Question

For the expression
b^2 −4by/2y^2 −by − 4y/b −2y
Find the domain.

Answer 1

The required domain is all real values of y, except y = 0 and y = \frac{b}{2}.

The given expression is

\frac{b^{2} - 4by}{2y^{2} - by} - \frac{4y}{b - 2y}

Let, x = \frac{b^{2} - 4by}{2y^{2} - by} - \frac{4y}{b - 2y}

Therefore, to find the domain of this function means to find those values of variable y for which the x-value exists.

It is clear that the denominator of the two terms in the expression must not be zero for x to exist.

So, {2y^{2} - by} \neq 0

⇒ y(2y - b) \neq 0

⇒ y \neq 0 , y \neq \frac{b}{2}

Again, b - 2y \neq 0

⇒ y \neq \frac{b}{2}

Therefore, the required domain is all real values of y, except y = 0 and y = \frac{b}{2}. (Answer)

Answer 2

In other words 5050 is the right answer.

Answer 3

use some parentheses to make clear what you mean
b^2 − 4by/2y^2 −by − 4y/b −2y
the LCD is 2by^2, so that is
[b^2(4by^2) - 4by(b) - by(4by^2) - 4y(4y^2) - 2y(4by^2)]/(2by^2)
= (4b^3y^2 - 4b^2y - 4b^2y^3 - 16y^3 - 8by^3)/(2by^2)

on the other hand,
(b^2 −4by)/(2y^2 −by) − 4y/(b −2y)
= (b^2-4by) / y(2y-b) + 4y/(y-2b)
= (b^2-4by + 4y^2) / y(2y-b)
= (2y-b)^2 / y(2y-b)
= (2y-b)/y

Answer 4

thanks