1. **State the problem:** Solve the equation $$1 - \frac{1}{1 - x} = \frac{2}{x^2 - 1}$$ for $x$. 2. **Rewrite the equation and identify denominators:** Note that $x^2 - 1 = (x-1)(x+1)$. 3. **Find a common denominator:** The denominators are $1 - x$ and $(x-1)(x+1)$. Since $1 - x = -(x-1)$, rewrite $\frac{1}{1-x} = -\frac{1}{x-1}$. 4. **Rewrite the equation:** $$1 + \frac{1}{x-1} = \frac{2}{(x-1)(x+1)}$$ 5. **Multiply both sides by $(x-1)(x+1)$ to clear denominators:** $$ (x-1)(x+1) \times 1 + (x-1)(x+1) \times \frac{1}{x-1} = (x-1)(x+1) \times \frac{2}{(x-1)(x+1)} $$ 6. **Simplify each term:** $$ (x-1)(x+1) + (x+1) = 2 $$ 7. **Expand $(x-1)(x+1)$:** $$ x^2 - 1 + x + 1 = 2 $$ 8. **Simplify:** $$ x^2 + x = 2 $$ 9. **Bring all terms to one side:** $$ x^2 + x - 2 = 0 $$ 10. **Factor the quadratic:** $$ (x+2)(x-1) = 0 $$ 11. **Solve for $x$:** $$ x = -2 \quad \text{or} \quad x = 1 $$ 12. **Check for restrictions:** The original denominators cannot be zero. - $1 - x \neq 0 \Rightarrow x \neq 1$ - $x^2 - 1 \neq 0 \Rightarrow x \neq \pm 1$ 13. **Exclude $x=1$ because it makes denominators zero.** 14. **Final solution:** $$ \boxed{x = -2} $$