1. **State the problem:** Solve the system of linear equations: $$6x - 5y = 8$$ $$-12x + 2y = 0$$ 2. **Identify the method:** We will use the elimination method to solve for $x$ and $y$. 3. **Eliminate one variable:** Multiply the second equation by $\frac{5}{2}$ to align the coefficients of $y$: $$\frac{5}{2} \times (-12x + 2y) = \frac{5}{2} \times 0$$ $$-30x + 5y = 0$$ 4. **Add the two equations:** $$6x - 5y = 8$$ $$-30x + 5y = 0$$ Adding: $$6x - 5y + (-30x + 5y) = 8 + 0$$ $$6x - 30x + (-5y + 5y) = 8$$ $$-24x + \cancel{-5y + 5y} = 8$$ Simplifies to: $$-24x = 8$$ 5. **Solve for $x$:** $$x = \frac{8}{-24} = \cancel{\frac{8}{-24}} = -\frac{1}{3}$$ 6. **Substitute $x$ back into one equation:** Use the second original equation: $$-12x + 2y = 0$$ Substitute $x = -\frac{1}{3}$: $$-12 \times \left(-\frac{1}{3}\right) + 2y = 0$$ $$4 + 2y = 0$$ 7. **Solve for $y$:** $$2y = -4$$ $$y = \frac{-4}{2} = -2$$ **Final answer:** $$x = -\frac{1}{3}, \quad y = -2$$