# How do you solve 15/(7b+5)=5/(2b+1)?

Jun 28, 2018

See a solution process below:

#### Explanation:

Because both sides of the equation are pure fractions we can reverse the fractions and rewrite the equation as:

$\frac{7 b + 5}{15} = \frac{2 b + 1}{5}$

Next, we can multiply each side of the equation by $\textcolor{red}{15}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{15} \times \frac{7 b + 5}{15} = \textcolor{red}{15} \times \frac{2 b + 1}{5}$

$\cancel{\textcolor{red}{15}} \times \frac{7 b + 5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}}} = \cancel{\textcolor{red}{15}} \textcolor{red}{3} \times \frac{2 b + 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}}}$

$7 b + 5 = \textcolor{red}{3} \left(2 b + 1\right)$

$7 b + 5 = \left(\textcolor{red}{3} \times 2 b\right) + \left(\textcolor{red}{3} \times 1\right)$

$7 b + 5 = 6 b + 3$

Now, subtract $\textcolor{red}{5}$ and $\textcolor{b l u e}{6 b}$ from each side of the equation to solve for $b$ while keeping the equation balanced:

$7 b - \textcolor{b l u e}{6 b} + 5 - \textcolor{red}{5} = 6 b - \textcolor{b l u e}{6 b} + 3 - \textcolor{red}{5}$

$\left(7 - \textcolor{b l u e}{6}\right) b + 0 = 0 - 2$

$1 b = - 2$

$b = - 2$