# How do you solve abs(32x-16)>32?

Nov 10, 2017

See a solution process below:

#### Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

$- 32 > 32 x - 16 > 32$

First, add $\textcolor{red}{16}$ to each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- 32 + \textcolor{red}{16} > 32 x - 16 + \textcolor{red}{16} > 32 + \textcolor{red}{16}$

$- 16 > 32 x - 0 > 48$

$- 16 > 32 x > 48$

Now, divide each segment by $\textcolor{red}{32}$ to solve for $x$ while keeping the system balanced:

$- \frac{16}{\textcolor{red}{32}} > \frac{32 x}{\textcolor{red}{32}} > \frac{48}{\textcolor{red}{32}}$

$- \frac{16}{\textcolor{red}{16 \times 2}} > \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{32}}} x}{\cancel{\textcolor{red}{32}}} > \frac{16 \times 3}{\textcolor{red}{16 \times 2}}$

$\frac{- \textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} 1}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{16}}} \times 2}} > \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{32}}} x}{\cancel{\textcolor{red}{32}}} > \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} \times 3}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{16}}} \times 2}}$

$- \frac{1}{2} > x > \frac{3}{2}$

Or

$x < - \frac{1}{2}$ and $x > \frac{3}{2}$

Or, in interval notation

$\left(\infty , - \frac{1}{2}\right)$ and $\left(\frac{3}{2} , + \infty\right)$