# How do you solve (y+2)/y = 1/(y-5)?

May 27, 2015

Provide $y \ne 0$ and $y \ne 5$ (either of which would result in an attempt to divide by zero)

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

can be re-written as
$\left(y + 2\right) \left(y - 5\right) = y$

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

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

using the quadratic formula for roots
#y = (-b+-sqrt(b^2-4ac)/(2a)

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

$= 2 \pm \sqrt{14}$