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Math Help - How to make "y" the subject of a given formula

  1. #1
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    How to make "y" the subject of a given formula

    Hi guys,

    Make y the subject of the formula: \dfrac{b}{g}=\sqrt{\dfrac{x+2y^{2}}{x-2y^{2}}}.

    I am having trouble with the above stated question. Please guide me, thanks a lot.
    Last edited by Chris L T521; March 31st 2011 at 11:45 AM. Reason: Reformatted the question.
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  2. #2
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    Yes, i had "transfer" the square root over to (b/g)^2 = (x+2y^2) / ( (x-2y^2) and then i am stuck at here because i don't know what to do with the "(b/g)^2".
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  3. #3
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    Hello, FailInMaths!

    Do not state the problem in the title . . .


    \text{Solve for }y\!:\;\;\dfrac{b}{g} \;=\;\sqrt{\dfrac{x+2y^2}{x-2y^2}}

    Square both sides: . \displaystyle \frac{b^2}{g^2} \;=\;\frac{x+2y^2}{x - 2y^2}

    . . . . . . . . b^2(x-2y^2) \;=\;g^2(x + 2y^2)

    . . . . . . . . b^2x - 2b^2y^2 \;=\;g^2x + 2g^2y^2

    . . . . . . 2g^2y^2 + 2b^2y^2 \;=\;b^2x = g^2x

    . . . . . . . 2(g^2+b^2)y^2 \;=\;x(b^2-g^2)

    . . . . . . . . . . . . . . y^2 \;=\;\dfrac{x(b^2-g^2)}{2(b^2+g^2}

    . . . . . . . . . . . . . . . y \;=\;\pm\sqrt{\dfrac{x(b^2-g^2)}{2(b^2+g^2)}}

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  4. #4
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    Ok, so you have

    \left(\dfrac{b}{g}\right)^{\!\!2}=\dfrac{x+2y^{2}}  {x-2y^{2}}.

    I'd recommend multiplying both sides by the denominator of the RHS. What do you get when you do that?
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, FailInMaths!

    Do not state the problem in the title . . .



    Square both sides: . \displaystyle \frac{b^2}{g^2} \;=\;\frac{x+2y^2}{x - 2y^2}

    . . . . . . . . b^2(x-2y^2) \;=\;g^2(x + 2y^2)

    . . . . . . . . b^2x - 2b^2y^2 \;=\;g^2x + 2g^2y^2

    . . . . . . 2g^2y^2 + 2b^2y^2 \;=\;b^2x = g^2x

    . . . . . . . 2(g^2+b^2)y^2 \;=\;x(b^2-g^2)

    . . . . . . . . . . . . . . y^2 \;=\;\dfrac{x(b^2-g^2)}{2(b^2+g^2}

    . . . . . . . . . . . . . . . y \;=\;\pm\sqrt{\dfrac{x(b^2-g^2)}{2(b^2+g^2)}}

    I was told to specify the question by one of the moderators in the past.
    Anyway thanks for the answer, I should change (b/g)^2 = (b^2)/(g^2).

    Quote Originally Posted by Ackbeet View Post
    Ok, so you have

    \left(\dfrac{b}{g}\right)^{\!\!2}=\dfrac{x+2y^{2}}  {x-2y^{2}}.

    I'd recommend multiplying both sides by the denominator of the RHS. What do you get when you do that?
    Thanks a lot for all the hints but i found out my problem already. It is to change (b/g)^2 = (b^2)/(g^2).

    Thank you everyone!

    -Thread Closed-
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  6. #6
    A Plied Mathematician
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    Quote Originally Posted by FailInMaths View Post
    I was told to specify the question by one of the moderators in the past.
    Anyway thanks for the answer, I should change (b/g)^2 = (b^2)/(g^2).



    Thanks a lot for all the hints but i found out my problem already. It is to change (b/g)^2 = (b^2)/(g^2).

    Thank you everyone!

    -Thread Closed-
    Glad you've figured out your problem. You do need to state the entire problem in the body of the post, but there's no rule against partially stating the problem in the title of the thread. The thread title should be at least one level more specific than the forum in which you post the thread. Saying, for example, "Algebra Problem", while posting in this subforum, would likely earn you an infraction for useless post title. On the other hand, "Solve for y" would be perfectly acceptable. At least, this is my current understanding of the rules.
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  7. #7
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    You might want to restore your original post (see rule 7)

    You can mark a thread as solved instead, which enables anyone with a similar problem to check this thread rather than make a new one
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