Set 9: Systems of Equations (Advanced)

Choice C is correct. Choice C is the correct answer. We'll multiply to eliminate $y$: first equation by 3, second by 4. Step 1: Multiply the equations: 15x + 12 y = 99$$8x - 12 y = 4 Step 2: Add: 23x = 103$x = \frac{103}{23}$$ Non-integer. - First: $5(5) + 4(2) = 25 + 8 = 33$ ✓ - Second: $2(5) - 3(2) = 10 - 6 = 4 \neq 1$ Step 1: Multiply: 15x + 12 y = 998x - 12 y = 16$$$ Step 2: Add: $$$23x = 115$x = 5 Step 3: Substitute: 5(5) + 4 y = 33$25+ 4 y = 33$y = 2 Strategic Tip: The LCM of 4 and 3 is 12, so we multiply to get $12y$ and $-12 y$. Choice A is incorrect because $5(6) + 4(0.75) = 33$ ✓, but $2(6) - 3(0.75) \neq 4$. Choice B is incorrect because $5(4) + 4(\frac{13}{4}) = 20 + 13 = 33$ ✓, but $2(4) - 3(\frac{13}{4}) \neq 4$. Choice D is incorrect because $5(3) + 4(4.5) = 15 + 18 = 33$ ✓, but $2(3) - 3(4.5) = -7.5 \neq 4$.