find y if ∣((y−3)/(y+1))∣<2 i mean modulus by ∣


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 33837 by NECx last updated on 25/Apr/18

$f i n d y i f ∣ \frac{y - 3}{y + 1} ∣< 2$ $i m e a n m o d u l u s b y ∣$
	Answered by Rasheed.Sindhi last updated on 26/Apr/18

	$f i n d y i f ∣ \frac{y - 3}{y + 1} ∣< 2$ $\pm \frac{y - 3}{y + 1} < 2$ $\pm (y - 3) < 2 (y + 1) \land y + 1 > 0 . . . . . . . A$ $or$ $\pm (y - 3) > 2 (y + 1) \land y + 1 < 0 . . . . . . . B$ $A$ $y - 3 < 2 y + 2 \land y > - 1 . . . . . . . . . . . . A 1$ $y > - 5 \land y > - 1$ $y > - 1$ $or$ $- y + 3 < 2 y + 2 \land y > - 1 . . . . . . . . . . A 2$ $3 y > 1 \land y > - 1$ $y > 1 / 3 \land y > - 1$ $y > 1 / 3$ $A : A 1 or A 2 : y > - 1 \lor y > 1 / 3$ $y > - 1$ $- - - - - - - - -$ $B$ $y - 3 > 2 y + 2 \land y < - 1 . . . . B 1$ $y < - 5 \land y < - 1$ $y < - 5$ $or$ $- y + 3 > 2 y + 2 \land y < - 1 . . . . B 2$ $y < 1 / 3 \land y < - 1$ $y < - 1$ $B : B 1 or B 2 : y < - 5 \lor y < - 1$ $y < - 1$ $- - - - - - - - -$ $Ans : A or B : y > - 1 or y < - 1$ $Some error is there .$
	Commented byMJS last updated on 26/Apr/18

	$- 2 < \frac{y - 3}{y + 1} < 2 \Rightarrow - 2 < \frac{y - 3}{y + 1} \land \frac{y - 3}{y + 1} < 2$ $y + 1 \neq 0 \Rightarrow y \neq - 1$ $1 . y + 1 < 0 \Rightarrow y < - 1$ $- 2 (y + 1) > y - 3 \land y - 3 > 2 (y + 1)$ $y < \frac{1}{3} \land y < - 5 \Rightarrow y < - 5$ $2 . y + 1 > 0 \Rightarrow y > - 1$ $- 2 (y + 1) < y - 3 \land y - 3 < 2 (y + 1)$ $y > \frac{1}{3} \land y > - 5 \Rightarrow y > \frac{1}{3}$ $y < - 5 \lor y > \frac{1}{3}$ $y \in] - \infty; - 5 [\cup] \frac{1}{3}; \infty [$ $. . . but anyway he meant m o d (\frac{y - 3}{y + 1}) < 2$
	Answered by MJS last updated on 26/Apr/18

	$\mod (5 / 3) = 2 because 1 \times 3 + 2 = 5$ $which definition shall we use for \mod (- 5 / 3) and$ $\mod (5 / - 3) ?$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum