# How do you solve 3 < absx – 5?

Aug 15, 2015

$x \in \left(- \infty , - 8\right) \cup \left(8 , + \infty\right)$

#### Explanation:

Start by isolating the modulus on one side of the inequality. You can do this by adding $5$ to both sides

$3 + 5 < | x | - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}$

$8 < | x |$

This is of course equivalent to

$| x | > 8$

Now, you need to take into account the fact that $x$ can be both positive or negative, which means that you get

• $x > 0 \implies | x | = x$

For positive values of $x$, the inequality will be

$x > 8$

• $x < 0 \implies | x | = - x$

For negative values of $x$, the inequality will be

$- x > 8$

Multiply both sides by $- 1$ to get $x$ on the left side of the inequality - do not forget that the sign of the inequality changes when you multiply or divide by a negative number

$- 1 \cdot \left(- x\right) \textcolor{red}{<} 8 \cdot \left(- 1\right)$

$x < - 8$

This means that your origininal inequality will be tru for any value of $x$ that is smaller than $- 8$ or bigger than $8$. In other words, you need $x$ to belong to two distinct intervals, $x < - 8$ and $x > 8$, which can be written as $x \in \left(- \infty , - 8\right) \cup \left(8 , + \infty\right)$