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    So I have this question:

    2y\sqrt 2 - 3 = \frac{5y}{\sqrt2} + 1

    So obv. the objective is to solve y

    I got:

    y = \frac{-4\sqrt2}{3}

    Here's my method:

    Q)

    2y\sqrt 2 - 3 = \frac{5y}{\sqrt2} + 1

    = 2y\sqrt 2 - 3 = \frac{5y\sqrt2}{2} + 1
    = 4 = 2y\sqrt2 - \frac{5y\sqrt2}{2}
    = 8 = -3y\sqrt2
    = y = \frac{-8}{3\sqrt2}
    = y = \frac{-8\sqrt2}{6}
    = y = \frac{-4\sqrt2}{3}

    But in the book it says that I shouldn't have a 3 in the denominator.

    Can someone tell me where I've gone wrong?
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    Well, post your working and we can show where it goes wrong. The book is right though.
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    (Original post by Swayum)
    Well, post your working and we can show where it goes wrong. The book is right though.
    Have done just
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    In the first step, when multiplying by root2 you have to take into account the negative 3 and +1 also.
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    (Original post by Potassium^2)
    In the first step, when multiplying by root2 you have to take into account the negative 3 and +1 also.
    Why the negative 3? That's on the other side, plus why do I have to take into account the +1? On other equations I have done the same method and got it right without taking into account other integers.
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    First of all, you're making my eyes bleed! You can't just put equals signs whereever you please! Don't stick one at the start of every line, especially when it's wrong (in particular, line 3 is wrong because all of that does not equal 4).

    Your actual mistake:

    (Original post by Dededex)
    = 4 = 2y\sqrt2 - \frac{5y\sqrt2}{2}
    = 8 = -3y\sqrt2
    You go wrong here. Try again.

    (to pick up on my first point again, here you appear to be saying that 4 = 8...)
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    (Original post by Potassium^2)
    In the first step, when multiplying by root2 you have to take into account the negative 3 and +1 also.
    He's not multiplying through by root2. Look again, the first line is fine.
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    (Original post by Dededex)
    So I have this question:

    2y\sqrt 2 - 3 = \frac{5y}{\sqrt2} + 1

    So obv. the objective is to solve y

    I got:

    y = \frac{-4\sqrt2}{3}

    Here's my method:

    Q)

    2y\sqrt 2 - 3 = \frac{5y}{\sqrt2} + 1

    = 2y\sqrt 2 - 3 = \frac{5y\sqrt2}{2} + 1
    = 4 = 2y\sqrt2 - \frac{5y\sqrt2}{2}
    = 8 = -3y\sqrt2
    = y = \frac{-8}{3\sqrt2}
    = y = \frac{-8\sqrt2}{6}
    = y = \frac{-4\sqrt2}{3}

    But in the book it says that I shouldn't have a 3 in the denominator.

    Can someone tell me where I've gone wrong?
    You dint multiply '2' to '2yroot2' in 3rd step when taking LCM
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    (Original post by Swayum)
    First of all, you're making my eyes bleed! You can't just put equals signs whereever you please! Don't stick one at the start of every line, especially when it's wrong (in particular, line 3 is wrong because all of that does not equal 4).

    Your actual mistake:



    You go wrong here. Try again.

    (to pick up on my first point again, here you appear to be saying that 4 = 8...)
    Is it when I move the denominator of 2 over? Sorry but I can't see it I need an explanation. The only thing I can think I'm doing wrong is my rearrangement.
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    (Original post by Dededex)
    Is it when I move the denominator of 2 over? Sorry but I can't see it I need an explanation. The only thing I can think I'm doing wrong is my rearrangement.
    When you multiply through by 2, you multiply the 4 on the LHS to get 8, you did this right.

    Then you multiply 2y*root2 by 2 to get 4y*root2

    Then you multiply -5y/2 * root2 by 2 to get -5y*root2.

    So you have 8 = 4y\sqrt2 - 5y\sqrt2...
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    (Original post by Swayum)
    When you multiply through by 2, you multiply the 4 on the LHS to get 8, you did this right.

    Then you multiply 2y*root2 by 2 to get 4y*root2

    Then you multiply -5y/2 * root2 by 2 to get -5y*root2.

    So you have 8 = 4y\sqrt2 - 5y\sqrt2...
    Oh yeah, sorry - I always get mixed up with denominators.

    Really sorry everyone, I know I'm a noob :rolleyes:
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    Ok the strategy is to make 'y' the subject, so shift all the terms containing 'y' to one side, and the rest to other then factorize out 'y'.
 
 
 
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