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

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Answer 1 · 2015-05-27T01:09:31

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

$(y+2)/y = 1/(y-5)$

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

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

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

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

$= (4+-sqrt(16+40))/2$

$= 2+- sqrt(14)$

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