It seems to be a problem for students, so here is another example:
Using superposition Theorem, it will be analysed without the second voltage source.
Analyse Vab due to voltage source V1(36V). The total resistance RT=6+4//(4+8) (visually)=6+4×12/16=6+3=9. The current IT=36V/9=4A.
The voltage dropped across R1=6×4=24V, so VR2=36-24=12V which is VR3+VR4. Using voltage divider to find Vab=VR4=12×8/12=8V.
Next is to find Vab due to V2(48V).
Same as before, RT=4+6//(4+8)=4+6×12/(6+12)=4+4=8 ohms. IT=-48V/8 ohms=-6A.
The the drop of R2=(-6)x4=-24V and VR1=-48-(-24)=-24V which is the same as VR3+VR4 because they are in parallel. Using voltage divider again to have Vab=VR4=-24×8/12=-16V.
Combining the two (superposition theorem says): Vab=8V-16V=-8V.