First, we label the nodes, define the directions of the currents, and define the voltages in accordance with the sign convention, as follows.

Then, using KVL and Ohm’s law, we have v_{x} = 3i_{x}, v_{y} = 6i_{x}, and
• KVL(1 → 2 → 3 → 1): −24 + v_{x} + v_{x} = 0 \longrightarrow v_{x} + v_{y} = 24V

Therefore, we have
3i_{x} + 6i_{y} = 24 \longrightarrow i_{x} + 2i_{y} = 8 A.
Furthermore, using KCL (see below for some details), we derive
• KCL(2):i_{x} − i_{y} − i_{z} = −2A.
Using v_{y} = v_{z} and i_{y} = v_{y}∕6 = v_{z}∕6 = i_{z}, we obtain
i_{x} − 2i_{y} = −2 A.
Finally, solving two equations with two unknowns, we get
i_{x} = 3 A.
In the above, we note that node 2 (as well as node 3) is defined simultaneously at two
intersections, and KCL is written accordingly by considering all entering and leaving currents, as follows.

This is a common practice in circuit analysis in order to reduce the number of equations.
Specifically, intersections without a component between them can be considered as a
single node to avoid writing redundant equations with redundant unknowns. On the
other hand, this not mandatory. For example, one can consider each intersection as a node, as follows.

In this case, we need to define a current i_{w} between nodes 2 and 4.Writing KCL at the nodes, we now have
• KCL(2): i_{x} − i_{y}− i_{w} = 0,
• KCL(4): i_{w} + 2 − i_{z} = 0.
Obviously, when these equations are combined (directly added), we arrive at the same
equation in the original solution,
i_{x} − i_{y} − i_{z} = −2 A.