We can solve the inequality:

$|x+1| - |2x - 1| < 1$

Step 1: Analyze the absolute value expressions

To break down this inequality, we consider the critical points where the expressions inside the absolute values change sign:

1. $|x+1| = x+1$ when $x \geq -1$, and $|x+1| = -(x+1)$ when $x < -1$.
2. $|2x-1| = 2x-1$ when $x \geq \frac{1}{2}$, and $|2x-1| = -(2x-1) = 1-2x$ when $x < \frac{1}{2}$.

These critical points split the number line into three intervals:

1. $x < -1$
2. $-1 \leq x < \frac{1}{2}$
3. $x \geq \frac{1}{2}$

We will now solve the inequality separately for each of these intervals.

Step 2: Solve in each interval

Case 1: $x < -1$

Here, both absolute values will be negative: $|x+1| = -(x+1) = -x-1, \quad |2x-1| = 1 - 2x$ Thus, the inequality becomes: $-(x+1) - (1-2x) < 1$ Simplify: $-x - 1 - 1 + 2x < 1$ $x - 2 < 1$ $x < 3$ Since this is for the region $x < -1$, the solution in this case is $-1 < x < 3$, but the restriction $x < -1$ means we take only $x < -1$.

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

In this interval, $|x+1| = x+1$ and $|2x-1| = 1 - 2x$. The inequality becomes: $(x+1) - (1-2x) < 1$ Simplify: $x + 1 - 1 + 2x < 1$ $3x < 1$ $x < \frac{1}{3}$ Since we are in the interval $-1 \leq x < \frac{1}{2}$, the solution in this case is $-1 \leq x < \frac{1}{3}$.

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

Here, both absolute values will be positive: $|x+1| = x+1, \quad |2x-1| = 2x - 1$ Thus, the inequality becomes: $(x+1) - (2x-1) < 1$ Simplify: $x + 1 - 2x + 1 < 1$ $-x + 2 < 1$ $-x < -1$ $x > 1$ Since $x \geq \frac{1}{2}$, the solution in this case is $x > 1$.

Step 3: Combine the solutions

The combined solution from all intervals is: $x < -1 \quad \text{or} \quad -1 \leq x < \frac{1}{3} \quad \text{or} \quad x > 1$ In interval notation, this is: $(-\infty, -1) \cup [-1, \frac{1}{3}) \cup (1, \infty)$

Final Answer:

$x \in (-\infty, -1) \cup [-1, \frac{1}{3}) \cup (1, \infty)$

Would you like further details on this or have any questions?

Here are some related questions you might find interesting:

1. How do you solve inequalities involving multiple absolute values?
2. What happens when an absolute value inequality has more than two critical points?
3. How do you graph the solution of absolute value inequalities?
4. Can we generalize this method for inequalities with three or more absolute value terms?
5. How would this process change if the inequality sign was $\leq$ instead of $<$?

Tip: Always check critical points where the absolute value expressions change to correctly split the intervals.