Question

question 12 (1 point)
solve the inequality.
$-2x + 6 \leq 12$
\bigcirc a \quad $x \geq -3$
\bigcirc b \quad $x \leq -3$
\bigcirc c \quad $x \leq -9$
\bigcirc d \quad $x \geq -9$

Explanation:

Step1: Subtract 6 from both sides

To isolate the term with \( x \), we subtract 6 from both sides of the inequality \( -2x + 6 \leq 12 \).
\[
-2x + 6 - 6 \leq 12 - 6
\]
Simplifying both sides, we get:
\[
-2x \leq 6
\]

Step2: Divide by -2 (and reverse inequality)

When we divide both sides of an inequality by a negative number, we must reverse the direction of the inequality sign. We divide both sides of \( -2x \leq 6 \) by -2.
\[
\frac{-2x}{-2} \geq \frac{6}{-2}
\]
Simplifying both sides, we get:
\[
x \geq -3
\]

Answer:

a. \( x \geq -3 \)