1. **Problem 1:** Nayeon won 40 lollipops and gave 2 to every student in her math class. She has at least 7 left. 2. **Write the inequality:** Let $x$ be the number of students. The total lollipops given away is $2x$. Since she has at least 7 left, the inequality is: $$40 - 2x \geq 7$$ 3. **Solve the inequality:** $$40 - 2x \geq 7$$ Subtract 40 from both sides: $$\cancel{40} - 2x - \cancel{40} \geq 7 - 40$$ $$-2x \geq -33$$ Divide both sides by $-2$ and reverse the inequality sign because dividing by a negative number reverses inequality: $$x \leq \frac{-33}{-2}$$ $$x \leq 16.5$$ Since $x$ must be a whole number, the maximum number of students is 16. --- 4. **Problem 2:** More than 450 students went on a field trip. Ten vans were filled, and 5 more students traveled in a car. 5. **Write the inequality:** Let $y$ be the number of students per van. Total students in vans is $10y$, plus 5 more in a car, total more than 450: $$10y + 5 > 450$$ 6. **Solve the inequality:** $$10y + 5 > 450$$ Subtract 5 from both sides: $$\cancel{10y} + 5 - 5 > 450 - 5$$ $$10y > 445$$ Divide both sides by 10: $$y > \frac{445}{10}$$ $$y > 44.5$$ Since $y$ must be whole, minimum number of students per van is 45. --- 7. **Problem 3:** Maloi spent 26 pesos on a magazine and five stationaries. The magazine costs 4. 8. **Write the inequality:** Let $s$ be the cost of each stationary. Total cost is magazine plus 5 stationaries, which should not exceed 26: $$4 + 5s \leq 26$$ 9. **Solve the inequality:** $$4 + 5s \leq 26$$ Subtract 4 from both sides: $$\cancel{4} + 5s - \cancel{4} \leq 26 - 4$$ $$5s \leq 22$$ Divide both sides by 5: $$s \leq \frac{22}{5}$$ $$s \leq 4.4$$ So, the maximum cost of each stationary is 4.4 pesos. **Final answers:** 1A. $40 - 2x \geq 7$ 1B. $x \leq 16$ 2A. $10y + 5 > 450$ 2B. $y > 44.5$, minimum $y=45$ 3A. $4 + 5s \leq 26$ 3B. $s \leq 4.4$