Let's solve the problem step by step.

Problem Breakdown:

• You invested a total of $11,000 in two accounts.
• One account earns 4% annual interest, and the other earns 6% annual interest.
• The total interest earned from both accounts in one year is $620.

We need to determine how much was invested in each account.

Let:

• $x$ be the amount invested in the 4% account.
• $y$ be the amount invested in the 6% account.

We have two key equations:

1. The total investment equation: $x + y = 11,000$
2. The total interest equation: $0.04x + 0.06y = 620$

Solve the system of equations:

1. From the first equation, express $y$ in terms of $x$: $y = 11,000 - x$

2. Substitute $y$ in the second equation: $0.04x + 0.06(11,000 - x) = 620$

3. Expand and simplify: $0.04x + 660 - 0.06x = 620$ $-0.02x + 660 = 620$

4. Solve for $x$: $-0.02x = 620 - 660$ $-0.02x = -40$ $x = \frac{-40}{-0.02} = 2,000$

So, $x = 2,000$.

5. Find $y$ using the first equation: $y = 11,000 - 2,000 = 9,000$

Final Answer:

• $2,000 was invested at 4% interest.
• $9,000 was invested at 6% interest.

Would you like more details or have any questions?

Here are five related questions to explore further:

1. How would the total interest change if the amount invested at 6% increased by $1,000?
2. What would be the interest earned if all $11,000 were invested at 5%?
3. How do you calculate the interest rate if the total interest and amounts invested are known?
4. How does the duration of investment impact the total interest earned?
5. What if the interest rates were compounded monthly instead of annually?

Tip: Always double-check your final equations to ensure accuracy in problem-solving, especially in investment-related questions.