1. We are given the price ratio of two vehicles as $8:7$. Let the original prices be $8x$ and $7x$ respectively. 2. After three years, the price of the first vehicle increases by 880, so its new price is $8x + 880$. 3. The price of the second vehicle increases by 10%, so its new price is $7x + 0.1 \times 7x = 7x + 0.7x = 7.7x$. 4. The problem states that the price ratio increases after these changes, so the new ratio is: $$\frac{8x + 880}{7.7x} > \frac{8}{7}$$ 5. Cross-multiplying, we get: $$(8x + 880) \times 7 > 8 \times 7.7x$$ 6. Expanding both sides: $$56x + 6160 > 61.6x$$ 7. Bringing all terms involving $x$ to one side: $$6160 > 61.6x - 56x = 5.6x$$ 8. Solving for $x$: $$x < \frac{6160}{5.6} = 1100$$ 9. Since $x$ is a positive number representing the price unit, $x$ must be less than 1100. 10. To find a specific value for $x$, consider the scenario where the ratio just starts increasing (equal case): $$\frac{8x + 880}{7.7x} = \frac{8}{7}$$ 11. Cross-multiplying: $$(8x + 880) \times 7 = 8 \times 7.7x$$ 12. Expand: $$56x + 6160 = 61.6x$$ 13. Rearranged: $$6160 = 61.6x - 56x = 5.6x$$ 14. Find $x$: $$x = \frac{6160}{5.6} = 1100$$ 15. Original prices: $$8x = 8 \times 1100 = 8800$$ $$7x = 7 \times 1100 = 7700$$ Final answer: The original prices were 8800 and 7700 respectively.