To determine which combination of food items and attractions Leo can afford with his $15 budget, we can plug in each option into the expression \(4a + 2b\) (where \(a\) is the number of food items and \(b\) is the number of attractions) and check if the total cost is less than or equal to $15.
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1 food item and 5 attractions: \[ 4(1) + 2(5) = 4 + 10 = 14 \] (Leo can afford this.)
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3 food items and 2 attractions: \[ 4(3) + 2(2) = 12 + 4 = 16 \] (Leo cannot afford this.)
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2 food items and 4 attractions: \[ 4(2) + 2(4) = 8 + 8 = 16 \] (Leo cannot afford this.)
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0 food items and 8 attractions: \[ 4(0) + 2(8) = 0 + 16 = 16 \] (Leo cannot afford this.)
The only combination that Leo can afford is 1 food item and 5 attractions.