A total of $6000 is invested: part at 5% and the remainder at 9% . How much is invested at each rate if the annual interest is $530?
Let x be the amount invested at 5%.
Therefore, the amount invested at 9% is 6000 - x.
The interest from the amount invested at 5% is x * 5% = 0.05x.
The interest from the amount invested at 9% is (6000 - x) * 9% = 0.09(6000 - x) = 540 - 0.09x.
The total interest is 0.05x + 540 - 0.09x = 530.
Combining like terms, we get -0.04x = -10.
Dividing both sides by -0.04, we get x = 250.
Thus, $2500 is invested at 5% and $3500 is invested at 9%. Answer: \boxed{2500, 3500}.