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.....
If x = 0, then
If x = 1 , then
If x = 2 , then
If x = 3 , then
If x = 4 , then
If x = 5 , then
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)