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

2 Answers

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

Sep 22, 2015

Answer:

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.