# How do you solve and write the following in interval notation: 1+ 20/(x + 1) ≤ 20/x?

May 10, 2017

The solution is $x \in \left[- 5 , 1\right) \cup \left(0 , 4\right]$

#### Explanation:

We cannot do crossing over.

So, we rewrite, simplify and factorise the inequality

$1 + \frac{20}{x + 1} \le \frac{20}{x}$

$\frac{x + 1 + 20}{x + 1} \le \frac{20}{x}$

$\frac{x + 1 + 20}{x + 1} - \frac{20}{x} \le 0$

$\frac{x \left(x + 21\right) - 20 \left(x + 1\right)}{x \left(x + 1\right)} \le 0$

$\frac{{x}^{2} + 21 x - 20 x - 20}{x \left(x + 1\right)} \le 0$

$\frac{{x}^{2} + x - 20}{x \left(x + 1\right)} \le 0$

$\frac{\left(x - 4\right) \left(x + 5\right)}{x \left(x + 1\right)} \le 0$

Let $f \left(x\right) = \frac{\left(x - 4\right) \left(x + 5\right)}{x \left(x + 1\right)}$

We can construct 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}$$- 5$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\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}$$+$$\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}$$+$

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

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

$\textcolor{w h i t e}{a a a a}$$x - 4$$\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}$$-$$\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}$$+$

$\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 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}$$+$$\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}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[- 5 , 1\right) \cup \left(0 , 4\right]$