Solve the circuit with the nodal analysis and determine $I_x$.

Solution
1) Identify all nodes in the circuit. Call the number of nodes $N$.
There are five nodes in the circuit:

Therefore $N=5$
2) Select a reference node
The best option is the node in the bottom because it is connected to both voltage sources.

3) Assign a variable for each node whose voltage is unknown.
There are four nodes beside the reference node:

Node III and Node IV are connected to the reference node through voltage sources. Therefore, their node voltages can be determined by the voltage sources.
$V_3=2I_x$ and $V_4=V_s=3V$.

$I_x$ is the current of $R_1$. The Ohm's law can be used to write $I_x$ in terms of the node voltages. Thus, $I_x=\frac{V_3-V_2}{R_1}=V_3-V_2$. Substituting $V_3=2I_x$, $I_x=2I_x-V_2 \to I_x=V_2$ (Eq. 1).
Therefore, $V_3=2V_2$.
4) Write down KCL equations.
We only need to write a KCL equation for Node I and Node II:
Node I: $I_s + \frac{V_1-V_3}{R_2} +\frac {V_1}{R_4}=0$. Substituting $V_3=2V_2$ and known variables,
$3V_1-2V_2=-2V$ (Eq. 2).

Node II: $-I_s+\frac{V_2-V_3}{R_1}+\frac{V2-V_4}{R_5}=0$. $\to -1+V_2-2V_2+\frac{V_2}{2}-\frac{3}{2}=0 \to V_2=-5V$.
Substituting in Eq. 2:
$V_1=-4V$.

Now, we need to determine the required quantities. Eq. 1 implies that
$I_x=V_2=-4A$.

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