Let's denote the price of an adult ticket as \( a \) and the price of a youth ticket as \( y \).
From the given information, we can set up the following equations based on the ticket prices:
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For Z's family: \[ 2a + 3y = 134 \quad \text{(1)} \]
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For Jack's family: \[ 3a + 2y = 146 \quad \text{(2)} \]
Now we have two equations with two unknowns. We can solve these equations simultaneously to find the values of \( a \) and \( y \).
To eliminate \( a \), we can multiply equation (1) by 3 and equation (2) by 2:
\[ 3(2a + 3y) = 3(134) \implies 6a + 9y = 402 \quad \text{(3)} \] \[ 2(3a + 2y) = 2(146) \implies 6a + 4y = 292 \quad \text{(4)} \]
Now, we can subtract equation (4) from equation (3):
\[ (6a + 9y) - (6a + 4y) = 402 - 292 \] \[ 9y - 4y = 110 \] \[ 5y = 110 \] \[ y = 22 \]
Now that we have found \( y \), which is the price of a youth ticket, we can substitute back to find the price of an adult ticket \( a \).
Using equation (1): \[ 2a + 3(22) = 134 \] \[ 2a + 66 = 134 \] \[ 2a = 134 - 66 \] \[ 2a = 68 \] \[ a = 34 \]
Thus, the price of an adult ticket is \( 34 \) and the price of a youth ticket is \( 22 \).
The final answer is: \[ \boxed{22} \]
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