https://file.aiquickdraw.com/mathbot/file/1726866147200zur0k83e.jpg

The inequality you've uploaded is:

\[
\left| \frac{x - 3}{x + 1} \right| < 2
\]

Let's solve this inequality step by step.

### Step 1: Remove the absolute value
The inequality \(\left| A \right| < B\) means that \(-B < A < B\). So, we can rewrite the inequality as two separate inequalities:

\[
-2 < \frac{x - 3}{x + 1} < 2
\]

### Step 2: Solve each part of the compound inequality

#### 1. Solve \(-2 < \frac{x - 3}{x + 1}\)

Multiply both sides by \((x + 1)\), but we need to consider two cases: \(x + 1 > 0\) and \(x + 1 < 0\).

1. **Case 1: \(x + 1 > 0\) (i.e., \(x > -1\))**

\[
-2(x + 1) < x - 3
\]

Simplify:

\[
-2x - 2 < x - 3
\]

Add \(2x\) to both sides:

\[
-2 < 3x - 3
\]

Add 3 to both sides:

\[
1 < 3x
\]

Divide by 3:

\[
x > \frac{1}{3}
\]

In this case, \(x > \frac{1}{3}\) and \(x > -1\), so the solution from this part is \(x > \frac{1}{3}\).

2. **Case 2: \(x + 1 < 0\) (i.e., \(x < -1\))**

Here, multiplying by \(x + 1\) reverses the inequality:

\[
-2(x + 1) > x - 3
\]

Simplify:

\[
-2x - 2 > x - 3
\]

Add \(2x\) to both sides:

\[
-2 > 3x - 3
\]

Add 3 to both sides:

\[
1 > 3x
\]

Divide by 3:

\[
x < \frac{1}{3}
\]

Since \(x < -1\), the solution for this case is simply \(x < -1\).

#### 2. Solve \(\frac{x - 3}{x + 1} < 2\)

Again, consider two cases based on \(x + 1\).

1. **Case 1: \(x + 1 > 0\) (i.e., \(x > -1\))**

Multiply both sides by \((x + 1)\):

\[
x - 3 < 2(x + 1)
\]

Simplify:

\[
x - 3 < 2x + 2
\]

Subtract \(x\) from both sides:

\[
-3 < x + 2
\]

Subtract 2 from both sides:

\[
-5 < x
\]

So the solution is \(x > -5\).

2. **Case 2: \(x + 1 < 0\) (i.e., \(x < -1\))**

Multiplying by \((x + 1)\) reverses the inequality:

\[
x - 3 > 2(x + 1)
\]

Simplify:

\[
x - 3 > 2x + 2
\]

Subtract \(x\) from both sides:

\[
-3 > x + 2
\]

Subtract 2 from both sides:

\[
-5 > x
\]

Thus, \(x < -5\).

### Step 3: Combine the results

We have two cases:
1. \(x > \frac{1}{3}\)
2. \(x < -1\)

The solution to the inequality is:

\[
x \in (-\infty, -1) \cup \left( \frac{1}{3}, \infty \right)
\]

Would you like more details on any specific part of the process or additional questions on this topic? Here are some related questions:

1. How do we handle inequalities involving absolute values?
2. What happens if the inequality were \(\left| \frac{x - 3}{x + 1} \right| \geq 2\)?
3. Could you explain the reasoning behind splitting the absolute value inequality?
4. What if the denominator caused a division by zero? How do we handle that case?
5. How would the solution change if the inequality were reversed to \(\left| \frac{x - 3}{x + 1} \right| > 2\)?

**Tip:** When solving inequalities involving absolute values, always remember to split the inequality into two cases (positive and negative) to cover all possible scenarios.

Learn how to solve the absolute value inequality |(x - 3) / (x + 1)| < 2 using algebraic methods. This problem covers key concepts such as solving absolute value inequalities, rational expressions, and handling cases with different conditions on variable denominators. Suitable for students in Grades 10-12.

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