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    (Original post by around)
    Or you can do what OP originally suggested (as none of those expressions are ever negative)
    The LHS is \frac{3x+1}{x-1} and when x=0 surely the expression is negative.

    Also if x=-0.2 the RHS is negative.
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    you will get (7x+1)(x+1)(x-1)>0 finally,then,look
    (1).suppose 7x+1>0,get x>-1/7,then(x+1)(x-1)>0;get x>1 & x<-1
    the intersection of x>-1/7 and x>1 & x<-1 will be x>1
    (2).suppose 7x+1<0,get x<-1/7,then(x+1)(x-1)<0;get -1<x<1
    the intersection of x<-1/7 and -1<x<1 will be -1<x<-1/7
    Union (1)and(2) will be the answer -1<x<-1/7 & x>1

    this inequality will be more diffcult when replace > with >=
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    (Original post by steve2005)
    The LHS is \frac{3x+1}{x-1} and when x=0 surely the expression is negative.

    Also if x=-0.2 the RHS is negative.
    Multiply by {x-1}^2 and {x+1}^2 ?
    I meant these.
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    (Original post by D-Day)
    Changing the sign only happens when dividing by a negative constant. Here you are multiplying by an expression.
    Not true. When you multiply through by variables you have to consider the cases when what you're multiplying by is negative. So, in this case, you'd have to consider cases where:
    1. x + 1 \ge 0 and x - 1 \ge 0, that is x \ge 1
    2. x + 1 \ge 0 and x - 1 &lt; 0, that is -1 \le x &lt; 1
    3. x + 1 &lt; 0 and x - 1 &lt; 0, that is x &lt; -1

    It makes no difference whether it's constant or not.
 
 
 
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