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    Solve for x : \dfrac {3x + 1}{x-1} > \dfrac {3x}{x+1}

    How would you go about solving this? Multiply by {x-1}^2 and {x+1}^2 ?
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    Christ, I am in awe of anyone who attempts maths during the holidays! I'm sorry but my brain just won't work between July and September:eek3:
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    Multiply by (x - 1)(x + 1), multiply out the brackets, things cancel, stick all the x's on one side.
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    (Original post by paddyman4)
    Multiply by (x - 1)(x + 1), multiply out the brackets, things cancel, stick all the x's on one side.
    x^2 - 1 is sometimes negative, which would reverse the sign of the inequality
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    (Original post by around)
    x^2 - 1 is sometimes negative, which would reverse the sign of the inequality
    Oh yeah, how embarrassing. So if I remember right then you do it how I said, but for the case of x>1 and then again for x<1.
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    Well ive subtract the RHS on the LHS then put them together and finally got :

     \dfrac{7x+1}{(x-1)(x+1)}&gt;0

    so if it i was = 0 id have x=1, x=-1 and x=-1/7

    so for > 0 what will the interval be?
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    Sugar , is this GCSE ?
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    easy:

    http://www91.wolframalpha.com/input/...%2F%28x%2B1%29
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    IB Higher Level, and i dont think we can solve using a GDC
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    (3x + 1)(x+1) &gt; (3x)(x-1)

    Cross multiply, collect terms, etc.`
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    (Original post by D-Day)
    (3x + 1)(x+1) &gt; (3x)(x-1)

    Cross multiply, collect terms, etc.`

    But we dont know if x+1 or x-1 is negative so would change the sign.
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    Ive solved a different way and got

     (x^2 -1)(7x+1)&gt;0

    but still unsure what the interval for x will be/
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    (Original post by Coded)
    But we dont know if x+1 or x-1 is negative so would change the sign.
    Changing the sign only happens when dividing by a negative constant. Here you are multiplying by an expression.
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    I got x&gt;-\frac {1}{7}
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    (Original post by Coded)
    Ive solved a different way and got

     (x^2 -1)(7x+1)&gt;0

    but still unsure what the interval for x will be/
    x^2 - 1 is positive for x > 1 or x < -1, and negative for -1 < x < 1. 7x + 1 is positive for x > -1/7, negative for x < -1/7. So in which intervals will their product be positive?

    (Original post by D-Day)
    I got x&gt;-\frac {1}{7}
    Disagree. Try x = 0.
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    ^ No, at D-Day (x=0 doesn't work!). The answer is, I believe, x>1 or -1<x<-1/7

    Just solve cases for positive and negative x!

    EDIT: Damn you GE! Always quicker!
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    (Original post by generalebriety)
    x^2 - 1 is positive for x > 1 or x < -1, and negative for -1 < x < 1. 7x + 1 is positive for x > -1/7, negative for x < -1/7. So in which intervals will their product be positive?
    im liking that, that you.

    so x > 1 or -1 < x < -1/7
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    (Original post by generalebriety)
    Disagree. Try x = 0.
    :facepalm:
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    You can't directly re-arrange it as an inequality with x as the subject because x-1 and x+1 can be either negative or positive

    You need to solve it as an equality to find the critical x values, then either draw a graph with both curves on it to find the intervals. (you could also use a table of values I think, but I've always used graphs for things like this so I don't forget to take asymptotes into account)
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    Or you can do what OP originally suggested (as none of those expressions are ever negative)
 
 
 
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