To solve for $V_a$ and $V_b$, we will solve the two simultaneous equations derived from the node voltages in what looks like a resistive circuit. These equations are:

For Node $a$: $\frac{V_a - 3}{4700} + \frac{V_a}{1200} + \frac{V_a - V_b}{5600} + \frac{V_a - V_b}{1707.22} = 0$

For Node $b$: $\frac{V_b - V_a}{1707.22} + \frac{V_b - V_a}{5600} + \frac{V_b}{1500} = 0$

Step 1: Simplify the Equations

For Node $a$:

We will combine the terms over a common denominator:

$\frac{V_a - 3}{4700} + \frac{V_a}{1200} + \frac{V_a - V_b}{5600} + \frac{V_a - V_b}{1707.22} = 0$

For Node $b$:

Similarly, combine terms for $V_b$:

$\frac{V_b - V_a}{1707.22} + \frac{V_b - V_a}{5600} + \frac{V_b}{1500} = 0$

Step 2: Substitute numerical values and solve

I will now solve the system of equations numerically.The solution for the node voltages is:

$V_a = 0.455 \, \text{V}$ $V_b = 0.243 \, \text{V}$

Would you like further details on how these were calculated, or do you have any specific questions?

Here are 5 related questions for further exploration:

1. How does the choice of resistances affect the node voltages?
2. Can this circuit be simplified using Thevenin's theorem?
3. How would adding another resistor between the nodes affect $V_a$ and $V_b$?
4. How do you calculate current in each branch once the node voltages are known?
5. What role does the 3V voltage source play in this circuit?

Tip: When solving systems of equations in circuit analysis, keep track of the reference (ground) node to simplify voltage assignments.