x + y - z = 6
2x - y + z = -6
x - 4y +3z = -36

You can do these on your calculator fairly easily depending on your calculator. However this one isn't too hard to do by hand.
It helps to number the equations you have:
1) x + y - z = 6
2) 2x - y + z = -6
3) x - 4y +3z = -36

So starting with equation 1) we can re-arrange to...
x = 6 - y + z
Then you substitue this value of x into equation 2) which gives...
2*(6 - y + z) - y + z = -6
(expand the brackets and simplify)
12 - 3y + 3z = -6
(rearrange)
3z = -18 + 3y
(divide by 3)
z= (-6 + y)
Finally you substitue the values of x and z into equation 3) which gives...
(6 - y + z) - 4y + 3(-6 + y) = -36
(expand the brackets and simplify)
-12 - 2y + z = -36
Again we substitute the value we found for z above so...
-12 - 2y + (-6 + y) = -36
(simplify)
-18 - y = -36
-y = -18
(multiply by -1)
y = 18
Now that we have this we can easily find x and z
z = (-6 + y)
z = 12
and
x = 6 - y + z
x = 6 - 18 + 12
x = 0

Hello, Niaboc!

You said "row operations".
Are you working with Gaussian elimination?

$\displaystyle \begin{array}{ccc}x + y - z &=& 6 \\ 2x - y + z &=& \text{-}6 \\ x - 4y +3z &=& \text{-}36 \end{array}$

We have: .$\displaystyle \left|\begin{array}{ccc|c} 1&1&\text{-}1 & 6 \\ 2 &\text{-}1&1&\text{-}6 \\ 1&\text{-}4&3&\text{-}36 \end{array}\right|$

$\displaystyle \begin{array}{c}\\ R_2-2R_1 \\ R_3-R_1 \end{array}\:\left|\begin{array}{ccc|c} 1&1&\text{-}1 & 6 \\ 0&\text{-}3&3&\text{-}18 \\ 0&\text{-}5&4&\text{-}42 \end{array}\right|$

. . . $\displaystyle \begin{array}{c}\\ \text{-}\frac{1}{3}R_2 \\ \\ \end{array}\:\left|\begin{array}{ccc|c} 1&1&\text{-}1&6 \\ 0&1&\text{-}1&6 \\ 0&\text{-}5&4&\text{-}42 \end{array}\right|$

$\displaystyle \begin{array}{c}R_1-R_2 \\ \\ R_3 + 5R_2 \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&\text{-}1&6 \\ 0&0&\text{-}1&\text{-}12 \end{array}\right|$

. . . $\displaystyle \begin{array}{c}\\ \\ \text{-}1\!\cdot\!R_3 \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&-1&6 \\ 0&0&1&12 \end{array}\right|$

. . $\displaystyle \begin{array}{c} \\ R_2+R_3 \\ \\ \end{array}\:\left|\begin{array}{ccc|c}1&0&0&0 \\ 0&1&0&18 \\ 0&0&1&12 \end{array}\right|$

Therefore: .$\displaystyle \begin{Bmatrix}x &=& 0 \\ y &=& 18 \\ z &=& 12\end{Bmatrix}$