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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??

Please help!!

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  • $\begingroup$ you can't multiple by a variable term both the sides to solve a inequality. @AtulMishra $\endgroup$ – Harsh Kumar Jul 9 '17 at 8:43
  • $\begingroup$ I don't believe it @HarshKumar $\endgroup$ – Atul Mishra Jul 9 '17 at 8:46
  • $\begingroup$ Just read the answer of Jean Marie... $\endgroup$ – Harsh Kumar Jul 9 '17 at 8:49
  • $\begingroup$ He said about the negatives your comment is about everything @HarshKumar $\endgroup$ – Atul Mishra Jul 9 '17 at 8:50
  • $\begingroup$ 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.) $\endgroup$ – Hans Lundmark Jul 9 '17 at 11:47
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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

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

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  • $\begingroup$ Ahh , I got that.... $\endgroup$ – Atul Mishra Jul 9 '17 at 8:46
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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: $x\in(-5,-2)\cup(-1,\infty)$

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

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

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