Apply Kirchhoff's Current Law (KCL) at Node V₁.

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):

$$3V_1 - V_2 = -20$$

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):

$$-3V_1 + 5V_2 = 180$$

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:

$$\boxed{V_1 = \frac{20}{3} \text{ V}}$$ $$\boxed{V_2 = 40 \text{ V}}$$