1. **State the problem:** Solve the inequality $$\left|\frac{1 - 3X}{5X - 2}\right| \geq 10$$. 2. **Recall the rule for absolute value inequalities:** For any expression $A$, $$|A| \geq k$$ means $$A \geq k \quad \text{or} \quad A \leq -k$$. 3. **Apply this to our problem:** $$\frac{1 - 3X}{5X - 2} \geq 10 \quad \text{or} \quad \frac{1 - 3X}{5X - 2} \leq -10$$. 4. **Solve the first inequality:** $$\frac{1 - 3X}{5X - 2} \geq 10$$ Multiply both sides by $5X - 2$, but remember to consider the sign of $5X - 2$ because it affects the inequality direction. - Case 1: $5X - 2 > 0 \Rightarrow X > \frac{2}{5}$ Multiply without changing inequality: $$1 - 3X \geq 10(5X - 2)$$ $$1 - 3X \geq 50X - 20$$ Bring all terms to one side: $$1 - 3X - 50X + 20 \geq 0$$ $$21 - 53X \geq 0$$ $$-53X \geq -21$$ Divide by -53 (negative, flip inequality): $$X \leq \frac{21}{53}$$ Combine with domain $X > \frac{2}{5} = 0.4$: $$\frac{2}{5} < X \leq \frac{21}{53} \approx 0.396$$ No values satisfy this because $\frac{21}{53} < \frac{2}{5}$ is false. - Case 2: $5X - 2 < 0 \Rightarrow X < \frac{2}{5}$ Multiply and flip inequality: $$1 - 3X \leq 10(5X - 2)$$ $$1 - 3X \leq 50X - 20$$ $$1 - 3X - 50X + 20 \leq 0$$ $$21 - 53X \leq 0$$ $$-53X \leq -21$$ $$X \geq \frac{21}{53}$$ Combine with domain $X < \frac{2}{5}$: $$X \geq \frac{21}{53} \quad \text{and} \quad X < \frac{2}{5}$$ Since $\frac{21}{53} \approx 0.396$ and $\frac{2}{5} = 0.4$, this interval is: $$0.396 \leq X < 0.4$$ 5. **Solve the second inequality:** $$\frac{1 - 3X}{5X - 2} \leq -10$$ Again consider sign of denominator. - Case 1: $5X - 2 > 0 \Rightarrow X > \frac{2}{5}$ Multiply without flipping: $$1 - 3X \leq -10(5X - 2)$$ $$1 - 3X \leq -50X + 20$$ $$1 - 3X + 50X - 20 \leq 0$$ $$-19 + 47X \leq 0$$ $$47X \leq 19$$ $$X \leq \frac{19}{47}$$ Combine with domain $X > \frac{2}{5} = 0.4$: $$\frac{2}{5} < X \leq \frac{19}{47} \approx 0.404$$ - Case 2: $5X - 2 < 0 \Rightarrow X < \frac{2}{5}$ Multiply and flip inequality: $$1 - 3X \geq -10(5X - 2)$$ $$1 - 3X \geq -50X + 20$$ $$1 - 3X + 50X - 20 \geq 0$$ $$-19 + 47X \geq 0$$ $$47X \geq 19$$ $$X \geq \frac{19}{47}$$ Combine with domain $X < \frac{2}{5}$: $$X \geq \frac{19}{47} \quad \text{and} \quad X < \frac{2}{5}$$ Since $\frac{19}{47} \approx 0.404$ and $\frac{2}{5} = 0.4$, no values satisfy this. 6. **Domain restriction:** The denominator $5X - 2 \neq 0 \Rightarrow X \neq \frac{2}{5}$. 7. **Combine all valid intervals:** From first inequality: $0.396 \leq X < 0.4$ From second inequality: $0.4 < X \leq 0.404$ 8. **Final solution:** $$\boxed{\left[\frac{21}{53}, \frac{2}{5}\right) \cup \left(\frac{2}{5}, \frac{19}{47}\right]}$$ This means $X$ is in the intervals $[0.396, 0.4)$ or $(0.4, 0.404]$, excluding $X=\frac{2}{5}$ where denominator is zero.