1. **State the problem:** Solve the equation $$\frac{V-9}{903} + \frac{V-15}{1195} + \frac{V}{822} = 0.$$\n\n2. **Identify the formula and approach:** We want to find the value of $V$ that satisfies the sum of these three fractions equal to zero. To do this, we find a common denominator and combine the fractions.\n\n3. **Find the least common denominator (LCD):** The denominators are 903, 1195, and 822. The LCD is their least common multiple (LCM).\n\n4. **Multiply both sides by the LCD to clear denominators:** Let $D = \mathrm{LCM}(903,1195,822)$. Multiply the entire equation by $D$:\n$$D \times \left(\frac{V-9}{903} + \frac{V-15}{1195} + \frac{V}{822}\right) = D \times 0$$\n\n5. **Simplify each term:**\n$$\frac{D}{903}(V-9) + \frac{D}{1195}(V-15) + \frac{D}{822}V = 0.$$\n\n6. **Let $A = \frac{D}{903}$, $B = \frac{D}{1195}$, and $C = \frac{D}{822}$. Then:**\n$$A(V-9) + B(V-15) + CV = 0.$$\n\n7. **Expand:**\n$$AV - 9A + BV - 15B + CV = 0.$$\n\n8. **Group terms with $V$ and constants:**\n$$V(A + B + C) - (9A + 15B) = 0.$$\n\n9. **Solve for $V$:**\n$$V = \frac{9A + 15B}{A + B + C}.$$\n\n10. **Calculate $D$, $A$, $B$, and $C$:**\n- Factor denominators:\n - $903 = 3 \times 7 \times 43$\n - $1195 = 5 \times 7 \times 34 = 5 \times 7 \times 2 \times 17$\n - $822 = 2 \times 3 \times 137$\n\n- LCM includes all prime factors at highest powers:\n - $2$, $3$, $5$, $7$, $17$, $43$, $137$\n\n- Calculate $D = 2 \times 3 \times 5 \times 7 \times 17 \times 43 \times 137 = 923430.$\n\n- Then:\n - $A = \frac{923430}{903} = 1023$\n - $B = \frac{923430}{1195} = 772$\n - $C = \frac{923430}{822} = 1123$\n\n11. **Substitute values:**\n$$V = \frac{9 \times 1023 + 15 \times 772}{1023 + 772 + 1123} = \frac{9207 + 11580}{2918} = \frac{20787}{2918}.$$\n\n12. **Simplify fraction:**\n- Divide numerator and denominator by their greatest common divisor (GCD). GCD(20787,2918) = 1 (since 2918 factors as $2 \times 1459$ and 20787 is not divisible by 2 or 1459).\n\nSo, $V = \frac{20787}{2918} \approx 7.12.$\n\n**Final answer:** $$V = \frac{20787}{2918} \approx 7.12.$$