Question

solve for q.
-2|q - 9| ≥ -6
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. use integers, fractions, or improper fractions in simplest form.

Explanation:

Step1: Divide both sides by -2 (reverse inequality)

When dividing an inequality by a negative number, the inequality sign flips. So we have:
$$\frac{-2|q - 9|}{-2} \leq \frac{-6}{-2}$$
Simplifying gives:
$$|q - 9| \leq 3$$

Step2: Solve the absolute - value inequality

The absolute - value inequality \(|a|\leq b\) (where \(b\geq0\)) is equivalent to \(-b\leq a\leq b\). So for \(|q - 9|\leq3\), we can write:
$$- 3\leq q - 9\leq3$$

Step3: Add 9 to all parts of the compound inequality

Add 9 to the left - hand side, the middle, and the right - hand side of the inequality:
Left - hand side: \(-3 + 9=6\)
Middle: \(q-9 + 9=q\)
Right - hand side: \(3 + 9 = 12\)
So we get \(6\leq q\leq12\)

Answer:

\(6\leq q\leq12\)