How do you solve and write the following in interval notation: 1/|x+5| >2?

Jul 27, 2016

$- \frac{11}{2} < x < - 5$ and $- 5 < x < = \frac{9}{2}$

Explanation:

First of all, we have to exclude the value when the inequality is undefined because the denominator is zero:
$x \ne - 5$

In all cases, except this, $| x + 5 | > 0$.

Both sides of an inequality can be multiplied by a POSITIVE number, leaving the sign of inequality as is.

Since $| x + 5 | > 0$ (after we have excluded value $x = - 5$), let's multiply left and right sides of our inequality by $| x + 5 |$ getting
$\frac{1 \cdot | x + 5 |}{| x + 5 |} > 2 \cdot | x + 5 |$
or
$1 > 2 \cdot | x + 5 |$

Next transformation is division of both sides of inequality by POSITIVE number $2$ preserving the sign of inequality:
$\frac{1}{2} > | x + 5 |$

Recall the definition of $| a |$:
$| a | = a$ for $a \ge 0$ and
$| a | = - a$ for $a < 0$

Consider now two cases (we excluded $x = - 5$):

Case 1. $x + 5 > 0$ or, equivalently, adding $- 5$ to both parts of inequality preserving the sign of inequality, $x > - 5$
Then $| x + 5 | = x + 5$ and our inequality looks like
$\frac{1}{2} > x + 5$
or, subtracting $5$ from both sides and retaining the sign of inequality, $x < - \frac{9}{2}$
We have to combine this with an inequality that defines our case,
$x > - 5$.
Both inequalities result in'
$- 5 < x < - \frac{9}{2}$

Case 2. $x + 5 < 0$ or, equivalently, adding $- 5$ to both parts of inequality preserving the sign of inequality, $x < - 5$
Then $| x + 5 | = - x - 5$ and our inequality looks like
$\frac{1}{2} \succ x - 5$
or, add $x$ to both sides and retaining the sign of inequality, $x + \frac{1}{2} > - 5$
or, subtracting $\frac{1}{2}$ from both sides,
$x > - \frac{11}{2}$
We have to combine this with an inequality that defines our case,
$x < - 5$.
Both inequalities result in'
$- \frac{11}{2} < x < - 5$

Final solution is
$- \frac{11}{2} < x < - 5$ and $- 5 < x < = \frac{9}{2}$

Here is an illustrative graph of a function $y = \frac{1}{|} x + 5 | - 2$. Notice, where it is positive.
graph{1/|x+5|-2 [-10, 2, -5, 5]}