1. The problem is to find the solution set of the inequality $$\frac{x-1}{x+2} > 0$$. 2. To solve inequalities involving rational expressions, we find where the numerator and denominator are zero and analyze the sign of the expression in each interval. 3. Set numerator zero: $$x-1=0 \Rightarrow x=1$$. 4. Set denominator zero: $$x+2=0 \Rightarrow x=-2$$ (excluded from domain). 5. The critical points divide the real line into intervals: $$(-\infty, -2), (-2, 1), (1, \infty)$$. 6. Test each interval: - For $$x < -2$$, pick $$x=-3$$: $$\frac{-3-1}{-3+2} = \frac{-4}{-1} = 4 > 0$$ (True). - For $$-2 < x < 1$$, pick $$x=0$$: $$\frac{0-1}{0+2} = \frac{-1}{2} = -0.5 < 0$$ (False). - For $$x > 1$$, pick $$x=2$$: $$\frac{2-1}{2+2} = \frac{1}{4} = 0.25 > 0$$ (True). 7. Since the inequality is strict, points where numerator or denominator is zero are excluded. 8. Therefore, the solution set is $$(-\infty, -2) \cup (1, \infty)$$. Final answer: $$\boxed{(-\infty, -2) \cup (1, \infty)}$$.