Question

A movie theater charges different prices for children and adults. On Friday, 10 adults and 25 children went to the movie and it cost $670. On Sunday 5 adults and 10 children went and it cost $290.

a) What equation would represent the cost of going to a movie on Friday? __________

Answer 1

b) What equation would represent the cost of going to a movie on Sunday? ___________

Answer 2

c) What would be the best algebraic method to solve given how the equations are set up? _______________

options:

Substitution because one of the equations already has a variable solved for

Elimination because the equations are already lined up nicely - you could just multiply the Sunday equation by -2 to eliminate the 'a' variable

Answer 3

What is the cost of an adult ticket?

Answer 4

Let \( a \) represent the price of an adult ticket and \( c \) represent the price of a child ticket.

On Friday, 10 adults and 25 children attended a movie, and the total cost was $670. The equation representing the cost of going to a movie on Friday can be written as:

\[
10a + 25c = 670
\]

Therefore, the equation is:

**10a + 25c = 670**.

Answer 5

On Sunday, 5 adults and 10 children attended a movie, and the total cost was $290. Using the variables \( a \) for the price of an adult ticket and \( c \) for the price of a child ticket, the equation representing the cost of going to a movie on Sunday can be written as:

\[
5a + 10c = 290
\]

Therefore, the equation is:

**5a + 10c = 290**.

Answer 6

The best algebraic method to solve the given equations is:

**Elimination because the equations are already lined up nicely - you could just multiply the Sunday equation by -2 to eliminate the 'c' variable.**

This method allows you to eliminate one of the variables easily, making it straightforward to solve for the other variable. After that, you can substitute back to find the value of the eliminated variable.

Answer 7

To find the cost of an adult ticket, we need to solve the system of equations derived from the data provided. The equations we have are:

1. \( 10a + 25c = 670 \) (from Friday)
2. \( 5a + 10c = 290 \) (from Sunday)

Let's first simplify the second equation by dividing all terms by 5:

\[
a + 2c = 58
\]

Now we can express \( a \) in terms of \( c \):

\[
a = 58 - 2c
\]

Next, we can substitute this expression for \( a \) into the first equation:

\[
10(58 - 2c) + 25c = 670
\]

Expanding this gives:

\[
580 - 20c + 25c = 670
\]

Combining like terms:

\[
580 + 5c = 670
\]

Now, subtract 580 from both sides:

\[
5c = 90
\]

Dividing by 5 gives:

\[
c = 18
\]

Now that we have the cost of a child ticket, we can substitute \( c = 18 \) back into the equation for \( a \):

\[
a = 58 - 2(18)
\]
\[
a = 58 - 36
\]
\[
a = 22
\]

Thus, the cost of an adult ticket is **$22**.