Solve the following inequality algebraically.
2, vertical bar, x, plus, 8, vertical bar, minus, 5, is greater than, 3
2∣x+8∣−5>3

To solve the inequality \( 2|x + 8| - 5 > 3 \) algebraically, follow these steps:

### Step 1: Isolate the absolute value expression
Start by isolating the absolute value expression on one side of the inequality.

\[
2|x + 8| - 5 > 3
\]

Add 5 to both sides:

\[
2|x + 8| > 8
\]

Next, divide both sides by 2:

\[
|x + 8| > 4
\]

### Step 2: Break the absolute value inequality into two cases
The inequality \( |x + 8| > 4 \) means that the expression inside the absolute value can be either greater than 4 or less than -4. So, we split this into two cases:

1. **Case 1:** \( x + 8 > 4 \)
2. **Case 2:** \( x + 8 < -4 \)

### Step 3: Solve each case

#### Case 1: \( x + 8 > 4 \)

Subtract 8 from both sides:

\[
x > -4
\]

#### Case 2: \( x + 8 < -4 \)

Subtract 8 from both sides:

\[
x < -12
\]

### Step 4: Combine the solutions
The solutions from the two cases give us the solution set:

\[
x > -4 \quad \text{or} \quad x < -12
\]

### Step 5: Write the solution in interval notation
The solution set is the union of the two intervals:

\[
x \in (-\infty, -12) \cup (-4, \infty)
\]

This is the algebraic solution to the inequality \( 2|x + 8| - 5 > 3 \).

Would you like more details or have any questions?

Here are five related questions for practice:

1. Solve the inequality \( 3|x - 5| + 2 > 11 \).
2. Solve the equation \( 4|x + 1| = 12 \).
3. What is the solution to \( |2x - 3| \leq 7 \)?
4. Solve the compound inequality \( |x + 6| < 2 \) and \( |x + 6| > -2 \).
5. Find the solution set for \( |3x + 9| \geq 15 \).

**Tip:** When solving absolute value inequalities, always consider both the positive and negative scenarios for the expression inside the absolute value.

Solve the following inequality algebraically: 2|x + 8| - 5 > 3

This guide explains how to solve the absolute value inequality 2|x + 8| - 5 > 3 step by step. Learn how to isolate the absolute value, split it into two cases, and solve the resulting inequalities. Ideal for students in Grades 9-11.

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