# How do you solve and write the following in interval notation: 2>8 - 3x>1/2?

Jun 20, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{8}$ from each segment of the system of inequalities to isolate the $x$ term while keeping the system balanced:

$2 - \textcolor{red}{8} > 8 - \textcolor{red}{8} - 3 x > \frac{1}{2} - \textcolor{red}{8}$

$- 6 > 0 - 3 x > \frac{1}{2} - \left(\frac{2}{2} \times \textcolor{red}{8}\right)$

$- 6 > - 3 x > \frac{1}{2} - \frac{\textcolor{red}{16}}{2}$

$- 6 > - 3 x > \frac{1 - \textcolor{red}{16}}{2}$

$- 6 > - 3 x > - \frac{15}{2}$

Next, divide each segment by $\textcolor{b l u e}{- 3}$ 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:

$\frac{- 6}{\textcolor{b l u e}{- 3}} \textcolor{red}{<} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{<} \frac{- \frac{15}{2}}{\textcolor{b l u e}{- 3}}$

$2 \textcolor{red}{<} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{<} \frac{15}{6}$

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

Or

$x > 2$; $x < \frac{5}{2}$

Or, in interval notation:

$\left(2 , \frac{5}{2}\right)$