what is the number of advance tickets sold? and the number of same-day tickets sold?

Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $35

 and same-day tickets cost $15

. For one performance, there were 35

 tickets sold in all, and the total amount paid for them was $925

. How many tickets of each type were sold?

Answer 1

Let:

1. x be the number of advance tickets sold, and
2. y be the number of same-day tickets sold.

We have two pieces of information:

1. The total number of tickets sold is 35: x + y = 35
2. The total amount paid for the tickets is $925. The cost of an advance ticket is $35, and the cost of a same-day ticket is $15: 35x + 15y = 925

We have the following system of equations:

x + y = 35 (1)

35x + 15y = 925 (2)

Solve this system of equations using Substitution or Elimination. Using Substitution:

Turn equation (1) into:

x = 35 - y

Substitute x = 35 - y for x in equation (2):

35(35 - y) + 15y = 925

y = 15

x = 35 - (15) = 20

The number of advance tickets sold is 20.

The number of same-day tickets sold is 15.

Hope this was helpful.

Answer 2

This type of question, where there are 2 unknown values:

The # of advance tickets sold

The # of same-day tickets sold

will require you to set up a system of 2 equations, using all given information.

Advance tickets cost $35 and same-day tickets cost $15.

Also, it is given that a total of 35 tickets are sold.

Then we also know that those 35 tickets sold produced a total amount of money collected of $925

Here's the 2 equations:

A + S = 35

35A + 15S = 925

A + S = 35

- S. - S

A = 35 - S

Substitute that in the second equation

35(35 - S) + 15S = 925

1165 - 35S + 15S = 925

1165 - 20S = 925

- 1165. - 1165

-20S = -240

÷ -20. ÷ -20

S = 12

12 same-day tickets were sold

Plug that into the 1st equation

A + S = 35

A + 12 = 35

- 12. - 12

A = 23 advance tickets sold