If `|3x + 2| lt 1`, then x belongs to the interval

Options

A. `(-1, -1//3)`
B. `[-1, -1//3]`
C. `(-oo, -1)`
D. `(-1//3, oo)`

Answer

To solve the inequality \( |3x + 2| < 1 \), we will follow these steps:

### Step 1: Remove the absolute value
The property of absolute values states that if \( |A| < B \), then \( -B < A < B \). Here, we can apply this property to our inequality:

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

### Step 2: Split into two inequalities
We can split the compound inequality into two separate inequalities:

1. \( 3x + 2 < 1 \)
2. \( 3x + 2 > -1 \)

### Step 3: Solve the first inequality
Let's solve the first inequality \( 3x + 2 < 1 \):

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

### Step 4: Solve the second inequality
Now, let's solve the second inequality \( 3x + 2 > -1 \):

\[ 3x > -1 - 2 \]
\[ 3x > -3 \]
\[ x > -1 \]

### Step 5: Combine the results
Now we combine the two results:

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

### Step 6: Write in interval notation
In interval notation, this can be expressed as:

\[ (-1, -\frac{1}{3}) \]

### Final Answer
Thus, the solution to the inequality \( |3x + 2| < 1 \) is:

\[ x \text{ belongs to the interval } (-1, -\frac{1}{3}) \]

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