How do you solve (7-2b)/2<3?

Feb 7, 2017

See the entire solution process below:

Explanation:

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

$\textcolor{red}{2} \times \frac{7 - 2 b}{2} < \textcolor{red}{2} \times 3$

$\cancel{\textcolor{red}{2}} \times \frac{7 - 2 b}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} < 6$

$7 - 2 b < 6$

Next, subtract $\textcolor{red}{7}$ from each side of the inequality to isolate the $b$ term while keeping the inequality balanced:

$7 - 2 b - \textcolor{red}{7} < 6 - \textcolor{red}{7}$

$7 - \textcolor{red}{7} - 2 b < - 1$

$0 - 2 b < - 1$

$- 2 b < - 1$

Now divide each side of the inequality by $\textcolor{b l u e}{- 2}$ to solve for $b$ while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative term we must reverse the inequality sign.

$\frac{- 2 b}{\textcolor{b l u e}{- 2}} \textcolor{red}{>} \frac{- 1}{\textcolor{b l u e}{- 2}}$

$\frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 2}}} b}{\cancel{\textcolor{b l u e}{- 2}}} \textcolor{red}{>} \frac{1}{2}$

$b > \frac{1}{2}$