# How do you solve abs(3x-4)>abs(x+6)?

Apr 23, 2017

Answer retracted by author due to errors.

#### Explanation:

Answer retracted by author due to errors.

Apr 23, 2017

#### Explanation:

Solve the equality.
Cut the number line at those solutions.
Test each piece of the number line.

$\left\mid 3 x - 4 \right\mid = \left\mid x + 6 \right\mid$ if and only if

$3 x - 4 = x + 6$ $\text{ }$ OR $\text{ }$ $3 x - 4 = - \left(x + 6\right)$

$2 x = 10$ $\text{ }$ OR $\text{ }$ $4 x = - 2$

$x = 5$ $\text{ }$ OR $\text{ }$ $x = - \frac{1}{2}$

Cut the number line at $- \frac{1}{2}$ and at $5$ and choose a test number in each piece of the line.

Test interval $\text{ }$ $\text{ }$ using $\text{ }$ $\text{ }$ True or false

$\left(- \infty , - \frac{1}{2}\right)$ $\text{ }$ $x = - 3$ $\text{ }$ $\left\mid - 13 \right\mid > \left\mid 3 \right\mid$ is true

$\left(- \frac{1}{2} , 5\right)$ $\text{ }$ $\text{ }$ $x = 0$ $\text{ }$ $\text{ }$ $\left\mid - 4 \right\mid > \left\mid 6 \right\mid$ is false

$\left(5 , \infty\right)$ $\text{ }$ $\text{ }$ $x = 6$ $\text{ }$ $\text{ }$ $\left\mid 14 \right\mid > \left\mid 12 \right\mid$ is true.

The solution set is

$\left(- \infty , - \frac{1}{2}\right) \cup \left(5 , \infty\right)$

Graphical solution

$\left\mid 3 x - 4 \right\mid > \left\mid x + 6 \right\mid$ when $\left\mid 3 x - 4 \right\mid - \left\mid x + 6 \right\mid$ is positive.

Here is the graph of $y = \left\mid 3 x - 4 \right\mid - \left\mid x + 6 \right\mid$

graph{abs(3x-4)-abs(x+6) [-11.79, 16.69, -7.2, 7.04]}

We can see that $y > 0$ for $x$ in $\left(- \infty , - \frac{1}{2}\right) \cup \left(5 , \infty\right)$

Apr 23, 2017

$\textcolor{red}{x \in \left(- \infty , - \frac{1}{2}\right) \cup \left(5 , \infty\right)}$

#### Explanation:

First, let us calculate roots/zeros of the expressions on both sides of the equation (i.e. the value(s) of $x$ for which these expressions become zero separately). The reason for doing this will become clear after a few steps.

1. $3 x - 4 = 0 \implies x = \frac{4}{3}$
2. $x + 6 = 0 \implies x = - 6$

Now let us put these points on a number line. [just for a ROUGH idea. No need to draw to scale.] Now observe that in the interval,
1. $\left(- \infty , - 6\right)$ value of both the expressions is negative, i.e. $3 x - 4 < 0$ and $x + 6 < 0$.
$\therefore$ in this interval, $| 3 x - 4 | = - \left(3 x - 4\right) = 4 - 3 x$ and $| x + 6 | = - \left(x + 6\right) = - x - 6$.

2. $\left[- 6 , \frac{4}{3}\right)$ , $3 x - 4$ is negative , i.e. $3 x - 4 < 0$ and $x + 6$ is positive i.e. $x + 6 \ge 0$.
$\therefore$ in this interval, $| 3 x - 4 | = - \left(3 x - 4\right) = 4 - 3 x$ and $| x + 6 | = \left(x + 6\right)$.

3. $\left[\frac{4}{3} , \infty\right)$ value of both expressions is positive, i.e. $3 x - 4 \ge 0$ and $x + 6 > 0$.
$\therefore$ in this interval, $| 3 x - 4 | = \left(3 x - 4\right)$ and $| x + 6 | = \left(x + 6\right)$.

Now let us solve this inequality using the above-obtained results:-

1. $\left(- \infty , - 6\right)$
$4 - 3 x > - x - 6$
$\implies 4 + 6 > 3 x - x$
$\implies 5 > x \implies x < 5$ $\therefore x \in \left(- \infty , 5\right)$ BUT the interval we are considering is $\left(- \infty , - 6\right)$.
Hence $\textcolor{b l u e}{x \in \left(- \infty , - 6\right) \cap \left(- \infty , 5\right)}$
$\implies \textcolor{red}{x \in \left(- \infty , - 6\right)}$

2. $\left[- 6 , \frac{4}{3}\right)$
$4 - 3 x > x + 6$
$\implies 4 - 6 > 3 x + x$
$\implies - \frac{1}{2} > x \implies x < - \frac{1}{2}$ $\therefore x \in \left(- \infty , - \frac{1}{2}\right)$ BUT our interval is $\left[- 6 , \frac{4}{3}\right)$.
Hence $\textcolor{b l u e}{x \in \left[- 6 , \frac{4}{3}\right) \cap \left(- \infty , - \frac{1}{2}\right)}$
$\implies \textcolor{red}{x \in \left[- 6 , - \frac{1}{2}\right)}$

3. $\left[\frac{4}{3} , \infty\right)$
$3 x - 4 > x + 6$
$\implies 3 x - x > 4 + 6$
$\implies x > 5$ $\therefore x \in \left(5 , \infty\right)$ BUT similar to previous two cases, $\textcolor{b l u e}{x \in \left[\frac{4}{3} , \infty\right) \cap \left(5 , \infty\right)}$
$\implies \textcolor{red}{x \in \left(5 , \infty\right)}$

$\therefore \textcolor{b l u e}{x \in \left(- \infty , - 6\right) \cup \left[- 6 , - \frac{1}{2}\right) \cup \left(5 , \infty\right)}$
$\implies \textcolor{red}{x \in \left(- \infty , - \frac{1}{2}\right) \cup \left(5 , \infty\right)}$