The school that Trevon goes to is selling tickets to a choral performance. On the first day of ticket sales the school sold 12 adult tickets and 4 student tickets for a total of $188. The school took in $171 on the second day by selling 6 adult tickets and 9 student tickets. What is the price each of one adult ticket and student ticket?

Let's assume the price of each adult ticket is "A" dollars and the price of each student ticket is "S" dollars.

According to the given information,
On the first day, 12 adult tickets and 4 student tickets were sold for a total of $188.
So, the equation can be formed as follows:
12A + 4S = 188 ...(1)

On the second day, 6 adult tickets and 9 student tickets were sold for a total of $171.
So, the equation can be formed as follows:
6A + 9S = 171 ...(2)

We have two equations (1) and (2) with two variables (A and S). We can solve these two equations to find the values of A and S.

To solve the equations, we can use the method of elimination or substitution.

Let's solve using the elimination method:

Multiply equation (1) by 3 and equation (2) by 2 to make the coefficients of S the same:
(3) 36A + 12S = 564
(4) 12A + 18S = 342

Now, subtract equation (4) from equation (3):
(3) - (4): (36A - 12A) + (12S - 18S) = 564 - 342
24A - 6S = 222

Divide the resulting equation by 6 to simplify:
4A - S = 37

Now, we have a new equation: 4A - S = 37 ...(5)

Let's solve equations (5) and (2) using the elimination method:

Multiply equation (5) by 9 and equation (2) by 4 to make the coefficients of S the same:
(5) 36A - 9S = 333
(2) 24A + 36S = 684

Now, subtract equation (2) from equation (5):
(5) - (2): (36A - 24A) + (-9S - 36S) = 333 - 684
12A - 45S = -351

Divide the resulting equation by -3 to simplify:
-4A + 15S = 117

Now, we have a new equation: -4A + 15S = 117 ...(6)

Now, we have two new equations:
(5) 4A - S = 37
(6) -4A + 15S = 117

Now, add equation (5) and equation (6):
(5) + (6): (4A - 4A) + (-S + 15S) = 37 + 117
14S = 154

Divide both sides of the equation by 14:
S = 154 / 14
S = 11

Substitute the value of S into equation (5):
4A - 11 = 37
4A = 37 + 11
4A = 48

Divide both sides of the equation by 4:
A = 48 / 4
A = 12

Therefore, the price of each adult ticket is $12 and the price of each student ticket is $11.

To solve this problem, we can set up a system of equations. Let's denote the price of an adult ticket as 'a' and the price of a student ticket as 's'.

From the information given, we can create two equations:

Equation 1: 12a + 4s = 188
This equation represents the sales on the first day, where 12 adult tickets and 4 student tickets were sold for a total of $188.

Equation 2: 6a + 9s = 171
This equation represents the sales on the second day, where 6 adult tickets and 9 student tickets were sold for a total of $171.

Now, we can solve this system of equations to find the values of 'a' and 's'.

Multiplying Equation 1 by 3, we get:
36a + 12s = 564

Next, multiplying Equation 2 by 2, we get:
12a + 18s = 342

Now, let's subtract Equation 2 from Equation 1 to eliminate 'a':
(36a + 12s) - (12a + 18s) = 564 - 342
24a - 6s = 222

Simplifying this equation, we have:
24a - 6s = 222

To further simplify, we can divide both sides of the equation by 6:
4a - s = 37

Now, we have a new equation:
4a - s = 37

From here, we can solve for 's' in terms of 'a':
s = 4a - 37

We can substitute this expression for 's' into Equation 1 to find 'a':

12a + 4(4a - 37) = 188
12a + 16a - 148 = 188
28a - 148 = 188
28a = 188 + 148
28a = 336
a = 336 / 28
a = 12

Now that we have the value of 'a', we can substitute it back into Equation 1 to find 's':

12(12) + 4s = 188
144 + 4s = 188
4s = 188 - 144
4s = 44
s = 44 / 4
s = 11

Therefore, the price of one adult ticket is $12, and the price of one student ticket is $11.

12a + 4s = 188

6a + 9s = 171

12a = 188 - 4s
3a = 47 - s

Substitute s-value for a-value in second equation.

2(47 - s) + 9s = 171

Solve for s. Put that value in the first equation to find a. Put both into the second equation to check.

I'll let you do the calculations. I hope this helps.