Set 7: Systems of Equations (Intermediate)

Choice B is correct. Choice B is the correct answer. I'll use elimination by multiplying the second equation by 3. Step 1: Multiply the second equation by 3: $6x - 3 y = 9$ Step 2: Add to the first equation: (x + 3 y) + (6 x - 3 y) = 13 + 9$7x = 22$x = \frac{22}{7} Non-integer. - First: $4+ 3(3) = 4 + 9 = 13$ ✓ - Second: $2(4) - 3 = 8 - 3 = 5 \neq 3$ Step 1: Multiply second by 3: $6x - 3 y = 15$ Step 2: Add: 7x = 28$x = 4$$ Step 3: Substitute: 4+ 3 y = 13y = 3$$ Solution: (4, 3)$Verification:$x + 3 y = 13✓ and $$2x - y = 5$$ ✓ Strategic Tip: With coefficient 1 on x, either substitution or elimination works well. Choice A is incorrect because $$2+ 3(\frac{11}{3}) = 13$$ ✓, but2(2) - \frac{11}{3} \neq 5. Choice C is incorrect because $$5+ 3(\frac{8}{3}) = 13$$ ✓, but 2(5) - \frac{8}{3} \neq 5. Choice D is incorrect because $$1+ 3(4) = 13$$ ✓, but 2(1) - 4 = -2 \neq 5$.