Why is a linear system of equations considered consistent when a solution contains 0/0? Suppose I have a system of equations as such, with $\alpha$ and $\beta$ being constants:
$$
x_1+5x_2+x_3=1\\
x_1+6x_2-x_3=1\\
2x_1+\alpha x_2-6x_3=\beta
$$
Suppose I want to calculate for which values of $\alpha$ and $\beta$ my system is consistent. I can do that by making a coëfficiënt matrix and using Gaussian elimination to derive the solution space, it being:
$$
x_1=1-\frac{11\beta -22}{2\alpha-28}\\
x_2=\frac{\beta -2}{\alpha -14}\\
x_3=\frac{\beta -2}{2\alpha -28}
$$
Then I can see that for the value $\alpha=14$ the denominator = 0, which is not possible. So, I can conclude that the system is only consistent for $\alpha\neq14$.
This is clear to me. However, my answer sheet says that $\alpha=14$ and $\beta=2$ is also an answer. Plugging these values into $x_1$, $x_2$ and $x_3$ gives these answers:
$$
x_1=1-\frac{0}{0}\\
x_2=\frac{0}{0}\\
x_3=\frac{0}{0}
$$
I thought $\frac{0}{0}$ was supposed to be undefined? How can it be then that using these values for $\alpha$ and $\beta$, we get a consistent system?
To try to figure it out, I checked the answer of $A\overrightarrow{x}=\overrightarrow{b}$, which gives me this (I think):
$$
\begin{bmatrix}1 & 5 & 1\\1 & 6 & -1\\2 & \alpha & -6\end{bmatrix}
\begin{bmatrix}1-\frac{0}{0}\\\frac{0}{0}\\\frac{0}{0}\end{bmatrix}
= 
\begin{bmatrix}1-\frac{0}{0}+5(1-\frac{0}{0})+1-\frac{0}{0}\\\frac{0}{0}+6\frac{0}{0}-\frac{0}{0}\\2\frac{0}{0}+\alpha\frac{0}{0}-6\frac{0}{0}\end{bmatrix}
$$
Which in turn implies that the equation below must be true, since the answer sheet says that the system is consistent:
$$
\begin{bmatrix}1-\frac{0}{0}+5(1-\frac{0}{0})+1-\frac{0}{0}\\\frac{0}{0}+6\frac{0}{0}-\frac{0}{0}\\2\frac{0}{0}+\alpha\frac{0}{0}-6\frac{0}{0}\end{bmatrix}
=
\begin{bmatrix}1\\1\\\beta\end{bmatrix}
$$
If I substitute $\frac{0}{0}$ for something else (lets say $\frac{0}{0}=p$), I get this:
$$
\begin{bmatrix}1-p+5(1-p)+1-p\\p+6p-p\\2p+\alpha p-6p\end{bmatrix}
=
\begin{bmatrix}7-7p\\6p\\\alpha p-4p\end{bmatrix}
=
\begin{bmatrix}1\\1\\\beta\end{bmatrix}
$$
I get this equation, which when looking at the first 2 components of the vectors already doesn't make any sense, since for $7-7p=1$ to make sense, $p$ would have to be $\frac{6}{7}$. But for $6p=1$, $p$ would have to be $\frac{1}{6}$.

I made a summary of the comments by @MaximilianJanish and @JustinBenfield in the comments under the answer.
After I calculate the solution set, I state that:

Then I can see that for the value α=14 the denominator = 0, which is not possible. So, I can conclude that the system is only consistent for α≠14.

This is actually not the full picture. When I calculate the solution set, I write it in a way where the solution only makes sense if $\alpha \neq14$, so I cannot conclude that there can be no answer for when $\alpha=14$. To check this case, I can plug $\alpha=14$ into the original set of equations, which when solved, gives me a value for $\beta$, namely $\beta=2$.
The final conclusion therefore is that the system is consistent for $\alpha \neq14$ and $\alpha=14, \beta=2$.
