How do you solve $- \frac{2}{x - 10} < 0$ $-2/(x-10)<0$ ?

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Answer 1 · 2017-09-22T12:17:29

Solution : $x > 10 or (10, \infty)$ $x > 10 or (10,oo)$

$- \frac{2}{x - 10} < 0 or \frac{2}{x - 10} > 0$ $- 2/(x-10) < 0 or 2/(x-10) > 0$ . Critical point is

$x - 10 = 0 or x = 10; x \neq 10$ $x -10=0 or x=10 ; x != 10$ since for $x = 10$ $x=10$ , the function

is undefined.

Sign change observation:

When $x < 10$ $x <10$ sign of $\frac{2}{x - 10}$ $2/(x-10)$ is $-; i . e < 0$ $- ; i.e <0$

When $x > 10$ $x >10$ sign of $\frac{2}{x - 10}$ $2/(x-10)$ is $+; i . e > 0$ $+ ; i.e > 0$

Solution : $x > 10 or (10, \infty)$ $x > 10 or (10,oo)$ [Ans]

How do you solve -2/(x-10)<0−2x−10<0-2/(x-10)<0?