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Math Help - Need help re - arraning this question

  1. #1
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    Need help re - arraning this question

    Make x the subject:

    Many thanks in advance.

    y = (2/3)zx

    And this one which I have already done but I need checking:

    y = (6x - 7)/3

    My answer:

    x = (3y + 7)/6
    Last edited by BomberAF; March 11th 2010 at 12:23 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by BomberAF View Post
    Make x the subject:

    Many thanks in advance.

    y = (2/3)zx

    And this one which I have already done but I need checking:

    y = (6x - 7)/3

    My answer:

    x = (3y + 7)/6
    y = \frac{6x-7}{3} gives x = \frac{3y+7}{6}

    Similarly
    y = \frac{2zx}{3} gives x = \frac{3y}{2z}
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  3. #3
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    Quote Originally Posted by harish21 View Post
    y = \frac{6x-7}{3} gives x = \frac{3y+7}{6}

    Similarly
    y = \frac{2zx}{3} gives x = \frac{3y}{2z}

    Thanks for the answer to the first question but you have got the second one incorrect:

    It is (2/3)yz, the y and z are not over the 3.

    Could you also check this one for me:

    y = (1/2)xz.

    The answer I get is:

    x = (2y)/z
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by BomberAF View Post
    Thanks for the answer to the first question but you have got the second one incorrect:

    It is (2/3)yz, the y and z are not over the 3.

    Could you also check this one for me:

    y = (1/2)xz.

    The answer I get is:

    x = (2y)/z
    Okay. For the first question, i thought zx was in the numerator.
    If zx is in the denominator, then your question becomes y = \frac{2}{3zx} which gives:
    x= \frac{2}{3yz}

    For y = \frac{1}{2xz} you get
    x = \frac{1}{2yz}



    hope that is what you are looking for!
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