A man invests $2,200 in three accounts that pay 6%, 8%, 9% in annual interest, respectively. he has three times as much invented at 9% as he does at 6%. if his total interest for year is $178, how much is invested at each rate?

6%: $200

8%: $1400

9%: $600

Step-by-step explanation:

Let x represent amount invested at a rate of 6% and y represent amount invested at a rate of 8%.

We have been given that a man has three times as much invented at 9% as he does at 6%. So amount invested at a rate of 9% would be .

We are also told that a man invests $2,200 in three accounts. We can represent this information in an equation as:

We are also told that his total interest for year is $178. We can represent this information in an equation as:

Upon substituting equation (1) in equation (2), we will get:

Therefore, an amount of $200 is invested at 6%.

Amount invested at a rate of 9% would be .

Therefore, an amount of $600 is invested at 9%.

Upon substituting in equation (1), we will get:

Therefore, an amount of $1400 is invested at 8%.

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure