Here’s a detailed breakdown of the problems shown in the image:

Problem 1

Solve the inequality and graph the solution set:
$|2x + 7| < 11$

Solution Steps

The absolute value inequality $|A| < B$ can be rewritten as:
$-B < A < B$
Here $A = 2x + 7$ and $B = 11$, so:
$-11 < 2x + 7 < 11$

1. Subtract 7 from all parts:
   $-11 - 7 < 2x < 11 - 7$
   $-18 < 2x < 4$

2. Divide through by 2:
   $-9 < x < 2$

Solution in Interval Notation

$x \in (-9, 2)$

Graph

On a number line:

• Open circles at $x = -9$ and $x = 2$.
• Shade the region between these points.

Problem 2

Solve the inequality and graph the solution set:
$|x + 1| \geq 2$

Solution Steps

The absolute value inequality $|A| \geq B$ can be rewritten as two cases:
$A \geq B \quad \text{or} \quad A \leq -B$

Here $A = x + 1$ and $B = 2$, so:

1. First case:
   $x + 1 \geq 2 \quad \implies \quad x \geq 1$

2. Second case:
   $x + 1 \leq -2 \quad \implies \quad x \leq -3$

Solution in Interval Notation

$x \in (-\infty, -3] \cup [1, \infty)$

Graph

On a number line:

• Closed circle at $x = -3$ and $x = 1$.
• Shade the regions $x \leq -3$ and $x \geq 1$.

Problem 3

Solve for $x$:
$|2x - 5| = 8$

Solution Steps

The absolute value equation $|A| = B$ can be rewritten as two cases:
$A = B \quad \text{or} \quad A = -B$

Here $A = 2x - 5$ and $B = 8$, so:

1. First case:
   $2x - 5 = 8 \quad \implies \quad 2x = 13 \quad \implies \quad x = \frac{13}{2} = 6.5$

2. Second case:
   $2x - 5 = -8 \quad \implies \quad 2x = -3 \quad \implies \quad x = -\frac{3}{2} = -1.5$

Solution

$x = 6.5 \quad \text{or} \quad x = -1.5$

Summary

1. Problem 1: $x \in (-9, 2)$
2. Problem 2: $x \in (-\infty, -3] \cup [1, \infty)$
3. Problem 3: $x = 6.5$ or $x = -1.5$

Would you like me to explain any part further or graph these solutions for clarity?

Related Questions:

1. How do you solve inequalities involving absolute values?
2. What is the difference between $|A| < B$ and $|A| > B$?
3. How can interval notation be converted to graphical representation?
4. Why do absolute value inequalities produce two cases?
5. How do we check if a solution is valid in an absolute value equation?

Tip: Always check absolute value solutions by substituting them back into the original equation to ensure they satisfy the conditions.