1. **State the problem:** Solve the system of linear equations: $$4x + 8y = 20$$ $$-x + 2y = -30$$ 2. **Formula and rules:** We can use the substitution or elimination method to solve this system. Here, we'll use elimination. 3. **Step 1: Multiply the second equation by 4 to align coefficients of $x$:** $$4(-x + 2y) = 4(-30)$$ $$-4x + 8y = -120$$ 4. **Step 2: Add the first equation and the new equation:** $$4x + 8y + (-4x + 8y) = 20 + (-120)$$ $$0x + 16y = -100$$ $$16y = -100$$ 5. **Step 3: Solve for $y$:** $$y = \frac{-100}{16}$$ Show cancellation: $$y = \frac{-\cancel{100}}{\cancel{16}} = \frac{-25}{4}$$ 6. **Step 4: Substitute $y = -\frac{25}{4}$ into the second original equation:** $$-x + 2\left(-\frac{25}{4}\right) = -30$$ $$-x - \frac{50}{4} = -30$$ $$-x - \frac{25}{2} = -30$$ 7. **Step 5: Add $\frac{25}{2}$ to both sides:** $$-x = -30 + \frac{25}{2}$$ Convert $-30$ to fraction with denominator 2: $$-30 = \frac{-60}{2}$$ So: $$-x = \frac{-60}{2} + \frac{25}{2} = \frac{-35}{2}$$ 8. **Step 6: Multiply both sides by $-1$ to solve for $x$:** $$x = \frac{35}{2}$$ **Final answer:** $$x = \frac{35}{2}, \quad y = -\frac{25}{4}$$