Question

On the first night, the Movie House Theater sold 7 adult tickets and 11 child tickets for $227. On the second night they took in $150 by selling 6 adult tickets and 6 child tickets. Find the price of an adult ticket and the price of a child ticket.
Let x = adult tickets y = child tickets

a
x=6 y = 12

b
x=12 y = 13

c
x=15 y =20

Answer 1

To find the prices of adult tickets and child tickets, we can set up a system of equations based on the information provided.

Let:
- \( a \) = price of an adult ticket
- \( c \) = price of a child ticket

From the first night, we have the equation:
\[ 7a + 11c = 227 \quad \text{(1)} \]

From the second night, we have the equation:
\[ 6a + 6c = 150 \quad \text{(2)} \]

We can simplify equation (2):
Divide all terms by 6:
\[ a + c = 25 \quad \text{(3)} \]

Now we will solve this system of equations using equation (3).

From equation (3), we can express \( c \) in terms of \( a \):
\[ c = 25 - a \]

Now we substitute \( c \) into equation (1):
\[ 7a + 11(25 - a) = 227 \]

Distribute the 11:
\[ 7a + 275 - 11a = 227 \]

Combine like terms:
\[ -4a + 275 = 227 \]

Now, isolate \( a \):
\[ -4a = 227 - 275 \]
\[ -4a = -48 \]
\[ a = 12 \]

Now we substitute \( a \) back into equation (3) to find \( c \):
\[ c = 25 - 12 \]
\[ c = 13 \]

Thus, the price of an adult ticket is $12, and the price of a child ticket is $13.

From the options you have given:
- b) \( x = 12 \) and \( y = 13 \) is the correct answer.