1. **State the problem:** Solve the system of equations: $$12x + 15y = 34$$ $$-6x + 5y = 3$$ 2. **Choose a method:** We will use the elimination method to eliminate one variable. 3. **Eliminate variable:** Multiply the second equation by 3 to align the coefficients of $y$: $$3(-6x + 5y) = 3 \times 3$$ $$-18x + 15y = 9$$ 4. **Subtract the first equation from this new equation:** $$(-18x + 15y) - (12x + 15y) = 9 - 34$$ $$-18x + 15y - 12x - 15y = -25$$ $$-30x = -25$$ 5. **Solve for $x$:** $$x = \frac{-25}{-30}$$ $$x = \frac{25}{30}$$ $$x = \frac{\cancel{25}}{\cancel{30}} = \frac{5}{6}$$ 6. **Substitute $x = \frac{5}{6}$ into the second original equation:** $$-6\left(\frac{5}{6}\right) + 5y = 3$$ $$-\cancel{6} \times \frac{5}{\cancel{6}} + 5y = 3$$ $$-5 + 5y = 3$$ 7. **Solve for $y$:** $$5y = 3 + 5$$ $$5y = 8$$ $$y = \frac{8}{5}$$ **Final answer:** $$x = \frac{5}{6}, \quad y = \frac{8}{5}$$