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 $3124 dollars. How much did he invest at each rate?
Your answer is:
Amount invested at 6% equals $ -
Amount invested at 10% equals $ -

Amount invested at 6% equals $28400

Amount invested at 10% equals $14200

Step-by-step explanation:

Let x represent the amount invested at 6%.

Let y represent the amount invested at 10%.

He puts twice as much in the lower-yielding account because it is less risky. This means that x = 2y

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest on the investment.

P represents the principal or amount invested

R represents interest rate

T represents the duration of the investment in years.

Considering the amount invested at 6%,

P = x

R = 6 %

T = 1 year

I = (x × 6 × 1)/100 = $0.06x

Considering the amount invested at 6%,

P = y

R = 10 %

T = 1 year

I = (y × 10 × 1)/100 = $0.1y

His annual interest is $3124. This means that

0.06x + 0.1y = 3124 - - - - - - - - - -1

Substituting x = 2y into equation 1, it becomes

0.06 × 2y + 0.1y = 3124

0.12y + 0.1y = 3124

0.22y = 3124

y = 3124/0.22

y = $14200

x = 2y = 2 × 14200

x = $28400

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)

the k side

step-by-step explanation:

plese mark mde brainliest!