OK, the first thing we do is combine the two north parallel 5 ohm resistors into one 2.5 ohm resistor. Then we realize that the 2 amp east branch will have 2 amps existing in it no matter what value the 5 ohm southeast resistance is. So we remove the 5 ohm southeast resistor in series with the 2 amp current source, and replace it with a zero ohm wire. Next we realize that the voltage across the 5 ohm south resistor in parallel with the 6 volt source going to be 6 volts no matter what the resistance value is. So we remove it completely from the circuit. Then we call on our friend Mr Thévenin to change the east branch to a 10 volt source in series with a 5 ohm resistor, and the west side branch to a 4 volt source in series with a 2.5 ohm resistor. Now we sum up the voltage sources around the outside loop. They are starting from the west side and going clockwise 4 volts + 10 volts + 6 volts = 20 volts. Then sum up all the resistances. They are starting from the west and going clockwise 2.5 ohms + 2.5 ohms + 5 ohms = 10 ohms . Therefore, the current around the loop is 2 amps and each of the 2 north resistors has 1 amp of current existing within it.
