# How do you solve \frac { 4} { v + 3} = \frac { 6} { v + 3} + 2?

Apr 19, 2017

See the entire solution process below:

#### Explanation:

First, subtract $\textcolor{red}{\frac{6}{v + 3}}$ from each side of the equation to consolidate the fractions:

$\frac{4}{v + 3} - \textcolor{red}{\frac{6}{v + 3}} = - \textcolor{red}{\frac{6}{v + 3}} + \frac{6}{v + 3} + 2$

$\frac{4 - 6}{v + 3} = 0 + 2$

$- \frac{2}{v + 3} = 2$

Next, multiply each side of the equation by $\textcolor{red}{\frac{1}{2}}$ to simplify the equation while keeping the equation balanced:

$\textcolor{red}{\frac{1}{2}} \cdot - \frac{2}{v + 3} = \textcolor{red}{\frac{1}{2}} \cdot 2$

$\cancel{\textcolor{red}{\frac{1}{2}}} \cdot - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{v + 3} = \frac{2}{2}$

$- \frac{1}{v + 3} = 1$

Then, multiply each side of the equation by $\textcolor{red}{v + 3}$ to eliminate the fraction while keeping the equation balanced:

$\textcolor{red}{v + 3} \cdot - \frac{1}{v + 3} = \textcolor{red}{v + 3} \cdot 1$

$\cancel{\textcolor{red}{v + 3}} \cdot - \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{v + 3}}}} = v + 3$

$- 1 = v + 3$

Subtract $\textcolor{red}{3}$ from each side of the equation to solve for $v$ while keeping the equation balanced:

$- 1 - \textcolor{red}{3} = v + 3 - \textcolor{red}{3}$

$- 4 = v + 0$

$- 4 = v$

$v = - 4$