The possible combinations of adult tickets and children tickets are : (0, 10), (1, 8), (2, 6), (3, 4), (4, 2), (5, 0)
<u><em>Explanation</em></u>
Suppose, the number of adult tickets is
and the number of children tickets is
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](https://tex.z-dn.net/?f=4x%2B2y%3D20%5C%5C%20%5C%5C%202y%3D%2020-4x%5C%5C%20%5C%5C%20y%3D%2010-2x)
If x = 0, then ![y= 10-2(0)= 10](https://tex.z-dn.net/?f=y%3D%2010-2%280%29%3D%2010)
If x = 1 , then ![y= 10-2(1)= 8](https://tex.z-dn.net/?f=y%3D%2010-2%281%29%3D%208)
If x = 2 , then ![y= 10-2(2)= 6](https://tex.z-dn.net/?f=y%3D%2010-2%282%29%3D%206)
If x = 3 , then ![y= 10-2(3)= 4](https://tex.z-dn.net/?f=y%3D%2010-2%283%29%3D%204)
If x = 4 , then ![y= 10-2(4)= 2](https://tex.z-dn.net/?f=y%3D%2010-2%284%29%3D%202)
If x = 5 , then ![y= 10-2(5)= 0](https://tex.z-dn.net/?f=y%3D%2010-2%285%29%3D%200)
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)