### Instruction: Find the sum of all $x$ that satisfy the equation $\frac{-9x}{x^2-1} = \frac{2x}{x+1} - \frac{6}{x-1}.$.### Response: This problem involves rational expressions, so I want to clear the fractions by multiplying both sides by a common denominator.
To find the common denominator, I need to factor the denominators of each term.
I notice that $x^2-1$ is a difference of squares, so I can write it as $(x+1)(x-1)$.
The other two denominators are already in factored form, so I can write them as $(x+1)$ and $(x-1)$.
The common denominator is the product of all the distinct factors, which is $(x+1)(x-1)$.
Multiplying both sides by $(x+1)(x-1)$, I get $-9x = 2x(x-1) - 6(x+1)$.
Expanding and simplifying, I get $-9x = 2x^2 - 8x - 6$.
Subtracting $-9x$ from both sides, I get $0 = 2x^2 + x - 6$.
This is a quadratic equation, which I can solve by factoring, using the quadratic formula, or completing the square.
I notice that this equation has a nice factorization, since $2x^2 + x - 6 = (2x-3)(x+2)$.
So I can set each factor equal to zero and get $x = \frac{3}{2}$ or $x = -2$.
These are the two values of $x$ that satisfy the equation.
To find the sum of these values, I just add them up: $\frac{3}{2} + (-2) = -\frac{1}{2}$.