A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $11000 dollars. How much did he invest at each rate?

amount in lower rate --- 2x dollar

amount at higher rate --- x dollars

.06(2x) + .10x = 11000
.12x + .10x = 11000
.22x = 11000
x = 11000/.22 = 50000

so 50000 invested at 10% and
100000 invested at 6%

To solve this problem, we can break it down into two equations. Let's assume that the amount invested in the lower-yielding account is x dollars. Since he puts twice as much in the lower-yielding account, the amount invested in the higher-yielding account is 2x dollars.

The formula for simple interest is: Interest = (Principal * Rate * Time) / 100
Let's calculate the interest earned from each account:

Interest from the lower-yielding account:
Principal = x dollars
Rate = 6%
Interest = (x * 6 * 1) / 100 = 6x / 100 = 0.06x dollars

Interest from the higher-yielding account:
Principal = 2x dollars
Rate = 10%
Interest = (2x * 10 * 1) / 100 = 20x / 100 = 0.2x dollars

Now, we can set up an equation using the given information:
0.06x + 0.2x = 11000 dollars

Combine like terms:
0.26x = 11000 dollars

Divide both sides by 0.26:
x = 11000 dollars / 0.26
x ≈ $42,307.69

Therefore, the man invested approximately $42,307.69 at a 6% interest rate and approximately 2 * $42,307.69 = $84,615.38 at a 10% interest rate.