The solution set of the inequation `0 lt |3x+ 1|lt (1)/(3)`, is

Options

A. `(-4//9, -2//9)`
B. `[-4//9, -2//9]`
C. `(-4//9, -2//9)-[-1//3]`
D. `[-4//9-2//9]-[-1//3]`

Answer

To solve the inequation \( 0 < |3x + 1| < \frac{1}{3} \), we will break it down into two parts:

1. **Understanding the absolute value inequality**:
- The first part is \( |3x + 1| > 0 \).
- The second part is \( |3x + 1| < \frac{1}{3} \).

### Step 1: Solve \( |3x + 1| > 0 \)

The expression \( |3x + 1| > 0 \) implies that \( 3x + 1 \neq 0 \).

- Therefore, we have:
\[ 3x + 1 \neq 0 \implies 3x \neq -1 \implies x \neq -\frac{1}{3} \]

### Step 2: Solve \( |3x + 1| < \frac{1}{3} \)

The expression \( |3x + 1| < \frac{1}{3} \) can be rewritten as:
\[ -\frac{1}{3} < 3x + 1 < \frac{1}{3} \]

Now, we will solve the two inequalities separately.

#### Part A: Solve \( 3x + 1 < \frac{1}{3} \)

1. Subtract 1 from both sides:
\[ 3x < \frac{1}{3} - 1 = \frac{1}{3} - \frac{3}{3} = -\frac{2}{3} \]
2. Divide by 3:
\[ x < -\frac{2}{9} \]

#### Part B: Solve \( -\frac{1}{3} < 3x + 1 \)

1. Subtract 1 from both sides:
\[ -\frac{1}{3} - 1 < 3x \implies -\frac{1}{3} - \frac{3}{3} < 3x \implies -\frac{4}{3} < 3x \]
2. Divide by 3:
\[ -\frac{4}{9} < x \]

### Step 3: Combine the results

From the two parts, we have:
1. \( x < -\frac{2}{9} \)
2. \( -\frac{4}{9} < x \)

This gives us the combined inequality:
\[ -\frac{4}{9} < x < -\frac{2}{9} \]

### Step 4: Exclude the value \( x = -\frac{1}{3} \)

We also have the condition from Step 1 that \( x \neq -\frac{1}{3} \). Since \( -\frac{1}{3} \) is within the interval \( -\frac{4}{9} < x < -\frac{2}{9} \), we need to exclude it from the solution set.

### Final Solution Set

Thus, the solution set of the inequation \( 0 < |3x + 1| < \frac{1}{3} \) is:
\[ x \in \left(-\frac{4}{9}, -\frac{2}{9}\right) \setminus \left\{-\frac{1}{3}\right\} \]