Doris invested some money at 7% and some money at 8%. She invested $6000 more at 8% than she did at 7%. Her total yearly interest from the two investments was $780. How much did Doris invest at each rate?

amount invested at 7% ---- x

amount invested at 8% = x+6000

solve for x:

.07x + .08(x+6000) = 780

To solve this problem, we can use a system of equations. Let's assume Doris invested x dollars at 7% and (x+6000) dollars at 8%.

We know that the interest from the 7% investment is 0.07x, and the interest from the 8% investment is 0.08(x+6000).

The problem also states that the total yearly interest from both investments is $780, so we can create the equation:
0.07x + 0.08(x+6000) = 780

Now, let's solve this equation step by step:

1. Distribute the 0.08:
0.07x + 0.08x + 480 = 780

2. Combine like terms on the left side:
0.15x + 480 = 780

3. Subtract 480 from both sides:
0.15x = 300

4. Divide both sides by 0.15:
x = 2000

So, Doris invested $2000 at 7% and $2000 + $6000 = $8000 at 8%.