Question

At the movie theater, child admission is $5.20 and adult admission is $8.60. On Friday, 132 tickets were sold for a total sales of $931.20. How many tickets of each category were sold that day?

Answer 1

For this case, we propose a system of two equations with two unknowns.

Let:

x: Represents the number of children's tickets

y: Represents the number of adult tickets

`5.20x + 8.60y = 931.20\\x + y = 132`

So, we clear x from the second equation:

x = 132-y

We replace in the first one.

`5.20 (132-y) + 8.60y = 931.20\\686.4-5.20y + 8.60y = 931.20\\-5.20y + 8.60y = 931.20-686.4\\3.4y = 244.8\\y = \frac {244.8} {3.4}\\y = 72`

72 adult tickets were sold.

`x = 132-y\\x = 132-72\\x = 60`

60 tickets for children were sold

Answer:

60 tickets for children

72 adult tickets

Answer 2

60 child tickets and 72 adult tickets

Explanation:

Let x denote the number of child tickets sold and y the number of adult tickets sold. The problem involves solving the following system of linear equations; x+y=132 and 5.2x+8.6y=931.2. Using technology to solve this simultaneous equation yields; x= 60, y =72