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

Question

I am trying to solve for the following inequality:

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

In the given answer,

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

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

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

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

+1 for showing thought and where the question is. It allows better answers, as you got. — Ross Millikan, Sep 10 '11 at 04:19
@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

score 4 · Accepted Answer · answered Sep 10 '11 at 02:48

4

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

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

answered Sep 10 '11 at 02:48

Ragib Zaman

1 Answers1