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    \frac{1}{2x+1}>\frac{x}{3x-2}.
    Anyone know how to solve this equality?
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    \frac{1}{2x+1}-\frac{x}{3x-2} = \frac{(3x-2)-x(2x+1)}{(2x+1)(3x-2)} = \frac{-2x^2+2x-2}{(2x+1)(3x-2)}>0

    \frac{x^2-x+1}{(2x+1)(3x-2)}<0

    Now solve i.e. find when the denominator is zero then find the appropriate inequality.
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    (Original post by Math12345)
    \frac{1}{2x+1}-\frac{x}{3x-2} = \frac{(3x-2)-x(2x+1)}{(2x+1)(3x-2)} = \frac{-2x^2+2x-2}{(2x+1)(3x-2)}>0

    \frac{x^2-x+1}{(2x+1)(3x-2)}<0

    Now solve i.e. find when the denominator is zero then find the appropriate inequality.
    Latex is being an absolute pain, must be to do with the site's recent technical difficulties.
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    (Original post by JustDynamite)
    \frac{1}{2x+1}>\frac{x}{3x-2}.
    Anyone know how to solve this equality?
    As math12345 suggested, then by completing the square in the numerator you will notice that it is always negative. So, you're interested in when the denominator is negative as well for the whole thing to be positive.
 
 
 
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