1. **State the problem:** We want to show that the recurring decimal $0.28\overline{13}$ equals the fraction $\frac{557}{1980}$. 2. **Express the decimal as a sum:** Let $x = 0.28\overline{13}$, which means $x = 0.28131313\ldots$. 3. **Separate the non-repeating and repeating parts:** Write $x$ as the sum of the non-repeating part and the repeating decimal: $$x = 0.28 + 0.00131313\ldots$$ 4. **Convert the repeating part to a fraction:** Let $y = 0.00131313\ldots$. Notice the repeating block "13" starts after three decimal places. Multiply $y$ by $100$ (since "13" has 2 digits) to shift the decimal two places: $$100y = 0.131313\ldots$$ 5. **Set up an equation for $y$: $$100y = 0.131313\ldots$$ $$y = 0.00131313\ldots$$ Subtract the second from the first: $$100y - y = 0.131313\ldots - 0.00131313\ldots$$ $$99y = 0.13$$ 6. **Convert $0.13$ to a fraction:** $$0.13 = \frac{13}{100}$$ So, $$99y = \frac{13}{100} \implies y = \frac{13}{9900}$$ 7. **Add the non-repeating part:** Recall $x = 0.28 + y = \frac{28}{100} + \frac{13}{9900}$. 8. **Find a common denominator and add:** The least common denominator of $100$ and $9900$ is $9900$. Convert $\frac{28}{100}$: $$\frac{28}{100} = \frac{28 \times 99}{100 \times 99} = \frac{2772}{9900}$$ So, $$x = \frac{2772}{9900} + \frac{13}{9900} = \frac{2785}{9900}$$ 9. **Simplify the fraction:** Divide numerator and denominator by 5: $$\frac{2785}{9900} = \frac{557}{1980}$$ **Final answer:** $$0.28\overline{13} = \frac{557}{1980}$$