To solve the problem, let's denote:
- \( x \) as the cost of an adult ticket.
- \( y \) as the cost of a child ticket.
Now we can formulate equations based on the information given.
a) What equation would represent the cost of going to a movie on Friday?
On Friday, 10 adults and 25 children went to the movie for a total of $670. The equation for this situation would be: \[ 10x + 25y = 670 \]
b) What equation would represent the cost of going to a movie on Sunday?
On Sunday, 5 adults and 10 children went to the movie for a total of $290. The equation for this situation would be: \[ 5x + 10y = 290 \]
c) What would be the best algebraic method to solve given how the equations are set up?
The best method to solve these equations is the substitution method or the elimination method. Given that both equations can be streamlined, elimination is particularly effective here due to the coefficients of \( y \).
d) What is the cost of an adult ticket?
First, let's simplify the second equation for easy elimination. We can divide the second equation by 5: \[ x + 2y = 58 \quad \text{(Equation 2)} \]
Now we have the following system of equations:
- \( 10x + 25y = 670 \) (Equation 1)
- \( x + 2y = 58 \) (Equation 2)
Next, let's multiply Equation 2 by 10 to align the coefficients of \( x \): \[ 10x + 20y = 580 \quad \text{(Equation 3)} \]
Now we can subtract Equation 3 from Equation 1: \[ (10x + 25y) - (10x + 20y) = 670 - 580 \] \[ 5y = 90 \] \[ y = 18 \]
Now that we have the value of \( y \) (the cost of a child ticket), we can substitute \( y \) back into Equation 2 to find \( x \): \[ x + 2(18) = 58 \] \[ x + 36 = 58 \] \[ x = 22 \]
Thus, the cost of an adult ticket is: \[ \boxed{22} \]