# How do you solve and write the following in interval notation:  3 abs(4-5x)<=9?

Jan 9, 2018

See a solution process below: $\left[\frac{1}{5} , \frac{7}{5}\right]$

#### Explanation:

First, divide each side of the inequality by $\textcolor{red}{3}$ to isolate the absolute value term while keeping the inequality balanced:

$\frac{3 \left\mid 4 - 5 x \right\mid}{\textcolor{red}{3}} \le \frac{9}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left\mid 4 - 5 x \right\mid}{\cancel{\textcolor{red}{3}}} \le 3$

$\left\mid 4 - 5 x \right\mid \le 3$

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.

$- 3 \le 4 - 5 x \le 3$

Next, subtract $\textcolor{red}{4}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- 3 - \textcolor{red}{4} \le 4 - \textcolor{red}{4} - 5 x \le 3 - \textcolor{red}{4}$

$- 7 \le 0 - 5 x \le - 1$

$- 7 \le - 5 x \le - 1$

Now, divide each segment by $\textcolor{b l u e}{- 5}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

$\frac{- 7}{\textcolor{b l u e}{- 5}} \textcolor{red}{\ge} \frac{- 5 x}{\textcolor{b l u e}{- 5}} \textcolor{red}{\ge} \frac{- 1}{\textcolor{b l u e}{- 5}}$

$\frac{7}{5} \textcolor{red}{\ge} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 5}}} x}{\cancel{\textcolor{b l u e}{- 5}}} \textcolor{red}{\ge} \frac{1}{5}$

$\frac{7}{5} \textcolor{red}{\ge} x \textcolor{red}{\ge} \frac{1}{5}$

Or

$x \ge \frac{1}{5}$; $x \le \frac{7}{5}$

Or, in interval notation

$\left[\frac{1}{5} , \frac{7}{5}\right]$