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9 March, 09:01

How can one halfx - 5 = one thirdx + 6 be set up as a system of equations? 2y + x = - 10 3y + x = 18 2y + 2x = - 10 3y + 3x = 18 2y - x = - 10 3y - x = 18 2y - 2x = - 10 3y - 3x = 18

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  1. 9 March, 09:24
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    2y-x = - 10

    3y-x = 18

    Step-by-step explanation:

    The correct option is. 2y-x = - 10, 3y-x = 18

    one half x - 5 = 1/2 (x-5)

    one third x + 6 = 1/3 * (x+6)

    1/2 x-5 = 1/3x+6 = y

    y = 1/2x-5

    y = 1/3x+6

    Now,

    y=x/2 - 5 equation 1

    y = x/3 + 6 equation 2

    By taking L. C. M of the first equation we get:

    y=x/2 - 5

    y = x-10/2

    Now multiply both terms by 2.

    2y=x-10

    2y-x = - 10

    Now lets solve second equation:

    Take L. C. M of the second equation:

    y = x/3 + 6

    y=x+18/3

    Multiply both sides by 3

    3y = x+18

    3y-x = 18

    Therefore the system of equations we get is:

    2y-x = - 10

    3y-x = 18 ...
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