1. **State the problem:** Solve the inequality $$\frac{2x - 5}{x + 3} \geq 1$$. 2. **Rewrite the inequality:** Subtract 1 from both sides to get a single rational expression: $$\frac{2x - 5}{x + 3} - 1 \geq 0$$ 3. **Find a common denominator and combine:** $$\frac{2x - 5}{x + 3} - \frac{x + 3}{x + 3} \geq 0$$ $$\Rightarrow \frac{2x - 5 - (x + 3)}{x + 3} \geq 0$$ 4. **Simplify the numerator:** $$2x - 5 - x - 3 = x - 8$$ So the inequality becomes: $$\frac{x - 8}{x + 3} \geq 0$$ 5. **Determine critical points:** The numerator is zero at $$x = 8$$, and the denominator is zero at $$x = -3$$ (which is a vertical asymptote and excluded from the domain). 6. **Analyze intervals:** The number line is divided into three intervals by the critical points: - $$(-\infty, -3)$$ - $$(-3, 8)$$ - $$(8, \infty)$$ 7. **Test each interval:** - For $$x < -3$$, pick $$x = -4$$: $$\frac{-4 - 8}{-4 + 3} = \frac{-12}{-1} = 12 \geq 0$$ (True) - For $$-3 < x < 8$$, pick $$x = 0$$: $$\frac{0 - 8}{0 + 3} = \frac{-8}{3} = -\frac{8}{3} < 0$$ (False) - For $$x > 8$$, pick $$x = 9$$: $$\frac{9 - 8}{9 + 3} = \frac{1}{12} > 0$$ (True) 8. **Include points where numerator is zero:** At $$x = 8$$, the expression equals zero, so include $$x = 8$$. 9. **Exclude points where denominator is zero:** At $$x = -3$$, the expression is undefined, so exclude $$x = -3$$. 10. **Final solution set:** $$x \leq -3 \text{ (excluding } x = -3\text{) or } x \geq 8$$ Since $$x = -3$$ is excluded, the solution is: $$(-\infty, -3) \cup [8, \infty)$$ **Answer choice matching this is:** c. $$\{x \leq -3, x \geq 8\}$$.