1. **State the problem:** Two classes, Class A and Class B, sell tickets for a movie night. Each ticket costs 5, and each class pays a 20 fee. Class B sells three times as many tickets as Class A. The total money collected by both classes is 170. We need to find how many tickets Class A sold and how many Class B sold. 2. **Define variables and expressions:** Let $x$ be the number of tickets sold by Class A. Class B sells $3x$ tickets. Total money collected by Class A: $$5x + 20$$ Total money collected by Class B: $$5(3x) + 20 = 15x + 20$$ 3. **Set up the equation for total money collected:** $$5x + 20 + 15x + 20 = 170$$ 4. **Simplify the equation:** $$5x + 15x + 20 + 20 = 170$$ $$20x + 40 = 170$$ 5. **Solve for $x$:** Subtract 40 from both sides: $$20x + \cancel{40} - \cancel{40} = 170 - 40$$ $$20x = 130$$ Divide both sides by 20: $$\frac{20x}{\cancel{20}} = \frac{130}{\cancel{20}}$$ $$x = \frac{130}{20} = 6.5$$ 6. **Interpret the result:** Class A sold 6.5 tickets, which is not possible since tickets must be whole numbers. This suggests either the problem allows fractional tickets or the total money collected is approximate. 7. **Calculate tickets sold by Class B:** $$3x = 3 \times 6.5 = 19.5$$ 8. **Context meaning:** Class A sold about 6 or 7 tickets, and Class B sold about 19 or 20 tickets. The total money collected is 170, including the 20 fee each class paid. **Final answer:** Class A sold 6.5 tickets (approximately 6 or 7), Class B sold 19.5 tickets (approximately 19 or 20).