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