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.