The inequality you provided is:

\[
|x - 2| \leq x + 1
\]

To solve this, we'll split the absolute value inequality into two cases, based on the definition of absolute value. The absolute value function results in two possible expressions, depending on whether \(x - 2\) is positive or negative.

### Case 1: \(x - 2 \geq 0\), meaning \(x \geq 2\)
For this case, we remove the absolute value, and the inequality becomes:
\[
x - 2 \leq x + 1
\]
Subtracting \(x\) from both sides:
\[
-2 \leq 1
\]
This is always true, so this condition holds for all \(x \geq 2\).

### Case 2: \(x - 2 < 0\), meaning \(x < 2\)
In this case, the absolute value will flip the sign of \(x - 2\), so we have:
\[
-(x - 2) \leq x + 1
\]
This simplifies to:
\[
- x + 2 \leq x + 1
\]
Now, adding \(x\) to both sides:
\[
2 \leq 2x + 1
\]
Subtracting 1 from both sides:
\[
1 \leq 2x
\]
Now, dividing both sides by 2:
\[
\frac{1}{2} \leq x
\]
Thus, for this case, \(x \geq \frac{1}{2}\).

### Final Solution:
The solution to the inequality is the intersection of both cases. From Case 1, we know the inequality holds for \(x \geq 2\). From Case 2, we know that the inequality holds for \(x \geq \frac{1}{2}\).

Therefore, the final solution is:

\[
x \in \left[ \frac{1}{2}, \infty \right)
\]

---

Do you have any questions or would you like more details?

### Related Questions:
1. How would this change if the inequality was \( |x - 2| > x + 1 \)?
2. What if the inequality was \( |x - 2| \leq -x + 1 \)?
3. How does the solution change if the expression inside the absolute value has more terms?
4. How do we solve inequalities with absolute values on both sides?
5. How can you graphically interpret the solution set?

### Tip:
When solving absolute value inequalities, always split them into two cases, one for when the expression inside the absolute value is non-negative, and one for when it is negative.

Learn how to solve the absolute value inequality |x - 2| ≤ x + 1 with a step-by-step approach. This solution covers key algebraic concepts like splitting the inequality into two cases and applying properties of absolute value. Ideal for students in Grades 9-11.

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