Solve the inequality 2 < |x + 1| <= 3.

Step-by-step solution

1. 1

   Split the sandwich into two separate conditions. A double inequality on an absolute value is really an intersection:

   $|x+1|>2\qquad\textbf{and}\qquad |x+1|\le3$

   These behave in opposite ways — "greater than" opens outward, "less than" closes inward — so they must be solved separately and then intersected.

2. 2

   Solve the greater-than part. $|A|>c$ (with $c>0$) means $A>c$ or $A<-c$:

   $x+1>2\ \text{or}\ x+1<-2\qquad\Longrightarrow\qquad x>1\ \text{or}\ x<-3$

   This is a union of two rays — the points far from $-1$.

3. 3

   Solve the at-most part. $|A|\le c$ means $-c\le A\le c$:

   $-3\le x+1\le3\qquad\Longrightarrow\qquad -4\le x\le2$

   This is a single band — the points near $-1$.

4. 4

   Intersect the two results, piece by piece. Take each ray against the band:

   • $x>1$ with $-4\le x\le2$ gives $1<x\le2$
   • $x<-3$ with $-4\le x\le2$ gives $-4\le x<-3$
5. 5

   Read the answer as a distance condition. $|x+1|$ is the distance from $x$ to $-1$, so the inequality asks for points more than $2$ but at most $3$ away — an annulus on the number line. Each side contributes one interval of length $1$:

   $[-4,-3)\cup(1,2]$

   Note the bracket pattern: closed at the outer ends (from $\le3$) and open at the inner ends (from the strict $>2$).

6. 6

   Verify by scanning. Testing $12{,}001$ exact rational values of $x$ on $[-6,6]$, the inequality $2<|x+1|\le3$ holds at exactly the points of $[-4,-3)\cup(1,2]$ and nowhere else ✓. Endpoint checks: at $x=-4$, $|x+1|=3$ ✓ included; at $x=-3$, $|x+1|=2$ ✗ excluded.