How do you solve (2x-4)/6>=-5x+2?

Jan 16, 2017

See the entire solution process below:

Explanation:

First step, multiply each side of the inequality by $\textcolor{red}{6}$ to eliminate the fraction:

$\textcolor{red}{6} \times \frac{2 x - 4}{6} \ge \textcolor{red}{6} \left(- 5 x + 2\right)$

$\cancel{\textcolor{red}{6}} \times \frac{2 x - 4}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}} \ge \left(\textcolor{red}{6} \times - 5 x\right) + \left(\textcolor{red}{6} \times 2\right)$

$2 x - 4 \ge - 30 x + 12$

Next step, add the necessary terms from each side of the inequality to isolate the $x$ terms on one side of the inequality and the constants on the other side of the inequality while keeping the inequality balanced.

$2 x - 4 + \textcolor{red}{4} + \textcolor{b l u e}{30 x} \ge - 30 x + 12 + \textcolor{red}{4} + \textcolor{b l u e}{30 x}$

$2 x + \textcolor{b l u e}{30 x} - 4 + \textcolor{red}{4} \ge - 30 x + \textcolor{b l u e}{30 x} + 12 + \textcolor{red}{4}$

$\left(2 + 30\right) x - 0 \ge 0 + 16$

$32 x \ge 16$

Last step, divide each side of the inequality by $\textcolor{red}{32}$ to solve for $x$ while keeping the inequality balanced:

$\frac{32 x}{\textcolor{red}{32}} \ge \frac{16}{\textcolor{red}{32}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{32}}} x}{\cancel{\textcolor{red}{32}}} \ge \frac{1}{2}$

$x \ge \frac{1}{2}$