The problem asks to solve the inequality $|10 - 3x| < 5$ and find one possible integer value of $x$.

Step 1: Break down the absolute value inequality

We know that if $|A| < B$, it implies $-B < A < B$. So, we can rewrite the inequality $|10 - 3x| < 5$ as: $-5 < 10 - 3x < 5$

Step 2: Solve the compound inequality

Part 1: Solve $-5 < 10 - 3x$

Subtract 10 from both sides: $-5 - 10 < -3x \quad \Rightarrow \quad -15 < -3x$ Now divide by -3 (remember to flip the inequality when dividing by a negative number): $5 > x \quad \Rightarrow \quad x < 5$

Part 2: Solve $10 - 3x < 5$

Subtract 10 from both sides: $-3x < 5 - 10 \quad \Rightarrow \quad -3x < -5$ Now divide by -3 (again, flip the inequality): $x > \frac{5}{3}$

Step 3: Combine the inequalities

From the two parts, we have: $\frac{5}{3} < x < 5$

Step 4: Identify integer values

The only integer value that satisfies this inequality is $x = 2$.

So, one possible integer value of $x$ is $\boxed{2}$.

Would you like more details or have any further questions? Here are five related questions:

1. How do you handle absolute value inequalities that involve greater than signs instead of less than?
2. What if the inequality involved a fraction instead of an integer? How would the solution change?
3. How do you graph the solution to absolute value inequalities on a number line?
4. Can there be multiple integer solutions to an absolute value inequality?
5. What happens when the absolute value inequality has a variable on both sides?

Tip: When solving absolute value inequalities, always remember to split the inequality into two parts and consider both the positive and negative scenarios.