Equation Exercise asked in math group ( Level -Secondary classes )

Exercise:

Solve in $\mathbb{R},\,\,\,\frac{3\left| 3x+1 \right|}{x+5}<3x$

Solution: we have $\frac{3\left| 3x+1 \right|}{x+5}<3x\Leftrightarrow \frac{3\left| 3x+1 \right|-3x\left( x+5 \right)}{x+5}<0$

Observe that, $x+5\ne 0\Leftrightarrow x\ne -5$

Let $f\left( x \right)=3\left| 3x+1 \right|-3x\left( x+5 \right)=3\left( \left| 3x+1 \right|-x\left( x+5 \right) \right)$ so $\frac{f\left( x \right)}{x+5}<0$

Observe that $x+5<0$ when $x<-5$

Now if $x>\frac{-1}{3}\Leftrightarrow f\left( x \right)=3\left( 3x+1-{{x}^{2}}-5x \right)=3\left( -{{x}^{2}}-2x+1 \right)=-3\left( {{x}^{2}}+2x-1 \right)$

But ${{x}^{2}}+2x-1=0\Leftrightarrow {{\left( x+1 \right)}^{2}}=2\Leftrightarrow x=-1\pm \sqrt{2}$

\[\begin{align}
  & \underline{\left. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,-5\,\,\,\,\,\,\,\,\,\,\,\,\,-1-\sqrt{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1+\sqrt{2}\,\,\,\,\,\,} \\
 & \underline{-3\left. \left( {{x}^{2}}+2x-1 \right)\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,-\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0-\,\,\,\,\,}\,\, \\
 & \underline{\,\,\,\,\left. \,x+5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,-\,\left. {} \right\|\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,+}\,\,\, \\
 & \left. \frac{-3\left( {{x}^{2}}+2x-1 \right)}{x+5}<0 \right|\,\,\,\,\,\,\,\,\,\,+\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,- \\
\end{align}\]

So ${{S}_{1}}=\left( -5;-1-\sqrt{2} \right)\cup \left( -1+\sqrt{2};+\infty  \right)$

Now if $x<-\frac{1}{3}\Leftrightarrow f\left( x \right)=3\left( -3x-1-{{x}^{2}}-5x \right)=3\left( -{{x}^{2}}-8x-1 \right)=-3\left( {{x}^{2}}+8x+1 \right)$

But ${{x}^{2}}+8x+1=0\Leftrightarrow {{\left( x+4 \right)}^{2}}=15\Leftrightarrow x=-4\pm \sqrt{15}$

\[\begin{align}
  & \underline{\left. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,-4-\sqrt{15}\,\,\,\,\,\,\,\,\,\,\,\,\,-5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4+\sqrt{15}\,\,\,\,\,\,} \\
 & \underline{-3\left. \left( {{x}^{2}}+8x+1 \right)\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,-\,\,0\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0-\,\,\,\,\,}\,\, \\
 & \underline{\,\,\,\,\left. \,x+5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,-\,0\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. 0 \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,+}\,\,\, \\
 & \left. \frac{-3\left( {{x}^{2}}+8x+1 \right)}{x+5}<0 \right|\,\,\,\,\,\,\,\,\,\,+\,\,\,0\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,- \\
\end{align}\]

So ${{S}_{2}}=\left( -4-\sqrt{15},-5 \right)\cup \left( -4+\sqrt{15},+\infty  \right)$

Thus $S={{S}_{1}}\cap {{S}_{2}}=\left( -4-\sqrt{15},-5 \right)\cup \left( -1+\sqrt{2};+\infty  \right)$