$$\left(\begin{array}{ccc|c} 2&2&1&4.5 \\ 4&6&3&12\\ 6&9&7&m \end{array}\right)$$ When I try to solve this, obviously always end up with this $m=x$ term. When you do this wth Gauss, you actually can reach to $y=(459-m)/90$ and $z=(2m-36)/5$, and once you plug $y$ and $z$ in the first equation you get $x=0.75$. Now, when you try to sustitute on the other equations to get $y$ and $z$, always end up with expression like $270=270$ I know it means the system is consistent but I can´t catch $y$ and $z$.
I also tried to do with cramer and this is what I get: $\det (M)= 10, \det(M_{11})= 0.75, \det(M_{12})= 51-2x, \det(M_{13})=4x-72$ so $x=\det(M_{11})/\det(M)= 0.75, y=(51-2m)/10, z= (2m-46)/2$
What I am missing? This is when I stuck. Thank´s for all kind of help.
Your solution is spot on with Gauss.
Yes, you always get $x = 0.75$ as fixed.
$y$ and $z$ have a "free" variable that you can set to "any' value you like. For example, choose $m = 0$ and $y = \dfrac{459}{90}$ and $z = -\dfrac{36}{5}$.
Now, when you sub those values in, of course each side should equate.
If you had shown this using Gaussian Elimination, you would have ended up with:
$$\left[\begin{array}{ccc|c} 1& 0& 0& 0.75\\ 0& 1& 0& 5.1 - 0.2m\\ 0& 0& 1& 0.4m-7.2 \end{array}\right]$$
It is best to leave the solution showing that you have 'free' variables.