1. **State the problem:** Solve the system of equations: $$\frac{1}{2}x + \frac{3}{5}y = 2$$ $$\frac{1}{6}x - \frac{1}{3}y = -10$$ 2. **Use substitution or elimination method. Here, we use elimination.** 3. Multiply the first equation by 6 and the second by 30 to clear denominators: $$6 \times \left(\frac{1}{2}x + \frac{3}{5}y\right) = 6 \times 2 \Rightarrow 3x + \frac{18}{5}y = 12$$ $$30 \times \left(\frac{1}{6}x - \frac{1}{3}y\right) = 30 \times (-10) \Rightarrow 5x - 10y = -300$$ 4. Multiply the first new equation by 5 to clear fraction: $$5 \times \left(3x + \frac{18}{5}y\right) = 5 \times 12 \Rightarrow 15x + 18y = 60$$ 5. Now system is: $$15x + 18y = 60$$ $$5x - 10y = -300$$ 6. Multiply second equation by 3: $$3 \times (5x - 10y) = 3 \times (-300) \Rightarrow 15x - 30y = -900$$ 7. Subtract first equation from this: $$\cancel{15x} - 30y - (\cancel{15x} + 18y) = -900 - 60$$ $$-30y - 18y = -960$$ $$-48y = -960$$ 8. Solve for $y$: $$y = \frac{-960}{-48} = 20$$ 9. Substitute $y=20$ into first original equation: $$\frac{1}{2}x + \frac{3}{5} \times 20 = 2$$ $$\frac{1}{2}x + 12 = 2$$ 10. Solve for $x$: $$\frac{1}{2}x = 2 - 12 = -10$$ $$x = -10 \times 2 = -20$$ **Final answer:** $$x = -20, \quad y = 20$$