<s>[INST] Linear equations with parameters 
For which $a$ the following had one solution, no solution, infinite solutions:
  $$\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a  & 4 \\
4a-8 & 16 & 4a-2  & a+14 \\
10a-20 & -2a+36 & 8a-4 & 2a+30 \\
\end{array}
\right]$$

I manage to come to this:
$$\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a  & 4 \\
0 & 2a+4 & 2a-2  & a+6 \\
0 & 3a+6 & 3a-4 & 2a+10 \\
\end{array}
\right]$$
Now I should try to create zero elements? 
 [/INST] What about that: divide the second row by $2$ and the third row by $3$
$$
\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a  & 4 \\
0 & 2a+4 & 2a-2  & a+6 \\
0 & 3a+6 & 3a-4 & 2a+10 \\
\end{array}
\right]=\
\left[
\begin{array}{ccc|c}
2a-4 & 6-a & a  & 4 \\
0 & a+2 & a-1  & \frac12 a+3 \\
0 & a+2 & a-\frac43 & \frac23a+\frac{10}{3} \\
\end{array}
\right]
$$
Is it clearer now what to do?
</s>