# How do you solve 8/(x-1)<1?

Dec 25, 2016

The answer is x in ] -oo,1 [ uu ] 9, +oo[

#### Explanation:

We cannot do crossing over

So,

$\frac{8}{x - 1} < 1$, $\implies$, $\frac{8}{x - 1} - 1 < 0$

$\frac{8 - \left(x - 1\right)}{x - 1} < 0$

$\frac{9 - x}{x - 1} < 0$

Let $f \left(x\right) = \frac{9 - x}{x - 1}$

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}$$1$$\textcolor{w h i t e}{a a a a}$$9$$\textcolor{w h i t e}{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}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$9 - x$$\textcolor{w h i t e}{a a a 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}$$-$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$-$

Therefore,

$f \left(x\right) < 0$, when x in ] -oo,1 [ uu ] 9, +oo[