The above is fine.
In the following, you choose to eliminate ##\ x \ ## in two equations, which is a fairly common approach. However, as you point out above, you can pick any variable and in the case of these three equations, equation [3] already has ##\ z \ ## eliminated. I'll say more on that later
†. Apparently, you intended to eliminate ##\ x\ ## by subtracting Eqn [1] from Eqn [4]. (As I stated in Post #2, the result you give as Eqn [5] is incorrect.)
Specifically, 12z − (−9z) → 21z, not 3z .
I find that I make fewer arithmetic errors, if I do the elimination by adding equations rather than subtracting them. So for your system of equations, get Eqn [4] by multiplying Eqn [2] by −4 .
Then add equations [1] and [4].
That gives: ##\ 0x - 7y - 21z = 0 \ ## for Eqn [5].
( Maybe divide this by −7 & use this instead. ##\ \ \ y + 3z = 0 \ ## , Eqn [5*]
Furthermore, this could also be written: ##\ 3z = -y \,.## Going even further, plug this back into Eqn[2] to get
##x + 2y + (-y) =0 ##
## x + y =0 \ ## ( Eqn [5**] )
Again, more on this later
† .
)
Again, an arithmetic (or algebra) mistake due to incorrect subtracting.
That should have given: −7y + 9z − 10 = 0 for Eqn [7]In writing the "system", it's customary to write a complete set of three equations. So, any of the original equations, Eqn [2] is recommended b/c the coefficient of x is 1 .
With the corrected equations, the system can be written as follows.
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ -7y + 9z &= 10 & \text{ [7*]} \nonumber
\end{align} \right. ##
Use elimination with Eqn's [5*] and [7*] to obtain an equation in which ## y ## is eliminated. Easily done by taking 7 times Eqn [5*] and adding Eqn [7*] to that.
You get ##\ 30z = 10 \ ## which simplifies to ##\ 3z = 1 \,. ##
The "system becomes:
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ 3z &= 1 & \text{ [8]} \nonumber
\end{align} \right. ##
The above system may be said to be in upper triangular form and the system can be solved by "back substitution".
† This post is way long enough. I'll do the "later" stuff in another post. Basically, it involves using the observation that eliminating ##z## from an additional equation ( using elimination on Eqn's [1] &[2] ) gets you a result with much less computation.