<s>[INST] Find the value of $a$ and $b$, so the linear system stays consistant I was wondering if someone could take a look at my solution and tell me if i'm wrong. I feel like my solution is incorrect.
Given
$ \left[
\begin{array}{ccc|c}
  1&1&2&2a-b+1\\
  2&1&1&3a-b+1\\
  1&2&5&3a-2b+6
\end{array}
\right] $ find the values of a and b so the system stays consistent. 
I tried to put the system into RREF. So, the first thing I did was R1-R2
$$ \left[
\begin{array}{ccc|c}
  -1&0&1&-a\\
  2&1&1&3a-b+1\\
  1&2&5&3a-2b+6
\end{array}
\right] $$
R2-2R1
$$ \left[
\begin{array}{ccc|c}
  -1&0&1&-a\\
  0&1&3&a-b+1\\
  1&2&5&3a-2b+6
\end{array}
\right] $$
R3+R1
$$ \left[
\begin{array}{ccc|c}
  -1&0&1&-a\\
  0&1&3&a-b+1\\
  0&2&6&2a-2b+6
\end{array}
\right] $$
R3-2R2
$$ \left[
\begin{array}{ccc|c}
  -1&0&1&-a\\
  0&1&3&a-b+1\\
  0&0&0&-a-4b+6
\end{array}
\right] $$
Lastly, -R1
$$ \left[
\begin{array}{ccc|c}
  1&0&-1&a\\
  0&1&3&a-b+1\\
  0&0&0&-a-4b+6
\end{array}
\right] $$
The last equations gives $0=-a-4b+6$. This means that $a$ and $b$ can be any values as long as it follows that equation. 
 [/INST] Multiplying the first equation by-2 and adding to the second we get
$$-x_2-3x_3=-a+b-1$$
and the first by -1 and adding to the third
$$x_2+3x_3=a-b+5$$
Can you finish?
</s>