1. **State the problem:** We have a bus fare initially at 5.50 per person, transporting 800 people daily. For every decrease of 0.05 in fare, 10 more people ride the bus. We define $x$ as the number of 0.05 decreases. 2. **Find the bus fare expression:** Each decrease reduces fare by 0.05, so the fare after $x$ decreases is: $$\text{Fare} = 5.50 - 0.05x$$ 3. **Find the number of people expression:** Each decrease increases riders by 10, so the number of people is: $$\text{People} = 800 + 10x$$ 4. **Find the daily revenue expression:** Revenue is fare times number of people: $$R(x) = (5.50 - 0.05x)(800 + 10x)$$ Expanding: $$R(x) = 5.50 \times 800 + 5.50 \times 10x - 0.05x \times 800 - 0.05x \times 10x$$ $$= 4400 + 55x - 40x - 0.5x^2$$ $$= 4400 + 15x - 0.5x^2$$ 5. **Find $x$ for revenue 4500:** Set $R(x) = 4500$: $$4400 + 15x - 0.5x^2 = 4500$$ Rearranged: $$-0.5x^2 + 15x + 4400 - 4500 = 0$$ $$-0.5x^2 + 15x - 100 = 0$$ Multiply both sides by -2 to clear decimals: $$x^2 - 30x + 200 = 0$$ Use quadratic formula: $$x = \frac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 200}}{2} = \frac{30 \pm \sqrt{900 - 800}}{2} = \frac{30 \pm \sqrt{100}}{2}$$ $$x = \frac{30 \pm 10}{2}$$ Two solutions: $$x = \frac{40}{2} = 20$$ $$x = \frac{20}{2} = 10$$ 6. **Domain for $x$:** - Fare must be non-negative: $5.50 - 0.05x \geq 0 \Rightarrow x \leq 110$ - Number of people must be non-negative: $800 + 10x \geq 0 \Rightarrow x \geq -80$ - Since $x$ counts decreases, it must be a whole number $\geq 0$ Thus, domain is: $$x \in \{0,1,2,\ldots,110\}$$ **Final answers:** - Fare: $5.50 - 0.05x$ - People: $800 + 10x$ - Revenue: $4400 + 15x - 0.5x^2$ - Revenue 4500 at $x=10$ or $x=20$ - Domain: $x$ integer, $0 \leq x \leq 110$