To determine which combinations of food items (costing $4 each) and attractions (costing $2 each) Leo can afford with his $15 budget, we can analyze the expression \(4a + 2b \leq 15\), where \(a\) is the number of food items and \(b\) is the number of attractions.
We can rearrange and analyze the expression:
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Max food items only: If Leo buys only food items, the maximum number of food items he can afford is: \[ 4a \leq 15 \implies a \leq \frac{15}{4} \implies a \leq 3.75 \] So, the maximum \(a\) is 3 (he can buy 3 food items).
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Max attractions only: If Leo buys only attractions, the maximum number of attractions he can afford is: \[ 2b \leq 15 \implies b \leq \frac{15}{2} \implies b \leq 7.5 \] So, the maximum \(b\) is 7 (he can buy 7 attractions).
Now, let's consider various combinations of \(a\) and \(b\) and check if they fit the budget:
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Combination 1: \(a = 0\), \(b = 7\) \[ 4(0) + 2(7) = 0 + 14 = 14 \quad \text{(within budget)} \]
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Combination 2: \(a = 1\), \(b = 5\) \[ 4(1) + 2(5) = 4 + 10 = 14 \quad \text{(within budget)} \]
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Combination 3: \(a = 2\), \(b = 3\) \[ 4(2) + 2(3) = 8 + 6 = 14 \quad \text{(within budget)} \]
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Combination 4: \(a = 3\), \(b = 0\) \[ 4(3) + 2(0) = 12 + 0 = 12 \quad \text{(within budget)} \]
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Combination 5: \(a = 1\), \(b = 6\) \[ 4(1) + 2(6) = 4 + 12 = 16 \quad \text{(over budget)} \]
These combinations show that Leo can afford various amounts of food items and attractions as long as he stays within the total cost of $15.
Please provide the specific combinations you would like to check, and we can evaluate them!
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