Let the cost of an adult ticket be \(x\) and the cost of a child ticket be \(y\). We can set up the following equations based on the information provided:
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From the first scenario (3 adults and 2 children): \[ 3x + 2y = 31 \]
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From the second scenario (2 adults and 3 children): \[ 2x + 3y = 29 \]
We can solve this system of equations to find the values of \(x\) and \(y\).
First, we will multiply the first equation by 3 and the second equation by 2 to eliminate \(y\):
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Multiply the first equation by 3: \[ 9x + 6y = 93 \]
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Multiply the second equation by 2: \[ 4x + 6y = 58 \]
Now, subtract the second equation from the first: \[ (9x + 6y) - (4x + 6y) = 93 - 58 \] This simplifies to: \[ 5x = 35 \] Thus, \[ x = 7 \]
Now substitute \(x = 7\) back into one of the original equations to find \(y\). We can use the first equation: \[ 3(7) + 2y = 31 \] This simplifies to: \[ 21 + 2y = 31 \] Subtract 21 from both sides: \[ 2y = 10 \] Thus, \[ y = 5 \]
Now we know the ticket prices are:
- Adult ticket: \(x = 7\)
- Child ticket: \(y = 5\)
Now we need to determine how many adults are in the group of 6 children that paid $58 for tickets. Let the number of adults be \(a\). The equation for the total cost becomes: \[ 7a + 6(5) = 58 \] This simplifies to: \[ 7a + 30 = 58 \] Subtract 30 from both sides: \[ 7a = 28 \] Dividing both sides by 7 gives: \[ a = 4 \]
Thus, the number of adults in the group is \( \boxed{4} \).
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