# How do you solve \frac { w - 1} { 5} = \frac { w + 2} { 2}?

Feb 23, 2017

See the entire solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{10}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{10} \times \frac{w - 1}{5} = \textcolor{red}{10} \times \frac{w + 2}{2}$

$\cancel{\textcolor{red}{10}} 2 \times \frac{w - 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}} = \cancel{\textcolor{red}{10}} 5 \times \frac{w + 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}$

$2 \left(w - 1\right) = 5 \left(w + 2\right)$

Next, eliminate the parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$\left(2 \times w\right) - \left(2 \times 1\right) = \left(5 \times w\right) + \left(5 \times 2\right)$

$2 w - 2 = 5 w + 10$

Then, subtract $\textcolor{red}{2 w}$ and $\textcolor{b l u e}{10}$ from each side of the equation to isolate the $w$ term while keeping the equation balanced:

$2 w - 2 - \textcolor{red}{2 w} - \textcolor{b l u e}{10} = 5 w + 10 - \textcolor{red}{2 w} - \textcolor{b l u e}{10}$

$2 w - \textcolor{red}{2 w} - 2 - \textcolor{b l u e}{10} = 5 w - \textcolor{red}{2 w} + 10 - \textcolor{b l u e}{10}$

$0 - 12 = 3 w + 0$

$- 12 = 3 w$

Now, divide each side of the equation by $\textcolor{red}{3}$ to solve for $w$ while keeping the equation balanced:

$\frac{- 12}{\textcolor{red}{3}} = \frac{3 w}{\textcolor{red}{3}}$

$- 4 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} w}{\cancel{\textcolor{red}{3}}}$

$- 4 = w$

$w = - 4$