Julia's school is selling tickets to a play. On the first day of ticket sales the school sold 6 adult tickets and 2 student tickets for a total of $40. The school took in $85 on the second day by selling 5 adult tickets and 12 student tickets. What is the price each of one adult ticket and one student ticket.
Let's assign variables to the prices of adult and student tickets.
Let's say the price of an adult ticket is 'a', and the price of a student ticket is 's'.
According to the information given on the first day:
6a + 2s = 40 ---(equation 1)
According to the information given on the second day:
5a + 12s = 85 ---(equation 2)
We can solve this system of equations to find the values of 'a' and 's'.
Step 1: Multiply equation 1 by 5 and equation 2 by 6 to eliminate 'a'.
5(6a + 2s) = 5(40) ---> 30a + 10s = 200 ---(equation 3)
6(5a + 12s) = 6(85) ---> 30a + 72s = 510 ---(equation 4)
Step 2: Subtract equation 3 from equation 4 to eliminate 'a'.
(30a + 72s) - (30a + 10s) = 510 - 200
30a + 72s - 30a - 10s = 310
62s = 310
s = 310/62
s = 5
Step 3: Substitute the value of 's' into equation 1 to solve for 'a'.
6a + 2(5) = 40
6a + 10 = 40
6a = 40 - 10
6a = 30
a = 30/6
a = 5
The price of an adult ticket is $5, and the price of a student ticket is also $5.
To find the price of one adult ticket and one student ticket, we can set up a system of equations based on the information given.
Let's assume the price of one adult ticket is A, and the price of one student ticket is S.
On the first day, the school sold 6 adult tickets and 2 student tickets for a total of $40. This can be expressed as:
6A + 2S = 40
On the second day, the school sold 5 adult tickets and 12 student tickets for a total of $85. This can be expressed as:
5A + 12S = 85
Now, we have a system of two equations:
6A + 2S = 40
5A + 12S = 85
There are several ways to solve this system of equations, but let's use the substitution method.
1. Rearrange the first equation to solve for A:
6A = 40 - 2S
A = (40 - 2S)/6
2. Substitute the value of A in the second equation:
5[(40 - 2S)/6] + 12S = 85
Now, let's solve for S:
(200 - 10S)/6 + 12S = 85
Multiply the entire equation by 6 to get rid of the fraction:
200 - 10S + 72S = 510
Combine like terms:
62S = 310
Divide both sides by 62:
S = 5
Now, substitute the value of S back into the first equation to solve for A:
6A + 2(5) = 40
6A + 10 = 40
6A = 30
A = 5
Therefore, the price of one adult ticket is $5, and the price of one student ticket is also $5.
6 a + 2 s = 40
5 a + 12 s = 85
multiply first one by 6
36 a + 12 s = 240
5 a + 12 s = 85
------------------ subtract
31 a = 155
a = $5
now find s