Question

type the correct answer in each bo
what is the solution set of this ineq
$2 < |x + 2| < 6$
$(-8, \square ) \cup ( \square , \square )$

Explanation:

Step1: Split compound inequality

Split 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:
$-6-2 < x < 6-2$
$\implies -8 < x < 4$

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

Apply absolute value rule: $x+2 < -2$ or $x+2 > 2$
Solve left part: $x < -2-2 \implies x < -4$
Solve right part: $x > 2-2 \implies x > 0$

Step4: Find intersection of solutions

Intersection of $-8 < x < 4$ with $x < -4$: $-8 < x < -4$
Intersection of $-8 < x < 4$ with $x > 0$: $0 < x < 4$
Combine the intervals.

Answer:

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