(1 point) 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 $9086 dollars. How much did he invest at each rate

$82436 is invested at 6% and $41218.2 is invested at 10%

Explanation:

We were told that a man invests his savings in two accounts,

Amount invested at 6 percent interest is ?

Amount invested at 10 percent interest is ?

Let us denote the amount invested at different percent as

X=the amount invested at 6%

Y is the amount invested at 10%

Then,

0.06X + 0.10Y = 9068

But from the question, we were told He puts twice as much in the lower-yielding account because it is less risky which means

X = 2Y

0.06(2Y) + 0.10Y = 9068

0.12Y + 0.10Y = 9068

0.22Y = 9068

Y= 9068/0.22

Y=41218.2

Then if we substitute Y into equation above we have

X= 2×41218.2

X=82436

Therefore, $82436 is invested at 6% and $41218.2 is invested at 10%

answer; ///i believe that the correct answer is (post conventional)