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Formula
Solve an expression involving the absolute value
Graph
$| 3 x + 1 | > 2$
$| 3 x + 1 | > 2$
Solution of inequality
$x < - 1 \text{ or } x > \dfrac { 1 } { 3 }$
$| 3x+1 | > 2$
$x < - 1$ or $x > \dfrac { 1 } { 3 }$
 Solve a solution to $x$
$| \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | > \color{#FF6800}{ 2 }$
 Divide the interval based on the value where the inside of the absolute value is 0 
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) > \color{#FF6800}{ 2 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right ) \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } > \color{#FF6800}{ 2 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right )$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) > \color{#FF6800}{ 2 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right ) \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } > \color{#FF6800}{ 2 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right )$
 Find the solution 
$\color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 1 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right ) \\ \color{#FF6800}{ x } > \color{#FF6800}{ \dfrac { 1 } { 3 } } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right )$
$\color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 1 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right ) \\ \color{#FF6800}{ x } > \color{#FF6800}{ \dfrac { 1 } { 3 } } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ - } \color{#FF6800}{ \frac { 1 } { 3 } } \right )$
 Make sure if the value is within the interval 
$\color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } > \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$\color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } > \color{#FF6800}{ \dfrac { 1 } { 3 } }$
 Find the union of sets of each interval 
$\color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 1 }$ or $\color{#FF6800}{ x } > \color{#FF6800}{ \dfrac { 1 } { 3 } }$
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Inequality
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