To solve the equation \( |x-1| + |2x-3| = |3x-4| \), we will analyze the absolute value expressions by considering different cases based on the critical points where the expressions inside the absolute values change their signs. The critical points are \( x = 1 \), \( x = \frac{3}{2} \) (or \( 1.5 \)), and \( x = \frac{4}{3} \) (or \( 1.333... \)).
### Step 1: Identify the intervals
The critical points divide the number line into the following intervals:
1. \( (-\infty, 1) \)
2. \( [1, \frac{4}{3}) \)
3. \( [\frac{4}{3}, \frac{3}{2}) \)
4. \( [\frac{3}{2}, \infty) \)
### Step 2: Solve for each interval
#### Case 1: \( x < 1 \)
In this interval, all expressions inside the absolute values are negative:
- \( |x-1| = -(x-1) = -x + 1 \)
- \( |2x-3| = -(2x-3) = -2x + 3 \)
- \( |3x-4| = -(3x-4) = -3x + 4 \)
Substituting these into the equation:
\[
-x + 1 - 2x + 3 = -3x + 4
\]
Simplifying:
\[
-3x + 4 = -3x + 4
\]
This is true for all \( x < 1 \). Therefore, the solution in this interval is:
\[
x \in (-\infty, 1)
\]
#### Case 2: \( 1 \leq x < \frac{4}{3} \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |2x-3| = -(2x-3) = -2x + 3 \)
- \( |3x-4| = -(3x-4) = -3x + 4 \)
Substituting these into the equation:
\[
x - 1 - 2x + 3 = -3x + 4
\]
Simplifying:
\[
-x + 2 = -3x + 4
\]
Rearranging gives:
\[
2x = 2 \implies x = 1
\]
Since \( x = 1 \) is within the interval, it is a valid solution.
#### Case 3: \( \frac{4}{3} \leq x < \frac{3}{2} \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |2x-3| = -2x + 3 \)
- \( |3x-4| = 3x - 4 \)
Substituting these into the equation:
\[
x - 1 - 2x + 3 = 3x - 4
\]
Simplifying:
\[
-x + 2 = 3x - 4
\]
Rearranging gives:
\[
4x = 6 \implies x = \frac{3}{2}
\]
Since \( x = \frac{3}{2} \) is at the boundary of the interval, it is a valid solution.
#### Case 4: \( x \geq \frac{3}{2} \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |2x-3| = 2x - 3 \)
- \( |3x-4| = 3x - 4 \)
Substituting these into the equation:
\[
x - 1 + 2x - 3 = 3x - 4
\]
Simplifying:
\[
3x - 4 = 3x - 4
\]
This is true for all \( x \geq \frac{3}{2} \). Therefore, the solution in this interval is:
\[
x \in [\frac{3}{2}, \infty)
\]
### Final Solution
Combining all the valid solutions from each case, we have:
\[
x \in (-\infty, 1) \cup \{1\} \cup \{\frac{3}{2}\} \cup [\frac{3}{2}, \infty)
\]
This can be simplified to:
\[
x \in (-\infty, 1] \cup [\frac{3}{2}, \infty)
\]
