You invested ​$12 comma 00012,000 in two accounts paying 3 %3% and 7 %7% annual​ interest, respectively. If the total interest earned for the year was $ 680 comma$680, how much was invested at each​ rate?

Let's denote the amount invested at 3% as $ x $, and the amount invested at 7% as $ y $.

We are given the following information:
1. The total investment is $12,000, so:
   \[
   x + y = 12,000
   \]
2. The total interest earned is $680. The interest from the 3% investment is $ 0.03x $, and the interest from the 7% investment is $ 0.07y $, so:
   \[
   0.03x + 0.07y = 680
   \]

Now, let's solve this system of equations:

### Step 1: Solve for one variable
From the first equation, solve for $ y $:
\[
y = 12,000 - x
\]

### Step 2: Substitute into the second equation
Substitute $ y = 12,000 - x $ into the second equation:
\[
0.03x + 0.07(12,000 - x) = 680
\]

### Step 3: Expand and simplify
Expanding the equation:
\[
0.03x + 0.07 \times 12,000 - 0.07x = 680
\]
\[
0.03x + 840 - 0.07x = 680
\]

### Step 4: Combine like terms
\[
-0.04x + 840 = 680
\]

### Step 5: Solve for $ x $
Subtract 840 from both sides:
\[
-0.04x = 680 - 840
\]
\[
-0.04x = -160
\]
Divide both sides by -0.04:
\[
x = \frac{-160}{-0.04} = 4,000
\]

So, $ x = 4,000 $. This means $4,000 was invested at 3%.

### Step 6: Find $ y $
Substitute $ x = 4,000 $ into the first equation:
\[
y = 12,000 - 4,000 = 8,000
\]

So, $8,000 was invested at 7%.

### Final Answer:
- $4,000 was invested at 3%.
- $8,000 was invested at 7%.

Would you like more details on any part of the solution, or do you have any further questions?

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Here are 5 related questions you might find interesting:
1. What would the total interest be if $12,000 was invested entirely at 7%?
2. How would the investments change if the total interest earned was $900?
3. What would happen if the interest rates were swapped?
4. How much would need to be invested at each rate to earn $1,000 in interest?
5. What is the impact on total interest if the rates are increased by 1% each?

**Tip:** When solving systems of equations, substitution is a helpful method when one equation is easily solvable for one variable.

You invested ​$12,000 in two accounts paying 3% and 7% annual​ interest, respectively. If the total interest earned for the year was $680, how much was invested at each​ rate?

Learn how to solve an investment problem by setting up and solving a system of linear equations. This problem involves investments of $12,000 at 3% and 7% interest rates and finding out how much was invested at each rate. Suitable for students in Grades 9-11.

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