Respuesta :
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 [tex]x[/tex] and the number of children tickets is [tex]y[/tex]
Given that, adult tickets cost 4 and children tickets cost 2 and Natalie sold 20 worth of tickets. So the equation will be.....
[tex]4x+2y=20\\ \\ 2y= 20-4x\\ \\ y= 10-2x[/tex]
If x = 0, then [tex]y= 10-2(0)= 10[/tex]
If x = 1 , then [tex]y= 10-2(1)= 8[/tex]
If x = 2 , then [tex]y= 10-2(2)= 6[/tex]
If x = 3 , then [tex]y= 10-2(3)= 4[/tex]
If x = 4 , then [tex]y= 10-2(4)= 2[/tex]
If x = 5 , then [tex]y= 10-2(5)= 0[/tex]
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)
