Usually I don't care about algebra and just focus on the functionality. However, below circuit form appeared so many times in last few days in op amps that I thought of memorizing it by heart. 

From my limited op amps study so far, it seems the entire op amp analysis boils down to figuring out the voltage at point \$\text{x}\$ shown in the diagram. 

Working the equations using superposition gave a nice symmetry:  

$$V(x) = \frac{V\_1R\_2 + V\_2R\_1}{R\_1+R\_2} = \frac{\left<V\_1,V\_2\right>\cdot \left<R\_2,R\_1\right>}{R\_1+R\_2}$$ 

If we think \$R\_2, R\_1\$ as weights, then \$V(x)\$ is the weighted average of \$V\_1,V\_2\$. It seems \$R\_2\$ controls \$V\_1\$ and \$R\_1\$ controls \$V\_2\$. This is beautiful and I'm dying to know ***if this symmetry has a nice physical explanation***. Thanks! 

![schematic](https://i.stack.imgur.com/2iK1x.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f2iK1x.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

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Few observations: 

1. When \$V\_1=V\_2=V\$ above formula gives \$V(x) = V\$. This means the entire branch including \$R\_1, R\_2\$ and every other point float to voltage \$V\$. Doesn't matter what \$R\_1, R\_2\$ are. This is used in instrumentation amplifiers to amplify AC while perfectly grounding common mode.
2. \$x\$ will be virtual ground when the average value is \$0\$