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How can the number 39 be divided into two parts in order that the sum of 2/3 of one part and 3/4 of the other part is 28?

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  1. I will assume that by "divided" you mean split up, and not actual division

    let one part be x, then the other is 39-x

    so (2/3)x + (3/4)(39-x) = 28
    times 12 , the LCD

    8x + 9(39-x) = 336
    8x + 351 - 9x = 336
    -x = -15
    x = 15

    so parts are 15 and 24

    check:
    (2/3)(15) = 10
    (3/4)(24) = 18 and their sum would be 28, as required

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  2. (2/3)x+(3/4)(39_x)=28
    8x+9(39_x)/12=28
    8x+9(39_x)=28 multiply by 12
    8x+351_9x=336
    _x=336_351
    _x=_15
    X=15

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