Number of solutions to max-min modulus equation

Note that $\max\{x, 2\} - \min\{x, 2\} = |x - 2|$.

The equation becomes: $|x - 2| = |x + 2| - |x - 2| \implies 2|x - 2| = |x + 2|$

Analyzing case by case:

1. $x \ge 2$: $2(x - 2) = x + 2 \implies x = 6$. (Valid, $6 \ge 2$)
2. $-2 \le x < 2$: $2(2 - x) = x + 2 \implies 4 - 2x = x + 2 \implies 3x = 2 \implies x = 2/3$. (Valid, $-2 \le 2/3 < 2$)
3. $x < -2$: $2(2 - x) = -(x + 2) \implies 4 - 2x = -x - 2 \implies x = 6$. (Invalid, since $6 \nless -2$)

Thus, there are 2 distinct real solutions: $x = 6$ and $x = 2/3$.