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.