I wonder if anyone can tell me why there is Vo=V1-6v in the solution on the 6th line in the PDF attached.
How come it doesn't include the 12k resistor and how come it's minus and not add?
Thanks

 

The answer is simple. They first write a nodal equation for node 1 (V1). And solve for V1 voltage.
And the voltage at the node 1 (V1) is 6V. But they need Vo voltage, the voltage across 12kΩ resistor.
The voltage on the left side of a 12kΩ resistor is V1 = 6V and one the right side the voltage is also equal to 6V (right voltage source).
Therefore the voltage across 12kΩ resistor is a potential differences between V1 and the right voltage source ---->Vo = V1 - Vsource = 6V - 6V = 0V

 

V1 is not indicated very precisely, it is between the lower Node junction (-) and the upper node junction (+). If you follow the loop to the right you have two elements, The 12K marked as having Vo across it and a 6v source.

By inspection of the voltage loop there are three voltages where:
V1 = Vo + 6V

Solve that for Vo and you get that negative 6V. To "see" why mark where V1 is in the circuit and follow around the loop: you see you enter V1 thru the + terminal and the 6V thru the negative, so one is added and the other subtracted.