# How do you solve \frac { 4} { v ^ { 2} - 3v } = \frac { 1} { 5- v }?

May 3, 2018

$v = - 5 , 4$

#### Explanation:

We start by cross-multiplying to get rid of the fractions.

$\frac{4}{{v}^{2} - 3 v} = \frac{1}{5 - v} \to 4 \left(5 - v\right) = {v}^{2} - 3 v$

This can be rearranged:

$4 \left(5 - v\right) = {v}^{2} - 3 v \to 20 - 4 v = {v}^{2} - 3 v \to {v}^{2} + v - 20 = 0$
$\to \left(v + 5\right) \left(v - 4\right) = 0 \to v = - 5 , 4$.

We've found two solutions, $v = - 5$ and $v = 4$.

May 3, 2018

$v = 4 \mathmr{and} - 5$

#### Explanation:

You'll want to start by cross multiplying.

$4 \left(5 - v\right) = 1 \left({v}^{2} - 3 v\right)$
$20 - 4 v = {v}^{2} - 3 v$

Have like terms on the same side.
$20 = {v}^{2} + v$

Now you can figure out in your head that $v = 4$, but if you use the quadratic formula, you'll get another possible solution.

${v}^{2} + v - 20 = 0$

$v = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in your values (a = 1, b = 1, c = -20), and you get:

$v = \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \cdot 1 \cdot - 20\right)}}{2 \cdot 1}$

Now let's simplify.

$v = \frac{- 1 \pm \sqrt{1 + 80}}{2}$

$v = \frac{- 1 \pm \sqrt{81}}{2}$

$v = \frac{- 1 \pm 9}{2}$

$\frac{- 1 + 9}{2} = \frac{8}{2} = 4$

$\frac{- 1 - 9}{2} = \frac{8}{2} = - 5$

So $v$ can be either $4$ or $- 5$