To solve this system of equations that represents the costs of adult and children tickets, both the elimination method and substitution method can be effective. However, given the structure of the equations, the elimination method might be simpler in this case.
Let's define:
- \( x \) = the cost of an adult ticket
- \( y \) = the cost of a children's ticket
From the information given, we can set up the following equations based on the ticket sales:
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From Friday's sales: \[ 10x + 25y = 670 \quad \text{(Equation 1)} \]
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From Sunday's sales: \[ 5x + 10y = 290 \quad \text{(Equation 2)} \]
Step 1: Simplify Equation 2
We can simplify Equation 2 by dividing all terms by 5: \[ x + 2y = 58 \quad \text{(Equation 2 simplified)} \]
Step 2: Eliminate \( x \)
Now we can use Equation 2 to express \( x \) in terms of \( y \): \[ x = 58 - 2y \]
Step 3: Substitute into Equation 1
Now, we substitute \( x \) in Equation 1: \[ 10(58 - 2y) + 25y = 670 \]
Expanding this gives: \[ 580 - 20y + 25y = 670 \]
Combining like terms results in: \[ 580 + 5y = 670 \]
Step 4: Solve for \( y \)
Now we move 580 to the other side: \[ 5y = 670 - 580 \] \[ 5y = 90 \] \[ y = 18 \]
Step 5: Substitute back to find \( x \)
Now that we have \( y \), we can substitute it back into the expression for \( x \): \[ x = 58 - 2(18) \] \[ x = 58 - 36 \] \[ x = 22 \]
Conclusion
The cost of an adult ticket (\( x \)) is $22. The cost of a children's ticket (\( y \)) is $18.
Both methods are valid, but the elimination method is often clearer when dealing with integers, as was done here.
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