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?

There were 40 advanced tickets and 15 same-day tickets sold

Step-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

Advanced tickets = 40

Same-day tickets = 15

Step-by-step explanation:

We know that the advance tickets cost $40 and same-day ticket costs $20.

Also, there were 55 tickets sold and the total amount paid for them was $1900.

Assuming to be the number of advance tickets and to be the number of same-day tickets, we can write the following equations:

--- (1)

--- (2)

Simplifying equation (2):

--- (3)

Now solving for A and S:

Substituting A from equation 1, in (3):

S = 15

Substituting this value of S in (1):

A = 40

0 is its x and y coordinate

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads