# How do you solve  |4k-2| =11?

May 23, 2017

See a solution process below:

#### Explanation:

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1)

$4 k - 2 = - 11$

$4 k - 2 + \textcolor{red}{2} = - 11 + \textcolor{red}{2}$

$4 k - 0 = - 9$

$4 k = - 9$

$\frac{4 k}{\textcolor{red}{4}} = - \frac{9}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} k}{\cancel{\textcolor{red}{4}}} = - \frac{9}{4}$

$k = - \frac{9}{4}$

Solution 2)

$4 k - 2 = 11$

$4 k - 2 + \textcolor{red}{2} = 11 + \textcolor{red}{2}$

$4 k - 0 = 13$

$4 k = 13$

$\frac{4 k}{\textcolor{red}{4}} = \frac{13}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} k}{\cancel{\textcolor{red}{4}}} = \frac{13}{4}$

$k = \frac{13}{4}$

The solutions are: $k = - \frac{9}{4}$ and $k = \frac{13}{4}$

May 23, 2017

$k \in \left\{- \frac{9}{4} , \frac{13}{4}\right\}$

#### Explanation:

The absolute value of $4 k - 2$ must be 11, so $4 k - 2$ must either be $+ 11$ or $- 11$.

To solve this, you need to break it into two equations and solve both of them.

Equation 1

$4 k - 2 = 11 \textcolor{w h i t e}{\text{XXX.}} \textcolor{w h i t e}{\frac{1}{x}}$ "4k-2" could be positive 11.
$\textcolor{w h i t e}{\text{XX-"4k = 13color(white)"XXXXX}}$ Add 2 to both sides.
$\textcolor{w h i t e}{\text{XXX"k=13/4color(white)"XXXX/}}$ Divide both sides by 4.

Equation 2

$4 k - 2 = - 11 \textcolor{w h i t e}{\text{X.}} \textcolor{w h i t e}{\frac{1}{x}}$ "4k-2" could be negative 11.
$\textcolor{w h i t e}{\text{XX-"4k=-9 color(white)"XXXX}}$Add 2 to both sides.
$\textcolor{w h i t e}{\text{XXX"k=-9/4color(white)"XXX}}$ Divide both sides by 4.

Therefore, our two solutions are $k = - \frac{9}{4}$ and $k = \frac{13}{4}$.

Since we technically have a "set" of solutions, we can write our solution more formally like this:

$k \in \left\{- \frac{9}{4} , \frac{13}{4}\right\}$

(This means that $k$ can be any element in the set containing $- \frac{9}{4}$ and $\frac{13}{4}$)