Find Io in the circuit in Fig. P3.7 using nodal analysis.

Image from: J. D. Irwin and R. M. Nelms, Basic engineering circuit analysis, 10th ed. Hoboken, NJ: John Wiley, 2011.
Solution:
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Let us label the nodes we will use to perform nodal analysis as follows:
We will now write a KCL equation at node V_1, the orange node:
Let us express these currents in terms of voltage and resistance using I=\dfrac{V}{R}.
\dfrac{V_1}{1k}+\dfrac{V_1-V_2}{2k}+4m=0\,\,\,\color{orange} {\text{(eq.1)}}
Now, we shift our focus to the second node, V_2 (purple node).
Again, we will write a KCL equation.
Expressing the currents in terms of voltage and resistance gives us:
\dfrac{V_2-V_1}{2k}+\dfrac{V_2}{2k}=6m\,\,\,\color{purple} {\text{(eq.2)}}
Solving equations 1 and 2 simultaneously gives us (see full steps):
V_2=5.6 v
We know I_0=I_2. Writing I_2 in terms of voltage and resistance gives us:
I_0=\dfrac{-0.8-5.6}{2k}
I_0=\dfrac{-6.4}{2k}
I_0=-3.2 mA
(A negative current value means the direction of the current flow is opposite to the direction shown in the circuit diagram.)
Final Answer: