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My solving . Tell me where I made the wrong move .

$ | 1 + {3\over x} | > 2 $

$ \implies 1 + {3\over x} > 2 $ or $ 1 + {3\over x} < -2 $

$ x \neq 0 $ and $ ( 1 + {3\over x} ) = 0 $ at $ x = -3 $

Case 1 :

When $ x \leq -3 $ $ \implies ( 1 + {3\over x} \geq 0 ) $

$ 1 + {3\over x} > 2 $

$ \implies {3\over x} > 1 $

but considering domain of x ( $ \leq -3 $ ) the above condition wont be true for any value of x .

but if I resume arithmetically :

$ {3\over x} > 1 \implies {x\over 3} < 1 \implies x < 3 $

!! I am confused .

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  • $\begingroup$ I noticed that 1 + 3/x > 0 for two intervals : $ x < -3 $ or $ R_{+} $ . $\endgroup$
    – Ricky
    Dec 14, 2015 at 14:35

3 Answers 3

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$| 1 + \frac{3}{x} |>2$ so you should solve $1+\frac{3}{x}>2$ for x>0, and you would obtain x<3 and solve $1+\frac{3}{x}<-2$ for $x <0$ and you would get $x>-1$.So $-1<x<3$.

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Hint: $$1+\frac{3}{x}>2\iff\frac{3-x}{x}>0\iff(3-x)x>0\text{ with }x\neq0\iff0<x<3$$

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  • $\begingroup$ When you have $ ( 3 - x ) / x > 0 $ , how can you multiply x to both sides and retain the same inequality ( here , $ > $ ) , without knowing if x is positive or negative since on multiplication of negative term to both side , the inequality must be changed to opposite sides. $\endgroup$
    – Ricky
    Dec 14, 2015 at 14:31
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    $\begingroup$ @Ricky Y.Fan didn't multiply by $x$ but rather by $x^2.$ $\endgroup$
    – coffeemath
    Dec 14, 2015 at 14:33
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    $\begingroup$ @coffeemath Sorry , missed that . Thanks for helping out . $\endgroup$
    – Ricky
    Dec 14, 2015 at 14:36
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I just do one of the cases for you. First, we note that

$$\eqalign{ & 1 + {3 \over x} \gt 2 \cr & {3 \over x} \gt 1 \cr} $$

Now, we want to multiply by $x$ but we should consider that whether the $x$ is positive or negative

$$\left\{ \matrix{ x \gt 0 \to 3 \gt x \to 0 \lt x \lt 3 \hfill \cr x < 0 \to 3 \lt x\,\,\text{impossible} \hfill \cr} \right.$$

You can do the other case similarly.

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