How do you solve 1/(x+1)>2/(x-1)?

Feb 20, 2017

$x < - 3$

Explanation:

$\frac{1}{x + 1} > \frac{2}{x - 1}$

$\frac{1}{x + 1} - \frac{2}{x - 1} > 0$

$\frac{\left(x - 1\right) - 2 \left(x + 1\right)}{{x}^{2} - 1} > 0$ [ $x \ne \pm 1$]

$- x - 3 > 0 \to x + 3 < 0$

$\therefore x < - 3$

Feb 20, 2017

The solution is x in ]-oo, -3[uu]-1,1[

Explanation:

We cannot do crossing over

Let's rearrange the equation

$\frac{1}{x + 1} > \frac{2}{x - 1}$

$\frac{2}{x - 1} - \frac{1}{x + 1} < 0$

$\frac{2 x + 2 - x + 1}{\left(x + 1\right) \left(x - 1\right)} < 0$

$\frac{x + 3}{\left(x + 1\right) \left(x - 1\right)} < 0$

Let $f \left(x\right) = \frac{x + 3}{\left(x + 1\right) \left(x - 1\right)}$

We can build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$ when x in ]-oo, -3[uu]-1,1[#