what values of b_2 result in solving this system? Solve the system of equations:
$$ \begin{pmatrix}
0  & 1 & 0 & 3  \\
0  & 2 & 0 & 6  \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\    x_3\\  x_4 \end{pmatrix}
 = \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$$
What value of $b_2$ results in this system having a solution?
I did my row reduction and came up with:
$$\begin{pmatrix}
0 & 1 & 0 & 3 & b_1\\
0 & 0 & 0 & 0 & 2b_1 - b_2\\
\end{pmatrix} $$
therefore, $2b_1=b_2$
Is it really that simple?
Any help is greatly appreciated!  

Your answer is correct and your reasoning is good (although I think most people would come up with $b_2-2b_1$ rather than $2b_1-b_2$ in the reduced row echelon form). However, in this particular case, there is a conceptually simpler solution (I emphasize "particular" because in general, your way of solving the problem is preferred). Note that
$$
\underbrace{\begin{pmatrix}0&1&0&3\\ 0&2&0&6\end{pmatrix}}_{A}
\begin{pmatrix}x_1 \\ x_2 \\ x_3\\ x_4\end{pmatrix}
=x_1\begin{pmatrix}0\\ 0\end{pmatrix}
+x_2\begin{pmatrix}1\\ 2\end{pmatrix}
+x_3\begin{pmatrix}0\\ 0\end{pmatrix}
+x_4\begin{pmatrix}3\\ 6\end{pmatrix}
=(x_2+3x_4)\begin{pmatrix}1\\ 2\end{pmatrix}.
$$
Therefore the system of linear equations is solvable if and only if $(b_1,b_2)^T$ is a scalar multiple of $(1,2)^T$, i.e. iff $b_2=2b_1$.
Using the language of linear algebra, $A\mathbf{x}=\mathbf{b}$ is solvable iff $\mathbf{b}$ lies in the column space of $A$. Now the column space of $A$ is the line spanned by $(1,2)^T$. Hence the result.
