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Algebra

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Sam invested $1950 part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% invest was 6$ more then twice the income from the 6% investment. how much did he invest at each rate

  • Algebra -

    let S be the amount at six percent, E be the amount at Eight percent.

    S+E=1950

    .06S=2*.08E+6

    can you handle if from here?

  • Algebra -

    Not really having trouble with this one for some reason

  • Algebra -

    well, s=1950-E
    so
    .06(1950-E)=2(.08E)+6
    now solve for E.
    then to get S,subtract E from 1950

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