# How do I correctly solve $| 1 + {3\over x} | > 2$?

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 .

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

$| 1 + \frac{3}{x} |>2$ 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 $-1<x<3$.

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

• 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. Dec 14, 2015 at 14:31
• @Ricky Y.Fan didn't multiply by $x$ but rather by $x^2.$ Dec 14, 2015 at 14:33
• @coffeemath Sorry , missed that . Thanks for helping out . Dec 14, 2015 at 14:36

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.