A drama club sold 120 child tickets and 80 adult tickets to a play a child ticket is 3 dollars less than an adult ticket The club sold a total of 640 dollars worth of tickets What were the prices?

The tickets are $5 adult and $2 child.

Here is the algebraic substitution.

A = adult ticket money

C = child ticket money

120C + 80A = 640

C = A-3 (three dollars less)

120 (A-3) + 80A = 640

120 A - 360 + 80A = 640

200 A = 1000

A = 5

C = A - 3 = 2

Checking your answer:

120(2) + 80 (5) = 240 + 400 = 640