Questions: A movie theater has a seating capacity of 139. The theater charges 5.00 for children, 7.00 for students, and 12.00 for adults. There are half as many adults as there are children. If the total ticket sales was 1004, How many children, students, and adults attended? children attended. students attended. adults attended.

Solution Steps

To solve this problem, we need to set up a system of linear equations based on the given information. Let's define the variables:

• \( C \) for the number of children
• \( S \) for the number of students
• \( A \) for the number of adults

We have the following information:

1. The total number of people is 139: \( C + S + A = 139 \)
2. The total ticket sales amount to $1004: \( 5C + 7S + 12A = 1004 \)
3. There are half as many adults as there are children: \( A = \frac{1}{2}C \)

We can use these equations to solve for \( C \), \( S \), and \( A \).

Let's define the variables for the number of children, students, and adults:

• Let \( c \) be the number of children.
• Let \( s \) be the number of students.
• Let \( a \) be the number of adults.

We are given three pieces of information that we can translate into equations:

1. The total seating capacity is 139.
2. The total ticket sales amount to \$1004.
3. There are half as many adults as there are children.

From these, we can write the following equations:

1. \( c + s + a = 139 \)
2. \( 5c + 7s + 12a = 1004 \)
3. \( a = \frac{1}{2}c \)

From the third equation, we have: \[ a = \frac{1}{2}c \]

Substitute \( a \) into the first and second equations:

1. \( c + s + \frac{1}{2}c = 139 \)
2. \( 5c + 7s + 12 \left(\frac{1}{2}c\right) = 1004 \)

Simplify the first equation: \[ c + s + \frac{1}{2}c = 139 \] \[ \frac{3}{2}c + s = 139 \] \[ 3c + 2s = 278 \quad \text{(Multiply by 2 to clear the fraction)} \]

Simplify the second equation: \[ 5c + 7s + 6c = 1004 \] \[ 11c + 7s = 1004 \]

We now have a system of two equations with two variables:

1. \( 3c + 2s = 278 \)
2. \( 11c + 7s = 1004 \)

We can use the method of elimination or substitution to solve this system. Let's use elimination.

Multiply the first equation by 7 and the second equation by 2 to align the coefficients of \( s \): \[ 7(3c + 2s) = 7(278) \] \[ 21c + 14s = 1946 \]

\[ 2(11c + 7s) = 2(1004) \] \[ 22c + 14s = 2008 \]

Subtract the first modified equation from the second: \[ (22c + 14s) - (21c + 14s) = 2008 - 1946 \] \[ c = 62 \]

Substitute \( c = 62 \) back into the first simplified equation: \[ 3(62) + 2s = 278 \] \[ 186 + 2s = 278 \] \[ 2s = 92 \] \[ s = 46 \]

Now, use \( c = 62 \) to find \( a \): \[ a = \frac{1}{2}c \] \[ a = \frac{1}{2}(62) \] \[ a = 31 \]

Final Answer