# What is the solution to the inequality abs(x-4)>3?

Aug 28, 2015

$x \in \left(- \infty , 1\right) \cup \left(7 , + \infty\right)$

#### Explanation:

You already have the modulus isolated on one side of the inequality, so you don't need to worry about that.

By definition, the absolute value of any real number will always be positive, regardless of the sign of said number.

This means that you need to take into account two scenarios, one in which $x - 4 \ge 0$ and one when $x - 4 < 0$.

• $x - 4 \ge 0 \implies | x - 4 | = x - 4$

The inequality becomes

$x - 4 > 3 \implies x > 7$

• $x - 4 < 0 \implies | x - 4 | = - \left(x - 4\right)$

This time, you get

$- \left(x - 4\right) > 3$

$- x + 4 > 3$

$- x > - 1 \implies x < 1$

This means that your solution set for this absolute value euqation will include any value of $x$ that is bigger than $7$ or smaller than $1$. $x = 7$ and $x = 1$ are not included in the solution set.

$x \in \left(- \infty , 1\right) \cup \left(7 , + \infty\right)$

For any value of $x \in \left[1 , 7\right]$, the inequality will not be true.