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.