The ball game took in $1,920 one Saturday. The number of $12 adult tickets was 15 more than twice the number of $5 child tickets. How many of each were sold?
adult tickets ---- x
children tickets --- 2x + 15
solve for x:
12x + 5(2x+15) = 1920
To solve this problem, let's represent the number of $12 adult tickets as 'a' and the number of $5 child tickets as 'c'.
We are given that the ball game took in $1,920, so we can write the equation:
12a + 5c = 1920
We are also given that the number of $12 adult tickets was 15 more than twice the number of $5 child tickets, which can be expressed as:
a = 2c + 15
Now we have a system of equations to solve:
Equation 1: 12a + 5c = 1920
Equation 2: a = 2c + 15
We can start by solving Equation 2 for 'a'.
a = 2c + 15
Next, we substitute this value for 'a' in Equation 1:
12(2c + 15) + 5c = 1920
Simplifying the equation:
24c + 180 + 5c = 1920
29c + 180 = 1920
29c = 1920 - 180
29c = 1740
Finally, we solve for 'c' by dividing both sides of the equation by 29:
c = 1740/29
c ≈ 60
Now that we have the value of 'c', we can substitute it back into Equation 2 to find 'a':
a = 2c + 15
a = 2(60) + 15
a = 135
Therefore, there were approximately 135 adult tickets and 60 child tickets sold.