We can re-write the given equation as
\[\frac{x - a + a}{x - a} + \frac{x - a - b + a}{x - a - b} = \frac{x - 2a + a}{x - 2a} + \frac{x - b + a}{x - b},\]so
\[1 + \frac{a}{x - a} + 1 + \frac{a}{x - a - b} = 1 + \frac{a}{x - 2a} + 1 + \frac{a}{x - b}.\]Then
\[\frac{1}{x - a} + \frac{1}{x - a - b} = \frac{1}{x - 2a} + \frac{1}{x - b}.\]Combining the fractions on each side, we get
\[\frac{2x - 2a - b}{(x - a)(x - a - b)} = \frac{2x - 2a - b}{(x - 2a)(x - b)}.\]Cross-multiplying, we get
\[(2x - 2a - b)(x - 2a)(x - b) = (2x - 2a - b)(x - a)(x - a - b),\]so
\[(2x - 2a - b)[(x - 2a)(x - b) - (x - a)(x - a - b)] = 0.\]This simplifies to $a(b - a)(2x - 2a - b) = 0.$  Therefore,
\[x = \boxed{\frac{2a + b}{2}}.\]
Solve for $x,$ where
\[\frac{x}{x - a} + \frac{x - b}{x - a - b} = \frac{x - a}{x - 2a} + \frac{x + a - b}{x - b}.\]Assume that $2a > x > b > a > 0.$
{'answer': '\\frac{2a + b}{2}', 'subject': 'Intermediate Algebra', 'level': 5, 'unique_id': 'test/intermediate_algebra/2011.json'}