# How do you solve (y+2) / y = 1 / (y-5) and find any extraneous solutions?

May 12, 2018

There are two solutions:

$y = 2 + \sqrt{14}$

$y = 2 - \sqrt{14}$

#### Explanation:

$\frac{y + 2}{y} = \frac{1}{y - 5}$

Then:

$1 \cdot y = \left(y + 2\right) \cdot \left(y - 5\right)$

So:

$y = {y}^{2} - 5 y + 2 y - 10 = {y}^{2} - 3 y - 10$

Unifying:

${y}^{2} - 4 y - 10 = 0$

As we know in:

$a \cdot {y}^{2} + b \cdot y + c = 0$ ,
$y = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$

So:

$y = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 1 \cdot \left(- 10\right)}}{2 \cdot 1}$

$y = \frac{4 \pm \sqrt{16 + 40}}{2} = \frac{4 \pm \sqrt{56}}{2} = 2 \pm \frac{\sqrt{56}}{2} =$

$y = 2 \pm \frac{\sqrt{14 \cdot 4}}{2} = 2 \pm \frac{2}{2} \cdot \sqrt{14} = 2 \pm \sqrt{14}$