The manager of a movie theater compared ticket
sales for two movies showing on the same night.
For the first movie, 150 adult tickets and
100 child tickets were sold for $3,087.50 in total
ticket sales. For the second movie, 200 adult
tickets and 150 child tickets were sold for
$4,287.50 in total ticket sales. What was the price
of each adult and each child ticket?

You have two revenue relations and want the price of each adult and child ticket. 150 adult and 100 child tickets sold for $3087.50; 200 adult and 150 child tickets sold for $4287.50.

Equations

Each of the revenue relations can be formulated as an equation. If x and y represent the price of each adult and child ticket, respectively, then ...

150x +100y = 3087.50

200x +150y = 4287.50

Solution

The solution can be found many ways. A graphing calculator (first attachment) shows the solution to be ...

adult price: $13.75
child price: $10.25

The calculator function that reduces a matrix to row-echelon form can be used on the augmented matrix of coefficients. That, too, gives the solution about as quickly as you can enter the coefficients. This is shown in the second attachment.

Subtracting twice the second equation from 3 times the first will eliminate the y-variable:

3(150x +100y) -2(200x +150y) = 3(3087.50) -2(4287.50)

50x = 687.50

x = 13.75 . . . . . . divide by 50

From the second equation, ...

y = (4287.50 -200(13.75))/150 = 1537.50/150

y = 10.25