1. **State the problem:** Solve the system of equations: $$\frac{x}{10} + \frac{y}{20} = 4$$ $$\frac{x}{20} - \frac{y}{5} = 1$$ 2. **Rewrite the equations to eliminate fractions:** Multiply the first equation by 20: $$20 \times \left(\frac{x}{10} + \frac{y}{20}\right) = 20 \times 4$$ $$2x + y = 80$$ Multiply the second equation by 20: $$20 \times \left(\frac{x}{20} - \frac{y}{5}\right) = 20 \times 1$$ $$x - 4y = 20$$ 3. **Use substitution or elimination method:** From the first equation: $$y = 80 - 2x$$ 4. **Substitute $y$ into the second equation:** $$x - 4(80 - 2x) = 20$$ $$x - 320 + 8x = 20$$ $$9x - 320 = 20$$ 5. **Solve for $x$:** $$9x = 20 + 320$$ $$9x = 340$$ $$x = \frac{340}{9}$$ 6. **Substitute $x$ back to find $y$:** $$y = 80 - 2 \times \frac{340}{9} = 80 - \frac{680}{9} = \frac{720}{9} - \frac{680}{9} = \frac{40}{9}$$ **Final answer:** $$x = \frac{340}{9}, \quad y = \frac{40}{9}$$