The party will cost $150, plus $10.50 per child and $5.50 per adult. There needs to be at least one adult and one child at the party, and the cost cannot exceed $300. (14.286, 0) is a viable solution for the above situation. Consider the ordered pair as (c,a) where c represents number of children and a represents number of adults at the party. (1 point) True False

To evaluate whether the ordered pair (14.286, 0) is a viable solution for the given cost situation, we need to analyze the constraints provided:

1. The total cost equation is given by: \[ \text{Total Cost} = 150 + 10.50c + 5.50a \] where \( c \) is the number of children and \( a \) is the number of adults.

2. The conditions specify:

   • There must be at least one adult (\( a \geq 1 \)).
   • There must be at least one child (\( c \geq 1 \)).
   • The total cost cannot exceed $300.

Substituting the ordered pair (14.286, 0):

• \( c = 14.286 \) (approximately 14 children)
• \( a = 0 \) (0 adults)

Now verifying the conditions:

• Total cost: \[ \text{Total Cost} = 150 + 10.50(14.286) + 5.50(0) = 150 + 150.003 = 300.003 \] This exceeds $300, which is not allowed.

• Number of adults: The number of adults \( a = 0 \), which violates the condition of having at least one adult.

• Number of children: The number of children can be roughly 14.286, which satisfies the condition of having at least one child.

Since \( a \) must be at least 1 and the cost exceeds $300, (14.286, 0) is not a viable solution.

Thus, the answer is False.