Showing that this set of equations have a unique solution 
I am stuck with the first part of this problem
This is what I tried : 

I put this set of equations to a matrix and got its reduced echleon form, which is :
$$\begin{array}{ccc|c}
1 & 4 & 12 & 5 \\
0 & a-8 & -12 & a-11 \\
0 & 0 & 2a-36 & -5 \\
\end{array}$$

As you can see it doesnt seem to me that it has a unique solution, and I don't know how to prove it because what I have been taught is that, when there is a unique solution it should look like this:
$$\begin{array}{ccc|c}
1 & 0 & 0 & a \\
0 & 1 & 0 & b \\
0 & 0 & 1 & c \\
\end{array}$$
And when there free variables, it has infinite solutions. This is what my teacher has taught me. So now, it is very clear that my reduced row echleon form matrix is not in the form (that was taught by definition) which is necessary to show it has a unique solution. I am lost, and I dont know how to proceed to prove this. Please help.

I only a high school student, so incase you answer this purely by using symbols (which is fair, because most of the users here are in  university) , it would be more helpful to me if you explain what you have done. 
 A: This is how I would do it: if $a$ isn't $8$ or $18$, then if you divide the 2nd row of your echelon form by $a - 8$ and the 3rd row by $2a - 36$ it'll look like $$\left[\begin{array}{ccc|c} 1 & 4 & 12 & 5 \\ 0 & 1 & \frac{-12}{a-8} & \frac{a-11}{a-8} \\ 0 & 0 & 1 & \frac{-5}{2a-36} \end{array}\right]$$ and then you can get the identity matrix on the left by adding $\frac{12}{a-8}$ times the 3rd row to the 2nd row, $-12$ times the 3rd row to the 1st row, and $-4$ times the new 2nd row to the 1st row, so you have a unique solution as long as $a$ isn't $8$ or $18$. I guess the rest of the problem is showing that you don't get a unique solution if $a = 8$ or $a = 18$, so you have a unique solution iff $a$ is not $8$ or $18$.
A: $$\left[\begin{array}{ccc|c} 1 & 4 & 12 & 5 \\ 0 & a-8 & -12 & a-11 \\ 0 & 0 & 2a-36 & -5 \end{array}\right]$$
$$R_2 \rightarrow R_2/{a-8}$$
$$R_3 \rightarrow R_3/{2a-36}$$
$$\left[\begin{array}{ccc|c} 1 & 4 & 12 & 5 \\ 0 & 1 & \frac{-12}{a-8} & \frac{a-11}{a-8} \\ 0 & 0 & 1 & \frac{-5}{2a-36} \end{array}\right]$$
$$R_2 \rightarrow R_2(R_3)+R_2$$
$$\left[\begin{array}{ccc|c} 1 & 4 & 12 & 5 \\ 0 & 1 & 0 & \frac{a-11}{a-8}\frac{-5}{2a-36} + \frac{12}{a-8}\\ 0 & 0 & 1 & \frac{-5}{2a-36} \end{array}\right]$$
If a=8 then $y=1/0$
If a=18 then $z=1/0$
If a is not 18 or 8 then it has a unique solution (it is the form your teacher taught).
A: Hint: The entries that depend on $a$ can be zero or nonzero depending on the value of $a$. What if $a-8$ is zero? What if $2a-36$ is zero? Try to substitute $a$ with values mentioned in the problem, and then continue row reduction.
A: You need a full triangular triangular matrix for the solutions to be unique, otherwise you get infinite solutions.   So,  if a=8 or a=18, your matrix collapses to a rank 2 matrix, and you no longer have unique solutions.  Otherwise, I think you should be good
