z& \end{pmatrix} &=& \dfrac{1}{35}& \begin{pmatrix} 14&-18&-5 \\ 0& 25&5\\ -7&4&5 \end{pmatrix} & \begin{pmatrix} 1& \\ 0& \\ 39& \end{pmatrix}$$

$$\dfrac{1}{35}& \begin{pmatrix} 35&0&0 \\ 0& 35&0\\ 0&0&35 \end{pmatrix} & \begin{pmatrix} x& \\ y& \\ z& \end{pmatrix} &=& \dfrac{1}{35}& \end{pmatrix} \begin{pmatrix} 14-0-195& \\ 0+0+195& \\ -7+0+195& \end{pmatrix}$$

$$\begin{pmatrix} x& \\ y& \\ z& \end{pmatrix} &=& \dfrac{1}{35}& \begin{pmatrix} -181& \\ 195& \\ 188& \end{pmatrix}$$

$$\begin{pmatrix} x& \\ y& \\ z& \end{pmatrix} &=& \dfrac{1}{35}& \begin{pmatrix} -181& \\ 195& \\ 188& \end{pmatrix}$$

the reason tha my answser and Gino's answers are different is because he assumed that =z in the middle equation was a mistype and he changed the z to a 2.

I left it as a z.

Gino's assumuption was obviously correct since his way the number worked out nicely and my way the numbers are horrible.  Even so, I have answered the question that was asked.