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Thomas invested $8,500 for one year. Part of the money was invested at6% and the rest at 9%. The total interest earned was $667.50. How much did Thomas invest at the 6% rate?

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  1. 2 June, 12:41
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    Okay so say that x=amount invested with 6% and y=amount invested with 9%

    x+y=8,500 so > x=8500-y

    6%=0.06 9%=0.09

    0.06x + 0.09y=667.5 (Substitute in x=8500-y so only numbers and y)

    0.06 (8500-y) + 0.09y=667.5 (expand brackets)

    510-0.06y+0.09y=667.5 (-510)

    0.03y=117.5 (/0.03)

    $3916.67=Y - >9%

    X=8500-Y

    x=$4583.33 - >6%
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