To solve the equation $|x + 2| + |x - 3| = 5$, we need to consider the different cases for $x$ because the absolute value function behaves differently depending on the sign of the expression inside.

Step-by-Step Solution:

1. Identify Critical Points: The expressions inside the absolute values change signs at $x = -2$ and $x = 3$. These are the critical points that we need to consider for our cases.

2. Set Up Cases:

   • Case 1: $x < -2$

     • In this range, both expressions inside the absolute values are negative.
     • $|x + 2| = -(x + 2) = -x - 2$
     • $|x - 3| = -(x - 3) = -x + 3$
     • The equation becomes: $-x - 2 + (-x + 3) = 5$ Simplify this: $-2x + 1 = 5$ Solve for $x$: $-2x = 4 \implies x = -2$ Since $x = -2$ is at the boundary of this case, we need to check if it satisfies the equation: $|(-2) + 2| + |(-2) - 3| = |0| + |-5| = 0 + 5 = 5$ This is true, so $x = -2$ is a solution.

   • Case 2: $-2 \leq x < 3$

     • In this range, $x + 2 \geq 0$ and $x - 3 < 0$.
     • $|x + 2| = x + 2$
     • $|x - 3| = -(x - 3) = -x + 3$
     • The equation becomes: $(x + 2) + (-x + 3) = 5$ Simplify this: $x - x + 2 + 3 = 5 \implies 5 = 5$ This is always true, so all $x$ in the interval $[-2, 3)$ are solutions.

   • Case 3: $x \geq 3$

     • In this range, both expressions inside the absolute values are positive.
     • $|x + 2| = x + 2$
     • $|x - 3| = x - 3$
     • The equation becomes: $(x + 2) + (x - 3) = 5$ Simplify this: $2x - 1 = 5$ Solve for $x$: $2x = 6 \implies x = 3$ Since $x = 3$ is at the boundary of this case, we need to check if it satisfies the equation: $|3 + 2| + |3 - 3| = |5| + |0| = 5 + 0 = 5$ This is true, so $x = 3$ is also a solution.

Conclusion:

The solutions to the equation $|x + 2| + |x - 3| = 5$ are:

$\boxed{x = -2, \, [-2, 3), \, x = 3}$

Or, in interval notation: $\boxed{[-2, 3]}$.

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

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Relative Questions:

1. How do you solve absolute value equations with more than two absolute value expressions?
2. What are some strategies for graphing piecewise functions involving absolute values?
3. How would the solution change if the equation was $|x + 2| - |x - 3| = 5$?
4. Can you show a different method for solving absolute value equations, such as graphing?
5. How does solving inequalities involving absolute values differ from solving absolute value equations?

Tip: When solving absolute value equations, always carefully consider the critical points where the expressions inside the absolute values change signs.