# How do you solve the inequality 2abs(7-x)+4>1 and write your answer in interval notation?

Jul 23, 2018

$\left[\frac{11}{2} , \frac{17}{2}\right]$

#### Explanation:

Isolate the absolute value (in red):
$2 \setminus \textcolor{red}{| 7 - x |} + 4 \setminus > 1$
$2 \setminus \textcolor{red}{| 7 - x |} \setminus > 1 - 4$
$\setminus \textcolor{red}{| 7 - x |} \setminus > \setminus \frac{- 3}{2}$

Expand absolute value:
$- \frac{3}{2} \setminus < 7 - x \setminus < - \left(- \frac{3}{2}\right)$
$- \frac{3}{2} \setminus < 7 - x \setminus < \frac{3}{2}$
$- \frac{3}{2} - 7 \setminus < - x \setminus < \frac{3}{2} - 7$

Note the sign change here. This happens when you divide by negatives.
$\frac{3}{2} + 7 \setminus \textcolor{\to m a \to}{\setminus >} x$ and $x \setminus \textcolor{\to m a \to}{\setminus >} - \frac{3}{2} + 7$
Effectively, this means $- \frac{3}{2} + 7 \setminus < x \setminus < \frac{3}{2} + 7$

Simplifying new inequality:
$- \frac{3}{2} + 7 \setminus < x \setminus < \frac{3}{2} + 7$
$- \frac{3}{2} + \frac{14}{2} \setminus < x \setminus < \frac{3}{2} + \frac{14}{2}$
$\frac{11}{2} \setminus < x \setminus < \frac{17}{2}$

In interval notation:
$\left[\frac{11}{2} , \frac{17}{2}\right]$ (brackets are used when you have $\setminus < , \setminus >$ signs).

Jul 23, 2018

The solution is $x \in \left(- \infty , + \infty\right)$

#### Explanation:

Let

$7 - x = 0$

$\implies$, $x = 7$

$\forall x \in \mathbb{R}$

$2 | 7 - x | > 0$

And

$2 | 7 - x | + 4 > 1$

The solution is

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

graph{2|7-x|+3 [-21.99, 29.33, -11.31, 14.36]}