# How do you solve 6/(3b+2)=3/5?

Feb 10, 2017

See the entire solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{5} \left(\textcolor{b l u e}{3 b + 2}\right)$ to eliminate the fraction while keeping the equation balanced:

$\textcolor{red}{5} \left(\textcolor{b l u e}{3 b + 2}\right) \times \frac{6}{3 b + 2} = \textcolor{red}{5} \left(\textcolor{b l u e}{3 b + 2}\right) \times \frac{3}{5}$

$\textcolor{red}{5} \cancel{\left(\textcolor{b l u e}{3 b + 2}\right)} \times \frac{6}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(3 b + 2\right)}}}} = \cancel{\textcolor{red}{5}} \left(\textcolor{b l u e}{3 b + 2}\right) \times \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}}$

$5 \times 6 = \left(3 b + 2\right) x 3$

$30 = \left(3 b \times 3\right) + \left(2 \times 3\right)$

$30 = 9 b + 6$

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

$30 - \textcolor{red}{6} = 9 b + 6 - \textcolor{red}{6}$

$24 = 9 b + 0$

$24 = 9 b$

Now, divide each side of the equation by $\textcolor{red}{9}$ to solve for $b$ while keeping the equation balanced:

$\frac{24}{\textcolor{red}{9}} = \frac{9 b}{\textcolor{red}{9}}$

$\frac{3 \times 8}{3 \times 3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} b}{\cancel{\textcolor{red}{9}}}$

$\frac{\cancel{3} \times 8}{\cancel{3} \times 3} = b$

$\frac{8}{3} = b$

$b = \frac{8}{3}$

Feb 10, 2017

$2 \frac{2}{3}$

#### Explanation:

$\frac{6}{3 b + 2} = \frac{3}{5}$

multiply both sides by $3 b + 2$

$\therefore {\cancel{3 b + 2}}^{1} / 1 \times \frac{6}{\cancel{3 b + 2}} ^ 1 = \frac{3}{5} \times \frac{3 b + 2}{1}$

$\therefore \frac{1}{1} \times \frac{6}{1} = \frac{9 b + 6}{5}$

$\therefore 6 = \frac{9 b + 6}{5}$

multiply both sides by $\frac{5}{1}$

$\therefore \frac{5}{1} \times 6 = \frac{9 b + 6}{\cancel{5}} ^ 1 \times {\cancel{5}}^{1} / 1$

$\therefore 30 = 9 b + 6$

$\therefore 9 b + 6 = 30$

$\therefore 9 b = 30 - 6$

$\therefore 9 b = 24$

$\therefore b = {\cancel{24}}^{8} / {\cancel{9}}^{3}$

$\therefore b = \frac{8}{3}$ or $2 \frac{2}{3}$

substitute $x = \frac{8}{3}$

$\therefore \frac{6}{3 \left(\frac{8}{3}\right) + 2} = \frac{3}{5}$

$\therefore \frac{6}{{\cancel{24}}^{8} / {\cancel{33}}^{1} + 2} = \frac{3}{5}$

$\therefore {\cancel{6}}^{3} / {\cancel{10}}^{5} = \frac{3}{5}$

$\therefore \frac{3}{5} = \frac{3}{5}$