1. **State the problem:** We have a total investment of 20000 split into two parts: one part invested at 11% simple interest for 2 years, and the other part invested at 8% simple interest for 3 years. The total interest earned from both investments is 4690. We need to find how much was invested at 11%. 2. **Define variables:** Let $x$ be the amount invested at 11%. Then, the amount invested at 8% is $20000 - x$. 3. **Recall the simple interest formula:** $$\text{Simple Interest} = P \times r \times t$$ where $P$ is principal, $r$ is rate (in decimal), and $t$ is time in years. 4. **Write the total interest equation:** Interest from 11% investment: $$I_1 = x \times 0.11 \times 2 = 0.22x$$ Interest from 8% investment: $$I_2 = (20000 - x) \times 0.08 \times 3 = 0.24(20000 - x)$$ Total interest: $$I_1 + I_2 = 4690$$ 5. **Set up the equation:** $$0.22x + 0.24(20000 - x) = 4690$$ 6. **Expand and simplify:** $$0.22x + 4800 - 0.24x = 4690$$ 7. **Combine like terms:** $$0.22x - 0.24x + 4800 = 4690$$ $$-0.02x + 4800 = 4690$$ 8. **Isolate $x$:** $$-0.02x = 4690 - 4800$$ $$-0.02x = -110$$ 9. **Divide both sides by -0.02:** $$x = \frac{-110}{-0.02}$$ Show cancellation: $$x = \frac{\cancel{-110}}{\cancel{-0.02}} = 5500$$ 10. **Answer:** The amount invested at 11% is **5500**.