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Try finding out the voltage across the 6k resistor.

 

You know the voltage at the 2k resistor. Between the 6k and the 2k there is clearly a voltage source, wonder what that could do with the voltage at the 6k resistor.

 

That is not correct, you got the minus signs in wrong places. Just look at it with some common sense, if on the right side you got 2V, the voltage source is 4V, and its plus sign is on the left side, how many volts have to be on the left side?

 

Depends how you draw the current loop for the KVL. If you go from bottom left through top to bottom right, then you have -I*R2 -4V -2V =0
If you go the other way, then 2V + 4V -I*R2 =0. You should not use symbols like V(r2) because it is ambiguous which way it is oriented and will get you confused, use current times resistance to get the voltage across a resistor.

 

R3 has 2k ohms and 2V across it, that is 1mA. R2 has 6k ohms and 6V across it, both currents are flowing towards ground.
That means that the sum of those two currents has to be flowing through R1, which makes the drop across R1 xV and which brings the Vs to yV.

 

No problem, you could also draw it up in ltspice for example and measure the circuit, then correct your method of applying kirchhoff equations so that it works.