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


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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 by MJS 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

	$\mod (5 / 3) = 2 because 1 \times 3 + 2 = 5$ $which definition shall we use for \mod (- 5 / 3) and$ $\mod (5 / - 3) ?$
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