If `|-3y+2|lt1`, what is one possible value of y?

Answer

To solve the inequality \( |-3y + 2| < 1 \), we can follow these steps:

### Step 1: Understand the Absolute Value Inequality
The absolute value inequality \( |a| < b \) can be rewritten as:
\[ -b < a < b \]
In our case, \( a = -3y + 2 \) and \( b = 1 \). Therefore, we can rewrite the inequality as:
\[ -1 < -3y + 2 < 1 \]

### Step 2: Break Down the Compound Inequality
Now, we will break this down into two separate inequalities:
1. \( -1 < -3y + 2 \)
2. \( -3y + 2 < 1 \)

### Step 3: Solve the First Inequality
Starting with the first inequality:
\[ -1 < -3y + 2 \]
Subtract 2 from both sides:
\[ -1 - 2 < -3y \]
This simplifies to:
\[ -3 < -3y \]
Now, divide both sides by -3. Remember, dividing by a negative number reverses the inequality:
\[ 1 > y \quad \text{or} \quad y < 1 \]

### Step 4: Solve the Second Inequality
Now, let's solve the second inequality:
\[ -3y + 2 < 1 \]
Subtract 2 from both sides:
\[ -3y < 1 - 2 \]
This simplifies to:
\[ -3y < -1 \]
Now, divide both sides by -3, reversing the inequality:
\[ y > \frac{1}{3} \]

### Step 5: Combine the Results
We now have two inequalities:
1. \( y < 1 \)
2. \( y > \frac{1}{3} \)

Combining these gives us:
\[ \frac{1}{3} < y < 1 \]

### Step 6: Find One Possible Value of \( y \)
From the range \( \frac{1}{3} < y < 1 \), we can choose any value that lies within this interval. A simple choice is:
\[ y = \frac{1}{2} \]

### Conclusion
One possible value of \( y \) that satisfies the inequality \( |-3y + 2| < 1 \) is:
\[ \boxed{\frac{1}{2}} \]