The solution set of the inequation `|(3)/(x)+1| gt 2,` is

Options

A. `(0, 3]`
B. `[-1, 0]`
C. `(-1, 0) cup (0, 3)`
D. none of these

Answer

To solve the inequation \( \left| \frac{3}{x} + 1 \right| > 2 \), we will break it down into two separate cases based on the definition of absolute value.

### Step 1: Set up the two cases
The absolute value inequality \( \left| A \right| > B \) can be rewritten as two separate inequalities:
1. \( A > B \)
2. \( A < -B \)

In our case, we have:
1. \( \frac{3}{x} + 1 > 2 \)
2. \( \frac{3}{x} + 1 < -2 \)

### Step 2: Solve the first inequality
Starting with the first inequality:
\[ \frac{3}{x} + 1 > 2 \]
Subtract 1 from both sides:
\[ \frac{3}{x} > 1 \]
Now, multiply both sides by \( x \) (note that we need to consider the sign of \( x \)):
- If \( x > 0 \):
\[ 3 > x \quad \Rightarrow \quad x < 3 \]
- If \( x < 0 \), the inequality reverses:
\[ 3 < x \quad \Rightarrow \quad \text{This is not possible since } x < 0. \]

Thus, from the first inequality, we have:
\[ 0 < x < 3 \]

### Step 3: Solve the second inequality
Now, we solve the second inequality:
\[ \frac{3}{x} + 1 < -2 \]
Subtract 1 from both sides:
\[ \frac{3}{x} < -3 \]
Multiply both sides by \( x \) (again considering the sign of \( x \)):
- If \( x > 0 \):
\[ 3 < -3x \quad \Rightarrow \quad x < -1 \quad \text{(not possible since } x > 0\text{)} \]
- If \( x < 0 \):
\[ 3 > -3x \quad \Rightarrow \quad x < -1 \]

Thus, from the second inequality, we have:
\[ x < -1 \]

### Step 4: Combine the results
Now, we combine the results from both inequalities:
1. From the first inequality, we have \( 0 < x < 3 \).
2. From the second inequality, we have \( x < -1 \).

The solution set for the inequation \( \left| \frac{3}{x} + 1 \right| > 2 \) is:
\[ (-\infty, -1) \cup (0, 3) \]

### Final Answer
The solution set of the inequation \( \left| \frac{3}{x} + 1 \right| > 2 \) is:
\[ (-\infty, -1) \cup (0, 3) \]