Let \( x \) be the cost of an adult ticket and \( y \) be the cost of a child ticket.
a) Cost equation for Friday
On Friday, 10 adults and 25 children went to the movie, costing a total of $670. The equation representing this situation is:
\[ 10x + 25y = 670 \]
b) Cost equation for Sunday
On Sunday, 5 adults and 10 children went to the movie, costing a total of $290. The equation representing this situation is:
\[ 5x + 10y = 290 \]
c) Best algebraic method to solve
The best method to solve these equations would likely be the substitution method or the elimination method, as both pairs are linear equations in two variables. The elimination method can be straightforward, especially given that both equations can be manipulated easily to eliminate one variable.
d) Finding the cost of an adult ticket
First, we can simplify the second equation. Dividing the entire equation by 5 gives:
\[ x + 2y = 58 \quad \text{(Equation 2)} \]
Now we have two equations to work with:
- \( 10x + 25y = 670 \quad \text{(Equation 1)} \)
- \( x + 2y = 58 \quad \text{(Equation 2)} \)
Next, we can express \( x \) from Equation 2:
\[ x = 58 - 2y \]
Now substitute this expression for \( x \) into Equation 1:
\[ 10(58 - 2y) + 25y = 670 \]
Expanding this:
\[ 580 - 20y + 25y = 670 \]
Combining like terms:
\[ 580 + 5y = 670 \]
Subtracting 580 from both sides:
\[ 5y = 90 \]
Dividing by 5 gives:
\[ y = 18 \]
Now we can substitute \( y \) back into the expression for \( x \):
\[ x = 58 - 2(18) \]
Calculating this:
\[ x = 58 - 36 = 22 \]
Conclusion
The cost of an adult ticket is \( \boxed{22} \) dollars.
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