# How do you solve 2( 8x - 1) \geq 0?

Mar 27, 2017

See the entire solution process below:

#### Explanation:

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

$\frac{2 \left(8 x - 1\right)}{\textcolor{red}{2}} \ge \frac{0}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(8 x - 1\right)}{\cancel{\textcolor{red}{2}}} \ge 0$

$8 x - 1 \ge 0$

Next, add $\textcolor{red}{1}$ to each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

$8 x - 1 + \textcolor{red}{1} \ge 0 + \textcolor{red}{1}$

$8 x - 0 \ge 1$

$8 x \ge 1$

Now, divide each side of the inequalityby $\textcolor{red}{8}$ to solve for $x$ while keeping the inequality balanced:

$\frac{8 x}{\textcolor{red}{8}} \ge \frac{1}{\textcolor{red}{8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} x}{\cancel{\textcolor{red}{8}}} \ge \frac{1}{8}$

$x \ge \frac{1}{8}$