What are we solving with row reduction? Why are row reductions so useful in linear algebra? It is easy to get lost in the mechanical solving of equations. I know I can get a matrix into reduced row echelon form. But what are the outcomes of this? What can this mean?

Well, Dan and muad have already told you everything, but let me stress the two main points (in my opinion):


*

*Row reduction is just the same as solving one equation for one unkown in terms of the others and then plugging the obtained expression into the remaining equations. (But without having to write down the unkowns all the time.) In a nutshell, row reduction = solving your system.

*Row reduction doesn't change the solution of the system of linear equations. (So the target in performing row reduction is to obtain a simpler system, with the same solutions as the original one, but such that solutions can be read directly from it.)


The second statement can be proved formally, but I think that, if I can convince you that (1) is true, you won't need this formal proof, will you? (Because, when you solve your system, you're sure that, if you don't make mistakes, the solution you arrive at is, well, the solution. You don't change the solution of a system when you perform your operations in order to solve it, do you? So, if row reduction = solving your system and you don't change the solutions of your system while solving it, then row reduction doesn't change the solutions of your system.)
So let's see how row reduction = solving your system. Say you're trying to solve the following system:
$$
\begin{align}
5x - 3y &= 3 \\
2x + 4y &= 2
\end{align}
$$
You realize that the second equation looks nicer than the first one, so you decide to start with it and interchange both equations:
$$
\begin{align}
2x + 4y &= 2    \\
5x - 3y &= 3 
\end{align}
\qquad \qquad \qquad \qquad \textbf{Step one}
$$
Then you solve the first equation for $x$. You divide by two your, now, first equation 
$$
x + 2y = 1
\qquad \qquad \qquad \qquad \textbf{Step two}
$$
solve it for $x$, $x = 1 -2y$, and plug it into the second one, that is $5(1-2y) - 3y = 3$. You obtain:
$$
-13y = -2
\qquad \qquad \qquad \qquad \textbf{Step three}
$$
Then, you solve the second equation for $y$, dividing by $-13$
$$
y = \frac{2}{13}
\qquad \qquad \qquad \qquad \textbf{Step four}
$$
and you perform back subtitution, that is you plug this $y$ into the first equation, $x = 1- 2\frac{2}{13}$, getting
$$
x = \frac{9}{13}
\qquad \qquad \qquad \qquad \textbf{Step five}
$$
So you know the solution of your system is:
$$
\begin{align}
x &= \frac{9}{13} \\
y &= \frac{2}{13}
\end{align}
$$
Now, we are going to do exactly the same, but with row reduction. Our system of equations is the same as its augmented matrix:
$$
\left(
\begin{array}{rr|r}
5 & -3 & 3 \\
2 & 4  & 2
\end{array}
\right)
$$
In Step one, we've interchanged both equations. Now, we interchange the two rows:
$$
\left(
\begin{array}{rr|r}
2 & 4  & 2 \\
5 & -3 & 3
\end{array}
\right)
\qquad \qquad \qquad \qquad \textbf{Step one}
$$
In Step two, we divided by two the first equation. Now, we divide by two the first row:
$$
\left(
\begin{array}{rr|r}
1 & 2  & 1 \\
5 & -3 & 3
\end{array}
\right)
\qquad \qquad \qquad \qquad \textbf{Step two}
$$
In Step three, we plugged $x = 1 -2y$ into the second equation. Now, we substract five times the first equation from the second:
$$
\left(
\begin{array}{rr|r}
1 & 2  & 1 \\
0 & -13 & -2
\end{array}
\right)
\qquad \qquad \qquad \qquad \textbf{Step three}
$$
In Step four, we divided the second equation by $-13$. Now, we divide the second row by $-13$:
$$
\left(
\begin{array}{rr|r}
1 & 2  & 1 \\
0 & 1 & \frac{2}{13}
\end{array}
\right)
\qquad \qquad \qquad \qquad \textbf{Step four}
$$
In Step five, we performed back substitution. Now, we substract $2$ times the second row from the first one:
$$
\left(
\begin{array}{rr|r}
1 & 0  & \frac{9}{13} \\
0 & 1  & \frac{2}{13}
\end{array}
\right)
\qquad \qquad \qquad \qquad \textbf{Step five}
$$
Now, the solution of your system is in the third column, because the system that corresponds to this matrix is:
$$
\begin{align}
x &= \frac{9}{13} \\
y &= \frac{2}{13}
\end{align}
$$
