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The problem asks to find which of the given values for \(x\) satisfy the inequality \(|1 - 3x| < 5\).

### Step 1: Solve the inequality

\[
|1 - 3x| < 5
\]

This absolute value inequality can be rewritten as:

\[
-5 < 1 - 3x < 5
\]

### Step 2: Solve the two parts of the inequality

1. \(-5 < 1 - 3x\)

Subtract 1 from both sides:

\[
-6 < -3x
\]

Now divide by \(-3\), remembering to flip the inequality sign:

\[
2 > x \quad \text{or} \quad x < 2
\]

2. \(1 - 3x < 5\)

Subtract 1 from both sides:

\[
-3x < 4
\]

Now divide by \(-3\), again flipping the inequality:

\[
x > -\frac{4}{3}
\]

### Step 3: Combine the solution

Thus, the combined solution is:

\[
-\frac{4}{3} < x < 2
\]

### Step 4: Check the given values

- I. \(x = -1\): Since \(-\frac{4}{3} \approx -1.33\), \(-1\) is within the interval \((-4/3, 2)\), so this satisfies the inequality.
- II. \(x = 1\): \(1\) is also within the interval \((-4/3, 2)\), so this satisfies the inequality.
- III. \(x = 2\): \(x = 2\) is not strictly less than 2, so it does not satisfy the inequality.

### Final Answer:

The values that satisfy the inequality are \(x = -1\) and \(x = 1\). Therefore, the correct answer is:

**D) I and II only**

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Would you like more details or have any questions?

Here are five related questions:
1. How do you solve absolute value inequalities in general?
2. What happens when there is an "equal to" sign in absolute value inequalities?
3. Can inequalities with absolute values be graphed on a number line?
4. How do you check if a solution to an inequality is correct?
5. How do you solve compound inequalities involving two absolute values?

**Tip**: Always remember to reverse the inequality sign when dividing or multiplying by a negative number in inequalities.

Which of the following values of x satisfies |1 - 3x| < 5?
I. -1
II. 1
III. 2

Learn how to solve the absolute value inequality |1 - 3x| < 5 and determine which of the values -1, 1, or 2 satisfy the inequality. This problem involves key concepts in algebra, including absolute value inequalities and solving linear inequalities. Suitable for students in Grades 9-11.

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