Solving simple parallel/series without Req
I'm going to try and answer my own question, hopefully I'm getting the gist of this.

Let's identify the current that flows through R\$_{1}\$, R\$_{2}\$, and R\$_{3}\$ as I\$_{1}\$, I\$_{2}\$, and I\$_{3}\$, respectively. We know by Kirchoff's current law that current flowing into a node = current flowing out of it, or:

\begin{equation*}
I_{1} = I_{2} + I_{3}
\end{equation*}

Next, from Kirchoff's voltage law, we know that the sum of voltages across any closed path is zero. This gives us two equations for each possible path the circuit has:

\begin{equation*}
5 - V_{1} - V_{2} = 0
\end{equation*}

\begin{equation*}
5 - V_{1} - V_{3} = 0
\end{equation*}

And substituting with Ohm's law (V = IR) to solve for current, which we already have in the first equation, we get:

\begin{equation*}
5 - 100I_{1} - 200I_{2} = 0
\end{equation*}

\begin{equation*}
5 - 100I_{1} - 300I_{3} = 0
\end{equation*}

This gives us three equations for three unknowns:

\begin{equation*}
I_{1} - I_{2} - I_{3} = 0
\end{equation*}
\begin{equation*}
100I_{1}\ + 200I_{2} = 5
\end{equation*}
\begin{equation*}
100I_{1}\ + 300I_{3} = 5
\end{equation*}

We can convert this to a matrix and solve by getting its inverse:

\begin{equation*}
\begin{pmatrix}
1 &amp; -1 &amp; -1 \\
100 &amp; 200 &amp; 0 \\
100 &amp; 0 &amp; 300
\end{pmatrix}
\begin{pmatrix}
I_{1} \\
I_{2} \\
I_{3}
\end{pmatrix}
=
\begin{pmatrix}
0 \\
5 \\
5
\end{pmatrix}
\end{equation*}

\begin{equation*}
\begin{pmatrix}
I_{1} \\
I_{2} \\
I_{3}
\end{pmatrix} = 
\begin{pmatrix}
0.545 &amp; 0.003 &amp; 0.002 \\
-0.272 &amp; 0.004 &amp; -0.001 \\
0.182 &amp; 0.001 &amp; 0.003
\end{pmatrix}
\begin{pmatrix}
0 \\
5 \\
5
\end{pmatrix}
\end{equation*}

Finally we get the currents for I:

\begin{equation*}
\begin{pmatrix}
I_{1} \\
I_{2} \\
I_{3}
\end{pmatrix}
=
\begin{pmatrix}
0.0227 \\
0.0136 \\
0.0090
\end{pmatrix}
\end{equation*}

Which is what you get at the top using \$R_{eq}\$, of 0.0227A.