Assign a potential of 0 to the bottom right of the circuit.
Assign a Potential of V to the point where V1 and the 4 and 5 ohm resistances mee
work out the currents through the 3 branches of the circuit.
for example the potential across the 5 ohm resistance is V - V3 so the current through it is [itex](V - V_3)/ 5[/itex]
Now you can apply Kirchhof's current law, and then solve for V. The currents can
then be found with the 3 equations you found worked out above.
I don't think that adding the 3 voltages is going to get you anywhere. It seems you
might mean the the superposition principle: If you calculate the currents with only V1 in the circuit and the other voltage sources replaces by a wire, and calculate the currents through the resistances, and then you repeat this with V2 and V3, you can then sum the 3 currents you found for each resistance to get the current through that resistance with all the voltage sources present. I don't think that's an efficient method here.