1. **State the problem:** We need to find the price of an adult's ticket and a child's ticket given two total fare amounts for different group sizes. 2. **Set variables:** Let $a$ be the price of an adult ticket and $c$ be the price of a child's ticket. 3. **Write equations from the problem:** - For 3 adults and 4 children: $$3a + 4c = 98$$ - For 2 adults and 3 children: $$2a + 3c = 69$$ 4. **Solve the system of equations:** Multiply the second equation by 3 and the first by 4 to align coefficients of $c$: $$4(3a + 4c) = 4 \times 98 \Rightarrow 12a + 16c = 392$$ $$3(2a + 3c) = 3 \times 69 \Rightarrow 6a + 9c = 207$$ 5. **Subtract the second from the first:** $$ (12a + 16c) - (6a + 9c) = 392 - 207 $$ $$ 6a + 7c = 185 $$ 6. **Express $a$ in terms of $c$:** $$ 6a = 185 - 7c $$ $$ a = \frac{185 - 7c}{6} $$ 7. **Substitute $a$ into one original equation, e.g., $2a + 3c = 69$:** $$ 2 \times \frac{185 - 7c}{6} + 3c = 69 $$ 8. **Multiply both sides by 6 to clear denominator:** $$ 2(185 - 7c) + 18c = 414 $$ $$ 370 - 14c + 18c = 414 $$ $$ 370 + 4c = 414 $$ 9. **Solve for $c$:** $$ 4c = 414 - 370 $$ $$ 4c = 44 $$ $$ c = 11 $$ 10. **Find $a$ using $a = \frac{185 - 7c}{6}$:** $$ a = \frac{185 - 7 \times 11}{6} = \frac{185 - 77}{6} = \frac{108}{6} = 18 $$ **Final answer:** - Adult ticket price $a = 18$ - Child ticket price $c = 11$