Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $40 and same-day tickets cost $20. For one performance, there were 55 tickets sold in all, and the total amount paid for them was $1900. How many tickets of each type were sold?

Answer:There were 40 advanced tickets and 15 same-day tickets soldStep-by-step explanation:* Lets change the story problem to equations and solve them- Let x is the number of the advance tickets- Let y is the number of the same-day tickets∵ There were 55 tickets sold∴ x + y = 55 ⇒ (1)- The cost of the advance ticket is $40- The cost of the same-day tickets is $20- The total amount paid for them was $1900∴ 40x + 20y = 1900 ⇒ (2)* Now lets solve the two equations (1) and (2) to find the number  of tickets of each type were sold- By using substitution method- From equation (1) ⇒ y = 55 - x - Substitute the value of y in equation (2)∴ 40x + 20(55 - x) = 1900 ⇒ open the bracket∴ 40x + 1100 - 20x = 1900 ⇒ collect the like term∴ 20x + 1100 = 1900 ⇒ subtract 1100 from both sides∴ 20x = 800 ⇒ divide both sides by 20∴ x = 40- Substitute this value of x in the equation of y∴ y = 55 - 40 = 15* There were 40 advance tickets and 15 same-day tickets sold