How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$ First I looked the x that doesnt belong to this function.
$$|4x -6| - 21 \neq 0$$
$$ x \neq \frac{-15}{4}$$ and $$ x \neq \frac{27}{4}$$
Then I found the roots of the x
$$x = 0$$
$$x = \frac{3}{2}$$
After I found the roots I wrote the inequality like this:
$$ x + 14|x| - 10 > 3|4x-6| - 63$$
$$ x + 14|x| > 3|4x-6| - 53$$
to finish we may write differents function for each values of x based on roots we found.
for $$ x<0$$ the function is $$ x - 14x > 3(6-4x) - 53$$
for $$x<\frac{3}{2}$$ we have $$ x + 14x > 3(6-4x) - 53$$ for $$x \geq \frac{3}{2}$$ the function is $$ x + 14x > 3(4x-6) - 53$$
But using this inequalities I couldnt find the solutions for x!! If we calculate that on wolfram we can see the solutions for x are $$ x < \frac{-15}{4}$$ and $$ x > \frac{27}{4}$$. Can anyone explain me why?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{{x + 14\verts{x} - 10 \over \verts{4x - 6} - 21} > 3:\ {\Large ?}}$.

$\ds{\Large x < 0:\ ?}$
\begin{align}
&3 < \left.{-13x - 10 \over -15 - 4x}\right\vert_{\ x\ \not=\ -15/4} \implies
3\pars{-15 - 4x}^{2} < \pars{-13x - 10}\pars{-15 - 4x}
\\[5mm] &\
\implies x^{2} - {125 \over 4}\,x - {525 \over 4} > 0
\implies \pars{x < -\,{15 \over 4}}\ \mbox{or}\ \pars{x > 35}
\end{align}
$$
\bbx{x < -\,{15 \over 4}}
$$

$\ds{\Large 0 \leq x < {3 \over 2}:\ ?}$
\begin{align}
&3 < {15x - 10 \over -15 - 4x} \implies
3\pars{-15 - 4x}^{2} < \pars{15x - 10}\pars{-15 - 4x} \implies
x^{2} + {545 \over 108}\,x + {525 \over 108} < 0
\\[5mm] &\
\implies \pars{-\,{15 \over 4} < x < -\,{35 \over 27}}
\end{align}
$$
\bbx{\mbox{There's not any solution in this case}}
$$

$\ds{\Large x \geq {3 \over 2}:\ ?}$
\begin{align}
&3 < \left.{15x - 10 \over 4x - 27}\right\vert_{\ x\ \not=\ -15/4} \implies
3\pars{4x - 27}^{2} < \pars{15x - 10}\pars{4x - 27}
\\[5mm] &\
\implies x^{2} + {203 \over 12}\,x - {639 \over 4} > 0
\implies \pars{x < -\,{71 \over 3}}\ \mbox{or}\ \pars{x > {27 \over 4}}
\end{align}
$$
\bbx{x > {27 \over 4}}
$$

$$
\bbox[15px,#ffd,border:2px groove navy]{\ds{\mbox{A solution}\ x \in
\mathbb{R}\setminus\bracks{-\,{15 \over 4},{27 \over 4}}}}
$$
A: ok, in the first case we assume that $$x\geq 0$$ and $$x\geq \frac{3}{2}$$ then our term is given by $$\frac{15x-10}{4x-27}$$ so we have $$x\geq \frac{3}{2}$$ and $$x\ne \frac{27}{4}$$
Further you must consider $$0\le x<\frac{3}{2}$$ and $$x<0$$ .Can you finish?
A: Your way is good, first consider


*

*$|4x-6|-21> 0$


and we have


*

*$x<\frac32\implies -4x+6-21> 0 \implies x<-\frac {15} 4$

*$x\ge\frac32\implies 4x-27> 0 \implies x>\frac {27} 4$


then


*

*$|4x-6|-21> 0$ for $x<-\frac {15} 4$ and $x>\frac {27} 4$

*$|4x-6|-21< 0$ for $-\frac {15} 4<x<\frac {27} 4$

*$|4x-6|-21= 0$ for $x=-\frac {15} 4$ and $x=\frac {27} 4$


Then consider the cases

1) $-\frac {15} 4<x<\frac {27} 4$



*

*$\frac{x+14|x|-10}{ |4x-6|-21 }>3\implies x+14|x|-10<3|4x-6|-63$


and


*

*$x<0 \implies -13x-10<3(-4x+6)-63$

*$0\le x<\frac32 \implies 15x-10<3(-4x+6)-63$

*$x\ge\frac32 \implies 15x-10<3(4x-6)-63$



2) $x<-\frac {15} 4$ and $x>\frac {27} 4$



*

*$\frac{x+14|x|-10}{ |4x-6|-21 }>3\implies x+14|x|-10>3|4x-6|-63$


and


*

*$x<-\frac {15} 4 \implies -13x-10<3(-4x+6)-63$

*$x>\frac {27} 4 \implies 15x-10<3(4x-6)-63$

A: Transform the inequality:
$$\frac{x+14|x|-10-3|4x-6|+63}{|4x-6|-21}>0.$$
Consider the $3$ cases:
$$1) \ \begin{cases} x<0 \\ \frac{x-35}{4x+15}>0\end{cases} \Rightarrow \begin{cases} x<0 \\ x<-\frac{15}{4} \ \ \text{or} \ \  x>35\end{cases} \Rightarrow \color{blue}{x<-\frac{15}{4}}.$$
$$2) \ \begin{cases} 0<x<\frac32 \\ \frac{27x+35}{-4x-15}>0\end{cases} \Rightarrow \begin{cases} 0<x<\frac32 \\ -\frac{15}{4}<x<-\frac{35}{27}\end{cases} \Rightarrow \emptyset.$$
$$3) \ \begin{cases} x>\frac32 \\ \frac{3x+71}{4x-27}>0 \end{cases} \Rightarrow \begin{cases} x>\frac32 \\ x<-\frac{71}{3} \ \ \text{or} \ \ x>\frac{27}{4}\end{cases} \Rightarrow \color{blue}{x>\frac{27}{4}}.$$
