1. **Problem Statement:** We are given a budget constraint and minimum purchase requirements for a pizza party. 2. **Define variables:** Let $x$ = number of large pizzas. Let $y$ = number of cases of soda. 3. **Write the system of inequalities:** - Each large pizza costs 10, each case of soda costs 3.5, and the total budget is 90: $$10x + 3.5y \leq 90$$ - The club wants at least 5 pizzas: $$x \geq 5$$ - The club wants at least 3 cases of soda: $$y \geq 3$$ 4. **Check if the club can buy 6 pizzas and 4 cases of soda:** Substitute $x=6$, $y=4$ into the budget inequality: $$10(6) + 3.5(4) = 60 + 14 = 74 \leq 90$$ This satisfies the budget constraint. Also, $6 \geq 5$ and $4 \geq 3$ satisfy the minimum requirements. So, yes, the club can buy 6 pizzas and 4 cases of soda. 5. **Find the maximum number of cases of soda if the club buys 7 pizzas:** Substitute $x=7$ into the budget inequality: $$10(7) + 3.5y \leq 90 \implies 70 + 3.5y \leq 90$$ Subtract 70 from both sides: $$3.5y \leq 20$$ Divide both sides by 3.5: $$y \leq \frac{20}{3.5} = \frac{20}{3.5} = \frac{200}{35} = \frac{40}{7} \approx 5.71$$ Since $y$ must be at least 3 and an integer (assuming whole cases), the maximum number of cases of soda is 5. **Final answers:** - System of inequalities: $$\begin{cases} 10x + 3.5y \leq 90 \\ x \geq 5 \\ y \geq 3 \end{cases}$$ - The club can buy 6 pizzas and 4 cases of soda. - Maximum cases of soda if buying 7 pizzas is 5.