Answer:
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
