1. **State the problem:** We need to find a two-digit number whose digits sum to 8 and when the digits are interchanged, the new number is 36 greater than the original. 2. **Define variables:** Let the tens digit be $x$ and the units digit be $y$. The original number is then $10x + y$. 3. **Write the equations:** - Sum of digits: $$x + y = 8$$ - Interchanged number is greater by 36: $$10y + x = 10x + y + 36$$ 4. **Simplify the second equation:** $$10y + x = 10x + y + 36$$ $$10y - y + x - 10x = 36$$ $$9y - 9x = 36$$ $$9(y - x) = 36$$ $$y - x = 4$$ 5. **Solve the system:** From the first equation: $$y = 8 - x$$ Substitute into the second: $$8 - x - x = 4$$ $$8 - 2x = 4$$ $$-2x = 4 - 8$$ $$-2x = -4$$ $$x = 2$$ 6. **Find $y$:** $$y = 8 - x = 8 - 2 = 6$$ 7. **Answer:** The original two-digit number is $$10x + y = 10 \times 2 + 6 = 26$$.