Step 1: Apply Kirchhoff's Current Law (KCL) at Node $V_1$. Assume currents leaving the node are positive. The current through the $2\Omega$ resistor to ground is $\frac{V_1 - 0}{2}$. The current through the $4\Omega$ resistor to $V_2$ is $\frac{V_1 - V_2}{4}$. The $5\text{A}$ current source is directed from $V_1$ to $V_2$, meaning $5\text{A}$ is leaving $V_1$. So, KCL at $V_1$: $$ \frac{V_1}{2} + \frac{V_1 - V_2}{4} + 5 = 0 $$ Multiply the entire equation by 4 to eliminate denominators: $$ 2V_1 + (V_1 - V_2) + 20 = 0 $$ $$ 3V_1 - V_2 + 20 = 0 $$ Equation (1): $$ 3V_1 - V_2 = -20 $$ Step 2: Apply Kirchhoff's Current Law (KCL) at Node $V_2$. Assume currents leaving the node are positive. The current through the $4\Omega$ resistor to $V_1$ is $\frac{V_2 - V_1}{4}$. The current through the $6\Omega$ resistor to ground is $\frac{V_2 - 0}{6}$. The $5\text{A}$ current source is directed from $V_1$ to $V_2$, meaning $5\text{A}$ is entering $V_2$. So, the current leaving $V_2$ due to this source is $-5\text{A}$. The $10\text{A}$ current source is directed upwards into $V_2$, meaning $10\text{A}$ is entering $V_2$. So, the current leaving $V_2$ due to this source is $-10\text{A}$. So, KCL at $V_2$: $$ \frac{V_2 - V_1}{4} + \frac{V_2}{6} - 5 - 10 = 0 $$ $$ \frac{V_2 - V_1}{4} + \frac{V_2}{6} - 15 = 0 $$ Multiply the entire equation by 12 (the least common multiple of 4 and 6) to eliminate denominators: $$ 3(V_2 - V_1) + 2V_2 - 180 = 0 $$ $$ 3V_2 - 3V_1 + 2V_2 - 180 = 0 $$ $$ -3V_1 + 5V_2 - 180 = 0 $$ Equation (2): $$ -3V_1 + 5V_2 = 180 $$ Step 3: Solve the system of linear equations. We have two equations: 1) $3V_1 - V_2 = -20$ 2) $-3V_1 + 5V_2 = 180$ Add Equation (1) and Equation (2): $$ (3V_1 - V_2) + (-3V_1 + 5V_2) = -20 + 180 $$ $$ 4V_2 = 160 $$ $$ V_2 = \frac{160}{4} $$ $$ V_2 = 40 \text{ V} $$ Substitute the value of $V_2$ into Equation (1): $$ 3V_1 - 40 = -20 $$ $$ 3V_1 = -20 + 40 $$ $$ 3V_1 = 20 $$ $$ V_1 = \frac{20}{3} \text{ V} $$ The node voltages are: $$ \boxed{V_1 = \frac{20}{3} \text{ V}} $$ $$ \boxed{V_2 = 40 \text{ V}} $$