To determine which combinations Leo can afford with his $15 budget using the expression \(4a + 2b\), we need to substitute the values of \(a\) (food items) and \(b\) (attractions) for each option provided.
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Zero food items and eight attractions:
- \(a = 0\), \(b = 8\)
- Cost = \(4(0) + 2(8) = 0 + 16 = 16\)
- Total: $16 (not affordable)
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Two food items and four attractions:
- \(a = 2\), \(b = 4\)
- Cost = \(4(2) + 2(4) = 8 + 8 = 16\)
- Total: $16 (not affordable)
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Three food items and two attractions:
- \(a = 3\), \(b = 2\)
- Cost = \(4(3) + 2(2) = 12 + 4 = 16\)
- Total: $16 (not affordable)
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One food item and five attractions:
- \(a = 1\), \(b = 5\)
- Cost = \(4(1) + 2(5) = 4 + 10 = 14\)
- Total: $14 (affordable)
The only combination that Leo can afford with his $15 budget is one food item and five attractions.