1. The problem is to solve the equation $$\left| \frac{9 - x^2}{x - 1} \right| = \frac{x^2 - 9}{x - 1}$$. 2. Notice that $$x^2 - 9 = (x - 3)(x + 3)$$ and $$9 - x^2 = -(x^2 - 9) = -(x - 3)(x + 3)$$. 3. Rewrite the equation as $$\left| \frac{-(x - 3)(x + 3)}{x - 1} \right| = \frac{(x - 3)(x + 3)}{x - 1}$$. 4. Simplify the absolute value: $$\left| -1 \cdot \frac{(x - 3)(x + 3)}{x - 1} \right| = \left| \frac{(x - 3)(x + 3)}{x - 1} \right|$$. 5. So the equation becomes $$\left| \frac{(x - 3)(x + 3)}{x - 1} \right| = \frac{(x - 3)(x + 3)}{x - 1}$$. 6. By definition of absolute value: $$|A| = A$$ means $$A \geq 0$$. 7. Therefore, $$\frac{(x - 3)(x + 3)}{x - 1} \geq 0$$. 8. Find the critical points where numerator or denominator is zero: $$x = 3, x = -3, x = 1$$. 9. Consider the intervals: $$(-\infty, -3), (-3,1), (1,3), (3, \infty)$$ and test sign of the expression in each. 10. Choose test points: - For $$x = -4$$ in $$(-\infty, -3)$$: numerator $$(-4-3)(-4+3) = (-7)(-1) = 7 > 0$$, denominator $$-4 - 1 = -5 < 0$$, so the expression is $$> 0 / < 0 = < 0$$. - For $$x=0$$ in $$(-3, 1)$$: numerator $$(0-3)(0+3) = (-3)(3) = -9 < 0$$, denominator $$0 - 1 = -1 < 0$$, expression $<0 / <0 = > 0$$. - For $$x=2$$ in $$(1,3)$$: numerator $$(2-3)(2+3) = (-1)(5) = -5 < 0$$, denominator $$2 - 1 = 1 > 0$$, expression $<0 / >0 = <0$$. - For $$x=4$$ in $$(3, \infty)$$: numerator $$(4-3)(4+3) = (1)(7) = 7 > 0$$, denominator $$4 - 1 = 3 >0$$, expression $>0 / >0 = >0$$. 11. So the expression is nonnegative on intervals $$[-3,1) \cup [3, \infty)$$. 12. We must exclude $$x=1$$ because denominator is zero. 13. Therefore, the solution set is $$x \in [-3,1) \cup [3, \infty)$$. 14. Final answer: $$\boxed{[-3,1) \cup [3, \infty)}$$.