One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman who has $140,900 to invest would like to have an income of $19,130 per year from her investments. How much should she invest at each rate?
10%    	$
18%    	$

Let's define the amounts the woman should invest at each rate as follows:

- Let $ x $ be the amount invested at 10%.
- Let $ y $ be the amount invested at 18%.

We know the following:
1. The total investment is $140,900, so:
   \[
   x + y = 140,900
   \]
2. The total income from both investments should be $19,130. The income from the 10% investment is $ 0.10x $, and the income from the 18% investment is $ 0.18y $. Thus:
   \[
   0.10x + 0.18y = 19,130
   \]

Now, we can solve this system of equations:

### Equation 1:
\[
x + y = 140,900
\]
Solve for $ y $:
\[
y = 140,900 - x
\]

### Equation 2:
\[
0.10x + 0.18y = 19,130
\]
Substitute $ y = 140,900 - x $ into the second equation:
\[
0.10x + 0.18(140,900 - x) = 19,130
\]
Now simplify and solve for $ x $:
\[
0.10x + 0.18 \times 140,900 - 0.18x = 19,130
\]
\[
0.10x + 25,362 - 0.18x = 19,130
\]
\[
-0.08x + 25,362 = 19,130
\]
\[
-0.08x = 19,130 - 25,362
\]
\[
-0.08x = -6,232
\]
\[
x = \frac{-6,232}{-0.08} = 77,900
\]

So, she should invest $77,900 at 10%.

To find $ y $, substitute $ x = 77,900 $ into the first equation:
\[
y = 140,900 - 77,900 = 63,000
\]

### Final answer:
- She should invest **$77,900** at 10%.
- She should invest **$63,000** at 18%.

Would you like further details or have any other questions?

Here are some related questions for further practice:
1. What if the total investment amount was $150,000? How would the investment amounts change?
2. How would the amounts change if the income requirement was $20,000 per year?
3. What would the total income be if the entire $140,900 was invested at 18%?
4. If the income requirement was reduced to $18,000, how much would she invest at each rate?
5. How would the amounts change if the risky investment paid 20% per year?

**Tip**: When solving systems of equations, substitution is useful when one variable can easily be expressed in terms of the other.

One safe investment pays 10% per year, and a more risky investment pays 18% per year. A woman who has $140,900 to invest would like to have an income of $19,130 per year from her investments. How much should she invest at each rate? 10% $ 18% $

Learn how to calculate how much to invest at 10% and 18% interest rates to achieve an annual income of $19,130. This problem involves solving a system of linear equations and covers algebraic concepts suitable for Grades 10-12.

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