The given equation is:

$|x - 1| + |3x - 2| = 2$

To solve this, we need to consider different cases based on the values of $x$ that make the expressions inside the absolute values change their sign. The critical points occur when the expressions inside the absolute values are equal to zero:

• $x - 1 = 0 \Rightarrow x = 1$
• $3x - 2 = 0 \Rightarrow x = \frac{2}{3}$

These critical points divide the real line into three intervals: $(-\infty, \frac{2}{3})$, $\left(\frac{2}{3}, 1\right)$, and $[1, \infty)$. We will analyze each interval separately.

Case 1: $x < \frac{2}{3}$

In this case, both $x - 1$ and $3x - 2$ are negative, so we have: $|x - 1| = -(x - 1) = 1 - x, \quad |3x - 2| = -(3x - 2) = 2 - 3x$ Thus, the equation becomes: $(1 - x) + (2 - 3x) = 2$ Simplifying: $3 - 4x = 2$ $3 - 2 = 4x$ $1 = 4x$ $x = \frac{1}{4}$ Since $\frac{1}{4}$ is in the interval $(-\infty, \frac{2}{3})$, this solution is valid.

Case 2: $\frac{2}{3} \leq x < 1$

In this case, $x - 1$ is negative and $3x - 2$ is positive, so we have: $|x - 1| = -(x - 1) = 1 - x, \quad |3x - 2| = 3x - 2$ Thus, the equation becomes: $(1 - x) + (3x - 2) = 2$ Simplifying: $1 - x + 3x - 2 = 2$ $2x - 1 = 2$ $2x = 3$ $x = \frac{3}{2}$ Since $\frac{3}{2}$ is not in the interval $\left[\frac{2}{3}, 1\right)$, this solution is not valid.

Case 3: $x \geq 1$

In this case, both $x - 1$ and $3x - 2$ are positive, so we have: $|x - 1| = x - 1, \quad |3x - 2| = 3x - 2$ Thus, the equation becomes: $(x - 1) + (3x - 2) = 2$ Simplifying: $4x - 3 = 2$ $4x = 5$ $x = \frac{5}{4}$ Since $\frac{5}{4}$ is greater than 1, this solution is valid.

Conclusion:

The solutions are $x = \frac{1}{4}$ and $x = \frac{5}{4}$.

Would you like any further details or clarifications on this?

Here are 5 related questions to deepen your understanding:

1. How do you determine where to break the absolute value equation into different cases?
2. Can we solve absolute value inequalities similarly?
3. What happens if both expressions inside the absolute value have the same critical point?
4. How would the solution change if one of the absolute values was negative?
5. Could a different approach like graphing help visualize the solution?

Tip: Always check that each solution satisfies the interval in which it's found to ensure it's valid!