# How do you solve  2(r-3)+6=14?

Mar 20, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{6}$ from each side of the equation to isolate the term with the parenthesis while keeping the equation balanced:

$2 \left(r - 3\right) + 6 - \textcolor{red}{6} = 14 - \textcolor{red}{6}$

$2 \left(r - 3\right) + 0 = 8$

$2 \left(r - 3\right) = 8$

Next, divide each side of the equation by $\textcolor{red}{2}$ to eliminate the parenthesis while keeping the equation balanced:

$\frac{2 \left(r - 3\right)}{\textcolor{red}{2}} = \frac{8}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \left(r - 3\right)}{\cancel{\textcolor{red}{2}}} = 4$

$r - 3 = 4$

Now, add $\textcolor{red}{3}$ from each side of the equation to solve for $r$ while keeping the equation balanced:

$r - 3 + \textcolor{red}{3} = 4 + \textcolor{red}{3}$

$r - 0 = 7$

$r = 7$