Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $30 and same-day tickets cost $40. For one performance, there were 45 tickets sold in all, and the total amount paid for them was $1550. How many tickets of each type were sold?

Answer 1

For this question, we will use the linear system of equations in order to find how many tickets were sold.

Let:

x = advance tickets

y = same-day tickets

x + y = 45

*this is because the total number of ticket solved was 45.

30x + 40y = 1550

*this is because the total amount of ticket sold was $1550

Using the linear system of equations, we can cancel out a variable. In this problem, let's try to cancel out x.

30x + 40y = 1550

x + y = 45

We multiply both sides of the second equation by -30 in order to cancel out x.

30x + 40y = 1550

-30x + 30y = 1359

This will leave us with

40y = 1550

30y = 1350

Subtract both equations and we will get

10y = 200

y = 20

Since we now have a value of y, we can now solve for the value of x.

Using x + y = 45;

x + (20) = 45

x = 45 - 20

x = 25

Going back, we know that

x = advance tickets

y = same-day tickets

Therefore, there are 25 advance tickets and 20 same-day tickets sold.