Let \( a \) be the price of an adult ticket and \( c \) be the price of a child's ticket. We have the following equations based on the given information:
-
For three adults and four children: \[ 3a + 4c = 136 \]
-
For two adults and three children: \[ 2a + 3c = 97 \]
Now we can solve this system of equations.
First, let's solve the second equation for \( a \): \[ 2a + 3c = 97 \] \[ 2a = 97 - 3c \] \[ a = \frac{97 - 3c}{2} \]
Next, substitute this expression for \( a \) into the first equation: \[ 3\left(\frac{97 - 3c}{2}\right) + 4c = 136 \] Multiply through by 2 to eliminate the fraction: \[ 3(97 - 3c) + 8c = 272 \] Distributing \( 3 \): \[ 291 - 9c + 8c = 272 \] Combining like terms: \[ 291 - c = 272 \] Solving for \( c \): \[ -c = 272 - 291 \] \[ -c = -19 \] \[ c = 19 \]
Now substitute \( c = 19 \) back into the expression for \( a \): \[ a = \frac{97 - 3(19)}{2} \] Calculating this: \[ a = \frac{97 - 57}{2} \] \[ a = \frac{40}{2} = 20 \]
Thus, the price of an adult ticket is \( \boxed{20} \) and the price of a child's ticket is \( \boxed{19} \).