We need to solve the equation $\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2 - b^2}$ for $x$ in terms of $a$ and $b$.

1. Problem Description

We need to solve the equation $\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2 - b^2}$ for $x$ in terms of $a$ and $b$.

2. Solution Steps

First, we can rewrite the equation as:

$\frac{a-x}{b-a} - \frac{a+x}{a+b} = \frac{2ax}{a^2 - b^2}$

Note that $a^2 - b^2 = (a-b)(a+b)$. Also, $b-a = -(a-b)$.

Then, we have:

$\frac{a-x}{-(a-b)} - \frac{a+x}{a+b} = \frac{2ax}{(a-b)(a+b)}$

Multiply both sides by $-(a-b)(a+b)$:

$(a-x)(a+b) + (a+x)(a-b) = -2ax$

Expanding the terms, we have:

$a^2 + ab - ax - bx + a^2 - ab + ax - bx = -2ax$

Simplifying the terms, we have:

$2a^2 - 2bx = -2ax$

$2a^2 = 2bx - 2ax$

$2a^2 = 2x(b-a)$

$a^2 = x(b-a)$

If $b-a \neq 0$, then we can divide by $b-a$ to solve for $x$:

$x = \frac{a^2}{b-a}$

If $b-a = 0$, then $b=a$. In that case, the original equation is undefined.

3. Final Answer

$x = \frac{a^2}{b-a}$ if $b \neq a$.

If $b=a$, the original equation is undefined.