A basketball team sells tickets that cost $10,520, or, for VIP seats, $30. The team has sold 3281 tickets overall. It has sold 180 more $20 tickets than $10 tickets. The
total sales are $63,600. How many tickets of each kind have been sold?
How many $10 tickets were sold?

The no. of three types of tickets = x, y, z

The team has sold 3281 tickets over all.
x + y + z = 3281

It has sold 180 more $20 tickets than $10 tickets.
y = x + 180

The total sales are $63,600.
10x + 20y + 30z = $63,600

x + (180+x) + z = 3281
2x + z = 3281 - 180
2x + z = 3101
and
10x + 20(x+180) + 30z = $63,600
10x + 20x + 3600 + 30z = $63,600
30x + 30z = 63,600 - 3600
30x + 30z = 60,000

Simplify, divide equation by 30
x + z = 2000

Using these two equations for elimination:
2x + z = 3101
x + z = 2000

x ($10 tickets) = 3101 - 2000 = 1101 tickets

NOTE: I would like to point out the fact that a ticket costs $10,520, but in the next few sentences it claims to be $10. I will use $10 as a more relevant price, so if it is really $10,520, just change the numbers and you’ll get the answer. I’m sorry if my answer is incorrect!