# How do you solve and write the following in interval notation: |8-3x|>5?

Apr 11, 2017

See the entire solution process below:

#### Explanation:

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

$- 5 > 8 - 3 x > 5$

First, subtract $\textcolor{red}{8}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- \textcolor{red}{8} - 5 > - \textcolor{red}{8} + 8 - 3 x > - \textcolor{red}{8} + 5$

$- 13 > 0 - 3 x > - 3$

$- 13 > - 3 x > - 3$

Now, divide each segment by $\textcolor{b l u e}{- 3}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative term we must reverse the inequality operators:

$\frac{- 13}{\textcolor{b l u e}{- 3}} \textcolor{red}{<} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{<} \frac{- 3}{\textcolor{b l u e}{- 3}}$

$\frac{13}{3} \textcolor{red}{<} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{<} 1$

$\frac{13}{3} \textcolor{red}{<} x \textcolor{red}{<} 1$

Or

$x > \frac{13}{3}$ and $x < 1$

Or, in interval form:

$\left(\frac{13}{3} , \infty\right)$ and $\left(- \infty , 1\right)$

Apr 11, 2017

$\left(- \infty , 1\right) \cup \left(\frac{13}{3} , + \infty\right)$

#### Explanation:

Inequalities of the form $| x | > a$ have solutions in the form.

$x < - a \textcolor{red}{\text{ or }} x > a$

$\Rightarrow 8 - 3 x < - 5 \textcolor{red}{\text{ OR }} 8 - 3 x > 5$

$\Rightarrow - 3 x < - 13 \textcolor{red}{\text{ OR }} - 3 x > - 3$

$\textcolor{w h i t e}{X X X} \Rightarrow x > \frac{13}{3} \textcolor{red}{\text{ OR }} x < 1$

[Remember to $\textcolor{b l u e}{\text{reverse signs}}$ when multiplying or dividing by a color(blue)"negative quantity"]

$\text{Solution is " x<1 " or } x > \frac{13}{3}$

$\text{Expressed in interval notation as}$

$\left(- \infty , 1\right) \cup \left(\frac{13}{3} , + \infty\right)$