Rational Inequality 80Ae84

1. **State the problem:** Solve the inequality $$\frac{2 - 2x}{5x + 3} \leq 0$$ and express the solution set in interval notation. 2. **Rewrite the inequality:** Define $$f(x) = \frac{2 - 2x}{5x + 3}$$ and solve $$f(x) \leq 0$$. 3. **Find zeros and undefined points:** - Numerator zero: $$2 - 2x = 0 \Rightarrow 2x = 2 \Rightarrow x = 1$$ - Denominator zero (undefined): $$5x + 3 = 0 \Rightarrow 5x = -3 \Rightarrow x = -\frac{3}{5}$$ 4. **Determine intervals:** The critical points split the real line into three intervals: - $$(-\infty, -\frac{3}{5})$$ - $$(-\frac{3}{5}, 1)$$ - $$(1, \infty)$$ 5. **Test signs of $$f(x)$$ on each interval:** - For $$x < -\frac{3}{5}$$, pick $$x = -1$$: $$f(-1) = \frac{2 - 2(-1)}{5(-1) + 3} = \frac{2 + 2}{-5 + 3} = \frac{4}{-2} = -2 < 0$$ - For $$-\frac{3}{5} < x < 1$$, pick $$x = 0$$: $$f(0) = \frac{2 - 0}{0 + 3} = \frac{2}{3} > 0$$ - For $$x > 1$$, pick $$x = 2$$: $$f(2) = \frac{2 - 4}{10 + 3} = \frac{-2}{13} < 0$$ 6. **Include points where $$f(x) = 0$$:** At $$x=1$$, numerator zero, so $$f(1) = 0$$, which satisfies $$\leq 0$$. 7. **Exclude points where $$f(x)$$ is undefined:** At $$x = -\frac{3}{5}$$, denominator zero, so $$f(x)$$ undefined, exclude this point. 8. **Write solution set:** $$(-\infty, -\frac{3}{5}) \cup [1, \infty)$$ 9. **Summary table:** | Interval | Sign of $$f(x)$$ | |---|---| | $$(-\infty, -\frac{3}{5})$$ | Negative $$< 0$$ | | $$(-\frac{3}{5}, 1)$$ | Positive $$> 0$$ | | $$(1, \infty)$$ | Negative $$< 0$$ | **Final answer:** $$\boxed{(-\infty, -\frac{3}{5}) \cup [1, \infty)}$$