The way I would approach this problem is to use Kirchhof's current law.
Define two different current loops in the circuit (say, i1 and i2 -- you can even use them the way the are already defined on your circuit diagram if you want to [or you an define your own if you prefer]). For each current loop, go through and add up the respective voltage drops, setting everything to 0.
You'll end up with 2 (simultaneous) equations and 2 unknowns. Solve for the unknowns using substitution, linear algebra, or whichever method you prefer to use for solving simultaneous equations.
VJ_1991 said:
is it possible to use voltage division in this circuit? if so, which battery will i use( 20 V or 32V)? or will i use the difference between the two batteries when calculating for voltage division?
There is another way to solve this problem using Thevenin equivalent circuits. After performing some Thevenin equivalent transformations, you can end up with some voltage dividers, so to speak (you'll have to do some work to get there though). Whatever the case, you can end up solving the problem that way. I've always preferred using Kirchhof's current law, but Thevienin equivalent approach works too.
[Edit: Technically, there are also Norton equivalent circuits/transformations that are sort of the same thing as Thevenin equivalent circuits/transformations, but in the other direction (above, I just grouped everything under the name Thevenin -- but technically there are both Thevenin and Norton equivalents). If you use these transformations to solve this problem, you can avoid solving simultaneous equations. But it might take more notebook paper, because you'll be redrawing the circuit a lot, and there is still a lot of math involved.]