Start with $\tag{1} \left[\matrix{ \color{maroon}{-4}&5&5\cr \color{darkgreen}{5}&-2&-4\cr 2&5&-6\cr }\ \ \Biggl |\ \ \matrix{-20\cr -1\cr -6}\right] $
The first phase is to eliminate (make 0) the first entries in rows two and three by adding multiples of row one to them.
But, to avoid fractions, you could also add a multiple of row one to a multiple of row two or three to effect this. Let's consider how to use this procedure to eliminate the $\color{darkgreen}{5}$ in row two: we need to multiply the $\color{maroon}{-4}$ in row 1 by something and the $\color{darkgreen}5$ in row two by something so that we we add these products, we obtain 0. What are the somethings?
A moment's reflection reveals that you may take the somethings to be $\color{darkgreen}5$ and $\color{maroon}4$. (There are other choices; but, multiplying each row by the coefficient of the leading entry of the other row and subtracting will always work, provided those coefficients are nonzero.)
So, to eliminate the 5 in row two, we may, and do, add five times row one to four times row two. To keep things simple, do not replace row one by five times itself.
In order to minimize the chance of making arithmetic errors, let's do the calculation on the side:
$5 \cdot\text{row}_1+4\cdot \text{row}_2$: $ \eqalign{ 5&({-}4\ \ \hphantom{ - }5 \ \ \hphantom{-}5\ \ {-}20)\cr +4&(\hphantom{-}5\ \ {-}2\ \ {-}4\ \ {-}1\hphantom{0})\cr & \ \ \cr &\ \ \phantom{(} }\qquad\Rightarrow \eqalign{ &({-}20\ \ \hphantom{ - }25 \ \ \hphantom{-}25\ \ {-}100)\cr + &(\hphantom{-}20\ \ {-}8\hphantom{3}\ \ {-}16\ \ {-}4\hphantom{0})\cr &\underline{\phantom{wwwwwwwwwwwwwww}}\cr &(\phantom{-} 0\phantom 5\ \ \hphantom{ - }17 \ \ \hphantom{-}9\ \ {-}104) } $
So $(1)$ becomes:
$\tag{2} \left[\matrix{ -4&5&5\cr 0&17&9\cr 2&5&-6\cr }\ \ \Biggl |\ \ \matrix{-20\cr -104\cr -6}\right] $
Now we want to eliminate the 2 in row three. To do this we add row one to twice row three:
We have $ \text{row}_1+2\cdot\text{row}_3$: $ \eqalign{ &({-}4\ \ \hphantom{ - }5 \ \ \hphantom{-}5\ \ {-}20)\cr +2&(\hphantom{-}2\ \ \phantom{-}5\ \ {-}6\ \ {-}6\hphantom{0})\cr & \ \ \cr &\ \ \phantom{(} }\qquad\Rightarrow \eqalign{ &({-}4\ \ \hphantom{ - }5\phantom{4} \ \ \hphantom{--}5\phantom{5}\ \ {-}20)\cr + &(\hphantom{-}4\ \ \phantom{-}10\hphantom{3}\ \ {-}12\ \ {-}12\hphantom{0})\cr &\underline{\phantom{wwwwwwwwwwwwwww}}\cr &(\phantom{-} 0\phantom 5\ \ 15 \ \ \ \ \ {-}7\ \ \ \ {-}32) } $
So $(2)$ becomes:
$\tag{3} \left[\matrix{ -4&5&5\cr 0&17&9\cr 0&15&-7\cr }\ \ \Biggl |\ \ \matrix{-20\cr -104\cr -32}\right] $
Yuch; ugly, but let's deal with it. We want to eliminate the 15 in row three now. Here we also work with row two; do not use row one to eliminate the 15 in row three; if you do this the 0 in row three will become nonzero...
So, we take 15 times row two and subtract 17 times row three. We have:
$15\cdot\text{row}_2+17\cdot\text{row}_3$: $ \eqalign{ 15&(\phantom{-}0\ \ \hphantom{ - }17 \ \ \hphantom{-}9\ \ {-}104)\cr -17&(\hphantom{-}0\ \ \phantom{-}15\ \ {-}7\ \ {-}32\hphantom{0})\cr & \ \ \cr &\ \ \phantom{(} }\qquad\Rightarrow \eqalign{ &(\phantom{-}0\ \ \hphantom{ - }235\phantom{4} \ \ \ \hphantom{-}135\phantom{5}\ \ {-}1560)\cr + &(\hphantom{-}0\ \ {-}235\hphantom{3}\ \ \ \phantom{-}119\ \ \hphantom{-}544\hphantom{0})\cr &\underline{\phantom{wwwwwwwwwwwwwww}}\cr &(\phantom{-} 0\phantom 0\ \ 0\phantom{35}\ \ \ \ \ \ \ \ \ 254 \ \ \ \ \ -1016) } $
So $(3)$ becomes: $\tag{4} \left[\matrix{ -4&5&5\cr 0&17&9\cr 0&0&254\cr }\ \ \Biggl |\ \ \matrix{-20\cr -104\cr -1016}\right] $
We may now solve by back substitution:
Solve for $z$ first.
From row three of $(4)$, $ z={-1016\over 254}=-4 $
Now solve for $y$.
From row two of $(4)$: $\eqalign{ & 17y+9z =-104\cr\iff& 17y+9(-4)=-104\cr\iff& 17y=-104+36\cr\iff &17y=-68\cr\iff &y=-4} $
Finally solve for $x$.
From row one of $(4)$: $ -4x+5y+5z=-20\iff-4x-20-20=-20\iff x=-5. $