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Question Number 153858 by liberty last updated on 11/Sep/21

Commented by liberty last updated on 11/Sep/21

${\begin{cases} x = ? \\ y = ? \\ z = ? \end{cases}$
	Answered by EDWIN88 last updated on 11/Sep/21
	${ ((((x+y)/(xyz))=(1/2)⇒(1/(yz))+(1/(xz))=(1/2))),(( ((y+z)/(xyz))=(5/6)⇒(1/(xz))+(1/(xy))=(5/6))),((((x+z)/(xyz))=(2/3)⇒(1/(yz))+(1/(xy))=(2/3))) :} ⇒2((1/(xy))+(1/(xz))+(1/(yz)))=2 ⇒(1/(xy))+(1/(xz))+(1/(yz))=1 → { (((1/(xy))=(1/2))),(((1/(xz))=(1/3))),(((1/(yz))=(1/6))) :} ⇒(1/((xyz)^2 )) = (1/(36)) →(1/(xyz)) =±(1/6)→ { ((z=±3)),((x=±1)),((y=±2)) :}$
	${\begin{cases} \frac{x + y}{x y z} = \frac{1}{2} \Rightarrow \frac{1}{y z} + \frac{1}{x z} = \frac{1}{2} \\ \frac{y + z}{x y z} = \frac{5}{6} \Rightarrow \frac{1}{x z} + \frac{1}{x y} = \frac{5}{6} \\ \frac{x + z}{x y z} = \frac{2}{3} \Rightarrow \frac{1}{y z} + \frac{1}{x y} = \frac{2}{3} \end{cases}$ $\Rightarrow 2 (\frac{1}{x y} + \frac{1}{x z} + \frac{1}{y z}) = 2$ $\Rightarrow \frac{1}{x y} + \frac{1}{x z} + \frac{1}{y z} = 1 \to {\begin{cases} \frac{1}{x y} = \frac{1}{2} \\ \frac{1}{x z} = \frac{1}{3} \\ \frac{1}{y z} = \frac{1}{6} \end{cases}$ $\Rightarrow \frac{1}{{(x y z)}^{2}} = \frac{1}{36}$ $\to \frac{1}{x y z} = \pm \frac{1}{6} \to {\begin{cases} z = \pm 3 \\ x = \pm 1 \\ y = \pm 2 \end{cases}$
	Commented by Tawa11 last updated on 12/Sep/21

	$Weldone sir$
	Answered by Rasheed.Sindhi last updated on 11/Sep/21
	$xyz=2(x+y)=(6/5)(y+z)=(3/2)(x+z) 20(x+y)=12(y+z)=15(x+z) 20(x+y)=12(y+z) ⇒20x+8y−12z=0 ⇒5x+2y−3z=0 12(y+z)=15(x+z) ⇒12y−15x−3z=0 ⇒4y−5x−z=0 20(x+y)=15(x+z) ⇒5x+20y−15z=0 ⇒x+4y−3z=0 { ((5x+2y−3z=0⇒z=(1/3)(5x+2y))),((−5x+4y−z=0⇒z=−5x+4y )),((x+4y−3z=0⇒z=(1/3)(x+4y))) :} (1/3)(5x+2y)=−5x+4y=(1/3)(x+4y) 5x+2y=−15x+12y=x+4y 5x+2y=−15x+12y ⇒20x−10y=0⇒2x−y=0 −15x+12y=x+4y 16x−8y=0⇒2x−y=0 (1/3)(5x+2y)=(1/3)(x+4y) ⇒4x−2y=0⇒2x−y=0⇒y=2x ⇒z=−5x+4(2x)=3x x=k,y=2k,z=3k Substituting in original equations: xyz=2(x+y)=(6/5)(y+z)=(3/2)(x+z) k.2k.3k=2(k+2k)=(6/5)(2k+3k)=(3/2)(k+3k) 6k^3 =6k=6k=6k⇒k=±1 x=±1,y=±2,z=±3$
	$x y z = 2 (x + y) = \frac{6}{5} (y + z) = \frac{3}{2} (x + z)$ $20 (x + y) = 12 (y + z) = 15 (x + z)$ $20 (x + y) = 12 (y + z)$ $\Rightarrow 20 x + 8 y - 12 z = 0$ $\Rightarrow 5 x + 2 y - 3 z = 0$ $12 (y + z) = 15 (x + z)$ $\Rightarrow 12 y - 15 x - 3 z = 0$ $\Rightarrow 4 y - 5 x - z = 0$ $20 (x + y) = 15 (x + z)$ $\Rightarrow 5 x + 20 y - 15 z = 0$ $\Rightarrow x + 4 y - 3 z = 0$ ${\begin{cases} 5 x + 2 y - 3 z = 0 \Rightarrow z = \frac{1}{3} (5 x + 2 y) \\ - 5 x + 4 y - z = 0 \Rightarrow z = - 5 x + 4 y \\ x + 4 y - 3 z = 0 \Rightarrow z = \frac{1}{3} (x + 4 y) \end{cases}$ $\frac{1}{3} (5 x + 2 y) = - 5 x + 4 y = \frac{1}{3} (x + 4 y)$ $5 x + 2 y = - 15 x + 12 y = x + 4 y$ $5 x + 2 y = - 15 x + 12 y$ $\Rightarrow 20 x - 10 y = 0 \Rightarrow 2 x - y = 0$ $- 15 x + 12 y = x + 4 y$ $16 x - 8 y = 0 \Rightarrow 2 x - y = 0$ $\frac{1}{3} (5 x + 2 y) = \frac{1}{3} (x + 4 y)$ $\Rightarrow 4 x - 2 y = 0 \Rightarrow 2 x - y = 0 \Rightarrow y = 2 x$ $\Rightarrow z = - 5 x + 4 (2 x) = 3 x$ $x = k, y = 2 k, z = 3 k$ $S u b s t i t u t i n g i n o r i g i n a l e q u a t i o n s :$ $x y z = 2 (x + y) = \frac{6}{5} (y + z) = \frac{3}{2} (x + z)$ $k . 2 k . 3 k = 2 (k + 2 k) = \frac{6}{5} (2 k + 3 k) = \frac{3}{2} (k + 3 k)$ $6 k^{3} = 6 k = 6 k = 6 k \Rightarrow k = \pm 1$ $x = \pm 1, y = \pm 2, z = \pm 3$
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