Suppose you invest money in two accounts. One of the accounts pay 6% annual interest, whereas the other pays 8% annual interest. If you have $6, 000 more invested at 8% than you invested at 6 %, how much do you have invested in each account if the total amount of interest you earn in a year is $1, 040

Let's assume the amount invested at 6% is $x.

According to the problem, the amount invested at 8% is " $6,000 more than the amount invested at 6%". So, the amount invested at 8% is $(x + $6,000).

The formula to calculate the interest earned from an investment is:

Interest = Principal x Rate x Time

In this case, we'll calculate the interest earned from each account separately.

The interest earned from the account with 6% interest is:

Interest_6% = x * 0.06

The interest earned from the account with 8% interest is:

Interest_8% = (x + $6,000) * 0.08

The total interest earned for the year is $1,040.

Therefore, we can write the equation:

Interest_6% + Interest_8% = $1,040

Substituting the values from above, we have:

x * 0.06 + (x + $6,000) * 0.08 = $1,040

Simplifying the equation, we distribute the 0.08:

0.06x + 0.08(x + $6,000) = $1,040

Now, we can solve for x:

0.06x + 0.08x + 0.08($6,000) = $1,040

0.14x + $480 = $1,040

Subtracting $480 from both sides:

0.14x = $1,040 - $480

0.14x = $560

Dividing both sides by 0.14:

x = $560 / 0.14

x = $4,000

So, the amount invested at 6% is $4,000.

The amount invested at 8% is $4,000 + $6,000 = $10,000.

Answer:
You have $4,000 invested at 6% and $10,000 invested at 8%.