# How do you solve and write the following in interval notation: 3/(x-1) - 4/x>=1?

Sep 5, 2017

The solution is $x \in \left[- 2 , 0\right) \cup \left(1 , 2\right]$

#### Explanation:

Let's rearrange the inequality

$\frac{3}{x - 1} - \frac{4}{x} \ge 1$

$\frac{3}{x - 1} - \frac{4}{x} - 1 \ge 0$

$\frac{3 x - 4 \left(x - 1\right) - x \left(x - 1\right)}{x \left(x - 1\right)} \ge 0$

$\frac{3 x - 4 x + 4 - {x}^{2} + x}{x \left(x - 1\right)} \ge 0$

$\frac{4 - {x}^{2}}{x \left(x - 1\right)} \ge 0$

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

Let $f \left(x\right) = \frac{\left(2 - x\right) \left(2 + x\right)}{x \left(x - 1\right)}$

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}$$- 2$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 + x$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\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}$$+$$\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}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{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}$$-$$\textcolor{w h i t e}{a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a}$color(white)(aa)-$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$2 - x$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a}$color(white)(aaaa)+$\textcolor{w h i t e}{a}$color(white)(aa)+$\textcolor{w h i t e}{a}$color(white)(aa)+$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{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}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$$| |$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a}$$-$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left[- 2 , 0\right) \cup \left(1 , 2\right]$

graph{3/(x-1)-4/x-1 [-35.57, 37.5, -25.57, 10.98]}