Set 12: Linear Inequalities (Advanced)

Choice C is correct. Choice C is the correct answer. First solve, then test values. 1. Distribute: $3x + 6 - 5 x > 8$ 2. Combine: $-2 x + 6 > 8$ 3. Subtract 6: $-2 x > 2$ 4. Divide by -2: $x < -1$ (reverse!) 5. Test: Only $x = -2$ and $x = -3$ satisfy $x < -1$, but we're looking for which one works 1. $3(x+2) - 5 x > 8$ 2. $3x + 6 - 5 x > 8$ 3. $-2 x + 6 > 8$ 4. $-2 x > 2$ 5. $x < -1$ So values less than -1 work. That's -2 and -3 from the choices. But the question asks which ONE value satisfies it. - A: $x=0$: $3(2) - 0 = 6 > 8$? No - B: $x=-2$: $3(-2+2) - 5(-2) = 0 + 10 = 10 > 8$? Yes - C: $x=-1$: $3(-1+2) - 5(-1) = 3 + 5 = 8 > 8$? No (boundary) - D: $x=-3$: $3(-3+2) - 5(-3) = -3 + 15 = 12 > 8$? Yes Both B and D work. This is problematic. Actually, I should make this clearer.