1. **State the problem:** Solve the inequalities: a. $3x - 5 \leq 7$ b. $x^2 + 6 > 42$ 2. **Solve part a:** Start with the inequality: $$3x - 5 \leq 7$$ Add 5 to both sides: $$3x \leq 7 + 5$$ $$3x \leq 12$$ Divide both sides by 3 (positive number, inequality direction stays the same): $$x \leq \frac{12}{3}$$ $$x \leq 4$$ 3. **Solve part b:** Start with the inequality: $$x^2 + 6 > 42$$ Subtract 6 from both sides: $$x^2 > 42 - 6$$ $$x^2 > 36$$ Recall that $x^2 > 36$ means $x$ is either greater than 6 or less than -6: $$x > 6 \quad \text{or} \quad x < -6$$ 4. **Summary of solutions:** - Part a: $x \leq 4$ - Part b: $x < -6$ or $x > 6$ 5. **Explanation:** For part a, we isolate $x$ by performing inverse operations while maintaining the inequality direction because we divide by a positive number. For part b, since $x^2$ is greater than a positive number, $x$ must lie outside the interval between $-6$ and $6$. The number line graph for part a shows a solid circle at 4 and shading to the left, representing $x \leq 4$.