Michael has $15 and wants to buy a combination of school lunches to feed at least three classmates. A sandwich costs $2, and hot lunch costs $3. This system of inequalities models the scenario: 2x + 3y ≤ 15 x + y ≥ 3 Choose a point in the solution set and interpret what it means in terms of the real-world context. (3 points)

For a point (3, 1), we have;

Michael will buy 3 sandwich and 1 hot meals  and such that the number of meals = 1 + 3 = 4 lunches

The cost of the 4 lunches is 2×3 + 3×1 = $9

Step-by-step explanation:

The given parameters are;

2·x + 3·y ≤ 15

x + y ≥ 3

Where:

x = The number of sandwiches Micheal will buy

y = The number of hot lunches that will be bought by Micheal

We plot the equations to obtain the possible range of values are;

x ≥ 3 - y

2×(3 - y) + 3×y ≤ 15

when x = 0, we have

2×0 + 3×y ≤ 15

y ≤ 5

Similarly we have;

2×x + 3×(3 - x) ≤ 15

when y = 0, we have

2×x + 3×0 ≤ 15

2×x  ≤ 15

x ≤ 10

Therefore, we have the possible solution as follows;

Either x =  and y =

x = 1 and y = 2

x = 2 and y = 1

x = 3 and y = 3

x = 4 and y = 1

x = 5 and y = 1

x = 6 and y = 1

Therefore, where the solution is the point (3, 1), we have;

Michael will buy 3 sandwich and 1 hot meals  and such that the number of meals = 1 + 3 = 4 lunches

The cost of the 4 lunches is 2×3 + 3×1 = $9.

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step-by-step explanation: