1. **State the problem:** Solve the equation $$\frac{x - 4}{2022} + \frac{x - 3}{2023} + \frac{x - 2}{2024} - 3 = 3$$ for $x$. 2. **Rewrite the equation:** Move the constant term on the left side to the right side: $$\frac{x - 4}{2022} + \frac{x - 3}{2023} + \frac{x - 2}{2024} = 3 + 3 = 6$$ 3. **Combine the fractions:** The denominators are different, so we find a common denominator or treat the sum as is. Since denominators are large and close, we can write the sum as: $$x\left(\frac{1}{2022} + \frac{1}{2023} + \frac{1}{2024}\right) - \left(\frac{4}{2022} + \frac{3}{2023} + \frac{2}{2024}\right) = 6$$ 4. **Calculate the sums of the fractions:** - Sum of coefficients of $x$: $$S_x = \frac{1}{2022} + \frac{1}{2023} + \frac{1}{2024}$$ - Sum of constants: $$S_c = \frac{4}{2022} + \frac{3}{2023} + \frac{2}{2024}$$ 5. **Rewrite the equation:** $$x S_x - S_c = 6$$ 6. **Solve for $x$:** $$x = \frac{6 + S_c}{S_x}$$ 7. **Approximate the sums:** - Calculate $S_x$: $$\frac{1}{2022} \approx 0.0004946, \quad \frac{1}{2023} \approx 0.0004944, \quad \frac{1}{2024} \approx 0.0004941$$ $$S_x \approx 0.0004946 + 0.0004944 + 0.0004941 = 0.0014831$$ - Calculate $S_c$: $$\frac{4}{2022} \approx 0.0019784, \quad \frac{3}{2023} \approx 0.0014829, \quad \frac{2}{2024} \approx 0.0009882$$ $$S_c \approx 0.0019784 + 0.0014829 + 0.0009882 = 0.0044495$$ 8. **Final calculation:** $$x = \frac{6 + 0.0044495}{0.0014831} = \frac{6.0044495}{0.0014831} \approx 4048.5$$ **Answer:** $$x \approx 4048.5$$