Answer:
Let's use a system of equations to solve the problem.
Let x be the number of regular admission tickets sold, and y be the number of fast pass tickets sold.
From the problem, we know that:
x + y = 104 (equation 1)
and
8x + 22y = 1280 (equation 2)
We can use equation 1 to solve for x in terms of y:
x = 104 - y
Substituting this expression for x into equation 2, we get:
8(104 - y) + 22y = 1280
Simplifying and solving for y, we get:
832 - 8y + 22y = 1280
14y = 448
y = 32
So the student sold 32 fast pass tickets.
To find the number of regular admission tickets sold, we can substitute y = 32 into equation 1 and solve for x:
x + 32 = 104
x = 72
So the student sold 72 regular admission tickets.
Therefore, the student sold 72 regular admission tickets and 32 fast pass tickets.