The school that Mei goes to is selling tickets to a play

Question

The school the Mei goes to is selling tickets to a play. On the first day of ticket sales the school sold 2 adult tickets and 1 child ticket for a total of $24. The school took in $132 on the second day by selling 14 adult tickets and 4 child tickets. Find the price of an adult ticket and the price of a child ticket.

I can’t quite figure out how to solve this by elimination, so if anyone could help that’d be great! Thanks!

Answer 1

Let x = cost of an adult ticket; y = cost of a child ticket

First Day:

(1) 2x + y = 24

Second Day:

(2) 14x +4y = 132

To use elimination we could multiply eqn. 1 by -7 (to match the magnitude of x terms)

or multiply eqn. 1 by -4 (to match the y terms) - LET'S USE THIS ONE

-8x - 4y = -96 (-4 x (Eqn. 1)

14x + 4y = 132 (Eqn. 2)

6x = 36

x = 6 (adult ticket price)

Use Eqn. 1 to find y:

2x + y = 24

2(6) + y = 24

12 + y = 24

y = 12 (child ticket price)

Adult ticket price: $6

Child ticket price: $12

Answer 2

Let's call the price of an adult ticket "a" and the price of a child ticket "c". We can set up a system of two equations based on the information given:

2a + 1c = 24 (equation 1)

14a + 4c = 132 (equation 2)

We can use any method to solve this system, such as substitution or elimination. Let's use elimination. We can multiply equation 1 by 4 to get:

8a + 4c = 96 (equation 3)

Now we can subtract equation 3 from equation 2 to eliminate the "c" variable:

14a + 4c = 132

(8a + 4c = 96)

6a = 36

Dividing both sides by 6 gives us:

a = 6

So an adult ticket costs $6. Now we can substitute this value back into equation 1 to find the price of a child ticket:

2a + 1c = 24

2(6) + c = 24

12 + c = 24

c = 12

So a child ticket costs $12.

Therefore, the price of an adult ticket is $6 and the price of a child ticket is $12.