# How do you solve the inequality abs(x+14)+3 > 17?

Sep 8, 2015

$x \in \left(- 00 , - 28\right) \cup \left(14 , + \infty\right)$

#### Explanation:

Start by isolating the modulus on one side of the inequality. You can do that by adding $- 3$ to both sides of the equation

$| x + 14 | + \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} > 17 - 3$

$| x + 14 | > 14$

Now, you know that the absolute value of a number will always return a positive value, regardless of the sign of said number.

This means that you need to take into account two possibilities, one that the expression inside the modulus is negative, and the other that the expression inside the modulus is positive.

• $x + 14 > 0 \implies | x + 14 | = x + 14$

The inequality becomes

$x + 14 > 14 \implies x > 0$

• $x + 14 < 0 \implies | x + 14 | = - \left(x + 14\right)$

This time, you have

$- \left(x + 14\right) > 14$

$- x - 14 > 14$

$- x > 28 \implies x < - 28$

So, in order for the original inequality to be true, you need $x$ to be bigger than $0$ or smaller than $\left(- 28\right)$, which means that the solution set will be $x \in \left(- 00 , - 28\right) \cup \left(0 , + \infty\right)$.