Suppose that there are two types of tickets to a show: advance and same-day. (rest in description)

Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $20 and same-day tickets cost $25. For one performance, there were 45 tickets sold in all, and the total amount paid for them was $1050. How many tickets of each type were sold?



Number of advance tickets sold:

Number of same-day tickets sold:

Answer 1

This is a system

x is the number of advanced tickets sold

y is the number of same day tickets sold

once you label your variables then write the equations

x+ y = 45 total tickets sold

20x + 25y = 1050 money taken in

multiply the top equation by -20 and the x will drop out

-20x - 20y = -900

20x + 25y = 1050. Add the equations result is

5y = 150, divide by 5 ,

y or same day tickets cost $30

substitute 30 into first equation

x + 30 = 45 therefore x = 15 😋

Answer 2

Advanced tickets = A = 20

Same Day Tickets = S = 25

A + S = 45

20*A + 25*S = 1050

A = 45 - S

20(45-S)+25*S=1050

900-20*S+25S=1050

5*S=150

S = 150/5 = 30

A + 30 = 45

A = 15

Number of advance tickets = 15

Number of Same Day tickets = 30

Answer 3

Let's say that the advanced tickets are labelled a and the same-day tickets are labelled s.

According to the question we can make a system of equations:

a + s = 45 (1) (adding up a and s gives us the total number of tickets)

20a + 25s = 1050 (2) (the number of each type of ticket multiplied by its cost and added together gives total cost)

We can manipulate the first equation by multiplying it by 20 to get: 20a +20s = 900 (3)

Subtracting (2) - (3)

20a -20a +25s -20s =1050 - 900

5s = 150

s = 30

Thus, a must be 45-s = 15

ANSWER

a=15

s=30