One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman who has $138,500 to invest would like to have an income of $18,730 per year from her investments. How much should she invest at each rate?
10%    	$ 
18%    	$ 


Let:
- $ x $ be the amount invested at 10% per year.
- $ 138,500 - x $ be the amount invested at 18% per year.

We know that the total income from the investments is $18,730. Therefore, the total interest earned from both investments can be expressed as:

\[
0.10x + 0.18(138,500 - x) = 18,730
\]

Now, solve for $ x $:

1. Distribute the 0.18:
\[
0.10x + 0.18(138,500) - 0.18x = 18,730
\]
\[
0.10x + 24,930 - 0.18x = 18,730
\]

2. Combine like terms:
\[
-0.08x + 24,930 = 18,730
\]

3. Subtract 24,930 from both sides:
\[
-0.08x = 18,730 - 24,930
\]
\[
-0.08x = -6,200
\]

4. Solve for $ x $:
\[
x = \frac{-6,200}{-0.08} = 77,500
\]

So, she should invest $77,500 at 10%. The remaining amount is:
\[
138,500 - 77,500 = 61,000
\]
She should invest $61,000 at 18%.

### Final answer:
- Amount invested at 10%: **$77,500**
- Amount invested at 18%: **$61,000**

Would you like more details or have any other questions? 

Here are 5 related questions to consider:
1. What would happen if the interest rates changed?
2. How would the solution change if she wanted a higher annual income?
3. What is the total interest earned over a 5-year period?
4. How would compounding interest affect this problem?
5. What other safe investment options are available with different interest rates?

**Tip:** In finance, diversifying investments between safe and risky options is a common strategy to balance risk and return.

One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman who has $138,500 to invest would like to have an income of $18,730 per year from her investments. How much should she invest at each rate?

Find out how to determine the optimal investment amounts at 10% and 18% interest rates to achieve an annual income of $18,730 from a $138,500 total investment. This solution involves algebraic concepts and linear equations, making it ideal for students in Grades 9-11.

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