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 original two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the units digit. 3. **Write equations from the problem:** - Sum of digits: $$x + y = 8$$ - Interchanged number is greater by 36: $$10y + x = 10x + y + 36$$ 4. **Simplify the second equation:** $$10y + x = 10x + y + 36$$ $$10y - y + x - 10x = 36$$ $$9y - 9x = 36$$ $$9(y - x) = 36$$ $$\cancel{9}(y - x) = \cancel{9}4$$ $$y - x = 4$$ 5. **Solve the system of equations:** From step 3 and 4: $$x + y = 8$$ $$y - x = 4$$ Add the two equations: $$x + y + y - x = 8 + 4$$ $$2y = 12$$ $$y = 6$$ Substitute $y=6$ into $x + y = 8$: $$x + 6 = 8$$ $$x = 2$$ 6. **Find the original number:** $$10x + y = 10 \times 2 + 6 = 26$$ **Final answer:** The two-digit number is **26**.