How do you simplify #(2y^2-4y)/(16y)#?

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Answer 1 · 2017-06-01T11:28:36

#frac(y - 2)(8)#; #y ne 0#

We have: #frac(2 y^(2) - 4 y)(16 y)#

Let's factorise the numerator:

#= frac(2 y (y - 2))(16 y)#

Simplifying:

#= frac(y - 2)(8)#

In the original fraction, the denominator is #16 y#.

To prevent division by zero, this denominator cannot be equal to zero:

#Rightarrow 16 y ne 0#

#Rightarrow y ne 0#

#therefore frac(y - 2)(8)#; #y ne 0#