solve the system of equations { ((3∣x−5∣+4=y)),((∣y−3∣=4x−12)) :}


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Question Number 67745 by Enock last updated on 31/Aug/19

$s o l v e t h e s y s t e m o f e q u a t i o n s {\begin{cases} 3 ∣ x - 5 ∣ + 4 = y \\ ∣ y - 3 ∣= 4 x - 12 \end{cases}$
	Answered by Rasheed.Sindhi last updated on 31/Aug/19

	${\begin{cases} 3 ∣ x - 5 ∣ + 4 = y \Rightarrow y ⩾ 4 \\ ∣ y - 3 ∣= 4 x - 12 \Rightarrow 4 x - 12 ⩾ 1 \Rightarrow x ⩾ \frac{13}{4} \end{cases}$ $(i) \to (ii) :$ $\Rightarrow∣ (3 ∣ x - 5 ∣ + 4) - 3 ∣= 4 x - 12$ $\Rightarrow (3 ∣ x - 5 ∣ + 4) - 3 = \pm (4 x - 12)$ $\Rightarrow 3 ∣ x - 5 ∣ + 1 = \pm (4 x - 12)$ $\Rightarrow 3 ∣ x - 5 ∣= \pm (4 x - 12) - 1$ $\Rightarrow∣ x - 5 ∣= \frac{\pm (4 x - 12) - 1}{3}$ $\Rightarrow x - 5 = \pm \frac{\pm (4 x - 12) - 1}{3}$ $\Rightarrow x = \frac{(4 x - 12) \mp 1}{3} + 5$ $\Rightarrow x = \frac{(4 x - 12) \mp 1 + 15}{3}$ $\Rightarrow x = \frac{4 x - 12 - 1 + 15}{3}, x = \frac{4 x - 12 + 1 + 15}{3}$ $3 x = 4 x + 2, 3 x = 4 x + 4$ $x = - 2, x = - 4$ $Both values contradict x ⩾ \frac{13}{4} \land y ⩾ 4$ $∴ The system has no solution .$
	Answered by mr W last updated on 31/Aug/19

	$c a s e 1 : x < 5 a n d y < 3$ ${\begin{cases} 3 (5 - x) + 4 = y \Rightarrow 3 x + y = 19 . . . (i) \\ 3 - y = 4 x - 12 \Rightarrow 4 x + y = 15 . . . (i i) \end{cases}$ $(i i) - (i) :$ $x = - 4$ $y = 19 + 12 = 31 w h i c h i s n o t < 3!$ $\Rightarrow n o s o l u t i o n i n c a s e 1 .$ $c a s e 2 : x < 5 a n d y ⩾ 3$ ${\begin{cases} 3 (5 - x) + 4 = y \Rightarrow 3 x + y = 19 . . . (i) \\ y - 3 = 4 x - 12 \Rightarrow 4 x - y = 9 . . . (i i) \end{cases}$ $(i) + (i i) :$ $7 x = 28 \Rightarrow x = 4 < 5$ $y = 19 - 12 = 7 ⩾ 3$ $(4, 7) i s a s o l u t i o n .$ $c a s e 3 : x ⩾ 5 a n d y < 3$ ${\begin{cases} 3 (x - 5) + 4 = y \Rightarrow 3 x - y = 11 . . . (i) \\ 3 - y = 4 x - 12 \Rightarrow 4 x + y = 15 . . . (i i) \end{cases}$ $(i) + (i i) :$ $7 x = 26 \Rightarrow x = \frac{26}{7} w h i c h i s n o t ⩾ 5!$ $y = 15 - \frac{4 \times 26}{7} = \frac{1}{7}$ $\Rightarrow n o s o l u t i o n i n c a s e 3 .$ $c a s e 4 : x ⩾ 5 a n d y ⩾ 3$ ${\begin{cases} 3 (x - 5) + 4 = y \Rightarrow 3 x - y = 11 . . . (i) \\ y - 3 = 4 x - 12 \Rightarrow 4 x - y = 9 . . . (i i) \end{cases}$ $(i i) - (i) :$ $x = - 2 w h i c h i s n o t ⩾ 5!$ $y = - 8 - 9 = - 17 w h i c h i s n o t ⩾ 3!$ $\Rightarrow n o s o l u t i o n i n c a s e 4 .$ $s u m m a r y :$ $x = 4, y = 7 i s t h e o n l y s o l u t i o n .$
	Commented by mr W last updated on 31/Aug/19

	Answered by Rasheed.Sindhi last updated on 31/Aug/19

	$A n O ther W ay$ ${\begin{cases} 3 ∣ x - 5 ∣ + 4 = y \Rightarrow y ⩾ 4 . . . . . . . (i) \\ ∣ y - 3 ∣= 4 x - 12 \Rightarrow 4 x - 12 ⩾ 1 \Rightarrow x ⩾ \frac{13}{4} . . (ii) \end{cases}$ $C ⊚ nditi ⊚ n for valid solution :$ $x ⩾ \frac{13}{4} \land y ⩾ 4$ $(i) \Rightarrow \pm 3 (x - 5) + 4 = y$ $3 (x - 5) + 4 = y . . . . . . . . . . (Ai)$ $3 x - y = 11$ $- 3 (x - 5) + 4 = y . . . . . . . . . . (Bi)$ $3 x + y = 19$ $(ii) \Rightarrow \pm (y - 3) = 4 x - 12$ $y - 3 = 4 x - 12 . . . . . . . (Aii)$ $4 x - y = 9$ $- (y - 3) = 4 x - 12 . . . . (Bii)$ $4 x + y = 15$ $Ai & Aii :$ $3 x - y = 11$ $4 x - y = 9$ $x = - 2, y = 17$ $Ai & Bii :$ $3 x - y = 11$ $4 x + y = 15$ $x = 26 / 7, y = 15 - 4 (26 / 7) = 1 / 7$ $Bi & Aii :$ $3 x + y = 19$ $4 x - y = 9$ $x = 4, y = 19 - 3 (4) = 7$ $Bi & Bii :$ $3 x + y = 19$ $4 x + y = 15$ $x = - 4, y = 19 - 3 (- 4) = 31$ $Only solution x = 4 & y = 7 meets$ $the condition : x ⩾ \frac{13}{4} & y ⩾ 4$ $Hence its only solution of the given$ $system .$
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