1. **State the problem:** Solve the equation $$\frac{12}{x^2-1} = 1$$ for $x$. 2. **Recall the formula and rules:** The denominator $x^2-1$ can be factored using the difference of squares formula: $$x^2-1 = (x-1)(x+1)$$. 3. **Rewrite the equation:** $$\frac{12}{(x-1)(x+1)} = 1$$ 4. **Multiply both sides by the denominator to clear the fraction:** $$12 = 1 \times (x-1)(x+1)$$ 5. **Expand the right side:** $$(x-1)(x+1) = x^2 - 1$$ So, $$12 = x^2 - 1$$ 6. **Add 1 to both sides:** $$12 + 1 = x^2$$ $$13 = x^2$$ 7. **Take the square root of both sides:** $$x = \pm \sqrt{13}$$ 8. **Check for restrictions:** The original denominator $x^2 - 1$ cannot be zero, so $x \neq \pm 1$. Since $\pm \sqrt{13}$ are not $\pm 1$, both solutions are valid. **Final answer:** $$x = \pm \sqrt{13}$$