Question

Ms. Burke invested ​$21 comma 000 in two​ accounts, one yielding 5​% interest and the other yielding 9​%. If she received a total of​ $1 comma 210 in interest at the end of the​ year, how much did she invest in each​ account?
Question content area bottom
Part 1
The amount invested at 5​% was ​$
  
enter your response here.

Answer 1

To solve the problem, let's define the amounts invested in each account:

Let:
- \( x \) be the amount invested at 5%
- \( 21000 - x \) be the amount invested at 9%

According to the problem, the total interest earned from both accounts is $1,210 at the end of the year. We can create an equation based on the interest earned from each account:

The interest earned from the 5% account is:
\[
0.05x
\]

The interest earned from the 9% account is:
\[
0.09(21000 - x)
\]

Now we can set up the equation based on the total interest:
\[
0.05x + 0.09(21000 - x) = 1210
\]

Now, we will solve the equation step by step.

1. Distribute \( 0.09 \):
\[
0.05x + 1890 - 0.09x = 1210
\]

2. Combine like terms:
\[
-0.04x + 1890 = 1210
\]

3. Subtract 1890 from both sides:
\[
-0.04x = 1210 - 1890
\]
\[
-0.04x = -680
\]

4. Divide both sides by -0.04:
\[
x = \frac{-680}{-0.04}
\]
\[
x = 17000
\]

So, Ms. Burke invested **$17,000** at 5%.

Now, to find the amount invested at 9%:
\[
21000 - x = 21000 - 17000 = 4000
\]

Thus, the amounts invested are as follows:
- Amount invested at 5%: **$17,000**
- Amount invested at 9%: **$4,000**

### Final Answer
The amount invested at 5% was **$17,000**.