Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $3280, and 128 people attended. How many tickets of each type were sold?

16  singles, 56  couples

Step-by-step explanation:

There's two linear equations that we can make: one for money and one for people. Let the number of single tickets be  s  and the number of couple tickets be  c .

We know that the amount of money we make is  $  =  20 s + 35 c = 2280

We also how many people can come  P  = 1 s + 2 c = 128

We know that both  s  are the same and both  c  are the same. We have two unknowns and two equations, so we can do some algebra to solve for each.

Take the first minus twenty times the second: 20 s + 35 c  =2280

− 20 s − 40 c = − 2560

− 5 c = − 280 ⇒ c = 56

Plugging this back into the second equation, s + 2 c = s + 2 ⋅ 56 = s + 112 = 128 ⇒ s = 16

found it xd it ok yt