1. **State the problem:** We have two intersecting lines forming four angles: $(3x + 33)^\circ$, $(16y + 10)^\circ$, $(25y - 8)^\circ$, and $(4x - 2)^\circ$. Opposite angles are equal. 2. **Use the property of vertical angles:** Vertical angles are equal, so: $$3x + 33 = 25y - 8$$ $$16y + 10 = 4x - 2$$ 3. **Rewrite the equations:** $$3x + 33 = 25y - 8 \implies 3x - 25y = -41$$ $$16y + 10 = 4x - 2 \implies 4x - 16y = 12$$ 4. **Solve the system of equations:** From the second equation: $$4x - 16y = 12$$ Divide both sides by 4: $$\cancel{4}x - \cancel{16}y = \cancel{12} \implies x - 4y = 3$$ So, $$x = 4y + 3$$ 5. **Substitute $x$ into the first equation:** $$3(4y + 3) - 25y = -41$$ $$12y + 9 - 25y = -41$$ $$-13y + 9 = -41$$ $$-13y = -50$$ $$y = \frac{-50}{-13} = \frac{50}{13}$$ 6. **Find $x$ using $y$:** $$x = 4\left(\frac{50}{13}\right) + 3 = \frac{200}{13} + 3 = \frac{200}{13} + \frac{39}{13} = \frac{239}{13}$$ 7. **Final answers:** $$x = \frac{239}{13} \approx 18.38$$ $$y = \frac{50}{13} \approx 3.85$$