Michael has $8 and wants to buy a combination of cupcakes and fudge to feed at least four siblings. Each cupcake costs $2, and each piece of fudge costs $1.
This system of inequalities models the scenario:
2x + y ≤ 8
x + y ≥ 4
Is the point (8, 10) included in the solution area for the system? Justify your answer mathematically. (3 points)
To determine if the point (8, 10) is included in the solution area for the system, we can substitute x = 8 and y = 10 into both inequalities and see if they are true.
2x + y ≤ 8 becomes 2(8) + 10 ≤ 8, which simplifies to 26 ≤ 8. This is false, so (8, 10) is not included in the solution area for this inequality.
x + y ≥ 4 becomes 8 + 10 ≥ 4, which simplifies to 18 ≥ 4. This is true, so (8, 10) is included in the solution area for this inequality.
Since (8, 10) is not included in the solution area for one of the inequalities but is included in the solution area for the other, it is not included in the solution area for the system of inequalities as a whole. Therefore, Michael cannot buy a combination of cupcakes and fudge that costs exactly $8 and feeds at least four siblings.
To determine if the point (8, 10) is included in the solution area for the system, we need to substitute the values x = 8 and y = 10 into both inequalities and check if the inequalities hold.
Substituting x = 8 and y = 10 into the first inequality:
2(8) + 10 ≤ 8
16 + 10 ≤ 8
26 ≤ 8
Since 26 is not less than or equal to 8, the first inequality is not satisfied.
Next, substituting x = 8 and y = 10 into the second inequality:
8 + 10 ≥ 4
18 ≥ 4
Since 18 is greater than or equal to 4, the second inequality is satisfied.
Therefore, since the point (8, 10) does not satisfy the first inequality, it is not included in the solution area for the system.