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 can break it down into two parts.

### Step 1: Break down the absolute value inequality
The inequality \( |3x + 1| < \frac{1}{3} \) implies two inequalities:
1. \( 3x + 1 < \frac{1}{3} \)
2. \( 3x + 1 > -\frac{1}{3} \)

### Step 2: Solve the first inequality
Let's solve the first inequality:
\[ 3x + 1 < \frac{1}{3} \]
Subtract 1 from both sides:
\[ 3x < \frac{1}{3} - 1 \]
Convert 1 to a fraction:
\[ 3x < \frac{1}{3} - \frac{3}{3} = -\frac{2}{3} \]
Now, divide by 3:
\[ x < -\frac{2}{9} \]

### Step 3: Solve the second inequality
Now, let's solve the second inequality:
\[ 3x + 1 > -\frac{1}{3} \]
Subtract 1 from both sides:
\[ 3x > -\frac{1}{3} - 1 \]
Convert 1 to a fraction:
\[ 3x > -\frac{1}{3} - \frac{3}{3} = -\frac{4}{3} \]
Now, divide by 3:
\[ x > -\frac{4}{9} \]

### Step 4: Combine the results
Now we have two inequalities:
1. \( x < -\frac{2}{9} \)
2. \( x > -\frac{4}{9} \)

Combining these gives:
\[ -\frac{4}{9} < x < -\frac{2}{9} \]

### Step 5: Consider the restriction from the absolute value
Since the original inequality is \( 0 < |3x + 1| \), we need to ensure that \( 3x + 1 \neq 0 \). This means:
\[ 3x + 1 \neq 0 \implies 3x \neq -1 \implies x \neq -\frac{1}{3} \]
Since \( -\frac{1}{3} \) is between \( -\frac{4}{9} \) and \( -\frac{2}{9} \), we exclude it from our solution set.

### Final Solution
Thus, the solution set of the inequation \( 0 < |3x + 1| < \frac{1}{3} \) is:
\[ -\frac{4}{9} < x < -\frac{2}{9}, \quad x \neq -\frac{1}{3} \]