A movie theater sells tickets.

Question

A movie theater sells tickets at different prices. Adults are charged $8.50 per ticket, and children are charged $5.50 per ticket.

If the theater sells 26 tickets for $194, how many adult tickets and how many child tickets were sold?

Answer 1

This is a "two equations, two unknowns" problem.

First lets find our two equations: Children are C and Adults are A

8.50A + 5.50C = 194.00 . COST EQUATION

A + C = 26 TICKETS EQUATION

based on these numbers substitution is the best method (rather than elimination). Since we need both A and C, it doesnt matter which one we do first.

A = 26-C

substituting in the first equation we get

8.5(26-C) + 5.5C = 194

distribute the 8.5

221-8.5C +5.5C = 194

combine like terms and solve for C

221-3C = 194

3C = 27

C = 9 so we sold 9 childrens tickets

to find the adult we can go back to our original second equation

A + C = 26 TICKETS EQUATION

subbing in 9 for C

we get A = 17

Answer 2

Lets say,

The number of adult tickets sold is A

The number of tickets sold for children is C

Since the total number of tickets sold is 26, A+C=26.......(1)

Charge per adult ticket=$8.50, Hence amount collected by selling adult tickets is 8.50*A

Charge per child ticket=$5.50, Hence amount collected by selling tickets for children is 5.50*C

We know that total amount earned by the theater is $194, 8.5A+5.5C=194......(2)

Solving equations 8.5*(1) and 2,

8.5A+ 8.5C=221...(3)

8.5A+5.5C=194...(2)

Subtract (3)-(2) and solve, A= 17 and C=9