Question

How do you simplify `(2-1/y)/(4-1/ y^2)`?

Answer 1

It is `(2-1/y)/(4-1/ y^2)=(2-1/y)/((2-1/y)*(2+1/y))=1/(2+1/y)`

Answer 2

Multiply both numerator and denominator by `y^2` and use the difference of square identity `a^2 - b^2 = (a-b)(a+b)` to find:

`(2-1/y)/(4-1/(y^2)) = y/(2y+1)`

with exclusions `y != 0` and `y != 1/2`

Explanation:

`f(y) = (2-1/y)/(4-1/(y^2)) = ((2y-1)y)/(4y^2-1) = ((2y-1)y)/((2y)^2 - 1^2)`

`= ((2y-1)y)/((2y-1)(2y+1)) = y/(2y+1)`

with exclusions `y != 0` and `y != 1/2`

Notice that if `y = 0` or `y = 1/2` then `f(y)` is undefined, but `y/(2y+1)` is defined. Hence the exclusions.