# How do you solve 9(4b - 1) = 2(9b + 3) ?

Apr 30, 2017

9(4b - 1) = 2(9b + 3)

Expanding the brackets we have:

36b - 9 = 18b + 6

Subtracting 18b we have:

18b - 9 = 6

Adding 9 we have:

18b = 15

Dividing by 18 we have:

$b = \frac{15}{18}$

or, $b = \frac{5}{6}$

:)>

Apr 30, 2017

See the solution process below: $b = \frac{5}{6}$

#### Explanation:

First, expand the terms in parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{9} \left(4 b - 1\right) = \textcolor{b l u e}{2} \left(9 b + 3\right)$

$\left(\textcolor{red}{9} \cdot 4 b\right) - \left(\textcolor{red}{9} \cdot 1\right) = \left(\textcolor{b l u e}{2} \cdot 9 b\right) + \left(\textcolor{b l u e}{2} \cdot 3\right)$

$36 b - 9 = 18 b + 6$

Next, add $\textcolor{red}{9}$ and subtract $\textcolor{b l u e}{18 b}$ from each side of the equation to isolate the $b$ term while keeping the equation balanced:

$36 b - 9 + \textcolor{red}{9} - \textcolor{b l u e}{18 b} = 18 b + 6 + \textcolor{red}{9} - \textcolor{b l u e}{18 b}$

$36 b - \textcolor{b l u e}{18 b} - 9 + \textcolor{red}{9} = 18 b - \textcolor{b l u e}{18 b} + 6 + \textcolor{red}{9}$

$\left(36 - \textcolor{b l u e}{18}\right) b - 0 = 0 + 15$

$18 b = 15$

Now, divide each side of the equation by $\textcolor{red}{18}$ to solve for $b$ while keeping the equation balanced:

$\frac{18 b}{\textcolor{red}{18}} = \frac{15}{\textcolor{red}{18}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{18}}} b}{\cancel{\textcolor{red}{18}}} = \frac{3 \times 5}{\textcolor{red}{3 \times 6}}$

$b = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 5}{\textcolor{red}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}} \times 6}}$

$b = \frac{5}{6}$