<s>[INST] Solve for $x$: $\frac{x-a+b}{x-a}+\frac{x-b}{x-2b}=\frac{x}{x-b}+\frac{x-a}{x-a-b}$ Solve for $x$: $\dfrac{x-a+b}{x-a}+\dfrac{x-b}{x-2b}=\dfrac{x}{x-b}+\dfrac{x-a}{x-a-b}$
$\Rightarrow \dfrac{(x-a)+b}{x-a}+\dfrac{(x-2b)+b}{x-2b}=\dfrac{(x-b)+b}{x-b}+\dfrac{(x-a-b)+b}{x-a-b} \ \ \ ...(1)$
$\Rightarrow 1+\dfrac{b}{x-a}+1+\dfrac{b}{x-2b}=1+\dfrac{b}{x-b}+1+\dfrac{b}{x-a-b} \ \ \ ...(2)$
$\Rightarrow \dfrac{b}{x-a}+\dfrac{b}{x-2b}=\dfrac{b}{x-b}+\dfrac{b}{x-a-b} \ \ \ ...(3)$
$\Rightarrow \dfrac{b(x-2b)+b(x-a)}{(x-a)(x-2b)}=\dfrac{b(x-a-b)+b(x-b)}{(x-b)(x-a-b)} \ \ \ ...(4)$
$\Rightarrow \dfrac{b(2x-2b-a)}{(x-a)(x-2b)}=\dfrac{b(2x-2b-a)}{(x-b)(x-a-b)} \ \ \ ...(5)$
$\Rightarrow b(2x-2b-a)(x-b)(x-a-b)=b(2x-2b-a)(x-a)(x-2b)\ \ \ ...(6)$
$\Rightarrow (x-a)(x-2b)=(x-b)(x-a-b) \ \ \ ...(7)$
$\Rightarrow x^2-2bx-ax+2ab=x^2-ax-bx-bx+ab+b^2 \ \ \ ...(8)$
$\Rightarrow 2ab=ab+b^2 \ \ \ ...(9)$
$\Rightarrow a=b \ \ \ $
So the result I got does not solve for $x$. Clearly something went wrong. I can't see it however because to me everything feels correct computationally. Thanks for the assistance.
 [/INST] From (6) to (7) you lost the solution $x=(2b+a)/2.$ And from (8) to (9) you lost $b=0.$ You must present the set of solutions $x$ in the case $b=0$ or $a$, and in the case $b\ne0,a.$ And in each case, exclude the values which would make some of the 4 denominators equal to $0.$
So you need to split in three cases:

*

*If $b\in\{0,a\},$ every $x$ distinct from $a$ and $2b$ is a solution.

*If $b\notin\{0,a\}$ and $a\notin\{0,2b\},$ the unique solution is $x=b+\frac a2.$

*If $b\ne0$ and $a\in\{0,2b\},$ there is no solution.

</s>