Question

For the school play, adult tickets cost 4 and children tickets cost 2. Natalie is working at the ticket counter and just sold 20 worth of tickets. what are all of the possible ticket combinations for 20 worth of tickets?

Answer 1

The possible combinations of adult tickets and children tickets are : (0, 10), (1, 8), (2, 6), (3, 4), (4, 2), (5, 0)

Explanation

Suppose, the number of adult tickets is $x$ and the number of children tickets is $y$

Given that, adult tickets cost 4 and children tickets cost 2 and Natalie sold 20 worth of tickets. So the equation will be.....

$4x+2y=20\\ \\ 2y= 20-4x\\ \\ y= 10-2x$

If x = 0, then $y= 10-2(0)= 10$

If x = 1 , then $y= 10-2(1)= 8$

If x = 2 , then $y= 10-2(2)= 6$

If x = 3 , then $y= 10-2(3)= 4$

If x = 4 , then $y= 10-2(4)= 2$

If x = 5 , then $y= 10-2(5)= 0$

If we take x value greater than 5 , then y will become negative which is not possible.

So, the possible combinations of adult tickets and children tickets are : (0, 10), (1, 8), (2, 6), (3, 4), (4, 2), (5, 0)