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A person has invested $6000 . Part of the money is invested at 3% and the remaining amout at 4% . The annual income from the two investment is $225.How much is invested at each rate?

  • math -

    amount invested at 3% ---- x
    amount invested at 4% --- 6000-x

    solve for x
    .03x + .04(6000-x) = 225

    I would multiply each of the 3 terms by 100 to get
    3x + 4(6000-x) = 22500

    I will let you finish it

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