# How do you solve the inequality 5+1/x>16/x?

Jan 18, 2017

The answer is x in ] -oo,0 [ uu ] 3, +oo[

#### Explanation:

Let's rewrite the inequality

$5 + \frac{1}{x} - \frac{16}{x} > 0$

$5 - \frac{15}{x} > 0$

$\frac{5 \left(x - 3\right)}{x} > 0$

Let $f \left(x\right) = \frac{5 \left(x - 3\right)}{x}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{0\right\}$

Now we can do 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}$$0$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\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}$$-$$\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 a}$$+$

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

Therefore,

$f \left(x\right) > 0$, when x in ] -oo,0 [ uu ] 3, +oo[