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The price of admission at a movie theater is $6 for an adult and $4 for a child. In one day, the movie theater sold 80 tickets and made $420. How many adults and how many children bought tickets to the movie theater that day?

x + y = 80,

6x + 4y = 420

Which is a solution of the system of equations, and what does it represent?
(20, 60); 20 adult tickets and 60 child tickets
(30, 50); 30 adult tickets and 50 child tickets
(40, 40); 40 adult tickets and 40 child tickets
(50, 30); 50 adult tickets and 30 child tickets

1 answer

To solve the system of equations, we have:

  1. \( x + y = 80 \) (Equation 1)
  2. \( 6x + 4y = 420 \) (Equation 2)

Where:

  • \( x \) is the number of adult tickets sold.
  • \( y \) is the number of child tickets sold.

Let's analyze the options given for their validity against these equations:

  1. Testing (20, 60):

    • \( x = 20 \) and \( y = 60 \)
    • From Equation 1: \( 20 + 60 = 80 \) (correct)
    • From Equation 2: \( 6(20) + 4(60) = 120 + 240 = 360 \) (not correct)

  2. Testing (30, 50):

    • \( x = 30 \) and \( y = 50 \)
    • From Equation 1: \( 30 + 50 = 80 \) (correct)
    • From Equation 2: \( 6(30) + 4(50) = 180 + 200 = 380 \) (not correct)

  3. Testing (40, 40):

    • \( x = 40 \) and \( y = 40 \)
    • From Equation 1: \( 40 + 40 = 80 \) (correct)
    • From Equation 2: \( 6(40) + 4(40) = 240 + 160 = 400 \) (not correct)

  4. Testing (50, 30):

    • \( x = 50 \) and \( y = 30 \)
    • From Equation 1: \( 50 + 30 = 80 \) (correct)
    • From Equation 2: \( 6(50) + 4(30) = 300 + 120 = 420 \) (correct)

The only solution that satisfies both equations is (50, 30).

This means that 50 adult tickets and 30 child tickets were sold that day at the movie theater.