Instead of applying KVL and KCL directly, we first reduce the circuit by considering series and parallel connections of resistors.We have
R_{eq1} = 8 + 16 = 24 Ω,

R_{eq2} = [8 + (6 ∥ 12)] ∥ R_{eq1} = 12 ∥ 24 = 8 Ω,

R_{eq3} = 24 + R_{eq2} = 24 + 8 = 32 Ω,

R_{eq4} = 4 + 32 ∥ R_{eq3} = 20 Ω.

Then, after finding the main current to be i_{x} = 3A, we trace back the circuit as follows.

Hence, we obtain
i_{x}=\frac{1}{2} A
v_{x} = 16i_{x} = 8 V.