Question
there are two lemonade stands that sell lemonade called stand A and stand B. The total cost for lemonade at Stand A is $15 (C) plus $2 per ice (I). The total cost for lemonade at Stand B is $17 (C) plus $4.50 per ice (I). Which Lemonade stand is cheaper. Lets use C for the total cost and I for the cost per ice.
Systems of linear equations can be used to solve many, but not all, real-world problems. When you chose the problem for your journal entry, how did you know that this problem could be solved using a system of linear equations? What specific characteristics does your problem have that made it a good choice? List at least three reasons why the problem you described in your journal prompt can be solved using a system of linear equations.
Systems of linear equations can be used to solve many, but not all, real-world problems. When you chose the problem for your journal entry, how did you know that this problem could be solved using a system of linear equations? What specific characteristics does your problem have that made it a good choice? List at least three reasons why the problem you described in your journal prompt can be solved using a system of linear equations.
Answers
GPT-4o mini
To analyze the costs at the two lemonade stands, we can establish equations based on the given information.
### Cost Equations
1. **Stand A:**
$$ C_A = 15 + 2I $$
- Where \( C_A \) is the total cost at Stand A, and \( I \) is the number of ice units.
2. **Stand B:**
$$ C_B = 17 + 4.5I $$
- Where \( C_B \) is the total cost at Stand B.
### Finding the Cheaper Option
To determine which stand is cheaper, we can set the two equations for costs equal to each other and solve for \( I \):
$$ 15 + 2I = 17 + 4.5I $$
Subtracting \( 2I \) from both sides gives us:
$$ 15 = 17 + 2.5I $$
Next, subtracting 17 from both sides leads to:
$$ -2 = 2.5I $$
Now, divide by 2.5:
$$ I = -\frac{2}{2.5} = -0.8 $$
Since \( I \) cannot be negative (as it represents the number of ice units), it indicates that Stand A is cheaper for all non-negative values of \( I \).
### Conclusion
- **Stand A’s cost:**
- If \( I = 0 \) (no ice), then \( C_A = 15 \).
- **Stand B’s cost:**
- If \( I = 0 \), then \( C_B = 17 \).
Thus, Stand A is cheaper than Stand B for any number of ice units.
### Characteristic Reasons for Using a System of Linear Equations
This problem can be effectively solved using a system of linear equations for several reasons:
1. **Two Variables:**
- There are two independent variables (the cost from each stand) that can be compared based on the number of ice units.
2. **Linear Relationships:**
- The costs at each stand can be described using linear equations, where the total cost increases linearly with the number of ice units added.
3. **Intersection Point:**
- The potential solution (where both costs are equal) can be found by setting the two equations equal to each other, indicating a point of intersection in a graphical representation.
Overall, the problem is a clear real-world scenario where comparing costs leads to establishing relations that fit the criteria for a system of linear equations.
### Cost Equations
1. **Stand A:**
$$ C_A = 15 + 2I $$
- Where \( C_A \) is the total cost at Stand A, and \( I \) is the number of ice units.
2. **Stand B:**
$$ C_B = 17 + 4.5I $$
- Where \( C_B \) is the total cost at Stand B.
### Finding the Cheaper Option
To determine which stand is cheaper, we can set the two equations for costs equal to each other and solve for \( I \):
$$ 15 + 2I = 17 + 4.5I $$
Subtracting \( 2I \) from both sides gives us:
$$ 15 = 17 + 2.5I $$
Next, subtracting 17 from both sides leads to:
$$ -2 = 2.5I $$
Now, divide by 2.5:
$$ I = -\frac{2}{2.5} = -0.8 $$
Since \( I \) cannot be negative (as it represents the number of ice units), it indicates that Stand A is cheaper for all non-negative values of \( I \).
### Conclusion
- **Stand A’s cost:**
- If \( I = 0 \) (no ice), then \( C_A = 15 \).
- **Stand B’s cost:**
- If \( I = 0 \), then \( C_B = 17 \).
Thus, Stand A is cheaper than Stand B for any number of ice units.
### Characteristic Reasons for Using a System of Linear Equations
This problem can be effectively solved using a system of linear equations for several reasons:
1. **Two Variables:**
- There are two independent variables (the cost from each stand) that can be compared based on the number of ice units.
2. **Linear Relationships:**
- The costs at each stand can be described using linear equations, where the total cost increases linearly with the number of ice units added.
3. **Intersection Point:**
- The potential solution (where both costs are equal) can be found by setting the two equations equal to each other, indicating a point of intersection in a graphical representation.
Overall, the problem is a clear real-world scenario where comparing costs leads to establishing relations that fit the criteria for a system of linear equations.