Question

Easy Question pls help. (Algebra II type problems)

This first answer is correct, but the second part is the domain of it and I don't know how to do it. Please help.

Answer 1

`\large\boxed{if\ x\\eq-3\ \wedge\ y\\eq0\ \wedge\ x\\eq y\ \wedge\ x\\eq -y}`

second

`\large\boxed{y\\eq0\ \wedge\ x\\eq-y\ (y\\eq-x)}`

Explanation:

We know that dividing by 0 is impossible.

Therefore, the denominator of the expression must be different from 0.

`\text{For}\\\\((x-y)^2)/(2xy+6y)*(4x+12)/(x^2-y^2)\\\\\\2xy+6y\\eq0\qquad(1)\\\\x^2-y^2\\eq0\qquad(2)`

`(1)\\2xy+6y\\eq0\qquad\text{distribute}\\\\2y(x+3)\\eq0\iff2y\\eq0\ \wedge\ x+3\\eq0\\\\2y\\eq0\qquad\text{divide both sides by 2}\\\boxed{y\\eq0}\\\\x+3\\eq0\qquad\text{subtract 3 from both sides}\\\boxed{x\\eq-3}\\==========================\\(2)\\x^2-y^2\\eq0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-y)(x+y)\\eq0\iff x-y\\eq0\ \wedge\ x+y\\eq0\\\boxed{x\\eq y}\ \wedge\ \boxed{x\\eq-y}`

`\text{For}\\\\(2(x-y))/(y(x+y))\\\\y(x+y)\\eq0\iff y\\eq0 \wedge\ x+y\\eq0\\\\ \boxed{y\\eq0}\ \wedge\ \boxed{x\\eq-y}`