# How do you solve  |-3y-2|+1 ≥22?

Apr 14, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{1}$ from each side of the inequality to isolate the absolute value term while keeping the inequality balanced:

$\left\mid - 3 y - 2 \right\mid + 1 - \textcolor{red}{1} \ge 22 - \textcolor{red}{1}$

$\left\mid - 3 y - 2 \right\mid + 0 \ge 21$

$\left\mid - 3 y - 2 \right\mid \ge 21$

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.

$- 21 \ge - 3 y - 2 \ge 21$

Next, add $\textcolor{red}{2}$ to each segment of the inequality system to isolate the $y$ term while keeping the system balanced:

$- 21 + \textcolor{red}{2} \ge - 3 y - 2 + \textcolor{red}{2} \ge 21 + \textcolor{red}{2}$

$- 19 \ge - 3 y - 0 \ge 23$

$- 19 \ge - 3 y \ge 23$

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

$\frac{- 19}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{- 3 y}{\textcolor{b l u e}{- 3}} \textcolor{red}{\le} \frac{23}{\textcolor{b l u e}{- 3}}$

$\frac{19}{3} \textcolor{red}{\le} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}} y}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{\le} - \frac{23}{3}$

$\frac{19}{3} \textcolor{red}{\le} y \textcolor{red}{\le} - \frac{23}{3}$

Or

$y \ge \frac{19}{3}$ and $y \le - \frac{23}{3}$