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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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+1 for showing thought and where the question is. It allows better answers, as you got. –  Ross Millikan Sep 10 '11 at 4:19
    
@jie Your last step has a typo. The first $(2x-3)$ factor should in fact be $(2x+3)$. –  Srivatsan Sep 10 '11 at 15:12
    
@Srivatsan Narayanan, thanks I corrected that –  Jiew Meng Sep 11 '11 at 0:10
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1 Answer 1

up vote 4 down vote accepted

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

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

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