Concepts behind augmented matrix Apparently the strategy for finding a general solution to the equation $Ax = v$, where $A$ is some $m\times n$ matrix, $x$ is an unknown vector of dimension n, and $v$ is not the zero vector, is to turn $v$ into an extra column of A (resulting in an augmented matrix, $aA$) and then simply proceed with the usual technique on $aA$ by putting $aA$ into row-echelon form. But why does this work? If I try to answer this question myself, the answer I give is because $Ax = v$ represents a system of equations, and moving $v$ from the right hand side of this system to the left does not change it. But in order to actually do that I think you'd have to subtract $v$ from both sides of the system, which would mean you should have $-v$ as the extra column in $aA$ and not $v$; but this is not how it works.
Where does my reasoning go wrong? What is the correct explanation for why performing row operations on $aA$ gives you the general solution to $A$?
 A: Using augmented matrices is just a nice way to encode simultaneous equations. And row operations are just a nice way to encode the operations we can do with equations, e.g. multiply both sides of an equation by a constant, add two equations together etc.
Take the example:
$$2x+2y=2\\
x-y=1.$$
This is a super simple example that you can solve in your head, but let's use this just to parse out what's going on behind the scenes of row-reducing an augmented matrix.
We could manipulate these equations as follows:

*

*Divide the first equation by $2$ so your equations are now

$$x+y=1\\
x-y=1.$$
2. Take the second equation, subtract it from the first and divide it by 2 so your equations are now
$$x+y=1\\
y=0.$$
3. Take the first equation, subtract it by the  second so your equations are now
$$x=1\\
y=0,$$
and we solved the simultaneous equations.
All the above can be encoded by row operations on an augmented matrix:
$$\begin{bmatrix} 2 &2 &|2\\1&-1&|1 \end{bmatrix}\overset{R1'=R1/2}{\longrightarrow} \begin{bmatrix} 1 &1 &|1\\1&-1&|1 \end{bmatrix}\overset{R2'=(R1-R2)/2}{\longrightarrow} \begin{bmatrix} 1 &1 &|1\\0&1&|0 \end{bmatrix}\overset{R1'=R1-R2}{\longrightarrow} \begin{bmatrix} 1 &0 &|1\\0&1&|0 \end{bmatrix}$$
