Can someone show me the work to do this equation

What is this equation called?

Solve for Y:

((y+2)/(4y)) - (1/2) = ((y-9)/(10y))

Y+2..... 1.......Y-9
------- - ------ = -------
4y........2.......10Y

Ignore the periods
It is all fractions. Is this a linear equation?? I'm not sure what type of equation this is.

((y+2)/(4y)) - (1/2) = (y-9)/(10y)
((y+2)/(4y)) - (1(2y)/(2(2y))) = (y-9)/(10y)
((y+2)/(4y)) - ((2y)/(4y)) = (y-9)/(10y)
((y+2)-2y)/(4y) = (y-9)/(10y)
(y+2-2y)/(4y) = (y-9)/(10y)
(-y+2)/(4y) = (y-9)/(10y)
(-y+2)(10y) = (y-9)(4y)
-10y^2+20y = 4y^2-36y
-10y^2+20y-4y^2+36y = 4y^2-36y-4y^2+36y
-14y^2+56y = 0
-14y(y-4) = 0
Either -14y=0 or y-4=0
Therefore y=0 or y=4

However, y=0 would make the denomina