A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 418 dollars. How much did he invest at each rate?

To solve this problem, we can use a system of equations.

Let's assume the amount invested in the account with 6% interest is x dollars. Since the man invested twice as much in the lower-yielding account, the amount invested in the account with 10% interest is 2x dollars.

The interest earned from the 6% account can be calculated as (6/100) * x = 0.06x dollars per year.
The interest earned from the 10% account can be calculated as (10/100) * 2x = 0.2x dollars per year.

The total interest earned per year is given as $418, so we can set up the equation:
0.06x + 0.2x = 418

Combining like terms, we get:
0.26x = 418

Now, we can solve for x by dividing both sides of the equation by 0.26:
x = 418 / 0.26
x ≈ 1607.69

So, the man invested approximately $1607.69 at a 6% interest rate. Since he invested twice as much in the 10% account, he invested approximately $3215.38 at a 10% interest rate.

if he has $x at 10%, then

.06(2x) + .10x = 418
.12x + .10x = 418
.22x = 418
...