# How do you solve 3/(b+2)=4/b?

Mar 18, 2018

See a solution process below:

#### Explanation:

First, because both sides of the equation are pure fractions we can flip each equation to rewrite the equation as:

$\frac{b + 2}{3} = \frac{b}{4}$

Next, multiply each side of the equation by $\textcolor{red}{12}$ to eliminate the fractions while keeping the equation balanced. $\textcolor{red}{12}$ is used because it is the lowest common denominator of the two fractions:

$\textcolor{red}{12} \times \frac{b + 2}{3} = \textcolor{red}{12} \times \frac{b}{4}$

$\cancel{\textcolor{red}{12}} \textcolor{red}{4} \times \frac{b + 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = \cancel{\textcolor{red}{12}} \textcolor{red}{3} \times \frac{b}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}$

$\textcolor{red}{4} \left(b + 2\right) = 3 b$

$\left(\textcolor{red}{4} \times b\right) + \left(\textcolor{red}{4} \times 2\right) = 3 b$

$4 b + 8 = 3 b$

Now, subtract $\textcolor{red}{8}$ and $\textcolor{b l u e}{3 b}$ from each side of the equation to solve for $b$ while keeping the equation balanced:

$4 b - \textcolor{b l u e}{3 b} + 8 - \textcolor{red}{8} = 3 b - \textcolor{b l u e}{3 b} - \textcolor{red}{8}$

$\left(4 - \textcolor{b l u e}{3}\right) b + 0 = 0 - \textcolor{red}{8}$

$1 b = - 8$

$b = - 8$

Mar 18, 2018

I get $b = - 8$

#### Explanation:

To solve equations with fractions use the rule of cross products. Each numerator is multiplied by the opposite denominator and the two products are equal.

$\setminus \frac{\textcolor{b l u e}{3}}{b + 2} = \setminus \frac{4}{\textcolor{b l u e}{b}}$

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

Now it's just an ordinary linear equation:

$3 b = 4 b + 8$

$- b = 8$

$b = - 8$

Check:

$\setminus \frac{3}{- 8 + 2} = \setminus \frac{3}{- 6} = - \setminus \frac{1}{2}$

$\setminus \frac{4}{- 8} = - \setminus \frac{1}{2}$