To solve the inequality \( |x+1| - |2x-1| < 1 \), we will need to consider different cases based on the expressions inside the absolute values. The critical points occur when the expressions inside the absolute values are zero, which are at \( x = -1 \) and \( x = \frac{1}{2} \).

### Step 1: Break the number line into intervals based on the critical points:
- \( x < -1 \)
- \( -1 \leq x < \frac{1}{2} \)
- \( x \geq \frac{1}{2} \)

### Step 2: Solve the inequality on each interval.

#### Case 1: \( x < -1 \)
In this region, \( x+1 \) is negative and \( 2x-1 \) is negative, so:
\[
|x+1| = -(x+1) = -x - 1
\]
\[
|2x-1| = -(2x-1) = -2x + 1
\]
Substitute these into the inequality:
\[
-(x+1) - (-(2x-1)) < 1
\]
Simplify:
\[
-x - 1 + 2x - 1 < 1
\]
\[
x - 2 < 1
\]
\[
x < 3
\]
In this case, \( x < 3 \) must also satisfy \( x < -1 \). Therefore, the solution in this region is \( x < -1 \).

#### Case 2: \( -1 \leq x < \frac{1}{2} \)
In this region, \( x+1 \) is non-negative and \( 2x-1 \) is negative, so:
\[
|x+1| = x+1
\]
\[
|2x-1| = -(2x-1) = -2x + 1
\]
Substitute these into the inequality:
\[
(x+1) - (-(2x-1)) < 1
\]
Simplify:
\[
x + 1 + 2x - 1 < 1
\]
\[
3x < 1
\]
\[
x < \frac{1}{3}
\]
So, the solution in this region is \( -1 \leq x < \frac{1}{3} \).

#### Case 3: \( x \geq \frac{1}{2} \)
In this region, both \( x+1 \) and \( 2x-1 \) are non-negative, so:
\[
|x+1| = x+1
\]
\[
|2x-1| = 2x-1
\]
Substitute these into the inequality:
\[
(x+1) - (2x-1) < 1
\]
Simplify:
\[
x + 1 - 2x + 1 < 1
\]
\[
-x + 2 < 1
\]
\[
-x < -1
\]
\[
x > 1
\]
Thus, the solution in this region is \( x > 1 \).

### Step 3: Combine all the solutions
- For \( x < -1 \), the solution is \( x < -1 \).
- For \( -1 \leq x < \frac{1}{2} \), the solution is \( -1 \leq x < \frac{1}{3} \).
- For \( x \geq \frac{1}{2} \), the solution is \( x > 1 \).

Thus, the final solution is:
\[
x \in (-\infty, -1) \cup \left[ -1, \frac{1}{3} \right) \cup (1, \infty)
\]

Would you like more details on any of these steps or a graph of the solution?

Here are some related questions you might find useful:
1. How do absolute value functions impact inequalities?
2. What is the general method for solving absolute value inequalities?
3. How do critical points divide the number line in inequalities?
4. Can you explain how to graph absolute value inequalities?
5. What are some special cases when solving inequalities involving absolute values?

**Tip:** Always check for cases by breaking down absolute value inequalities into multiple intervals!

Learn how to solve the absolute value inequality |x+1| - |2x-1| < 1 with a detailed step-by-step solution. This problem covers key algebraic concepts like absolute value, inequalities, and piecewise functions. Suitable for students in Grades 9-11.

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