So the best way to answer this without just giving you the answer it to solve a similar question so with that in mind, I will use this circuit.
There are other methods out there but I'm going to use the Branch-Current method (look up loop current method).
Step 1 is to define the branch currents and the direction that they are acting in.
Steb 2 Define the voltages in terms of the currents
[tex]V_{ab} = 2I_1[/tex]
[tex]V_{bd} = 8I_3[/tex]
[tex]V_{cb} = 4I_2[/tex]
Step 3 using Kirchoffs Voltage law we get
[tex]ƩV_{abda} = 0 = V_{ab} + V_{bd} + V_{da}[/tex]
[tex]ƩV_{bcdb} = 0 = V_{cb} + V_{bd} + V_{dc}[/tex]
These can be re written as
[tex]ƩV_{abda} = V_{ab} + V_{bd} + V_{da} = 2I_1 + 8I_3 - 32[/tex]
[tex]ƩV_{bcdb} = V_{cb} + V_{bd} + V_{dc} = 4I_2 + 8I_3 - 20[/tex]
then finally You use Kichchoffs current laws at node b to get your final equation.
[tex]ƩI_b = 0 = I_1 + I_2 - I_3[/tex]
this gives the 3 simultaneous equations
[tex]I_1 + I_2 - I_3= 0[/tex]
[tex]2I_1 + 8I_3 = 32[/tex]
[tex]4I_2 + 8I_3 = 20[/tex]
Solving gives the currents [itex]I_1 = 4[/itex], [itex]I_2 = -1[/itex] and [itex]I_3= 3[/itex].
Hope that helps you.