1. **State the problem:** Solve the system of linear equations using the elimination method: $$5x + 4y = -14$$ $$3x + 6y = 6$$ 2. **Goal:** Eliminate one variable by making the coefficients of either $x$ or $y$ the same (or opposites) in both equations. 3. **Eliminate $x$:** Multiply the first equation by 3 and the second equation by 5 to align the coefficients of $x$: $$3(5x + 4y) = 3(-14) \Rightarrow 15x + 12y = -42$$ $$5(3x + 6y) = 5(6) \Rightarrow 15x + 30y = 30$$ 4. **Subtract the first new equation from the second:** $$\cancel{15x} + 30y - (\cancel{15x} + 12y) = 30 - (-42)$$ $$30y - 12y = 30 + 42$$ $$18y = 72$$ 5. **Solve for $y$:** $$y = \frac{72}{18} = 4$$ 6. **Substitute $y=4$ into one original equation to find $x$:** Using the first equation: $$5x + 4(4) = -14$$ $$5x + 16 = -14$$ $$5x = -14 - 16$$ $$5x = -30$$ 7. **Solve for $x$:** $$x = \frac{-30}{5} = -6$$ **Final answer:** $$x = -6, \quad y = 4$$