To solve this problem, we can set up variables for the ticket prices and create equations based on the information given.
Let:
- \( x \) = price of an adult ticket
- \( y \) = price of a children's ticket
a) Equation for the cost of going to a movie on Friday
On Friday, 10 adults and 25 children attended, costing $670. This can be represented by the equation: \[ 10x + 25y = 670 \]
b) Equation for the cost of going to a movie on Sunday
On Sunday, 5 adults and 10 children attended, costing $290. This can be represented by the equation: \[ 5x + 10y = 290 \]
c) Best algebraic method to solve the system of equations
The best algebraic method for this set of equations is the substitution method or the elimination method. In this case, the elimination method might be particularly effective since both equations can be manipulated to eliminate \( y \).
To use elimination, we can simplify the second equation by dividing every term by 5: \[ x + 2y = 58 \]
Now we have a system:
- \( 10x + 25y = 670 \)
- \( x + 2y = 58 \)
d) Solve for the costs
Now we can use the substitution method. From equation 2: \[ x + 2y = 58 \implies x = 58 - 2y \]
Substituting this value of \( x \) into equation 1: \[ 10(58 - 2y) + 25y = 670 \] \[ 580 - 20y + 25y = 670 \] \[ 580 + 5y = 670 \] \[ 5y = 670 - 580 \] \[ 5y = 90 \implies y = 18 \]
Now that we have \( y \), substitute it back to find \( x \): \[ x = 58 - 2(18) \] \[ x = 58 - 36 = 22 \]
Thus, the cost of an adult ticket is: \[ \boxed{22} \]
The cost of a children's ticket is: \[ \boxed{18} \]
In conclusion, an adult ticket costs $22 and a children's ticket costs $18.
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