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​ $1 comma 210 in interest at the end of the​ year, how much did she invest in each​ account?
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Part 1
The amount invested at 5​% was ​$
  
17,000.
Part 2
The amount invested at 9​% was ​$
  
enter your response here.

Answer 1

Let \( x \) be the amount invested at 5%, and \( y \) be the amount invested at 9%. According to the problem, we have the following two equations:

1. The total investment:
\[
x + y = 21,000
\]

2. The total interest earned:
\[
0.05x + 0.09y = 1,210
\]

Now, we can solve this system of equations step by step.

### Step 1: Solve for \( y \)
From the first equation, we can express \( y \) in terms of \( x \):
\[
y = 21,000 - x
\]

### Step 2: Substitute into the interest equation
Substituting \( y \) into the second equation:
\[
0.05x + 0.09(21,000 - x) = 1,210
\]

### Step 3: Distribute and simplify
\[
0.05x + 1,890 - 0.09x = 1,210
\]
\[
-0.04x + 1,890 = 1,210
\]

### Step 4: Isolate \( x \)
Subtract \( 1,890 \) from both sides:
\[
-0.04x = 1,210 - 1,890
\]
\[
-0.04x = -680
\]

### Step 5: Solve for \( x \)
Dividing both sides by -0.04:
\[
x = \frac{-680}{-0.04} = 17,000
\]

### Step 6: Find \( y \)
Now substitute \( x \) back into the equation for \( y \):
\[
y = 21,000 - 17,000 = 4,000
\]

### Final Answer
The amount invested at 5% was **$17,000**, and the amount invested at 9% was **$4,000**.