# How do you solve  4/3r - 3 < r + 2/3 - 1/3r?

May 3, 2017

See the solution process below:

#### Explanation:

First, multiply each side of the inequality by $\textcolor{red}{3}$ to eliminate the fractions while keeping the inequalitybalanced:

$\textcolor{red}{3} \left(\frac{4}{3} r - 3\right) < \textcolor{red}{3} \left(r + \frac{2}{3} - \frac{1}{3} r\right)$

$\left(\textcolor{red}{3} \cdot \frac{4}{3} r\right) - \left(\textcolor{red}{3} \cdot 3\right) < \left(\textcolor{red}{3} \cdot r\right) + \left(\textcolor{red}{3} \cdot \frac{2}{3}\right) - \left(\textcolor{red}{3} \cdot \frac{1}{3} r\right)$

$\left(\cancel{\textcolor{red}{3}} \cdot \frac{4}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} r\right) - 9 < 3 r + \left(\cancel{\textcolor{red}{3}} \cdot \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right) - \left(\cancel{\textcolor{red}{3}} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} r\right)$

$4 r - 9 < 3 r + 2 - 1 r$

Next, group and combine like terms on the left side of the inequality:

$4 r - 9 < 3 r - 1 r + 2$

$4 r - 9 < \left(3 - 1\right) r + 2$

$4 r - 9 < 2 r + 2$

Then, add $\textcolor{red}{9}$ and subtract $\textcolor{b l u e}{2 r}$ from each side of the inequality to isolate the $r$ term while keeping the inequality balanced:

$- \textcolor{b l u e}{2 r} + 4 r - 9 + \textcolor{red}{9} < - \textcolor{b l u e}{2 r} + 2 r + 2 + \textcolor{red}{9}$

$\left(- \textcolor{b l u e}{2} + 4\right) r - 0 < 0 + 11$

$2 r < 11$

Now, divide each side of the inequality by $\textcolor{red}{2}$ to solve for $r$ while keeping the inequality balanced:

$\frac{2 r}{\textcolor{red}{2}} < \frac{11}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} r}{\cancel{\textcolor{red}{2}}} < \frac{11}{2}$

$r < \frac{11}{2}$