1. **State the problem:** We need to find a two-digit number whose digits sum to 9, and nine times this number equals twice the number formed by reversing its digits. 2. **Define variables:** Let the two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit. 3. **Write the conditions:** - Sum of digits: $x + y = 9$ - Relation between numbers: $9(10x + y) = 2(10y + x)$ 4. **Solve the system:** From the first equation: $y = 9 - x$ Substitute into the second: $$9(10x + y) = 2(10y + x)$$ $$9(10x + 9 - x) = 2(10(9 - x) + x)$$ $$9(9x + 9) = 2(90 - 10x + x)$$ $$81x + 81 = 180 - 18x + 2x$$ $$81x + 81 = 180 - 16x$$ Bring all terms to one side: $$81x + 16x = 180 - 81$$ $$97x = 99$$ 5. **Find $x$:** Since $x$ must be an integer digit (1 to 9), check integer solutions near $x = \frac{99}{97} \approx 1.02$. Try $x=1$: $y = 9 - 1 = 8$ Check the relation: $9(10*1 + 8) = 9*18 = 162$ $2(10*8 + 1) = 2*81 = 162$ The condition holds. 6. **Answer:** The number is $\boxed{18}$.