# How do you graph and solve: color(white)("d") 4-|5/3y+4|<2/5?

Jan 29, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{4}$ from each side of the inequality to isolate the absolute value function while keeping the inequality balanced:

$4 - \textcolor{red}{4} - \left\mid \frac{5}{3} y + 4 \right\mid < \frac{2}{5} - \textcolor{red}{4}$

$0 - \left\mid \frac{5}{3} y + 4 \right\mid < \frac{2}{5} - \left(\frac{5}{5} \times \textcolor{red}{4}\right)$

$- \left\mid \frac{5}{3} y + 4 \right\mid < \frac{2}{5} - \frac{20}{5}$

$- \left\mid \frac{5}{3} y + 4 \right\mid < - \frac{18}{5}$

Next, multiply each side of the inequality by $\textcolor{red}{- 1}$ to make each side positive while keeping the inequality balanced. However, because we are multiplying an inequality by a negative number we must reverse the inequality operator.

$\textcolor{red}{- 1} \times - \left\mid \frac{5}{3} y + 4 \right\mid > \textcolor{red}{- 1} \times - \frac{18}{5}$

$\left\mid \frac{5}{3} y + 4 \right\mid > \frac{18}{5}$

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.

$- \frac{18}{5} > \frac{5}{3} y + 4 > \frac{18}{5}$

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

$- \frac{18}{5} - \textcolor{red}{4} > \frac{5}{3} y + 4 - \textcolor{red}{4} > \frac{18}{5} - \textcolor{red}{4}$

$- \frac{18}{5} - \left(\frac{5}{5} \times \textcolor{red}{4}\right) > \frac{5}{3} y + 0 > \frac{18}{5} - \left(\frac{5}{5} \times \textcolor{red}{4}\right)$

$- \frac{18}{5} - \frac{20}{5} > \frac{5}{3} y > \frac{18}{5} - \frac{20}{5}$

$- \frac{38}{5} > \frac{5}{3} y > - \frac{2}{5}$

Now, multiply each segment by $\frac{\textcolor{red}{3}}{\textcolor{b l u e}{5}}$ to solve for $y$ while keeping the system balanced:

$\frac{\textcolor{red}{3}}{\textcolor{b l u e}{5}} \times - \frac{38}{5} > \frac{\textcolor{red}{3}}{\textcolor{b l u e}{5}} \times \frac{5}{3} y > \frac{\textcolor{red}{3}}{\textcolor{b l u e}{5}} \times - \frac{2}{5}$

$- \frac{114}{25} > \frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{5}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{5}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} y > - \frac{6}{25}$

$- \frac{114}{25} > y > - \frac{6}{25}$

Or

$y < - \frac{114}{25}$; $y > - \frac{6}{25}$

Or, in interval notation

$\left(- \infty , - \frac{114}{25}\right)$; $\left(- \frac{6}{25} , + \infty\right)$