1. **State the problem:** We need to find the nodal voltages $V_1$ and $V_2$ in the given circuit using nodal analysis. 2. **Identify nodes and reference:** Let the bottom node be the reference (ground) with voltage 0 V. 3. **Write nodal equations:** - At node $V_1$: The currents leaving $V_1$ through the 5 $\Omega$ resistor, 3 $\Omega$ resistor, and the connection to $V_2$ must sum to zero. $$\frac{V_1 - 10}{5} + \frac{V_1 - V_2}{3} = 0$$ Here, $V_1 - 10$ is the voltage across the 5 $\Omega$ resistor (since the left side is connected to 10 V source). - At node $V_2$: The currents leaving $V_2$ through the 3 $\Omega$ resistor and the $\frac{4}{3}$ $\Omega$ resistor connected to 20 V source must sum to zero. $$\frac{V_2 - V_1}{3} + \frac{V_2 - 20}{\frac{4}{3}} = 0$$ 4. **Simplify equations:** - Equation 1: $$\frac{V_1 - 10}{5} + \frac{V_1 - V_2}{3} = 0$$ Multiply both sides by 15 (LCM of 5 and 3): $$3(V_1 - 10) + 5(V_1 - V_2) = 0$$ $$3V_1 - 30 + 5V_1 - 5V_2 = 0$$ $$8V_1 - 5V_2 = 30$$ - Equation 2: $$\frac{V_2 - V_1}{3} + \frac{V_2 - 20}{\frac{4}{3}} = 0$$ Note that dividing by $\frac{4}{3}$ is multiplying by $\frac{3}{4}$: $$\frac{V_2 - V_1}{3} + \frac{3}{4}(V_2 - 20) = 0$$ Multiply both sides by 12 (LCM of 3 and 4): $$4(V_2 - V_1) + 9(V_2 - 20) = 0$$ $$4V_2 - 4V_1 + 9V_2 - 180 = 0$$ $$13V_2 - 4V_1 = 180$$ 5. **Solve the system of equations:** From equation 1: $$8V_1 - 5V_2 = 30$$ From equation 2: $$-4V_1 + 13V_2 = 180$$ Multiply equation 1 by 4: $$32V_1 - 20V_2 = 120$$ Multiply equation 2 by 8: $$-32V_1 + 104V_2 = 1440$$ Add the two equations: $$(32V_1 - 20V_2) + (-32V_1 + 104V_2) = 120 + 1440$$ $$84V_2 = 1560$$ $$V_2 = \frac{1560}{84} = \frac{130}{7} \approx 18.57\text{ V}$$ Substitute $V_2$ back into equation 1: $$8V_1 - 5 \times \frac{130}{7} = 30$$ $$8V_1 = 30 + \frac{650}{7} = \frac{210}{7} + \frac{650}{7} = \frac{860}{7}$$ $$V_1 = \frac{860}{7 \times 8} = \frac{860}{56} = \frac{215}{14} \approx 15.36\text{ V}$$ 6. **Final answer:** $$V_1 = \frac{215}{14} \approx 15.36\text{ V}, \quad V_2 = \frac{130}{7} \approx 18.57\text{ V}$$