As in the previous examples, we label the nodes, define the directions of the currents, and define the voltages in accordance with the sign convention.

In order to simplify the solution, we again do not define voltage variables separately and
write all equations in terms of currents.ApplyingKVL andKCL consecutively,we obtain
• KVL(2 → 4 → 2): 4i_{y} − 1i_{w} = 0 \longrightarrow i_{w} = 4i_{y},
• KCL(4):i_{y} + i_{w} − i_{z} = 0\longrightarrow i_{z} = 5i_{y},

• KCL(3): i_{z} − i_{x} − 2 = 0\longrightarrow i_{x} = 5iy − 2,
• KVL(1 → 2 → 4 → 3 → 1): 2i_{x} + 4i_{y} + 6iz − 20 = 0 \longrightarrow i_{y} = 6∕11A.
Finally, we have  i_{x} = 30∕11 − 2 = 8∕11A  and v_{x} = 16∕11V.