How do you solve 1/9 - 2/(3b) = 1/18?

Jul 18, 2017

See a solution process below:

Explanation:

Step 1) Subtract $\textcolor{red}{\frac{1}{9}}$ from each side of the equation to isolate the $b$ term while keeping the equation balanced:

$- \textcolor{red}{\frac{1}{9}} + \frac{1}{9} - \frac{2}{3 b} = - \textcolor{red}{\frac{1}{9}} + \frac{1}{18}$

$0 - \frac{2}{3 b} = - \left(\frac{2}{2} \times \textcolor{red}{\frac{1}{9}}\right) + \frac{1}{18}$

$- \frac{2}{3 b} = - \frac{2}{18} + \frac{1}{18}$

$- \frac{2}{3 b} = - \frac{1}{18}$

Step 2) multiply each side of the equation by $\textcolor{red}{b}$ to move the $b$ term to the numerator while keeping the equation balanced:

$\textcolor{red}{b} \times - \frac{2}{3 b} = - \frac{1}{18} \times \textcolor{red}{b}$

$\cancel{\textcolor{red}{b}} \times - \frac{2}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{b}}}} = \frac{- 1}{18} \times \textcolor{red}{b}$

$\frac{- 2}{3} = \frac{- 1}{18} b$

Step 3) Multiply each side of the equation by $\textcolor{red}{- 18}$ to solve for $b$ while keeping the equation balanced:

$\textcolor{red}{- 18} \times \frac{- 2}{3} = \textcolor{red}{- 18} \times \frac{- 1}{18} b$

$- 6 \cancel{\textcolor{red}{- 18}} \times \frac{- 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = - 1 \cancel{\textcolor{red}{- 18}} \times \frac{- 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{18}}}} b$

$12 = b$

$b = 12$