Alex bought 9 tickets for $30. Adult tickets cost $4 and children's cost $2. How many children's tickets did he purchase?

A*4+C*2=30

A+C=9

double the second equation
2A+2C=18, now subtract from eq 1

4A-2A=12
A=6
C=3

Let a = adult tickets

c = children's tickets

a + c = 9
4a + 2c = 30

a = 9 - c

4(9 - c) + 2c = 30
36 - 4c + 2c = -6
-2c = -6
c = 3

To solve this problem, we can use algebra. Let's assume that Alex bought x adult tickets and y children's tickets.

According to the information given, the total number of tickets Alex bought is 9. Therefore, we can write our first equation as:

x + y = 9 (Equation 1)

We are also told that the total cost of the tickets was $30. The cost of each adult ticket is $4, and the cost of each children's ticket is $2. So, we can write our second equation as:

4x + 2y = 30 (Equation 2)

We now have a system of two equations (Equation 1 and Equation 2) that we can solve simultaneously.

Using the substitution method, we can solve Equation 1 for x:

x = 9 - y

Now, we substitute this value of x in Equation 2:

4(9 - y) + 2y = 30

Expanding the equation and simplifying, we get:

36 - 4y + 2y = 30

Combining like terms, we have:

-2y + 36 = 30

Subtracting 36 from both sides, we get:

-2y = 30 - 36
-2y = -6

Dividing both sides by -2, we have:

y = -6 / -2
y = 3

Therefore, Alex purchased 3 children's tickets.