To solve the equation $|x - 3| + |x + 3| = a$, we need to analyze the behavior of the absolute values depending on the value of $x$. This involves splitting the equation into cases based on the points where the expressions inside the absolute values change sign, which are $x = 3$ and $x = -3$.

Step 1: Case Analysis

Case 1: $x \geq 3$

In this case, both $x - 3 \geq 0$ and $x + 3 \geq 0$, so we can drop the absolute values: $|x - 3| + |x + 3| = (x - 3) + (x + 3) = 2x$ Thus, for $x \geq 3$, we have: $2x = a$ or $x = \frac{a}{2}$ This solution is valid only if $x \geq 3$, which means $\frac{a}{2} \geq 3$, or equivalently, $a \geq 6$.

Case 2: $-3 \leq x < 3$

In this case, $x - 3 \leq 0$ and $x + 3 \geq 0$. So we rewrite the absolute values as follows: $|x - 3| + |x + 3| = -(x - 3) + (x + 3) = 6$ Thus, for $-3 \leq x < 3$, we have: $6 = a$ This solution is valid only if $-3 \leq x < 3$, so $a$ must be exactly 6.

Case 3: $x < -3$

In this case, both $x - 3 \leq 0$ and $x + 3 \leq 0$. So we rewrite the absolute values as follows: $|x - 3| + |x + 3| = -(x - 3) - (x + 3) = -2x$ Thus, for $x < -3$, we have: $-2x = a$ or $x = -\frac{a}{2}$ This solution is valid only if $x < -3$, which means $-\frac{a}{2} < -3$, or equivalently, $a > 6$.

Summary of Solutions

1. If $a > 6$, then $x = \frac{a}{2}$ or $x = -\frac{a}{2}$.
2. If $a = 6$, then $x$ is any value in the interval $-3 \leq x < 3$.
3. If $a < 6$, there are no solutions.

Would you like a more detailed explanation of any specific case?

Here are some related questions:

1. How would the equation change if $a$ were a function of $x$?
2. What would happen if we modified the values inside the absolute values, say to $|x - 2|$ and $|x + 4|$?
3. How would graphing $|x - 3| + |x + 3|$ help us understand the solution set?
4. What are the implications of absolute value equations in real-world scenarios?
5. How would you approach solving $|x - 3| - |x + 3| = a$?

Tip: Always check for critical points when working with absolute values to simplify cases.