# How do you solve \frac { 5x - 8} { 2} \geq \frac { 4x - 2} { 3}?

Jun 11, 2017

See a solution process below:

#### Explanation:

First, multiply each side of the inequality by $\textcolor{red}{6}$ to eliminate the factions while keeping the inequality balanced. $\textcolor{red}{6}$ is the lowest common denominator of the two fractions:

$\textcolor{red}{6} \times \frac{5 x - 8}{2} \ge \textcolor{red}{6} \times \frac{4 x - 2}{3}$

$\cancel{\textcolor{red}{6}} 3 \times \frac{5 x - 8}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \ge \cancel{\textcolor{red}{6}} 2 \times \frac{4 x - 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}$

$3 \left(5 x - 8\right) \ge 2 \left(4 x - 2\right)$

Next, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{3} \left(5 x - 8\right) \ge \textcolor{b l u e}{2} \left(4 x - 2\right)$

$\left(\textcolor{red}{3} \times 5 x\right) - \left(\textcolor{red}{3} \times 8\right) \ge \left(\textcolor{b l u e}{2} \times 4 x\right) - \left(\textcolor{b l u e}{2} \times 2\right)$

$15 x - 24 \ge 8 x - 4$

Then, add $\textcolor{red}{24}$ and subtract $\textcolor{b l u e}{8 x}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

$- \textcolor{b l u e}{8 x} + 15 x - 24 + \textcolor{red}{24} \ge - \textcolor{b l u e}{8 x} + 8 x - 4 + \textcolor{red}{24}$

$\left(- \textcolor{b l u e}{8} + 15\right) x - 0 \ge 0 + 20$

$7 x \ge 20$

Now, divide each side of the inequality by $\textcolor{red}{7}$ to solve for $x$ while keeping the inequality balanced:

$\frac{7 x}{\textcolor{red}{7}} \ge \frac{20}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} \ge \frac{20}{7}$

$x \ge \frac{20}{7}$