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.