To formulate the spending equation for Darnell, we start with the cost of salad and pizza:
Let:
- \( S \) = number of bowls of salad
- \( P \) = number of slices of pizza
- \( E \) = total amount spent (in dollars)
The equation for total spending \( E \) can be represented as:
\[ E = 5S + 3P \]
We want to rearrange this equation to express \( P \) in terms of \( E \) and \( S \):
\[ 3P = E - 5S \]
Dividing both sides by 3 gives us:
\[ P = \frac{E - 5S}{3} \]
Next, using \( E = 30.00 \) (Darnell's budget), we can rewrite the equation to find \( P \):
\[ P = \frac{30 - 5S}{3} \]
Now, we can fill out the table using this equation by substituting different values for \( S \) (number of salad bowls) and calculating the corresponding \( P \) (number of pizza slices) values.
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When \( S = 0 \): \[ P = \frac{30 - 5(0)}{3} = \frac{30}{3} = 10 \]
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When \( S = 1 \): \[ P = \frac{30 - 5(1)}{3} = \frac{30 - 5}{3} = \frac{25}{3} \approx 8.33 \quad \text{(not a whole number, so can't purchase this} \]
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When \( S = 2 \): \[ P = \frac{30 - 5(2)}{3} = \frac{30 - 10}{3} = \frac{20}{3} \approx 6.67 \quad \text{(not a whole number)} \]
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When \( S = 3 \): \[ P = \frac{30 - 5(3)}{3} = \frac{30 - 15}{3} = \frac{15}{3} = 5 \]
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When \( S = 4 \): \[ P = \frac{30 - 5(4)}{3} = \frac{30 - 20}{3} = \frac{10}{3} \approx 3.33 \quad \text{(not a whole number)} \]
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When \( S = 5 \): \[ P = \frac{30 - 5(5)}{3} = \frac{30 - 25}{3} = \frac{5}{3} \approx 1.67 \quad \text{(not a whole number)} \]
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When \( S = 6 \): \[ P = \frac{30 - 5(6)}{3} = \frac{30 - 30}{3} = \frac{0}{3} = 0 \]
Now, let's summarize the valid combinations into a table:
| Budget (Dollars) | Salad (Bowls) | Pizza (Slices) | |-------------------|----------------|-----------------| | 30.00 | 0 | 10 | | 30.00 | 3 | 5 | | 30.00 | 6 | 0 |
These are the valid combinations Darnell can choose within his budget of $30.00. Note that not every increment of salad will yield a whole number of pizza slices based on the budget constraint.
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