You can do this using loop analysis:
Let the current in the upper loop be I1.
Let the current in the lower loop be I2.
Taking the loop on the top:
20 + 5I1 + 10 + 10(I1 - I2) = 0 => 15I1 - 10I2 = -30 ........... (1)
Taking the loop on the bottom:
20I2 + 10(I2 - I1) - 10 + 10I2 + 30 = 0 => -10I1 + 40I2 = -20 ......... (2)
Solving (1) and (2):
I1 = -2.8 A I2 = -1.2 A
So, I1 you are asking for is simply I1 = -2.8 A
I2 you are asking for is I2 = -1.2 A
I3 = I2 - I1 = 1.6A