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

Question

Genesis Allen · Accepted 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# #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.

Eli Assouline · Answer

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

Eva Adamson · Answer

undefined To simplify the expression (2-1/y)/(4-1/y^2), we can start by finding a common denominator for the fractions. The common denominator is y^2.

Next, we can simplify the numerator and denominator separately.

For the numerator, we multiply 2 by y/y to get 2y, and then subtract 1/y.

For the denominator, we multiply 4 by y^2/y^2 to get 4y^2, and then subtract 1/y^2.

Now, we have (2y - 1/y) / (4y^2 - 1/y^2).

To simplify further, we can multiply the numerator and denominator by y to eliminate the fractions.

This gives us (2y^2 - 1) / (4y^2 - y^2).

Simplifying the denominator, we have (2y^2 - 1) / (3y^2).

Therefore, the simplified expression is (2y^2 - 1) / (3y^2).