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

1. Problem Description

The problem asks us to solve the given equation for $x$ and then find the real number $a$ such that $x > 2$. The equation is

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

2. Solution Steps

First, we factor the denominators:

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

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

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

$x-1 = x-1$

So, the equation can be written as:

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

We multiply both sides of the equation by $3(x-1)(x+1)$ to clear the denominators. We assume $x \ne 1$ and $x \ne -1$:

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

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

$4a^2(x-1) + 9a+9 = 2a^2 + 9(x+1)$

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

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

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

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

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

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

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

We are given $x>2$. So, $\frac{3a}{2a+3} > 2$.

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

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

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

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

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

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

We also need to check the condition $2a-3 \ne 0$, so $a \ne \frac{3}{2}$. Since $\frac{3}{2}$ is not in the range $-6 < a < -\frac{3}{2}$, we don't need to exclude any values.

Also, we need to make sure that $x \ne 1$ and $x \ne -1$.

If $x=1$, $\frac{3a}{2a+3}=1$, so $3a=2a+3$, so $a=3$. This is not in the range $-6 < a < -\frac{3}{2}$.

If $x=-1$, $\frac{3a}{2a+3}=-1$, so $3a=-2a-3$, so $5a=-3$, so $a=-\frac{3}{5}$. This is not in the range $-6 < a < -\frac{3}{2}$.

3. Final Answer

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