Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. ticket sales totaled $2280, and 128 people attended. how many tickets of each type were sold?

Let x be the number of single tickets sold.

Let y be the number of couple tickets sold.

So

The total number of people attended is x + 2y.

The total money earned from selling the tickets is 20x + 35y.

As 128 people attended, setting x + 2y equal to 128, we have:

x + 2y = 128 ... (1)

As the total revenue from selling tickets is $2,280, setting 20x + 35y equal to 2280, we have:

20x + 35y = 2280 ... (2)

Next solve the simultaneous equations (1) and (2).

From equation (1), we get x = 128 - 2y.

Substituting it into equation (2), we have:

20 (128 - 2y) + 35y = 2280

Solving for y, we obtain:

2560 - 40y + 35y = 2280

-40y + 35y = 2280 - 2560

-5y = - 280

y = (-280) / (-5)

y = 56

Substituting y = 56 into x = 128 - 2y, we have

x = 128 - 2*56 = 16

So 16 single and 56 couple tickets were sold.