$$ \left\vert x-3\right\vert $$$$ +\left\vert x+3\right\vert=\alpha $$

To solve the equation 
\[
\left| x - 3 \right| + \left| x + 3 \right| = \alpha,
\]
we need to consider the absolute values, which change depending on the value of $ x $. 

Let's analyze the function piece by piece by dividing it into three cases, depending on the intervals of $ x $:

### Case 1: $ x \geq 3 $
In this interval:
- $ |x - 3| = x - 3 $
- $ |x + 3| = x + 3 $

Substitute these into the equation:
\[
(x - 3) + (x + 3) = \alpha.
\]
Simplify:
\[
2x = \alpha \Rightarrow x = \frac{\alpha}{2}.
\]
For this solution to be valid in this interval, $ \frac{\alpha}{2} \geq 3 $, so $ \alpha \geq 6 $.

### Case 2: $ -3 \leq x < 3 $
In this interval:
- $ |x - 3| = 3 - x $
- $ |x + 3| = x + 3 $

Substitute these into the equation:
\[
(3 - x) + (x + 3) = \alpha.
\]
Simplify:
\[
6 = \alpha.
\]
So in this case, $ \alpha = 6 $, which is a fixed value. Any $ x $ in the interval $ -3 \leq x < 3 $ will satisfy this.

### Case 3: $ x < -3 $
In this interval:
- $ |x - 3| = 3 - x $
- $ |x + 3| = -x - 3 $

Substitute these into the equation:
\[
(3 - x) + (-x - 3) = \alpha.
\]
Simplify:
\[
-2x = \alpha \Rightarrow x = -\frac{\alpha}{2}.
\]
For this solution to be valid in this interval, $ -\frac{\alpha}{2} < -3 $, so $ \alpha > 6 $.

### Summary of Solutions
- If $ \alpha = 6 $, then $ x $ can be any value in $ -3 \leq x < 3 $.
- If $ \alpha > 6 $, then $ x = \frac{\alpha}{2} $ or $ x = -\frac{\alpha}{2} $.

Would you like further details or have any questions?

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Here are some related questions to explore this concept further:
1. What happens if we change the constant $3$ to another number?
2. How would the solution change if we had $\left| x - 3 \right| - \left| x + 3 \right| = \alpha$?
3. How do absolute values affect equations in other forms, such as inequalities?
4. Can we visualize the solution to this equation on a graph?
5. How would the solution change if we had three absolute terms?

**Tip:** When solving equations with absolute values, divide the equation into cases based on where the expressions inside absolute values change from positive to negative.

Explore how to solve the absolute value equation |x - 3| + |x + 3| = α by analyzing cases based on values of x. This solution involves piecewise functions, examining three cases where x ≥ 3, -3 ≤ x < 3, and x < -3. Suitable for Grades 9-11.

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