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

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Answer 1 · 2015-09-22T18:10:55

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

Answer 2 · 2015-09-22T20:02:13

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 $y != 1/2$

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

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

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.

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