To solve the equation \( |x+2| + |x-3| = 5 \), we need to consider the different cases where the absolute value expressions change, based on the critical points \( x = -2 \) and \( x = 3 \).

### Case 1: \( x < -2 \)
- In this interval, \( x+2 < 0 \) and \( x-3 < 0 \).
- So, \( |x+2| = -(x+2) = -x-2 \) and \( |x-3| = -(x-3) = -x+3 \).

The equation becomes:
\[
-x-2 + -x+3 = 5
\]
\[
-2x + 1 = 5
\]
\[
-2x = 4 \quad \Rightarrow \quad x = -2
\]
However, \( x = -2 \) is not in the interval \( x < -2 \), so there is no solution in this case.

### Case 2: \( -2 \leq x < 3 \)
- In this interval, \( x+2 \geq 0 \) and \( x-3 < 0 \).
- So, \( |x+2| = x+2 \) and \( |x-3| = -(x-3) = -x+3 \).

The equation becomes:
\[
x+2 + -x+3 = 5
\]
\[
5 = 5
\]
This is always true for any \( x \) in the interval \( -2 \leq x < 3 \).

### Case 3: \( x \geq 3 \)
- In this interval, \( x+2 \geq 0 \) and \( x-3 \geq 0 \).
- So, \( |x+2| = x+2 \) and \( |x-3| = x-3 \).

The equation becomes:
\[
x+2 + x-3 = 5
\]
\[
2x - 1 = 5
\]
\[
2x = 6 \quad \Rightarrow \quad x = 3
\]
\( x = 3 \) is in the interval \( x \geq 3 \), so this solution is valid.

### Final Answer:
The solutions to the equation are \( x = -2 \) and \( x = 3 \).

Would you like more details or have any questions about this solution?

Here are five related questions you might consider:
1. How would the solution change if the equation were \( |x+2| + |x-3| = 7 \)?
2. What happens if the equation is \( |x+2| - |x-3| = 1 \)?
3. How can we graph the function \( |x+2| + |x-3| \) and what does it look like?
4. What are the key properties of absolute value functions that were used here?
5. How would you solve \( |2x+1| + |x-4| = 6 \)?

**Tip:** Always check if a solution lies within the interval assumed during its calculation, as not all solutions derived from algebraic manipulation might be valid for the original equation.

Learn how to solve the absolute value equation |x+2| + |x-3| = 5 step-by-step with detailed solutions. Understand the behavior of absolute value functions and solve for critical points. Suitable for high school students in Grades 9-12.

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