Apply Thevenin Theorem to a circuit with 2 sources?
Well, if you want to find the Thevenin equivalent, I would suggest using source transformations.


In this case, start with the 3A current source and 16 ohm resistor.  This pair forms a Norton equivalent source.
Transform it to a Thevenin equivalent by replacing them with a series combination of a voltage source and 16 ohm resistor. The voltage source will need to be I*R volts, in this case 48 volts.
Then combine the now series 16 ohm and 4 ohm resistors into a 20 ohm resistor.
Then transform both sources to current sources. The 48 volt source in series with the 20 ohm resistor becomes a 2.4 A source in parallel with a 20 ohm resistor. Likewise the 12 volt source and 5 ohm resistor become a 2.4 A source in parallel with a 5 ohm resistor.
Now combine the sources and resistors - two parallel 2.4 A sources are one 4.8 A source and 20 ohms in parallel with 5 ohms is 4 ohms.
Transform this back to a Thevenin equivalent - 19.2 volts in series with 4 ohms.
Combine the 4 ohm resistor with the 1 ohm resistor.
Then transform back again to a Norton equivalent - a 3.84 amp source in parallel with 5 ohms.
Now combine the 5 ohm and 10 ohm resistors to get a 3.33 ohm resistor.
Transform the source back to get a 12.8 volt source in series with a 3.33 ohm resistor.