Diagonal Wheatstone-bridge (not balanced!)

I can think of two ways to do this one:

1) Apply a fictitious voltage source and use linear algebra to solve for the node voltages, source current, and equivalent resistance
2) Apply successive delta-wye transforms and simplify. I have outlined this approach below. Pardon the eccentric color schemes.

$\text{edges}=\{1\leftrightarrow 2,1\leftrightarrow 3,1\leftrightarrow 4,2\leftrightarrow 3,3\leftrightarrow 4,2\leftrightarrow 5,3\leftrightarrow 5,4\leftrightarrow 5\}$

$\text{weights}=\{2,1,6,1,1,6,1,2\}$

$\text{resistance}=\text{With}\left[\left\{\Gamma =\text{PseudoInverse}\left[\text{With}\left[\{\text{wam}=\text{WeightedAdjacencyMatrix}[\text{\$\#\$1}]\},\text{DiagonalMatrix}\left[\text{Tr}\text{/@}\text{wam}^T\right]-\text{wam}\right]\right]\right\}, \\ \text{Outer}[\text{Plus},\text{Diagonal}[\Gamma ],\text{Diagonal}[\Gamma ]]-\Gamma -\Gamma ^T\right]\&;$

$g=\text{PlanarGraph}\left[\text{edges},\text{EdgeWeight}\to \text{Table}\left[\frac{1}{r},\{r,\text{weights}\}\right],\text{VertexLabels}\to \text{Automatic},\text{EdgeLabels}\to \text{EdgeWeight}\right]$

$\left( \begin{array}{ccccc} 0 & \frac{13}{14} & \frac{75}{112} & \frac{33}{28} & \frac{5}{4} \\ \frac{13}{14} & 0 & \frac{75}{112} & \frac{5}{4} & \frac{33}{28} \\ \frac{75}{112} & \frac{75}{112} & 0 & \frac{75}{112} & \frac{75}{112} \\ \frac{33}{28} & \frac{5}{4} & \frac{75}{112} & 0 & \frac{13}{14} \\ \frac{5}{4} & \frac{33}{28} & \frac{75}{112} & \frac{13}{14} & 0 \\ \end{array} \right)$