Electronic – Solving this circuit using node analysis

Without dependent (voltage) sources one could normally define a matrix equation to solve for the values in the network:
[Z][I] = [V]

In this case though we must add in dependent voltage sources
[Z][I] = [V] + [k][I]

Apologies, I renamed I3 above as I2 below, per the usual custom of ordering currents left to right as shown in sketch.

From this diagram, we extract the matrix equation suggested above.

This rearranges as
[Z][I] - [k][I] = [V]

Rearranging again [I] = ([Z]-[k])^-1 · [V]

Solution of the matrix math is standard matrix math left to the reader. It produces an equation for each of the 3 currents defined in the sketch as a function of the single independent voltage.

This gives the currents as a function of the one independent voltage.

At this point
Va = (I1 - I2)·R2

And Vb = (I2 - I3)·R4

The classic method looks like this

^{simulate this circuit – Schematic created using CircuitLab}

Step one Calculate Thevenin's equivalent impedance \$R_{TH}\$. Short all voltage sources and open circuit current sources.

As you can see \$R_{TH} = R1+R2 = 12\Omega\$

Step two We need to find the \$V_{TH}\$ voltage for this circuit:

^{simulate this circuit}

$$V_{TH} = V1 + R1 \cdot I1 = 12V + 12V = 24V$$

Hence, the Thevenin equivalent circuit will look like this:

^{simulate this circuit}

So we can find \$ V_{AB} = 24V * \frac{4\Omega}{12\Omega+4\Omega} = 6V \$ and additional \$IR_3 = \frac{6V}{4\Omega} = 1.5A\$

As for the nodal equation

The corrent equation looks like this

$$\frac{V_X}{R_2+R_3} + \frac{V_X - V1}{R_1} - I_1 = 0$$