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. (4 points)

Question

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

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

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

d) What is the cost of an adult ticket?



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Answer 1

Let's denote the price of an adult ticket as \( a \) and the price of a child ticket as \( c \).

a) Friday's Equation

On Friday, 10 adults and 25 children went to the movie for a total cost of $670. We can represent this situation with the following equation: \[ 10a + 25c = 670 \]

b) Sunday's Equation

On Sunday, 5 adults and 10 children went to the movie for a total cost of $290. We can represent this situation with this equation: \[ 5a + 10c = 290 \]

c) Best Algebraic Method

Given how the equations are set up, the best algebraic method to use for solving this system of equations is the substitution method or the elimination method. Both methods are viable, but in this case, elimination might be more straightforward since we have clear coefficients for both \( a \) and \( c \).

d) Cost of an Adult Ticket

To solve for \( a \) and \( c \), we can use the elimination method.

First, let's simplify the second equation from part (b): \[ 5a + 10c = 290 \quad \text{(divide everything by 5)} \] This simplifies to: \[ a + 2c = 58 \quad \text{(Equation 2)} \]

Now we have the following system of equations:

1. \( 10a + 25c = 670 \) (Equation 1)
2. \( a + 2c = 58 \) (Equation 2)

Next, we can multiply Equation 2 by 10 to align it with Equation 1: \[ 10a + 20c = 580 \quad \text{(Equation 3)} \]

Now we can subtract Equation 3 from Equation 1: \[ (10a + 25c) - (10a + 20c) = 670 - 580 \] This simplifies to: \[ 5c = 90 \] Now, solving for \( c \): \[ c = 18 \]

Now that we have \( c \), we can substitute it back into Equation 2 to find \( a \): \[ a + 2(18) = 58 \] \[ a + 36 = 58 \] \[ a = 58 - 36 \] \[ a = 22 \]

Thus, the cost of an adult ticket is: \[ \boxed{22} \]

Summary of Results:

• Price of an adult ticket \( a = 22 \)
• Price of a child ticket \( c = 18 \)

Answer 2

choose one method

Answer 3

Let's choose the elimination method to solve the system of equations.

Given Equations:

1. \( 10a + 25c = 670 \) (Equation 1)
2. \( 5a + 10c = 290 \) (Equation 2)

Step 1: Simplify Equation 2

First, we can simplify Equation 2 by dividing everything by 5: \[ a + 2c = 58 \quad \text{(Equation 2 simplified)} \]

Step 2: Convert Equation 2 for Elimination

Now we express Equation 2 in a form compatible for elimination:

• Multiply the simplified Equation 2 by 10 to match the coefficient of \( a \) in Equation 1: \[ 10a + 20c = 580 \quad \text{(Equation 3)} \]

Step 3: Subtract Equation 3 from Equation 1

Now we can subtract Equation 3 from Equation 1: \[ (10a + 25c) - (10a + 20c) = 670 - 580 \]

This gives us: \[ 5c = 90 \]

Step 4: Solve for \( c \)

Dividing both sides by 5: \[ c = 18 \]

Step 5: Substitute back to find \( a \)

Now we substitute \( c = 18 \) back into the simplified Equation 2: \[ a + 2(18) = 58 \] \[ a + 36 = 58 \] Subtract 36 from both sides: \[ a = 58 - 36 \] \[ a = 22 \]

Final Results

Thus, the cost of an adult ticket is: \[ \boxed{22} \] And the cost of a child ticket is: \[ \boxed{18} \]

The elimination method effectively helped us find the ticket prices step-by-step by reducing one variable at a time.