Question

1. |x + 11| - 8 ≤ -3
2. -5|x - 3| - 5 < 15

8|x + 4| + 10 < 2

Explanation:

Step1: Isolate absolute value (Q14)

Add 8 to both sides:
$$|x+11| - 8 + 8 \leq -3 + 8$$
$$|x+11| \leq 5$$

Step2: Rewrite without absolute value (Q14)

Use absolute value inequality rule:
$$-5 \leq x+11 \leq 5$$

Step3: Solve for x (Q14)

Subtract 11 from all parts:
$$-5 - 11 \leq x \leq 5 - 11$$
$$-16 \leq x \leq -6$$

Step4: Isolate absolute value (Q15)

Add 5 to both sides:
$$-5|x-3| - 5 + 5 < 15 + 5$$
$$-5|x-3| < 20$$

Step5: Divide by negative (reverse inequality)

$$|x-3| > \frac{20}{-5}$$
$$|x-3| > -4$$

Step6: Analyze absolute value (Q15)

Absolute value is always non-negative, so $|x-3| \geq 0 > -4$ is true for all real x.

Answer:

1. $\boldsymbol{-16 \leq x \leq -6}$
2. All real numbers ($\boldsymbol{x \in \mathbb{R}}$)