SOLUTION: You invest $6000 in two accounts paying 6% and 9% annual interest. At the end of the year, the accounts earn the same interest. How mush was invested at each rate? I can't figure

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Question 405425: You invest $6000 in two accounts paying 6% and 9% annual interest. At the end of the year, the accounts earn the same interest. How mush was invested at each rate?
I can't figure out how you find the total. I set the problem up like this. I know I am missing how to get the total amount
.06x + (6000-x).09 = What? How do I find this? I can solve the problems when given the total. Can you show me how to get the total
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi, Good work setting up the .06x and .09($6000-x)

"At the end of the year, the accounts earn the 'same' interest"

Writing as we Read:

    .06x =.09($6000-x)

solving for x

    .06x + .09x = .09*$6000

            x = .09*$6000/.15

            x = $3600, is invested at 6%.  $2400 invested at 9%   ($6000-$3600)



CHECKING our Answer***

   .06*$3600 = .09*$2400