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

Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be

4x + 2y = 20

There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.

When x=0,
4(0) + 2y = 20
y = 10

When x=1,
4(1) + 2y = 20
y = 8

When x=2,
4(2) + 2y = 20
y = 6

When x=3,
4(3) + 2y= 20
y = 4

When x = 4,
4(4) + 2y = 20
y = 2

When x = 5,
4(5) + 2y = 20
y = 0

When x = 6,
4(6) + 2y = 20
y = -2

A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:

   Number of adult tickets             Number of children tickets
                  0                                                   10
                  1                                                    8
                  2                                                    6
                  3                                                    4
                  4                                                    2
                  5                                                    0

It would be 15 m how i got the answer: (20x20) + b^2 = (25x25)400 + b^2 = 625625-400=225square that and get 15hope this : )