A basketball team sells tickets that cost $10, $20, or, for VIP seats, $30. The team has sold 3296 tickets overall. It has sold 304 more $20 tickets than $10 tickets. The total sales are $65,180. How many tickets of each kind have been sold?

1022 of $10 tickets, 1326 of $20 tickets, 948 of VIP $30 tickets

Step-by-step explanation:

We can set-up a system of equations to find the number of each kind of ticket. We know there are the cheap tickets c, the medium tickets m and the VIP tickets v. Since 3296 tickets were purchased, then c+m+v=3296.

We also know that 304 more medium priced tickets have been sold then cheap tickets. We write 304+c=m.

Lastly, we know that a total of $65,180 was sold. We write 10c+20m+30v=65,180.

We will solve by substituting one equation into the other. We substitute m=304+c into c+m+v=3296. Simplify and isolate the variable v.

c+m+v=3296

c + (304+c) + v=3296

c+304+c+v=3296

2c+304-304+v=3296-304

2c-2c+v=2992-2c

v=2992-2c

We will now substitute this into the remaining equation with our first substitution.

10c+20m+30v=65,180

10c+20 (304+c) + 30 (2992-2c) = 65,180

10c+6080+20c+89760-60c=65180

-30c+95840=65180

-30c+95840-95840=65180-95840

-30c=-30660

c=1022

This means 1,022 tickets purchased were $10 tickets. We substitute this equation back into m=304+c to find m.

m=304+1022

m=1326

This means 1,326 were $20 tickets.

Lastly we know 948 VIP tickets were bought because 1022+1326+948=3296