Question

Adult movie tickets cost $4.00 more than tickets for children. Henry bought 3 tickets for adults and 9 tickets for children, and the total cost was $54.00. The situation can be represented by a system of equations.

{y=x+49x+3y=54

What is the cost of one adult movie ticket?

A. 

$11.50

B. 

$7.50

C. 

$8.50

D. 

$3.50

Answer 1

Let \( x \) be the price of a children's ticket, and \( y \) be the price of an adult ticket.

From the problem statement, we have two equations:

1. The relationship between the prices:
\[
y = x + 4
\]

2. The total cost of the tickets purchased:
\[
3y + 9x = 54
\]

Now, we can substitute the first equation into the second to solve for \( x \):

Substituting \( y = x + 4 \) into the second equation:
\[
3(x + 4) + 9x = 54
\]

Now, distribute the \( 3 \):
\[
3x + 12 + 9x = 54
\]

Combine like terms:
\[
12x + 12 = 54
\]

Subtract \( 12 \) from both sides:
\[
12x = 42
\]

Now, divide by \( 12 \):
\[
x = 3.50
\]

Now that we have the price of a children's ticket, we can find the price of an adult ticket using the first equation:
\[
y = x + 4 = 3.50 + 4 = 7.50
\]

Thus, the cost of one adult movie ticket is:
\[
\boxed{7.50}
\]

So the answer is **B. $7.50**.