The problem asks us to solve the equation for $x$: $\frac{4a^2}{3x+3} + \frac{3a+3}{x^2-1} = \frac{2a^2}{3x^2-3} + \frac{3}{x-1}$ Then, we need to determine the real number $a$ such that $x > 2$.

1. Problem Description

The problem asks us to solve the equation for $x$:

$\frac{4a^2}{3x+3} + \frac{3a+3}{x^2-1} = \frac{2a^2}{3x^2-3} + \frac{3}{x-1}$

Then, we need to determine the real number $a$ such that $x > 2$.

2. Solution Steps

First, we factor the denominators:

$3x+3 = 3(x+1)$

$x^2-1 = (x-1)(x+1)$

$3x^2-3 = 3(x^2-1) = 3(x-1)(x+1)$

Substitute these into the equation:

$\frac{4a^2}{3(x+1)} + \frac{3a+3}{(x-1)(x+1)} = \frac{2a^2}{3(x-1)(x+1)} + \frac{3}{x-1}$

Multiply both sides of the equation by $3(x-1)(x+1)$ to eliminate the denominators:

$3(x-1)(x+1) \left[ \frac{4a^2}{3(x+1)} + \frac{3a+3}{(x-1)(x+1)} \right] = 3(x-1)(x+1) \left[ \frac{2a^2}{3(x-1)(x+1)} + \frac{3}{x-1} \right]$

$4a^2(x-1) + 3(3a+3) = 2a^2 + 3(3(x+1))$

$4a^2x - 4a^2 + 9a + 9 = 2a^2 + 9x + 9$

$4a^2x - 4a^2 + 9a = 2a^2 + 9x$

$4a^2x - 9x = 6a^2 - 9a$

$x(4a^2 - 9) = 6a^2 - 9a$

$x(4a^2 - 9) = 3a(2a - 3)$

If $4a^2 - 9 \neq 0$, then

$x = \frac{3a(2a-3)}{4a^2-9} = \frac{3a(2a-3)}{(2a-3)(2a+3)} = \frac{3a}{2a+3}$

Now, we want to find $a$ such that $x > 2$.

$\frac{3a}{2a+3} > 2$

$\frac{3a}{2a+3} - 2 > 0$

$\frac{3a - 2(2a+3)}{2a+3} > 0$

$\frac{3a - 4a - 6}{2a+3} > 0$

$\frac{-a-6}{2a+3} > 0$

$\frac{a+6}{2a+3} < 0$

Consider two cases:

Case 1: $a+6 < 0$ and $2a+3 > 0$

$a < -6$ and $a > -\frac{3}{2}$

This case is impossible.

Case 2: $a+6 > 0$ and $2a+3 < 0$

$a > -6$ and $a < -\frac{3}{2}$

So, $-6 < a < -\frac{3}{2}$

Also, we need to check the condition $4a^2 - 9 \neq 0$, which means $a \neq \pm \frac{3}{2}$. Since $a < -\frac{3}{2}$ in our solution, we have $a \neq -\frac{3}{2}$.

Therefore, the solution is $-6 < a < -\frac{3}{2}$.

3. Final Answer

$-6 < a < -\frac{3}{2}$