Solving the inequality $\frac{|x + 3| + x}{x + 2} > 1$

I have an inequality which is as follows

If $x\in \mathbb R$ , solve the following inequality for $x$ $$\frac{|x+3|+x}{x+2}\gt 1$$

I rearranged and got this $$|x+3|\gt2$$

solving for it gives me $x\in(-\infty,-5)\cup (-1,\infty)$

But putting $-6$ in original inequality is making the statement false

I told it to my friend and he got $x\in R-[-2,-1]-\{-5\}$

And he is right ,

But I want to know why am I wrong and how to approach the correct answer??

• you can't multiple by a variable term both the sides to solve a inequality. @AtulMishra – Harsh Kumar Jul 9 '17 at 8:43
• I don't believe it @HarshKumar – Atul Mishra Jul 9 '17 at 8:46
• Just read the answer of Jean Marie... – Harsh Kumar Jul 9 '17 at 8:49
• He said about the negatives your comment is about everything @HarshKumar – Atul Mishra Jul 9 '17 at 8:50
• To be more precise, you cannot multiply by an expression whose sign you don't know. (But it would be fine to multiply by $x^2+1$, for example, since it's always $\ge 1$ and hence positive.) – Hans Lundmark Jul 9 '17 at 11:47

You error comes from the fact that, in your "rearranging" process, you have (maybe without noticing it) had to multiply both sides of your inequality by $(x+2)$ but, doing that, you have implicitly assumed that $x+2>0$. If, on the contrary, $x+2<0$, you have to reverse the inequality symbol, i.e., your inequality becomes

$$(x+2)\left(\frac{|x+3|+x}{x+2}\right)\lt (x+2) \ \ \iff \ \ |x+3|+x \lt x+2 \ \ \iff \ \ |x+3| \lt 2$$

• Ahh , I got that.... – Atul Mishra Jul 9 '17 at 8:46

I like to make it easy to myself by first solving:

$$f(x)=\frac{|x+3|+x}{x+2}-1=0$$

We get:

$$|x+3|+x=x+2$$ $$|x+3|=2$$ $$x=-5 \lor x=-1$$

Furthermore, $x$ cannot be $-2$ else we would divide by $0$. $f(x)$ is a continuous function except at $x=-2$. By simply choosing some values for $x$ we get:

$$\begin{array}{c|lcr} x & &-5 && -2 && -1 \\ \hline f(x) & -&0&+ & \text{X}&- & 0&+\\ \end{array}$$

Therefore: $x\in(-5,-2)\cup(-1,\infty)$

• I think it's the best way here. +1 – Michael Rozenberg Jul 9 '17 at 12:39

your inequality is equivalent to $$\frac{|x+3|-2}{x+2}>0$$ now we do case work: 1)$$x\geq -3$$ then we get $$\frac{x+1}{x+2}>0$$ a) if $x+2>0$ then we have $x+1>x+2$ which is impossible b) if $x+2<0$ then $x+1<x+2$ which is true. and we get $$-3\le x<-2$$ can you finish? you can not multiply be $x+2$ because this can be negative

• But what was mistake in my process @Dr.SonnhardGraubner sir? – Atul Mishra Jul 9 '17 at 8:45