1. The problem asks which method eliminates one of the variables in the system of equations: Equation A: $5x + 9y = 12$ Equation B: $4x - 3y = 8$ 2. To eliminate a variable, we multiply the equations by constants so that the coefficients of either $x$ or $y$ become opposites and then add the equations. 3. Check each method: (1) Multiply equation A by $-\frac{1}{3}$: $$-\frac{1}{3} \times (5x + 9y) = -\frac{5}{3}x - 3y$$ Add to equation B: $$(4x - 3y) + (-\frac{5}{3}x - 3y) = (4 - \frac{5}{3})x + (-3y - 3y) = \frac{12}{3}x - \frac{5}{3}x - 6y = \frac{7}{3}x - 6y$$ The $y$ terms do not cancel, so no elimination. (2) Multiply equation B by 3: $$3 \times (4x - 3y) = 12x - 9y$$ Add to equation A: $$(5x + 9y) + (12x - 9y) = 17x + 0y$$ The $y$ terms cancel, so $y$ is eliminated. (3) Multiply equation A by 2: $$2 \times (5x + 9y) = 10x + 18y$$ Multiply equation B by -6: $$-6 \times (4x - 3y) = -24x + 18y$$ Add: $$(10x + 18y) + (-24x + 18y) = -14x + 36y$$ No variable eliminated. (4) Multiply equation B by 5: $$5 \times (4x - 3y) = 20x - 15y$$ Multiply equation A by 4: $$4 \times (5x + 9y) = 20x + 36y$$ Add: $$(20x - 15y) + (20x + 36y) = 40x + 21y$$ No variable eliminated. 4. Therefore, method (2) eliminates variable $y$. Final answer: Method (2)