1. **State the problem:** Solve the inequality $$\frac{3x - 5}{x + 1} \geq 1$$. 2. **Rewrite the inequality:** Subtract 1 from both sides to get a single rational expression: $$\frac{3x - 5}{x + 1} - 1 \geq 0$$. 3. **Find a common denominator and simplify:** $$\frac{3x - 5}{x + 1} - \frac{x + 1}{x + 1} \geq 0$$ $$\Rightarrow \frac{3x - 5 - (x + 1)}{x + 1} \geq 0$$ $$\Rightarrow \frac{3x - 5 - x - 1}{x + 1} \geq 0$$ $$\Rightarrow \frac{2x - 6}{x + 1} \geq 0$$. 4. **Factor the numerator:** $$\frac{2(x - 3)}{x + 1} \geq 0$$. 5. **Analyze critical points:** The numerator is zero at $$x = 3$$, and the denominator is zero at $$x = -1$$ (excluded from domain). 6. **Determine sign intervals:** - For $$x < -1$$: numerator $$2(x-3) < 0$$, denominator $$x+1 < 0$$, so fraction is $$\frac{negative}{negative} = positive$$. - For $$-1 < x < 3$$: numerator $$2(x-3) < 0$$, denominator $$x+1 > 0$$, fraction is $$\frac{negative}{positive} = negative$$. - For $$x > 3$$: numerator $$2(x-3) > 0$$, denominator $$x+1 > 0$$, fraction is $$\frac{positive}{positive} = positive$$. 7. **Include points where fraction equals zero:** At $$x=3$$, fraction equals zero, so include $$x=3$$. 8. **Exclude points where denominator is zero:** At $$x=-1$$, fraction is undefined, so exclude $$x=-1$$. 9. **Write solution:** $$(-\infty, -1) \cup [3, \infty)$$. **Final answer:** $$x \in (-\infty, -1) \cup [3, \infty)$$.