Solving linear system after Gaussian elimination If I had an initial $2 \times 3$ matrix $A$ and a $2 \times 1$ constant vector $b$, such that I have $A$ and $b$ of $Ax=b$:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$$
$$b = \begin{pmatrix} 7 \\ 8 \end{pmatrix}$$
Such that in augmented form it appears as such;
$$\left(\begin{array}{ccc|c} 1 & 2 & 3 & 7 \\ 4 & 5 & 6 & 8 \end{array}\right)$$
And valid operations were performed such that it was in $r.e.f.$ (the values here are just examples):
$$\left(\begin{array}{ccc|c} 1 & -3 & 2 & 11 \\ 0 & 1 & -9 & 2 \end{array}\right)$$
How does one solve this system, if there is not a row that only contain: 
$$\left(\begin{array}{ccc|c} 0 & 0 & 1 & N \end{array}\right)$$
...so that you could back solve the entire system?

Lets take a step back from the linear algebra to more basic algebra.
$x -3y  +2z = 11\\
0x  +1y -9z = 2$
Since you have two equations and 3 unknowns there will be a "line of solutions."
You can set $z = 0$ and find a pair of $x,y$ that will solve the equation.
i.e.
$x - 3y = 11\\
0x  +1y = 2$
$(17,2,0)$ is a solution.
Then set $z = 1$
$x - 3y +2= 11\\
0x  +1y -9= 2$
$(42, 11, 1)$ is also a solution
the full line of solutions is $(17,2,0) + (25,9,1) t$
It is worth noting that:
$\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 17\\2\\0\end{bmatrix} = \begin{bmatrix} 11\\2\end{bmatrix}$
and 
$\begin{bmatrix} 1&-3&2\\0&1&-9\end{bmatrix}\begin{bmatrix} 29\\9\\1\end{bmatrix} = \bf 0$
