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?
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?