# How do you solve abs(3 + 2x)>abs(4 - x)?

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#### Explanation

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#### Explanation:

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Apr 24, 2018

The solution is $x \in \left(- \infty , - 7\right) \cup \left(\frac{1}{3} , + \infty\right)$

#### Explanation:

This is an inequality with absolute values

$| 3 + 2 x | > | 4 - x |$

$| 3 + 2 x | - | 4 - x | > 0$

Let $f \left(x\right) = | 3 + 2 x | - | 4 - x |$

$\left\{\begin{matrix}3 + 2 x \ge 0 \\ 4 - x \ge 0\end{matrix}\right.$, $\iff$, $\left\{\begin{matrix}x \ge - \frac{3}{2} \\ x \le 4\end{matrix}\right.$

The sign chart is as follows :

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a a a}$$- \frac{3}{2}$$\textcolor{w h i t e}{a a a a a a a a a a a}$$4$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$3 + 2 x$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$4 - x$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$color(white)(aaaaa)+$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$| 3 + 2 x |$$\textcolor{w h i t e}{a a a a a}$$- 3 - 2 x$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$3 + 2 x$$\textcolor{w h i t e}{a a a a a a}$$3 + 2 x$

$\textcolor{w h i t e}{a a a a}$$| 4 - x |$$\textcolor{w h i t e}{a a a a a a a a}$$4 - x$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)4-x$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$- 4 + x$

In the interval $\left(- \infty , - \frac{3}{2}\right)$,

$f \left(x\right) = - 3 - 2 x - 4 + x = - 7 - x$

$f \left(x\right) > 0$, $\implies$, $- 7 - x > 0$, $\implies$, $x < - 7$

In the interval $\left[- \frac{3}{2} , 4\right]$,

$f \left(x\right) = 3 + 2 x - 4 + x = - 1 + 3 x$

$f \left(x\right) > 0$, $\implies$, $- 1 + 3 x > 0$, $x > \frac{1}{3}$

In the interval $\left(4 , + \infty\right)$,

$f \left(x\right) = 3 + 2 x + 4 - x = 7 + x$

$f \left(x\right) > 0$, $\implies$, $7 + x > 0$, $\implies$, $x < - 7$

This result is not valid since $x \notin$(4, +oo)

The solution is

$x \in \left(- \infty , - 7\right) \cup \left(\frac{1}{3} , + \infty\right)$

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