Set 8: Systems of Equations (Intermediate)

Choice B is correct. Choice B is the correct answer. To eliminate $y$, multiply the second equation by 2. Step 1: Multiply the second equation by 2: $10x - 4 y = 22$ Step 2: Add to the first equation: (3 x + 4 y) + (10 x - 4 y) = 27 + 22$13x = 49$x = \frac{49}{13} Non-integer. - First: $3(5) + 4(3) = 15 + 12 = 27$ ✓ - Second: $5(5) - 2(3) = 25 - 6 = 19 \neq 11$ Step 1: Multiply second by 2: $10x - 4 y = 38$ Step 2: Add: $$$13x = 65x = 5$$ Step 3: Substitute: $$3(5) + 4 y = 2715+ 4 y = 27y = 3$$ Strategic Tip: Multiplying by 2 turned the -2 y$into$-4 y$to match the$+4 y$in the first equation. Choice A is incorrect because$3(4) + 4(\frac{15}{4}) = 12 + 15 = 27$✓, but$5(4) - 2(\frac{15}{4}) = 20 - 7.5 = 12.5 \neq 19$. Choice C is incorrect because$3(3) + 4(4.5) = 9 + 18 = 27$✓, but$5(3) - 2(4.5) = 15 - 9 = 6 \neq 19$. Choice D is incorrect because$3(6) + 4(2.25) = 18 + 9 = 27$ ✓, but multiplication in second fails.