To find a viable solution for the number of student and adult tickets that can be sold, we can set up the equation based on the ticket prices and total revenue goal:
Let \( x \) be the number of student tickets sold at $5.00 each, and \( y \) be the number of adult tickets sold at $8.00 each. The equation representing the total revenue from ticket sales is:
\[ 5x + 8y = 1575 \]
Now let's evaluate each of the options provided to see if they satisfy this equation, ensuring that both \( x \) and \( y \) are non-negative integers.
A) 125 student tickets and 115 adult tickets \[ 5(125) + 8(115) = 625 + 920 = 1545 \] This doesn't meet the total of $1,575.
B) 371 student tickets and -35 adult tickets Student tickets are fine, but adult tickets cannot be negative. Not a viable solution.
C) 81 student tickets and 146.25 adult tickets Adult tickets must be whole numbers. Therefore, 146.25 is not a viable solution.
D) 115 student tickets and 125 adult tickets \[ 5(115) + 8(125) = 575 + 1000 = 1575 \] This meets the total of $1,575 exactly.
Thus, the only viable solution for the number of student tickets and adult tickets that can be sold is:
D) 115, 125