1. The problem is to solve the equation $$\frac{x}{8!} + \frac{x}{9!} = \frac{1}{10!}$$ for $x$. 2. Recall the factorial values and the property that $n! = n \times (n-1)!$. Here, $9! = 9 \times 8!$ and $10! = 10 \times 9!$. 3. Rewrite the equation with a common denominator to combine the left side: $$\frac{x}{8!} + \frac{x}{9!} = \frac{x}{8!} + \frac{x}{9 \times 8!} = \frac{x}{8!} + \frac{x}{9 \times 8!} = \frac{9x}{9 \times 8!} + \frac{x}{9 \times 8!} = \frac{9x + x}{9 \times 8!} = \frac{10x}{9 \times 8!}$$ 4. The equation becomes: $$\frac{10x}{9 \times 8!} = \frac{1}{10!}$$ 5. Since $10! = 10 \times 9 \times 8!$, substitute this in the denominator on the right: $$\frac{10x}{9 \times 8!} = \frac{1}{10 \times 9 \times 8!}$$ 6. Multiply both sides by $9 \times 8!$ to eliminate denominators: $$10x = \frac{9 \times 8!}{10 \times 9 \times 8!} = \frac{1}{10}$$ 7. Simplify the right side: $$10x = \frac{1}{10}$$ 8. Divide both sides by 10: $$x = \frac{1}{10 \times 10} = \frac{1}{100}$$ 9. Therefore, the solution is: $$x = \frac{1}{100}$$