Let \( s \) be the price of a student ticket and \( a \) be the price of an adult ticket.
From the information provided in the table, we can set up the following equations based on the ticket sales for both classes:
For the Algebra I class: \[ 32s + 8a = 164 \quad (1) \]
For the Geometry class: \[ 27s + 4a = 120.50 \quad (2) \]
We will solve this system of equations step by step.
Step 1: Simplify Equation (1)
We can simplify Equation (1) by dividing the entire equation by 4: \[ 8s + 2a = 41 \quad (3) \]
Step 2: Simplify Equation (2)
We can also simplify Equation (2) by dividing the entire equation by 4: \[ 6.75s + a = 30.125 \quad (4) \]
Step 3: Solve for \( a \) in terms of \( s \) from Equation (4) \[ a = 30.125 - 6.75s \quad (5) \]
Step 4: Substitute Equation (5) into Equation (3)
Now substitute \( a \) from Equation (5) into Equation (3): \[ 8s + 2(30.125 - 6.75s) = 41 \] Distribute the 2: \[ 8s + 60.25 - 13.5s = 41 \] Combine like terms: \[ -5.5s + 60.25 = 41 \] Subtract 60.25 from both sides: \[ -5.5s = 41 - 60.25 \] \[ -5.5s = -19.25 \] Divide by -5.5: \[ s = \frac{-19.25}{-5.5} = 3.5 \]
Step 5: Use \( s \) to find \( a \)
Now, substitute \( s = 3.5 \) back into Equation (5): \[ a = 30.125 - 6.75(3.5) \] Calculate \( 6.75 \times 3.5 \): \[ 6.75 \times 3.5 = 23.625 \] Now calculate \( a \): \[ a = 30.125 - 23.625 = 6.5 \]
Final Answer: Student tickets cost \( \mathbf{3.50} \) each and adult tickets cost \( \mathbf{6.50} \) each.