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

1. Problem Description

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

2. Solution Steps

First, we factor the denominators:

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

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

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

The equation becomes:

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

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

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

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

$4a^2x - 4a^2 + 9a + 9 - 2a^2 - 9x - 9 = 0$

$4a^2x - 6a^2 + 9a - 9x = 0$

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

$x = \frac{6a^2 - 9a}{4a^2 - 9}$

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

If $2a - 3 \neq 0$, then we can simplify to:

$x = \frac{3a}{2a+3}$

We are given that $x > 2$. Therefore,

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

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

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

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

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

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

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

The critical points are $a = -6$ and $a = -\frac{3}{2}$.

We test the intervals:

1. $a < -6$: Choose $a=-7$. $\frac{-7+6}{2(-7)+3} = \frac{-1}{-11} = \frac{1}{11} > 0$. This interval does not satisfy the inequality.

2. $-6 < a < -\frac{3}{2}$: Choose $a=-2$. $\frac{-2+6}{2(-2)+3} = \frac{4}{-1} = -4 < 0$. This interval satisfies the inequality.

3. $a > -\frac{3}{2}$: Choose $a=0$. $\frac{0+6}{2(0)+3} = \frac{6}{3} = 2 > 0$. This interval does not satisfy the inequality.

So we have $-6 < a < -\frac{3}{2}$.

Also we must have $2a-3 \neq 0$, so $a \neq \frac{3}{2}$. Since $\frac{3}{2}$ is not in the interval $(-6, -\frac{3}{2})$, this condition does not affect our result.

3. Final Answer

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