# How do you solve -4x-1>3 or x+2>7?

Sep 11, 2015

$x < - 1 \mathmr{and} x > 5$
$\textcolor{w h i t e}{\text{XXX}}$OR
$\left\mid x - 2 \right\mid > 3$

#### Explanation:

Part 1
$\textcolor{w h i t e}{\text{XXX}} - 4 x - 1 > 3$
add $4 x - 3$ to both sides
$\textcolor{w h i t e}{\text{XXX}} - 4 > 4 x$
divide by $4$
$\textcolor{w h i t e}{\text{XXX}} - 1 > x$
or, equivalently
$\textcolor{w h i t e}{\text{XXX}} x < - 1$

Part 2
$\textcolor{w h i t e}{\text{XXX}} x + 2 > 7$
subtract $2$ from both sides
$\textcolor{w h i t e}{\text{XXX}} x > 5$

Note that for $x < - 1$ or $x > 5$
$\textcolor{w h i t e}{\text{XXX}} x$ must be further that $2$ the midpoint between $\left(- 1\right)$ and $\left(5\right)$ (i.e. from $3$)
So
$\textcolor{w h i t e}{\text{XXX}} \left\mid x - 2 \right\mid > 3$