what interval includes all possible values of x, where -3(6 - 2x) ≥ 4x + 12?\n(-∞, -3\n-3, ∞)\n(-∞, 15\n15, ∞)

Explanation:

Step1: Expand the left - hand side

$-3(6 - 2x)=-18 + 6x$. So the inequality becomes $-18 + 6x\geq4x + 12$.

Step2: Move the terms with $x$ to one side

Subtract $4x$ from both sides: $6x-4x-18\geq4x - 4x+12$, which simplifies to $2x-18\geq12$.

Step3: Isolate the variable term

Add 18 to both sides: $2x-18 + 18\geq12 + 18$, resulting in $2x\geq30$.

Step4: Solve for $x$

Divide both sides by 2: $\frac{2x}{2}\geq\frac{30}{2}$, so $x\geq15$.

Answer:

D. $[15,\infty)$