To solve the inequality \( |2x + 1| > 5 \), we can break it down into two separate cases because of the absolute value.

### Step 1: Definition of absolute value inequality

For any expression \( |A| > B \), where \( B > 0 \), this means:
\[
A > B \quad \text{or} \quad A < -B
\]
In our case, \( A = 2x + 1 \) and \( B = 5 \). Thus, the inequality becomes:

1. \( 2x + 1 > 5 \)
2. \( 2x + 1 < -5 \)

### Step 2: Solve both cases

#### Case 1: \( 2x + 1 > 5 \)
Subtract 1 from both sides:
\[
2x > 4
\]
Now, divide by 2:
\[
x > 2
\]

#### Case 2: \( 2x + 1 < -5 \)
Subtract 1 from both sides:
\[
2x < -6
\]
Now, divide by 2:
\[
x < -3
\]

### Step 3: Combine the solutions
The solution is:
\[
x > 2 \quad \text{or} \quad x < -3
\]

In interval notation, this is:
\[
(-\infty, -3) \cup (2, \infty)
\]

Would you like more details or have any questions?

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**Related Questions:**

1. How do you solve absolute value inequalities where the inequality is \( \leq \)?
2. What is the geometric interpretation of \( |x - a| > b \)?
3. How do you handle inequalities with two absolute value terms?
4. What is the difference between solving \( |A| > B \) and \( |A| < B \)?
5. Can you explain how to graph the solution set of \( |2x + 1| > 5 \)?

**Tip:** Always check the boundary points when solving absolute value inequalities to ensure you don't miss any critical intervals.

Learn how to solve the absolute value inequality |2x + 1| > 5 with a step-by-step breakdown. This problem involves key algebraic concepts, including the absolute value inequality theorem. Suitable for students in Grades 9-10.

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