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Solve an expression involving the absolute value
Answer
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$y = | 2 x - 1 | - \left ( | x + 5 | \right )$
$y = 21$
$x$Intercept
$\left ( - \dfrac { 4 } { 3 } , 0 \right )$, $\left ( 6 , 0 \right )$
$y$Intercept
$\left ( 0 , - 4 \right )$
$\begin{array} {l} x = - 15 \\ x = 27 \end{array}$
$ $ Solve a solution to $ x$
$| \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \color{#FF6800}{ - } \left ( | \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } | \right ) = \color{#FF6800}{ 21 }$
$ $ Divide the interval based on the value where the inside of the absolute value is 0 $ $
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) \\ \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right )$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) \\ \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right )$
$ $ Find the solution $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 25 } } { \color{#FF6800}{ 3 } } } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right )$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 25 } } { \color{#FF6800}{ 3 } } } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \right )$
$ $ Make sure if the value is within the interval $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \\ \color{#FF6800}{ x } \in \emptyset \left ( \text{Do not have the solution} \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \\ \color{#FF6800}{ x } \in \emptyset \left ( \text{Do not have the solution} \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 }$
$ $ Find the union of sets of each interval $ $
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \end{array}$
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