How do i solve this problem. a person invested $8000 for 1 year, part at 4%, part at 11%, and the remainder at 13%. the total annual income from these investments was $882. the amount of money invested at 13% was $600 more than the invested at 4% and 11% combined. find the amount invested at each rate.

The amounts invested at 4%, 11% and 13% are $1200, $2500 and $4300 respectively.

Step-by-step explanation:

Let the amount invested at 4% be x and the amount invested at 11% be y.

Total amount of investment is $8000, so the amount invested at 13% is (8000-x-y).

The total annual income from these investments was $882.

               ..... (1)

The amount of money invested at 13% was $600 more than the invested at 4% and 11% combined.

                       ..... (2)

Solve equation (1) and (2) by using elimination method.

Therefore amounts invested at 4%, 11% and 13% are $1200, $2500 and $4300 respectively.

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

what golf ball

step-by-step explanation: