We are given the equation:

$\frac{2}{x^2 - 1} + \frac{1}{1 - x} = 0$

Step 1: Factor the denominator on the left side

Notice that $x^2 - 1$ is a difference of squares, which can be factored as: $x^2 - 1 = (x - 1)(x + 1)$ Thus, the equation becomes: $\frac{2}{(x - 1)(x + 1)} + \frac{1}{1 - x} = 0$

Step 2: Simplify the second term

Since $1 - x = -(x - 1)$, we rewrite the second term as: $\frac{1}{1 - x} = \frac{-1}{x - 1}$ Now the equation becomes: $\frac{2}{(x - 1)(x + 1)} - \frac{1}{x - 1} = 0$

Step 3: Combine the terms

To combine the two fractions, we need a common denominator. The common denominator is $(x - 1)(x + 1)$. Rewrite the second term with this denominator: $\frac{-1}{x - 1} = \frac{-1(x + 1)}{(x - 1)(x + 1)} = \frac{-(x + 1)}{(x - 1)(x + 1)}$ Now the equation becomes: $\frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} = 0$

Step 4: Combine the numerators

Since the denominators are the same, we combine the numerators: $\frac{2 - (x + 1)}{(x - 1)(x + 1)} = 0$ Simplifying the numerator: $2 - (x + 1) = 2 - x - 1 = 1 - x$ Thus, the equation becomes: $\frac{1 - x}{(x - 1)(x + 1)} = 0$

Step 5: Solve the equation

For this fraction to equal zero, the numerator must be zero: $1 - x = 0$ Solving for $x$: $x = 1$

Step 6: Check for extraneous solutions

We should check if $x = 1$ makes any of the denominators zero. In the original equation, the denominators are $x^2 - 1 = (x - 1)(x + 1)$ and $1 - x$. If $x = 1$, both denominators $(x - 1)$ and $1 - x$ become zero, which leads to division by zero. Therefore, $x = 1$ is not a valid solution.

Conclusion

Since $x = 1$ leads to division by zero, there are no valid solutions to this equation.

Would you like any further details or have questions on this process?

Here are five related questions you might explore:

1. How do you solve rational equations with more than two terms?
2. What are extraneous solutions, and how do they occur?
3. How do you factor differences of squares in polynomial expressions?
4. Can you always simplify the expression $\frac{1}{1 - x} = \frac{-1}{x - 1}$?
5. How do you solve equations involving rational expressions that have higher powers of $x$?

Tip: Always check for extraneous solutions in rational equations by substituting the solutions back into the original denominators.