Intermidiate algebra

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a woman ivests in two bank accounts, one with a 3% percent interest and the other paying 9% simple interest per year.She puts twice as much in the lower account. Her annual interest is 7110. How much did she invest in each account?

  • Intermidiate algebra -

    higher account --- x
    lower account ---- 2x

    .09x + .03(2x) = 7110
    .15x = 7110
    x = 47400

    $47400 at 9% and $94800 at 3%

    check:
    .09(47400) = 94800(.03) = indeed 7110

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