1. We are given the equation $263 + 441 = 714$ and asked to determine the base in which this equation is true. 2. Let's represent the digits in base $b$ and convert them to decimal. 3. Convert $263_b$ to decimal: $$2b^2 + 6b + 3$$ 4. Convert $441_b$ to decimal: $$4b^2 + 4b + 1$$ 5. Convert $714_b$ to decimal: $$7b^2 + b + 4$$ 6. According to the problem, the sum should satisfy: $$263_b + 441_b = 714_b$$ 7. Substitute decimal equivalents: $$ (2b^2 + 6b + 3) + (4b^2 + 4b + 1) = 7b^2 + b + 4$$ 8. Simplify the left side: $$6b^2 + 10b + 4 = 7b^2 + b + 4$$ 9. Rearrange to isolate terms: $$6b^2 + 10b + 4 - 7b^2 - b - 4 = 0$$ 10. Simplify: $$-b^2 + 9b = 0$$ 11. Factor: $$b(-b + 9) = 0$$ 12. Thus, $b = 0$ or $b = 9$. Since a base cannot be zero, the base used is $9$. 13. To verify, check digits are valid in base 9: digits are from 0 to 8. But digit 7 and 6 appear, which are valid, however digit 7 and digit 7 appear in 714 which is fine since digits are less than 9. Therefore, the base is 9.