Question

svlc algebra 1a - standard (15260)
linear equations and inequalities
what is the solution to $-4(8 - 3x) \geq 6x - 8$
options: $x \geq -\frac{4}{3}$, $x \leq 4$, $x \leq -\frac{4}{3}$, $x \geq 4$

Explanation:

Step1: Distribute -4

First, we distribute the -4 across the terms inside the parentheses: $-4(8 - 3x) = -32 + 12x$. So the inequality becomes $-32 + 12x \geq 6x - 8$.

Step2: Subtract 6x from both sides

Subtract $6x$ from both sides to get the x terms on one side: $-32 + 12x - 6x \geq 6x - 8 - 6x$. This simplifies to $-32 + 6x \geq -8$.

Step3: Add 32 to both sides

Add 32 to both sides to isolate the term with x: $-32 + 6x + 32 \geq -8 + 32$. This simplifies to $6x \geq 24$.

Step4: Divide by 6

Divide both sides by 6 to solve for x: $\frac{6x}{6} \geq \frac{24}{6}$. This gives $x \geq 4$.

Answer:

$x \geq 4$ (corresponding to the option "x ≥ 4")