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. 

$8.50

B. 

$3.50

C. 

$7.50

D. 

$11.50

Answer 1

Let's denote the cost of one children's ticket as \( x \) and the cost of one adult ticket as \( y \).

From the problem, we know that:
1. The cost of an adult ticket is $4.00 more than a children's ticket:
\[
y = x + 4
\]

2. Henry bought 3 adult tickets and 9 children's tickets for a total of $54.00, which gives us the second equation:
\[
3y + 9x = 54
\]

Now, we can substitute the expression for \( y \) from the first equation into the second equation.

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

Expanding this gives:
\[
3x + 12 + 9x = 54
\]

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

Subtracting 12 from both sides:
\[
12x = 42
\]

Dividing both sides by 12:
\[
x = 3.5
\]

Now substitute \( x \) back into the equation for \( y \):
\[
y = x + 4 = 3.5 + 4 = 7.5
\]

Thus, the cost of one adult movie ticket \( y \) is \( 7.50 \).

The correct answer is:
C. $7.50