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I am trying to solve for the following inequality:

$\frac{12}{2x-3}<1+2x$

In the given answer,

$\frac{12}{2x-3}-(1+2x)<0$

$\frac{-(2x+3)(2x-5)}{2x-3}<0 \rightarrow \textrm{ How do I get to this step?}$

$\frac{(2x+3)(2x-5)}{2x-3}>0$

$(2x+3)(2x-5)(2x-3)>0 \textrm{ via multiply both sides by }(2x-3)^2$

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    @Srivatsan Narayanan, thanks I corrected that2011-09-11

1 Answers 1

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$ \frac{12}{2x-3} - (1-2x) = \frac{12 - (1+2x)(2x-3) }{2x-3} = \frac{ 12 - (2x-3+4x^2-6x)}{2x-3} $

$= - \frac{4x^2-4x-15}{2x-3} = - \frac{(2x+3)(2x-5)}{2x-3} $