Let's assume that Larry Mach invested x amount of dollars at a 6% annual interest rate.
So, the amount he invested at a 4% annual interest rate would be (21,000 - x) dollars.
Now, we'll calculate the interest earned from both accounts.
The interest earned from the amount invested at a 6% annual interest rate would be: 0.06x.
The interest earned from the amount invested at a 4% annual interest rate would be: 0.04(21,000 - x).
According to the given information, the total interest earned was $1060, so we can set up the following equation:
0.06x + 0.04(21,000 - x) = 1060
Now, let's solve this equation to find the value of x:
0.06x + 0.04(21,000) - 0.04x = 1060
0.06x + 840 - 0.04x = 1060
0.02x + 840 = 1060
0.02x = 1060 - 840
0.02x = 220
x = \frac{220}{0.02}
Now, let's calculate the value of x:
x = \frac{220}{0.02} = 11,000
Therefore, Larry Mach invested $11,000 at a 6% annual interest rate, and the remaining amount (21,000 - 11,000 = $10,000) at a 4% annual interest rate.
Answer: Larry Mach invested $11,000 at a 6% annual interest rate and $10,000 at a 4% annual interest rate.