We are tasked with solving the inequality $|x + 1| - |x + 2| \leq 2$. To solve this, we need to break the absolute values into different cases based on the critical points where the expressions inside the absolute values change sign.

Step 1: Find critical points

The critical points occur where the arguments of the absolute values are equal to zero. These are: $x + 1 = 0 \quad \Rightarrow \quad x = -1$ $x + 2 = 0 \quad \Rightarrow \quad x = -2$ These critical points divide the number line into three regions:

• Region 1: $x < -2$
• Region 2: $-2 \leq x < -1$
• Region 3: $x \geq -1$

Step 2: Solve in each region

Region 1: $x < -2$

In this region, both $x + 1$ and $x + 2$ are negative, so: $|x + 1| = -(x + 1) \quad \text{and} \quad |x + 2| = -(x + 2)$ Substitute into the inequality: $-(x + 1) - (-(x + 2)) \leq 2$ Simplify: $-(x + 1) + (x + 2) \leq 2$ $-x - 1 + x + 2 \leq 2$ 1 \leq 2 \quad \text{(true for all x)} Thus, in this region, the inequality holds for all $x < -2$.

Region 2: $-2 \leq x < -1$

Here, $x + 1$ is negative and $x + 2$ is non-negative, so: $|x + 1| = -(x + 1) \quad \text{and} \quad |x + 2| = x + 2$ Substitute into the inequality: $-(x + 1) - (x + 2) \leq 2$ Simplify: $-x - 1 - x - 2 \leq 2$ $-2x - 3 \leq 2$ $-2x \leq 5$ $x \geq -\frac{5}{2}$ Since $-\frac{5}{2} \approx -2.5$, this condition is automatically satisfied in this region because $x \geq -2$. Therefore, the inequality holds for $-2 \leq x < -1$.

Region 3: $x \geq -1$

In this region, both $x + 1$ and $x + 2$ are non-negative, so: $|x + 1| = x + 1 \quad \text{and} \quad |x + 2| = x + 2$ Substitute into the inequality: $(x + 1) - (x + 2) \leq 2$ Simplify: $x + 1 - x - 2 \leq 2$ -1 \leq 2 \quad \text{(true for all x)} Thus, in this region, the inequality holds for all $x \geq -1$.

Step 3: Combine the results

The inequality holds for all $x$, as it is true in each of the three regions.

Final Answer:

$\boxed{x \in \mathbb{R}}$ The inequality is satisfied for all real numbers $x$.

Would you like more details on any step or have any further questions?

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Related Questions:

1. What happens if we change the inequality to $|x+1| - |x+2| > 2$?
2. How do absolute values affect inequalities in general?
3. What if the terms inside the absolute values were more complex (e.g., quadratics)?
4. How can you graph the solution for an inequality involving absolute values?
5. Can this technique be used for inequalities involving more absolute values?

Tip: Always start by identifying the critical points of absolute values to split your inequality into manageable cases.