Symbol

# Calculator search results

Formula
Solve an expression involving the absolute value
Graph
$y = | 2 x - 1 | - | x + 5 |$
$y = 21$
$x$Intercept
$\left ( - \dfrac { 4 } { 3 } , 0 \right )$, $\left ( 6 , 0 \right )$
$y$Intercept
$\left ( 0 , - 4 \right )$
$| 2x-1 | - | x+5 | = 21$
$\begin{array} {l} x = - 15 \\ x = 27 \end{array}$
 Solve a solution to $x$
$| \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \color{#FF6800}{ - } | \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } | = \color{#FF6800}{ 21 }$
 Divide the interval based on the value where the inside of the absolute value is 0 
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { 1 } { 2 } } \right ) \\ \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { 1 } { 2 } } \right )$
$\color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { 1 } { 2 } } \right ) \\ \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ 21 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { 1 } { 2 } } \right )$
 Find the solution 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 3 } } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { 1 } { 2 } } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { 1 } { 2 } } \right )$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \left ( \text{However (or only)} \color{#FF6800}{ x } < \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 3 } } \left ( \text{However (or only)} \color{#FF6800}{ - } \color{#FF6800}{ 5 } \leq \color{#FF6800}{ x } < \color{#FF6800}{ \frac { 1 } { 2 } } \right ) \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \left ( \text{However (or only)} \color{#FF6800}{ x } \geq \color{#FF6800}{ \frac { 1 } { 2 } } \right )$
 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 
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ 27 } \end{array}$
Solution search results
$|x+5|=2x-9$
$|x+5|=2x-9$
$|2x-3|<|x+5|$