How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$

Question

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?

Answer 1

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?

Answer 2

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

• $x<-\frac {15} 4 \implies -13x-10<3(-4x+6)-63$
• $x>\frac {27} 4 \implies 15x-10<3(4x-6)-63$

Answer 3

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

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{x + 14\verts{x} - 10 \over \verts{4x - 6} - 21} > 3:\ {\Large ?}}$.

$\ds{\Large x < 0:\ ?}$ \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} $$ \bbx{x < -\,{15 \over 4}} $$

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$\ds{\Large 0 \leq x < {3 \over 2}:\ ?}$ \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} $$ \bbx{\mbox{There's not any solution in this case}} $$

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$\ds{\Large x \geq {3 \over 2}:\ ?}$ \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} $$ \bbx{x > {27 \over 4}} $$

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$$ \bbox[15px,#ffd,border:2px groove navy]{\ds{\mbox{A solution}\ x \in \mathbb{R}\setminus\bracks{-\,{15 \over 4},{27 \over 4}}}} $$