1. **State the problem:** Solve the system of linear equations: $$-4x + 5y = 1$$ $$-x - 5y = 19$$ 2. **Add the two equations to eliminate $y$: ** $$(-4x + 5y) + (-x - 5y) = 1 + 19$$ $$-4x - x + 5y - 5y = 20$$ $$-5x + \cancel{5y} - \cancel{5y} = 20$$ 3. **Simplify and solve for $x$: ** $$-5x = 20$$ $$x = \frac{20}{-5}$$ $$x = -4$$ 4. **Substitute $x = -4$ into one of the original equations to find $y$: ** Using $$-x - 5y = 19$$: $$-(-4) - 5y = 19$$ $$4 - 5y = 19$$ 5. **Isolate $y$: ** $$-5y = 19 - 4$$ $$-5y = 15$$ $$y = \frac{15}{-5}$$ $$y = -3$$ 6. **Final answer:** $$x = -4, \quad y = -3$$ The solution to the system is $(-4, -3)$.