# 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 $x\in (-5,-2)$.

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?

• If you are unsure which of two regions to use, just try an example. For instance, $1$ is in the union but not the intersection. Is $1$ a solution? Continuing along those lines, $-10$ is in the union...is it a solution? As to your second question, since you do not show any of your work we can't tell you where you are going wrong. – lulu Aug 12 '17 at 11:20
• Worth noting: as a guess, you may run into difficulties when you multiply both sides by $x+2$. Remember that you need to change the inequality when $x+2<0$. – lulu Aug 12 '17 at 11:25

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}

• Thanks so much! I got it! Finally! – Tanuj Aug 12 '17 at 14:28
• Great, glad to hear that! – Hans Lundmark Aug 12 '17 at 14:58

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?

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: $$(-5,-2)\cup(-1,+\infty).$$

• Wow that is great! Can you explain me what exactly is an interval method and how did you do it? – Tanuj Aug 12 '17 at 11:50
• @user38227 We need to draw the $x$- axes, and to put there points $-5$, $-3$ and $-1$. The sign of the function $f(x)=\frac{|x+3|+x}{x+2}-1$ on $(-1,+\infty)$ is $+$, which gives signs on all four intervals: $-,+,-,+$ because we have no double points. Since we need $+$, we just can write the answer. – Michael Rozenberg Aug 12 '17 at 11:55


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, $\ds{{\verts{\color{#f00}{-3} + 3} + \pars{\color{#f00}{-3}} \over \color{#f00}{-3} + 2} = 3 \color{#f00}{>} 1}$.