1. **State the problem:** Solve the inequality $$\frac{1}{x} + \frac{1}{x+1} > \frac{3}{2}$$ for $x$. 2. **Find a common denominator and combine the left side:** The common denominator is $x(x+1)$. $$\frac{1}{x} + \frac{1}{x+1} = \frac{x+1}{x(x+1)} + \frac{x}{x(x+1)} = \frac{x+1+x}{x(x+1)} = \frac{2x+1}{x(x+1)}$$ 3. **Rewrite the inequality:** $$\frac{2x+1}{x(x+1)} > \frac{3}{2}$$ 4. **Bring all terms to one side:** $$\frac{2x+1}{x(x+1)} - \frac{3}{2} > 0$$ 5. **Find a common denominator for the entire expression:** $$\frac{2(2x+1)}{2x(x+1)} - \frac{3x(x+1)}{2x(x+1)} > 0$$ 6. **Combine the fractions:** $$\frac{2(2x+1) - 3x(x+1)}{2x(x+1)} > 0$$ 7. **Expand numerator:** $$2(2x+1) = 4x + 2$$ $$3x(x+1) = 3x^2 + 3x$$ 8. **Substitute back:** $$\frac{4x + 2 - (3x^2 + 3x)}{2x(x+1)} > 0$$ 9. **Simplify numerator:** $$4x + 2 - 3x^2 - 3x = -3x^2 + x + 2$$ 10. **Rewrite inequality:** $$\frac{-3x^2 + x + 2}{2x(x+1)} > 0$$ 11. **Factor numerator:** $$-3x^2 + x + 2 = -(3x^2 - x - 2)$$ 12. **Factor quadratic inside parentheses:** $$3x^2 - x - 2 = (3x + 2)(x - 1)$$ 13. **Rewrite numerator:** $$-(3x + 2)(x - 1)$$ 14. **Rewrite entire inequality:** $$\frac{-(3x + 2)(x - 1)}{2x(x+1)} > 0$$ 15. **Analyze critical points:** The expression is undefined or zero at $x = -1, 0, -\frac{2}{3}, 1$. 16. **Determine sign intervals:** Test values in intervals divided by these points: - $(-\infty, -1)$ - $(-1, -\frac{2}{3})$ - $(-\frac{2}{3}, 0)$ - $(0, 1)$ - $(1, \infty)$ 17. **Sign analysis:** - For $x < -1$, numerator negative (since $-(3x+2)(x-1)$), denominator positive (since $2x(x+1)$ with both factors negative, product positive), fraction negative. - For $-1 < x < -\frac{2}{3}$, numerator positive, denominator negative, fraction negative. - For $-\frac{2}{3} < x < 0$, numerator negative, denominator negative, fraction positive. - For $0 < x < 1$, numerator negative, denominator positive, fraction negative. - For $x > 1$, numerator positive, denominator positive, fraction positive. 18. **Solution set:** Inequality holds where fraction $> 0$, so $$x \in \left(-\frac{2}{3}, 0\right) \cup (1, \infty)$$ 19. **Exclude points where denominator is zero:** $x \neq -1, 0$. **Final answer:** $$\boxed{\left(-\frac{2}{3}, 0\right) \cup (1, \infty)}$$