Let's define the variables and create the system of linear inequalities based on the problem description:
Variables:
- Let \( x \) be the number of cookies Jace buys.
- Let \( y \) be the number of brownies Jace buys.
Inequalities:
- Jace needs at least 8 cookies: \[ x \geq 8 \]
- Jace needs at least 4 brownies: \[ y \geq 4 \]
- Each cookie costs $3 and each brownie costs $5. Jace can spend a maximum of $60: \[ 3x + 5y \leq 60 \]
Thus, the system of inequalities is:
- \( x \geq 8 \)
- \( y \geq 4 \)
- \( 3x + 5y \leq 60 \)
Now, let's evaluate each potential solution:
-
10 cookies and 5 brownies: \[ x = 10, , y = 5 \]
- Check the inequalities: \[ 10 \geq 8 \quad (\text{true}) \] \[ 5 \geq 4 \quad (\text{true}) \] \[ 3(10) + 5(5) = 30 + 25 = 55 \leq 60 \quad (\text{true}) \]
- Conclusion: This is a viable solution.
-
8.5 cookies and 4 brownies: \[ x = 8.5, , y = 4 \]
- Check the inequalities: \[ 8.5 \geq 8 \quad (\text{true}) \] \[ 4 \geq 4 \quad (\text{true}) \] \[ 3(8.5) + 5(4) = 25.5 + 20 = 45.5 \leq 60 \quad (\text{true}) \]
- Conclusion: This is a viable solution.
-
20 cookies and 0 brownies: \[ x = 20, , y = 0 \]
- Check the inequalities: \[ 20 \geq 8 \quad (\text{true}) \] \[ 0 \geq 4 \quad (\text{false}) \] \[ 3(20) + 5(0) = 60 + 0 = 60 \quad (\text{true}) \]
- Conclusion: This is a nonviable solution due to failing the brownie requirement.
Summary of Solutions:
- 10 cookies and 5 brownies: viable solution
- 8.5 cookies and 4 brownies: viable solution
- 20 cookies and 0 brownies: nonviable solution