<s>[INST] Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$? Suppose we have this system of equations:
$$\begin{align}
2x+6y-3z&=10\tag{1} \\
5x+2y-1z&=12\tag{2}
\end{align}$$
Why does solving as follows give an incorrect solution?
$$\begin{align}
2x+6y-3z-10&=5x+2y-1z-12 \\
-3x+4y-2z&=-2\tag{3}
\end{align}$$
According to this equation, $(0,0,1)$ is a solution. However, when I plug this into the two starting equations, the result is inconsistent.
What is wrong with the logic I used in finding the solution?
By the way, I was only looking for a single solution of the two starting equations, not a general solution. I know how to use matrices to solve the system, but I am still curious as to what is wrong with the approach I demonstrated above.
 [/INST] \begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align}
Matrix form: $AX=b$
$A=\begin{pmatrix} 2 & 6&-3\\5&2&-1\end{pmatrix}$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}
$
$b=\begin{pmatrix} 10\\12\end{pmatrix}$
Solve : $[A|b]=\begin{pmatrix} 2 & 6&-3&|10\\5&2&-1&|12\end{pmatrix}$
Then you get the solution :
$X=\{c\begin{pmatrix} 0\\1\\2\end{pmatrix}+\begin{pmatrix} 2\\1\\0\end{pmatrix}:c\in\Bbb{R} \}$

Now back to the original problem :
$-3x+4y-2z=-2$
represent a plane inside the $3$ dimensional space.But the solution of the original system lies on a line.
Also the line lies on the plane $-3x+4y-2z=-2$.
Hence any solution of the original system are also the solution of $-3x+4y-2z=-2$.

\begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align} $$\overset{(1')=(1)-(2)}\downarrow$$
\begin{align}
-3x+4y-2z&=-2 \tag {1'} \\
5x+2y-z&=12 \tag {2'}
\end{align}

Now solve the system $(1') , (2')$ and you will get the original solutions as both systems are equivalent system.This is the way how elimination works!
</s>