1. **State the problem:** Ellie invested some money in a bank account with simple interest at 14% per annum. We know the amounts at the end of 2010 and 2015, and we want to find the year she invested the money. 2. **Define variables:** Let $P$ be the principal (initial amount invested), $r = 0.14$ the interest rate per year, and $t$ the number of years from the investment year to 2010. 3. **Use the simple interest formula:** The amount $A$ after $t$ years is given by $$A = P(1 + rt)$$ 4. **Set up equations:** - At the end of 2010: $$871.20 = P(1 + 0.14t)$$ - At the end of 2015 (which is 5 years later): $$1179.20 = P(1 + 0.14(t + 5))$$ 5. **Express $P$ from the first equation:** $$P = \frac{871.20}{1 + 0.14t}$$ 6. **Substitute $P$ into the second equation:** $$1179.20 = \frac{871.20}{1 + 0.14t} \times (1 + 0.14(t + 5))$$ 7. **Simplify the right side:** $$1179.20 = 871.20 \times \frac{1 + 0.14t + 0.14 \times 5}{1 + 0.14t} = 871.20 \times \frac{1 + 0.14t + 0.7}{1 + 0.14t} = 871.20 \times \frac{1.7 + 0.14t}{1 + 0.14t}$$ 8. **Multiply both sides by denominator:** $$1179.20 (1 + 0.14t) = 871.20 (1.7 + 0.14t)$$ 9. **Expand both sides:** $$1179.20 + 1179.20 \times 0.14 t = 871.20 \times 1.7 + 871.20 \times 0.14 t$$ $$1179.20 + 165.088 t = 1481.04 + 121.968 t$$ 10. **Group like terms:** $$165.088 t - 121.968 t = 1481.04 - 1179.20$$ $$43.12 t = 301.84$$ 11. **Solve for $t$:** $$t = \frac{301.84}{43.12} \approx 7$$ 12. **Interpret $t$:** $t$ is the number of years from the investment year to 2010, so the investment year is $$2010 - 7 = 2003$$ **Final answer:** Ellie invested the money in the year **2003**.