# How do you solve and write the following in interval notation: -1 + | 1/3 - x/2 | > -11/12?

Aug 25, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{1}$ to each side of the inequality to isolate the absolute value function while keeping the inequality balanced:

$\textcolor{red}{1} - 1 + \left\mid \frac{1}{3} - \frac{x}{2} \right\mid > \textcolor{red}{1} - \frac{11}{12}$

$0 + \left\mid \frac{1}{3} - \frac{x}{2} \right\mid > \left(\frac{12}{12} \times \textcolor{red}{1}\right) - \frac{11}{12}$

$\left\mid \frac{1}{3} - \frac{x}{2} \right\mid > \frac{12}{12} - \frac{11}{12}$

$\left\mid \frac{1}{3} - \frac{x}{2} \right\mid > \frac{1}{12}$

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

$- \frac{1}{12} > \frac{1}{3} - \frac{x}{2} > \frac{1}{12}$

Next, subtract $\textcolor{red}{\frac{1}{3}}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$- \textcolor{red}{\frac{1}{3}} + - \frac{1}{12} > - \textcolor{red}{\frac{1}{3}} + \frac{1}{3} - \frac{x}{2} > - \textcolor{red}{\frac{1}{3}} + \frac{1}{12}$

$\left(\frac{4}{4} \times - \textcolor{red}{\frac{1}{3}}\right) + - \frac{1}{12} > 0 - \frac{x}{2} > \left(\frac{4}{4} \times - \textcolor{red}{\frac{1}{3}}\right) + \frac{1}{12}$

$- \frac{4}{12} + - \frac{1}{12} > - \frac{x}{2} > - \frac{4}{12} + \frac{1}{12}$

$- \frac{5}{12} > - \frac{x}{2} > - \frac{3}{12}$

$- \frac{5}{12} > - \frac{x}{2} > - \frac{1}{4}$

Now, multiply each segment by $\textcolor{b l u e}{- 2}$ to solve for $x$ while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

$\textcolor{b l u e}{- 2} \times - \frac{5}{12} \textcolor{red}{<} \textcolor{b l u e}{- 2} \times - \frac{x}{2} \textcolor{red}{<} \textcolor{b l u e}{- 2} \times - \frac{1}{4}$

$\frac{10}{12} \textcolor{red}{<} - \cancel{\textcolor{b l u e}{2}} \times - \frac{x}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}} \textcolor{red}{<} \frac{2}{4}$

$\frac{5}{6} \textcolor{red}{<} x \textcolor{red}{<} \frac{1}{2}$

Or

$x > \frac{5}{6}$ and $x < \frac{1}{2}$

Or, in interval notation:

$\left(- \infty , \frac{1}{2}\right) \text{ and } \left(\frac{5}{6} , + \infty\right)$