The problem is to solve the equation for $x$, where $a$, $b$, and $c$ are real numbers: $\frac{a}{c} - \frac{ax}{cx-1} = \frac{c}{a} - \frac{cx}{ax-1}$.

1. Problem Description

The problem is to solve the equation for $x$, where $a$, $b$, and $c$ are real numbers:

$\frac{a}{c} - \frac{ax}{cx-1} = \frac{c}{a} - \frac{cx}{ax-1}$.

2. Solution Steps

The given equation is:

$\frac{a}{c} - \frac{ax}{cx-1} = \frac{c}{a} - \frac{cx}{ax-1}$

First, we rearrange the terms:

$\frac{a}{c} - \frac{c}{a} = \frac{ax}{cx-1} - \frac{cx}{ax-1}$

$\frac{a^2 - c^2}{ac} = \frac{ax(ax-1) - cx(cx-1)}{(cx-1)(ax-1)}$

$\frac{a^2 - c^2}{ac} = \frac{a^2x^2 - ax - c^2x^2 + cx}{acx^2 - ax - cx + 1}$

$\frac{a^2 - c^2}{ac} = \frac{(a^2 - c^2)x^2 - (a-c)x}{acx^2 - (a+c)x + 1}$

If $a=c$, then the left side is

0. Also, the numerator of the right side is $(a^2-c^2)x^2 - (a-c)x = 0$. Then $x=0$ is a solution. If $a=c$, then the equation becomes $\frac{a}{a} - \frac{ax}{ax-1} = \frac{a}{a} - \frac{ax}{ax-1}$ which simplifies to $1 - \frac{ax}{ax-1} = 1 - \frac{ax}{ax-1}$, meaning any $x$ such that $ax-1 \ne 0$ is a solution. In other words, any $x \ne \frac{1}{a}$ is a solution.

If $a \ne c$ and $a \ne -c$, then $a^2 - c^2 \ne 0$. Then

$\frac{a+c}{ac} = \frac{(a+c)x^2 - x}{acx^2 - (a+c)x + 1}$. Since $a+c \ne 0$:

$\frac{1}{ac} = \frac{x^2 - \frac{1}{a+c}x}{acx^2 - (a+c)x + 1}$

$acx^2 - (a+c)x + 1 = acx^2 - \frac{ac}{a+c}x$

$-(a+c)x + 1 = - \frac{ac}{a+c}x$

$1 = (a+c - \frac{ac}{a+c})x$

$1 = (\frac{(a+c)^2 - ac}{a+c})x$

$1 = (\frac{a^2+2ac+c^2 - ac}{a+c})x$

$1 = (\frac{a^2+ac+c^2}{a+c})x$

$x = \frac{a+c}{a^2 + ac + c^2}$

If $a = -c$, then $a^2-c^2 = 0$ and $a-c = -2c \ne 0$. So the equation becomes $0 = \frac{-2cx}{acx^2 - 0x + 1}$, which means $x = 0$.

3. Final Answer

If $a=c$, any $x \ne \frac{1}{a}$ is a solution.

If $a = -c$, $x = 0$.

If $a^2 \ne c^2$, then $x = \frac{a+c}{a^2 + ac + c^2}$.