# How do you solve | x - 3 | + 2 >7?

Apr 17, 2016

$x \in \left(- \infty , - 2\right) \cup \left(8 , + \infty\right)$

#### Explanation:

The first thing to do here is isolate the modulus on one side of the inequality by subtracting $2$ from both sides

$| x - 3 | + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} > 7 - 2$

$| x - 3 | > 5$

Now, because you're dealing with an absolute value inequality, you must account for two possible cases

$\textcolor{w h i t e}{a}$

• color(purple)(ul(color(black)(x - 3 >= 0 implies |x- 3| = x-3))

When this is the case, the inequality becomes

$x - 3 > 5$

$x > 8$

$\textcolor{w h i t e}{a}$

• color(purple)(ul(color(black)(x - 3 <0 implies | x -3| = -(x-3)))

This time, you will have

$- \left(x - 3\right) > 5$

$- x + 3 > 5$

$- x > 2 \implies x < - 2$

So, you need to have $x < - 2$ and $x > 8$, which means that the solution interval for this inequality will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left(- \infty , - 2\right) \cup \left(8 , + \infty\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$