# How do you solve and write the following in interval notation: (x+1) / (x-1) <=2?

the interval is $\left[3 , \infty\right)$

#### Explanation:

the given inequlity is $\frac{x + 1}{x - 1} \le 2$ $\Rightarrow \left(x + 1\right) \le 2 \left(x - 1\right) \Rightarrow \left(x + 1\right) \le 2 x - 2$ $\Rightarrow \left(x - 2 x + 1 + 2\right) \le 0 \Rightarrow \left(- x + 3\right) \le 0$ $\Rightarrow \left(x - 3\right) \ge 0 \Rightarrow x \ge 3$
$\therefore x \in \left[3 , \infty\right)$

Nov 1, 2017

The solution is $x \in \left(- \infty , 1\right) \cup \left[3 , + \infty\right)$

#### Explanation:

We cannot do crossing over

Let's rewrite the expression

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

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

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

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

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

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

Let's build a 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 a a a}$$1$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$3 - x$$\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}$$0$$\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}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , 1\right) \cup \left[3 , + \infty\right)$

graph{(x+1)/(x-1)-2 [-12.66, 12.65, -6.33, 6.33]}