Absolute Value Inequalities 6E6Ae6

1. **Problem statement:** Solve the absolute value inequalities and express the solutions in interval notation. --- ### (a) Solve $\left|\frac{9-2x}{5}\right| + 4 \leq 7$ 2. **Isolate the absolute value:** $$\left|\frac{9-2x}{5}\right| + 4 \leq 7 \implies \left|\frac{9-2x}{5}\right| \leq 7 - 4$$ $$\left|\frac{9-2x}{5}\right| \leq 3$$ 3. **Recall the rule for absolute value inequalities:** If $|A| \leq c$ where $c \geq 0$, then $-c \leq A \leq c$. 4. **Apply the rule:** $$-3 \leq \frac{9-2x}{5} \leq 3$$ 5. **Multiply all parts by 5 to clear the denominator:** $$-3 \times 5 \leq 9 - 2x \leq 3 \times 5$$ $$-15 \leq 9 - 2x \leq 15$$ 6. **Subtract 9 from all parts:** $$-15 - 9 \leq -2x \leq 15 - 9$$ $$-24 \leq -2x \leq 6$$ 7. **Divide all parts by -2, remembering to reverse inequality signs because dividing by a negative:** $$\frac{-24}{-2} \geq x \geq \frac{6}{-2}$$ $$12 \geq x \geq -3$$ 8. **Rewrite the solution in standard interval notation:** $$[-3, 12]$$ --- ### (b) Solve $|2x - 13| - 3 > 4$ 9. **Isolate the absolute value:** $$|2x - 13| - 3 > 4 \implies |2x - 13| > 4 + 3$$ $$|2x - 13| > 7$$ 10. **Recall the rule for absolute value inequalities:** If $|A| > c$ where $c > 0$, then $A < -c$ or $A > c$. 11. **Apply the rule:** $$2x - 13 < -7 \quad \text{or} \quad 2x - 13 > 7$$ 12. **Solve each inequality separately:** - For $2x - 13 < -7$: $$2x < -7 + 13$$ $$2x < 6$$ $$x < 3$$ - For $2x - 13 > 7$: $$2x > 7 + 13$$ $$2x > 20$$ $$x > 10$$ 13. **Write the solution in interval notation:** $$(-\infty, 3) \cup (10, \infty)$$ --- ### Final answers: - (a) $x \in [-3, 12]$ - (b) $x \in (-\infty, 3) \cup (10, \infty)$ --- ### Number line representation: - For (a), shade the segment from $-3$ to $12$ including endpoints. - For (b), shade all values less than $3$ and all values greater than $10$, excluding endpoints.