Set 16: Systems of Equations (Intermediate)

Choice D is correct. Choice D is the correct answer. We can use elimination by multiplying the second equation by 2. Step 1: Multiply the second equation by 2: $6x + 2 y = 28$ Step 2: Add to the first equation: (4 x - 2 y) + (6 x + 2 y) = 18 + 28$10x = 46$x = 4.6 Non-integer. - First: $4(4) - 2(-1) = 16 + 2 = 18$ ✓ - Second: $3(4) + (-1) = 12 - 1 = 11 \neq 14$ Step 1: Multiply second by 2: $6x + 2 y = 22$ Step 2: Add: $$$10x = 40x = 4$$ Step 3: Substitute: $$4(4) - 2 y = 1816- 2 y = 18-2 y = 2y = -1$$ Strategic Tip: Multiplying by 2 creates opposite coefficients for easy elimination. Choice A is incorrect because 4(5) - 2(1) = 18$✓, but$3(5) + 1 = 16 \neq 11$. Choice B is incorrect because$4(3) - 2(-3) = 18$✓, but$3(3) + (-3) = 6 \neq 11$. Choice C is incorrect because$4(6) - 2(3) = 18$✓, but$3(6) + 3 = 21 \neq 11$.