# How do you solve -3v - \frac { 5} { 2} = \frac { 7} { 2} v - \frac { 4} { 3}?

Jun 22, 2017

$v = \frac{23}{39}$

#### Explanation:

First, get rid of the fractions by multiplying everything by a number which a multiple of both $2$ and $3$, say $6$:

$\textcolor{red}{6} \times \left(- 3 v - \frac{5}{2}\right) = \textcolor{red}{6} \times \left(\frac{7}{2} v - \frac{4}{3}\right)$

Distribute the $6$ through the parentheses on both sides

$6 \times \left(- 3 v\right) - 6 \times \left(\frac{5}{2}\right) = 6 \times \left(\frac{7}{2} v\right) - 6 \times \left(\frac{4}{3}\right)$

$- 18 v - 3 \times 5 = 3 \times 7 v - 2 \times 4$

$- 18 v - 15 = 21 v - 8$

Add $18 v$ to both sides

$- 18 v \textcolor{red}{+ 18 v} - 15 = 21 v \textcolor{red}{+ 18 v} - 8$

$15 = 39 v - 8$

Add $8$ to both sides, to the variable $v$ "more alone"

$15 \textcolor{red}{+ 8} = 39 v - 8 \textcolor{red}{+ 8}$

$23 = 39 v$

Divide both sides by $39$ to isolate the $v$

$\frac{23}{\textcolor{red}{39}} = \frac{\cancel{39} v}{\textcolor{red}{\cancel{39}}}$

This leaves

$v = \frac{23}{39}$

Because $23$ is a prime number, this fraction is reduced to lowest terms.