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?
 A: One approach is to find the (potential) "switchover" points. Note that one rational function can only become bigger than another after either intersecting or after a hole or vertical asymptote (of one or both). (Why?) There our no holes, and our asymptotes occur at $x=0$ and $x=-\frac32$. Intersection occurs when $x\neq 0$, $x\neq-\frac32$ and $$\frac{2x-13}{2x+3}=\frac{15}x$$ $$2x^2-13x=30x+45$$ $$2x^2-43x-45=0$$ $$2x^2-45x+2x-45=0$$ $$(2x-45)(x+1)=0.$$ Thus, our other possible switchover points are $x=-1$ and $x=\frac{45}2$.
Now, check the inequality on the intervals $\left(-\infty,-\frac32\right)$, $\left(-\frac32,-1\right)$, $(-1,0)$, $\left(0,\frac{45}2\right)$, and $\left(\frac{45}2,\infty\right)$ by using test points in each interval. Note that for the first interval a sign chart would readily tell us that the desired inequality fails (without need for a test point), since $\cfrac{2x-13}{2x+3}$ is positive and $\cfrac{15}x$ is negative.
A: Hint: From here:
$$\frac{2x^2 - 43x - 45}{2x(x + \frac{3}{2})} \lt 0$$
you could say that the fraction is less than zero if and only if either (1) the numerator is positive and the denominator is negative or (2) the numerator is negative and the denominator is positive. So for case (2), you solve:
$$
2x^2 - 43x - 45 <0 \quad \text{and}\quad 2x(x +\frac{3}{2}) > 0.
$$
You might already know this, but the way you solve each of these two inequalities is by finding possible roots. So for example for the $2x(x+ \frac{3}{2}) > 0$, the roots are $x = 0$ and $x = -\frac{3}{2}$. Now you just need to find the (constant) signs of the expression $2x(x+\frac{3}{2})$ on the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, 0)$, and $(0, \infty)$.
A: After simplifying you have $\frac{2x^2-43x-45}{x(2x+3)}=\frac{(2x-45)(x+1)}{x(2x+3)}<0$. So we consider the function on the intervals  $x<-\frac{3}{2}$, $x \in (-3/2,-1)$, $x \in (-1,0)$, $x \in (0,\frac{45}{2})$, $x>\frac{45}{2}$. At $x=-2$ the function is positive and the sign alternates  in each successive region.
Thus the function is negative for $x \in (-3/2,-1)$ and $x \in (0,\frac{45}{2})$.
A: 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 $$x \in (-\frac{3}{2}, -1) \cup (0, \frac{45}{2}).$$
A: There is at least one incorrect "solution" above. 
The usual trick is to multiply both sides by something positive. So multiply both sides by $x^2(2x+3)^2$. Next, eliminate common factors in the numerators and denominators, bring everything over to the left hand side and then factorise. The inequality reduces to 
$x(2x-45)(2x+3)(x+1) < 0 \, .$
You need to find the values of $x$ for which an odd number of $x$, $2x-45$, $2x+3$ and $x+1$ are negative. So clearly $x = 0$, $x = 45/2$, $x = -3/2$ and $x = -1$ play an important role: these are the values for which the factors change sign. The final result, contrary to Iuli's "solution", is: $-3/2 < x < -1$ or $0 < x < 45/2$.
To see why Iuli's "solution" is incorrect, let $x = -1/2 \in (-3/2,45/2) \backslash \{0\}$, yet
$-7 \not< -30 \, .  $
