Typical Absolute value inequality $$
\text{How to solve}\quad
\frac{\left\vert\,{x + 3}\,\right\vert + x}{x+2} > 1\quad{\large ?}.
$$ 
I tried and wrote two cases, once opening the mod as it is and then the other case opening the mod with a negative sign. 
I got the two cases as : $x\in (-\infty,-2)\cup (-1,\infty)$ and $x\in (-5,-2)$.
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The problem is I don't know whether to take union or intersection. Also, the answer I get is different from what's given in the book. Where am I going wrong? What's the best (errorless) way you would handle such problems with the modulus?
Thanks for your effort. 
 A: note that $$\frac{|x+3|+x}{x+2}=\frac{|x+3|-2}{x+2}>0$$
If $$x\geq -3$$ we get $$\frac{x+1}{x+2}>0$$
if $x>-2$ then we can multiply by $x+2$ and we get $$x>-1$$
if $x<-2$ then we get by multiplication with $x+2>0$ the solution set $$x<-1$$
Can you do the rest?
A: We need to solve $$\frac{|x+3|+x}{x+2}-1>0$$ or
$$\frac{|x+3|-2}{x+2}>0.$$
Now, $x+2=0$ for $x=-2$ and $|x+3|=2$ for $x=-1$ or $x=-5$, which by the intervals method gives the answer:
$$(-5,-2)\cup(-1,+\infty).$$
A: Instead of just "mindlessly" dividing into cases, write out the logical connectives "and" and "or", and everything should (hopefully) take care of itself:
$$
\begin{aligned}
\frac{|x+3|+x}{x+2}>1
\quad
\iff
\quad
& \Biggl( x+3 \ge 0 \quad\text{and}\quad \frac{(x+3)+x}{x+2}>1 \Biggr)
\\
&
\text{or} \quad
\Biggl( x+3 < 0 \quad\text{and}\quad \frac{-(x+3)+x}{x+2}>1 \Biggr)
\\[1em]
\iff \quad
& \dots
\end{aligned}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{x \not= -2}$:

\begin{align}
{\verts{x + 3} + x \over x + 2} > 1 & \implies
\pars{\verts{x + 3} + x}\pars{x + 2} > \pars{x + 2}^{2}
\\[3mm] & \implies
\bbx{\pars{\verts{x + 3} - 2}\pars{x + 2} > 0}
\\[3mm] & \implies
\left\{\begin{array}{lcl}
\ds{\pars{x + 5}\pars{x + 2} < 0} & \text{if} & \ds{x < -3}
\\[3mm]
\ds{\pars{x + 1}\pars{x + 2} > 0} & \text{if} & \ds{x > -3}
\end{array}\right.
\\[3mm] & \implies
\left\{\begin{array}{lcl}
\ds{\pars{-5 < x < -2}} & \text{if} & \ds{x < -3}
\\[3mm]
\ds{\pars{x < -2\quad \text{or}\quad x > -1}} & \text{if} & \ds{x > -3}
\end{array}\right.
\\[3mm] & \implies
\left\{\begin{array}{lcl}
\ds{\pars{-5 < x < -3}} & \text{if} & \ds{x < -3}
\\[3mm]
\ds{\pars{-3 < x < -2}\quad \text{or}\quad x > -1} & \text{if} & \ds{x > -3}
\end{array}\right.
\end{align}

Clearly, the solution is given by
  $\ds{\bbx{x \in \pars{-5,-2}\cup\pars{-1,\infty}}}$.

  Note that $\ds{x = -3}$ satisfies the initial inequality: Namely,
  $\ds{{\verts{\color{#f00}{-3} + 3} + \pars{\color{#f00}{-3}} \over \color{#f00}{-3} + 2} = 3 \color{#f00}{>} 1}$.

