# How do you solve (q - 12) 3 <5q + 2?

Feb 4, 2017

See the entire solution process below:

#### Explanation:

First, expand the terms on the left side of the inequality:

$\left(3 \times q\right) - \left(3 \times 12\right) < 5 q + 2$

$3 q - 36 < 5 q + 2$

Next, add $\textcolor{red}{36}$ and subtract $\textcolor{b l u e}{5 q}$ from each side of the inequality to isolate the $q$ term while keeping the inequality balanced:

$3 q - 36 + \textcolor{red}{36} - \textcolor{b l u e}{5 q} < 5 q + 2 + \textcolor{red}{36} - \textcolor{b l u e}{5 q}$

$3 q - \textcolor{b l u e}{5 q} - 36 + \textcolor{red}{36} < 5 q - \textcolor{b l u e}{5 q} + 2 + \textcolor{red}{36}$

$\left(3 - 5\right) q - 0 < 0 + 38$

$- 2 q < 38$

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

$\frac{- 2 q}{\textcolor{b l u e}{- 2}} \textcolor{red}{>} \frac{38}{\textcolor{b l u e}{- 2}}$

$\frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 2}}} q}{\cancel{\textcolor{b l u e}{- 2}}} \textcolor{red}{>} - 19$

$q > - 19$