Question

question 3 of 5
type the correct answer in each box. use numerals instead of words.
what is the solution set of this inequality?
$2 < |x + 2| < 6$
$(-8, \square) \cup (\square, \square)$

Explanation:

Step1: Split the compound inequality

Split $2 < |x + 2| < 6$ into two inequalities:
$|x + 2| > 2$ and $|x + 2| < 6$

Step2: Solve $|x + 2| < 6$

Apply absolute value rule: $-6 < x + 2 < 6$
Subtract 2 from all parts:
$-6 - 2 < x < 6 - 2$
$-8 < x < 4$

Step3: Solve $|x + 2| > 2$

Apply absolute value rule: $x + 2 < -2$ or $x + 2 > 2$
For $x + 2 < -2$: subtract 2, $x < -4$
For $x + 2 > 2$: subtract 2, $x > 0$

Step4: Find intersections

Intersection of $-8 < x < 4$ and $x < -4$: $-8 < x < -4$
Intersection of $-8 < x < 4$ and $x > 0$: $0 < x < 4$

Answer:

$(-8, -4) \cup (0, 4)$