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if $9,000 is to be invested, part at 13% and the rest at 8% simple interest , how much should be invested at each rate so that the total annual return will be the same as $9,000 invested at 9% set up as a system of linear equation

asked by jota
  1. It's a linear partition problem, since both investments are paid simple interest.
    Amount invested at 8%
    =$9000*(13-9)/(13-8)
    =$7200
    Amount invested at 13%
    =$9000*(9-8)/(13-8)
    =$1800

    Alternatively, using algebra,
    Let x=amount invested at 8%,
    Then (9000-x)=amount invested at 13%
    and the interest obtained should be the same as $9000 invested at 9%:
    x*(8/100)+(9000-x)*(13/100)=9000*(9/100)
    Isolate x,
    x((13-8)/100)=9000(13-9)/100
    x=9000*(13-9)/(13-8)=7200 as before.

    posted by MathMate

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