find the set of values of x for which y is real if y=(((x−2)(x−1))/(x+2)) , x≠−2, x∈R


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Question Number 63534 by Rio Michael last updated on 05/Jul/19

$f i n d t h e s e t o f v a l u e s o f x f o r w h i c h y i s r e a l i f$ $y = \frac{(x - 2) (x - 1)}{x + 2}, x \neq - 2, x \in R$
	Answered by MJS last updated on 05/Jul/19
	$the answer is already given: x∈R\{−2}$
	$the answer is already given : x \in R ∖ {- 2}$
	Commented by Rio Michael last updated on 05/Jul/19

	$w h y i s i t s o ?$
	Commented by Prithwish sen last updated on 05/Jul/19

	$for x = - 2 y is undefined$
	Commented by MJS last updated on 05/Jul/19
	$∀x∈R: (x−2)(x−1)∈R ∀t∈R: (t/0) is not defined ∀t∈R: (t/(x+2))∈R ⇔ x+2≠0 ⇔ x≠−2 ∀x∈R\{−2}: (((x−2)(x−1))/(x+2))∈R$
	$\forall x \in R : (x - 2) (x - 1) \in R$ $\forall t \in R : \frac{t}{0} is not defined$ $\forall t \in R : \frac{t}{x + 2} \in R \Leftrightarrow x + 2 \neq 0 \Leftrightarrow x \neq - 2$ $\forall x \in R ∖ {- 2} : \frac{(x - 2) (x - 1)}{x + 2} \in R$
	Commented by Rio Michael last updated on 11/Jul/19

	$s o t h e v a l u e o f x f o r w h i c h y i s r e a l i s t h e v a l u e o f x f o r w h i c h$ $y i s u n d e f i n e d ?$
	Commented by MJS last updated on 12/Jul/19
	$∀x∈R\{−2}: (((x−2)(x−1))/(x+2))∈R means ∀ for all x∈R\{−2} x real but x≠−2 : the following is true (((x−2)(x−1))/(x+2))∈R this expression is real$
	$\forall x \in R ∖ {- 2} : \frac{(x - 2) (x - 1)}{x + 2} \in R means$ $\forall for all$ $x \in R ∖ {- 2} x real but x \neq - 2$ $: the following is true$ $\frac{(x - 2) (x - 1)}{x + 2} \in R this expression is real$
	Commented by Rio Michael last updated on 12/Jul/19

	$i g e t i t n o w .$
	Commented by Rio Michael last updated on 12/Jul/19

	$m e a n s i f g i v e n s o m e t h i n g l i k e$ $y = (x + 3) (x + 6) t o f i n d t h e v a l u e o f x f o r w h i c h y i s r e a l$ $x \in R ?$
	Commented by MJS last updated on 12/Jul/19
	$yes y=(x−3)(x+5) x∈R ⇒ y∈R y=(1/((x−3)(x+5))) x∈R\{−5, 3} because y is not defined for these values y=((x−3)/(x+5)) x∈R\{−5}$
	$yes$ $y = (x - 3) (x + 5)$ $x \in R \Rightarrow y \in R$ $y = \frac{1}{(x - 3) (x + 5)}$ $x \in R ∖ {- 5, 3} because y is not defined for$ $these values$ $y = \frac{x - 3}{x + 5}$ $x \in R ∖ {- 5}$
	Commented by Rio Michael last updated on 12/Jul/19

	$t h a n k s s o m u c h$
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