Math club purchased 3 adult tickets and 12 student tickets for $262.50. Physics club purchased 4 adult tickets and 8 student tickets for $230.00. How much do the adult and student tickets cost?

Nevermind guys, I solved it.

For future strugglers:

equation A: 3a+12s=262.5
equation B: 4a+8s=230

multiply equation B so that a variable can be negated when you add the two equations together. In this case I chose variable (S) so i multiplied equation B by -1.5 so that -3a=-82.5. Solving for a would get a=27.5

Now you just have to plug in a for one of the two original equations and solve for (s).

3(27.5)+12s=262.5
s=15

3x + 12y =262.5 4x + 8y =230

Solve simultaneously x =$24.25 , y =$16.67. . But the question says how much do they cost ansa =4(24.25)=$97 Dat x and y =8(16.67)=$133 this means that child cost =97 and adult cost is $133

To find the cost of adult and student tickets, let's assign some variables to the unknowns.

Let's say the cost of an adult ticket is "A" dollars and the cost of a student ticket is "S" dollars.

From the given information, we have two equations:

Equation 1: 3A + 12S = 262.50
Equation 2: 4A + 8S = 230.00

Now, we can solve this system of equations using the method of substitution:

From Equation 1, we can rewrite it as A = (262.50 - 12S)/3

Substituting this value of A into Equation 2, we get:

4((262.50 - 12S)/3) + 8S = 230.00

Multiplying through by 3 to clear the fraction:

4(262.50 - 12S) + 24S = 690.00

Distributing:

1050.00 - 48S + 24S = 690.00

Combining like terms:

-24S = -360.00

Dividing both sides by -24:

S = 15.00

Now that we have the value of S, we can substitute it back into Equation 1 to find A:

3A + 12(15.00) = 262.50

3A + 180.00 = 262.50

Subtracting 180.00 from both sides:

3A = 82.50

Dividing both sides by 3:

A = 27.50

Therefore, the cost of an adult ticket is $27.50, and the cost of a student ticket is $15.00.