# How do you solve and write the following in interval notation: 3 |x + 5| + 6>=15?

Jun 8, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{6}$ from each side of the inequality to isolate the absolute value term while keeping the inequality balanced:

$3 \left\mid x + 5 \right\mid + 6 - \textcolor{red}{6} \ge 15 - \textcolor{red}{6}$

$3 \left\mid x + 5 \right\mid + 0 \ge 9$

$3 \left\mid x + 5 \right\mid \ge 9$

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

$\frac{3 \left\mid x + 5 \right\mid}{\textcolor{red}{3}} \ge \frac{9}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \left\mid x + 5 \right\mid}{\cancel{\textcolor{red}{3}}} \ge 3$

$\left\mid x + 5 \right\mid \ge 3$

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.

$- 3 \ge x + 5 \ge 3$

Now, subtract $\textcolor{red}{5}$ from each segment of the system of inequalities to solve for $x$ while keeping the system balanced:

$- 3 - \textcolor{red}{5} \ge x + 5 - \textcolor{red}{5} \ge 3 - \textcolor{red}{5}$

$- 8 \ge x + 0 \ge - 2$

$- 8 \ge x \ge - 2$

Or

$x \le - 8$ and $x \ge - 2$

Or, in interval notation:

$\left(- \infty , - 8\right]$ and $\left[- 2 , \infty\right)$