jmc

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, $$x=\boxed{\frac{2a+b}{2}}.$$