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

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Part 1
The amount invested at 5​% was ​$
  
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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.