<s>[INST] Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$. 
But how do I get the last step? Or am I already dead wrong?
 [/INST] you have to resolve this inequality depending $x$ : $$\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$$
that is equivalent with: $\displaystyle \frac{x(2x-13)-15(2x+3)}{x(2x+3)} \lt 0$ so: 
$\displaystyle \frac{2x^2-13x-30x-45}{2x^2+3x} \lt 0$ and now the simple form is :
$$\frac{2x^2-43x-45}{2x^2+3x} \lt 0$$
Now we have two equation . first: $2x^2-43x-45=0$ with : $\displaystyle x_{1,2}=-1 \mbox{and} \frac{45}{2}$
and second $2x^2+3x=0$ with $\displaystyle x_{1,2}=-\frac{3}{2} \mbox{and} 0$. 
For that inequality to be satisfied it must: $\displaystyle 2x^2-43x-45 \lt0$ and $\displaystyle 2x^2+3x \gt 0$  (1) OR $2x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$  (2). 
I attach a draw. it is easy now to view the solutions 

for first case: $2x^2-43x-45 \lt 0$ when $\displaystyle x \in (-1, \frac{45}{2})$ and $2x^2+3x \gt 0$ equivalent with $x\in (-\infty, -\frac{3}{2}) \cup (0, \infty)$. Now the solution for this case is the intersection :
$$ (-1, \frac{45}{2}) \cap( (-\infty, -\frac{3}{2}) \cup (0, \infty))=(0, \frac{45}{2}) $$ and now the answer for second case is: $x^2-43x-45 \gt 0$ and $2x^2+3x \lt 0$. We have for first inequality $x\in (-\infty, -1) \cup (\frac{45}{2}, \infty)$ and for second inequality we have $x \in (-\frac{3}{2},0)$. So the answer for second case is :
$$(-\frac{3}{2},0) \cap((-\infty, -1) \cup (\frac{45}{2}, \infty))=(-\frac{3}{2},-1)$$
And the final answer is $$x \in (-\frac{3}{2}, -1) \cup (0, \frac{45}{2}).$$
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