algebra

Janet invested $26,000, part at 6% and part at 3%. If the total interest at the end of the year is $1,080, how much did she invest at 6%

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  1. let x = the portion invested at 6%
    let y = the portion invested at 3%

    x + y = 26000
    0.06x + 0.03y = 1080

    Solve for x.

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