How do you solve 6abs(1-5x)-9=57?

Jan 21, 2017

See the entire solution process below:

$x = - 2$ and $x = \frac{12}{5}$

Explanation:

We need to isolate the absolute value term. Therefore, first we will add $\textcolor{red}{9}$ to each side of the equation to start to isolate the absolute value term while keeping the equation balanced:

$6 \left\mid 1 - 5 x \right\mid - 9 + \textcolor{red}{9} = 57 + \textcolor{red}{9}$

$6 \left\mid 1 - 5 x \right\mid - 0 = 66$

$6 \left\mid 1 - 5 x \right\mid = 66$

Next, we will divide each side of the equation by $\textcolor{red}{6}$:

$\frac{6 \left\mid 1 - 5 x \right\mid}{\textcolor{red}{6}} = \frac{66}{\textcolor{red}{6}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} \left\mid 1 - 5 x \right\mid}{\cancel{\textcolor{red}{6}}} = 11$

$\left\mid 1 - 5 x \right\mid = 11$

The absolution value function takes any positive or negative term and transforms it into its positive form. Therefore we must find two solutions - one for the negative and one form the positive form of what the absolute value term is equated to.

Solution 1):

$1 - 5 x = 11$

$1 - 5 x - \textcolor{red}{1} = 11 - \textcolor{red}{1}$

$1 - \textcolor{red}{1} - 5 x = 10$

$0 - 5 x = 10$

$- 5 x = 10$

$\frac{- 5 x}{\textcolor{red}{- 5}} = \frac{10}{\textcolor{red}{- 5}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} x}{\cancel{\textcolor{red}{- 5}}} = - 2$

$x = - 2$

Solution 2):

$1 - 5 x = - 11$

$1 - 5 x - \textcolor{red}{1} = - 11 - \textcolor{red}{1}$

$1 - \textcolor{red}{1} - 5 x = - 12$

$0 - 5 x = - 12$

$- 5 x = - 12$

$\frac{- 5 x}{\textcolor{red}{- 5}} = - \frac{12}{\textcolor{red}{- 5}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} x}{\cancel{\textcolor{red}{- 5}}} = \frac{12}{5}$

$x = \frac{12}{5}$

The solution to this problem is $x = - 2$ and $x = \frac{12}{5}$