1. **Problem 11a:** Write inequalities for the ticket sales constraints. - Let $x$ be the number of adult tickets sold. - Let $y$ be the number of child tickets sold. The school wants to make at least 3000 dollars: $$10x + 5y \geq 3000$$ The auditorium can hold at most 400 people: $$x + y \leq 400$$ 2. **Problem 11b:** Graph each inequality separately. - For the money inequality $10x + 5y \geq 3000$, rewrite as: $$5y \geq 3000 - 10x$$ $$y \geq 600 - 2x$$ - For the tickets inequality $x + y \leq 400$, rewrite as: $$y \leq 400 - x$$ 3. **Problem 11c:** Solve the system algebraically by substitution. We solve the system: $$\begin{cases} 10x + 5y = 3000 \\ x + y = 400 \end{cases}$$ From the second equation: $$y = 400 - x$$ Substitute into the first: $$10x + 5(400 - x) = 3000$$ $$10x + 2000 - 5x = 3000$$ $$5x + 2000 = 3000$$ $$5x = 1000$$ $$x = \cancel{\frac{5x}{5}}{\frac{1000}{5}} = 200$$ Then: $$y = 400 - 200 = 200$$ 4. **Problem 12:** Solve the system by substitution. Given: $$\begin{cases} y = 5x - 7 \\ y = 4x - 5 \end{cases}$$ Set equal: $$5x - 7 = 4x - 5$$ $$5x - 4x = -5 + 7$$ $$x = 2$$ Substitute back: $$y = 5(2) - 7 = 10 - 7 = 3$$ Solution: $(x,y) = (2,3)$ 5. **Problem 13:** Solve the system by substitution. Given: $$\begin{cases} y = -3x + 12 \\ y = 2x - 8 \end{cases}$$ Set equal: $$-3x + 12 = 2x - 8$$ $$-3x - 2x = -8 - 12$$ $$-5x = -20$$ $$x = \cancel{\frac{-5x}{-5}}{\frac{-20}{-5}} = 4$$ Substitute back: $$y = 2(4) - 8 = 8 - 8 = 0$$ Solution: $(x,y) = (4,0)$