A movie theater sold 120 tickets for a show. Adult tickets cost $10 each and child tickets cost $6 each. If the total revenue was $980, how many adult and how many child tickets were sold?

Answer 1

10a +6c = 980

a + c = 120

6a +6c = 720

subtract last from 1st equation to get

4a = 260

a = 260/4 = 65 adult tickets

c = 120 -65 = 55 child tickets

check the solution

10(65) +6(55) = 650 +330 = 980 It works

or use a graphing calculator

graph the lines x+y=120 and 10x+6y = 980

find the intersection point (x,y) = (65,55)

Answer 2

This is a typical format for a system of linear equations question.

First, decide on variables to represent your unknowns: choose one variable for the number of adult tickets, and another variable for the number of child tickets.

Then, you will have to create two equations based on the information you're given.

1. The number or adult tickets and the number of child tickets add up to 120.
2. $10 x the number of adult tickets and $6 x the number of child tickets must add up to $980.

Once you have your two equations using two variables, you can combine them into a single equation with a single variable using either the substitution method or elimination method. Solve for that variable, and plug that value into one of your equations to solve for the other variable.

Good luck!

Answer 3

Can you solve for c and determine the value of a?

Answer 4

Let x = adult tickets, y = child tickets.

Set up relevant equations:

x + y = 120 (for total tickets sold)

10x + 6y = 980 (for total dollar amount sold, or revenue collected)

Solve:

x + y = 120 ⟹ y = 120 − x

10x + 6*(120 − x) = 980 (y substitution made)

10x + 720 − 6x = 980

4x = 260 ⟹ x = 65, 

y = 55

So, 65 adult and 55 child tickets were sold.