We again label the nodes, from 1 to 4, and define the current directions. Node 2 is defined as the combination of three intersection points

Applying KCL at nodes 3 and 2, we obtain
• KCL(3): 2 + i_{w} − 4 = 0 \longrightarrow i_{w} = 2,
• KCL(2): i_{s} − i_{z} − i_{w} + 4 + i_{y} = 0 \longrightarrow i_{s} + i_{y} − i_{z} = −2.
Then, using KVL, we derive
• KVL(1 → 2 → 4 → 1): 4i_{s} + 7i_{z} = 10,
• KVL(2 → 3 → 4 → 2): 20i_{w} − 2 × 6 − 7i_{z} = 0 \longrightarrow i_{z}= 4A.

Using the updated information, we can also find i_{s}, as well as i_{y} as
4i_{s} = 10 − 7i_{z} = −18 \longrightarrow i_{s} = −9∕2 A,
i_{y} = −2 − i_{s} + i_{z} = −2 + 9∕2 + 4 = 13∕2 A.
Finally, we apply KVL(1 → 2 → 1) to find v_{x} as
−v_{x} + 2i_{y} − 4i_{s} = 0 \longrightarrow v_{x} = 2i_{y} − 4i_{s} = 13 + 18 = 31 V.