# How do you solve \frac { ( - x + 8) } { 2} \geq 3?

Sep 25, 2017

See a solution process below

#### Explanation:

First, multiply each side of the inequality by $\textcolor{red}{2}$ to eliminate the fraction while keeping the inequality balanced:

$\textcolor{red}{2} \times \frac{- x + 8}{2} \ge \textcolor{red}{2} \times 3$

$\cancel{\textcolor{red}{2}} \times \frac{- x + 8}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \ge 6$

$- x + 8 \ge 6$

Next, subtract $\textcolor{red}{8}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

$- x + 8 - \textcolor{red}{8} \ge 6 - \textcolor{red}{8}$

$- x + 0 \ge - 2$

$- x \ge - 2$

Now, multiply each side of the inequality by $\textcolor{b l u e}{- 1}$ to solve for $x$ while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative number we must reverse the inequality operator:

$\textcolor{b l u e}{- 1} \times - x \textcolor{red}{\le} \textcolor{b l u e}{- 1} \times - 2$

$x \textcolor{red}{\le} 2$