To solve the equation $|x - 2| + |x - 3| = 3$, we need to consider different cases based on the values of $x$ that affect the absolute value expressions. The absolute value function changes behavior depending on whether the argument inside the absolute value is positive or negative.

Step 1: Identify critical points

The expressions $|x - 2|$ and $|x - 3|$ change their behavior at $x = 2$ and $x = 3$. Thus, we will break the solution into three cases based on the value of $x$:

1. $x < 2$
2. $2 \leq x < 3$
3. $x \geq 3$

Step 2: Solve the cases

Case 1: $x < 2$

In this case, both $x - 2$ and $x - 3$ are negative, so we can rewrite the absolute values: $|x - 2| = -(x - 2) = -x + 2 \quad \text{and} \quad |x - 3| = -(x - 3) = -x + 3$ The equation becomes: $(-x + 2) + (-x + 3) = 3$ Simplifying: $-2x + 5 = 3$ $-2x = 3 - 5$ $-2x = -2$ $x = 1$ Since $x = 1$ is less than 2, this solution is valid for this case.

Case 2: $2 \leq x < 3$

In this range, $x - 2 \geq 0$ and $x - 3 < 0$, so we have: $|x - 2| = x - 2 \quad \text{and} \quad |x - 3| = -(x - 3) = -x + 3$ The equation becomes: $(x - 2) + (-x + 3) = 3$ Simplifying: $1 = 3$ This is a contradiction, so there is no solution in this case.

Case 3: $x \geq 3$

In this case, both $x - 2 \geq 0$ and $x - 3 \geq 0$, so we have: $|x - 2| = x - 2 \quad \text{and} \quad |x - 3| = x - 3$ The equation becomes: $(x - 2) + (x - 3) = 3$ Simplifying: $2x - 5 = 3$ $2x = 3 + 5$ $2x = 8$ $x = 4$ Since $x = 4$ is greater than 3, this solution is valid for this case.

Step 3: Combine solutions

From the three cases, the solutions are:

• $x = 1$ from Case 1.
• $x = 4$ from Case 3.

Thus, the solutions to the equation are: $\boxed{x = 1 \text{ or } x = 4}$

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Would you like further details or clarifications on any part of the solution?

Here are some related questions to explore further:

1. How do you solve equations involving absolute values with more than two terms?
2. What is the general approach for solving inequalities with absolute values?
3. How do absolute value equations behave when there is a variable inside multiple absolute values?
4. Can you solve $|x - a| + |x - b| = c$ where $a$, $b$, and $c$ are arbitrary constants?
5. What happens if the right-hand side of an absolute value equation is negative?

Tip: When solving absolute value equations, it is crucial to break the problem down into intervals determined by the points where the expressions inside the absolute values change sign.