Solve the equation. x^2 − (y − z)^2 = 10 ... (i) y^2 − (z − x)^2 = 5 ... (ii) z^2 − (x − y)^2 = 2 ... (iii)


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Question Number 78694 by TawaTawa last updated on 19/Jan/20

$Solve the equation .$ $x^{2} - {(y - z)}^{2} = 10 . . . (i)$ $y^{2} - {(z - x)}^{2} = 5 . . . (ii)$ $z^{2} - {(x - y)}^{2} = 2 . . . (iii)$
Commented by john santu last updated on 20/Jan/20
$(i)+(ii)+(iii) x^2 +y^2 +z^2 −{(x−y)^2 +(y−z)^2 +(z−x)^2 }=17 2xy+2xz+2yz−(x^2 +y^2 +z^2 )=17 (iv) (∗) (x+y+z)^2 = x^2 +y^2 +z^2 +2xy+2xz+2yz x^2 +y^2 +z^2 =(x+y+z)^2 −2xy−2xz−2yz (iv)4xy+4xz+4yz−(x+y+z)^2 =17$
$(i) + (i i) + (i i i)$ $x^{2} + y^{2} + z^{2} - {{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}} = 17$ $2 x y + 2 x z + 2 y z - (x^{2} + y^{2} + z^{2}) = 17 (i v)$ $(*) {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 x y + 2 x z + 2 y z$ $x^{2} + y^{2} + z^{2} = {(x + y + z)}^{2} - 2 x y - 2 x z - 2 y z$ $(i v) 4 x y + 4 x z + 4 y z - {(x + y + z)}^{2} = 17$
Commented by TawaTawa last updated on 20/Jan/20

$God bless you sir$
	Answered by Rasheed.Sindhi last updated on 20/Jan/20

	$x^{2} - {(y - z)}^{2} = 10 . . . (A)$ $y^{2} - {(z - x)}^{2} = 5 . . . (B)$ $z^{2} - {(x - y)}^{2} = 2 . . . (C)$ $A / B : \frac{(x - y + z) (x + y - z)}{(y - z + x) (y + z - x)} = \frac{10}{5} = 2$ $x - y + z = 2 (y + z - x)$ $x - y + z = 2 y + 2 z - 2 x)$ $3 x - 3 y - z = 0 . . . . . . . . $ $A / C : \frac{(x - y + z) (x + y - z)}{(z - x + y) (z + x - y)} = \frac{10}{2} = 5$ $\frac{x + y - z}{z - x + y} = 5$ $x + y - z = 5 z - 5 x + 5 y$ $6 x - 4 y - 6 z = 0$ $3 x - 2 y - 3 z = 0 . . . . . . . . . $ $B / C : \frac{(y - z + x) (y + z - x)}{(z - x + y) (z + x - y)} = \frac{5}{2}$ $5 (z + x - y) = 2 (y - z + x)$ $3 x - 7 y + 3 z = 0 . . . . . . * $ $, * * & * * * are simultaneous linear equations$ $in 3 variables and can be solved$ $easily :$ $[\begin{matrix} 3 x - 3 y - z = 0 \\ 3 x - 2 y - 3 z = 0 \\ 3 x - 7 y + 3 z = 0 \end{matrix}]$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir .$
	Answered by jagoll last updated on 20/Jan/20

	$(i) (x + y - z) (x - y + z) = 10$ $(ii) (y + z - x) (y - z + x) = 5$ $(iii) (z + x - y) (z - x + y) = 2$ $(i) = (ii) \times (iii)$ $(x + y - z) (x - y + z) = (y + z - x) (y - z + x) (z + x - y) (z - x + y)$
	Commented by jagoll last updated on 20/Jan/20

	$(z + y - x) (z - x + y) = 1$ $\Rightarrow z + y - x = 1 \Rightarrow z + x - y = 2 \Rightarrow y - z + x = 5$
	Commented by mr W last updated on 20/Jan/20

	$y o u r a p p r o c h i s n o t ‘ ‘ {e x a c t}^{″} m a t h e m a t i c a l l y,$ $i t h i n k .$ $w h a t i f t h e e q u a t i o n s w e r e e . g . :$ $(i) (x + y - z) (x - y + z) = 7$ $(ii) (y + z - x) (y - z + x) = 11$ $(iii) (z + x - y) (z - x + y) = 2 ?$ $t h e n y o u c a n n o t s i m p l y g u e s s t h e v a l u e s$ $f o r x + y - z e t c . a n d r e a c h t h e c o r r e c t$ $s o l u t i o n b y a c c i d e n t .$
	Commented by jagoll last updated on 20/Jan/20

	$see the result of equation (i) = (ii) \times (iii)$
	Commented by jagoll last updated on 20/Jan/20

	$from the result of that equation$ $, we get a value (z + y - x)) (z - x + y) = 1$
	Commented by jagoll last updated on 20/Jan/20

	$that finally we get$ $- x + y + z = 1, x - y + z = 2, x + y - z = 5$
	Commented by jagoll last updated on 20/Jan/20

	${don}^{'} t play that guess sir . there$ $is process, which i shorten the$ $process$
	Commented by mr W last updated on 20/Jan/20

	$m a y b e i {d i d n}^{'} t u n t e r s t a n d y o u r m e t h o d$ $c o r r e c t l y . i t h o u g h t y o u j u s t g u e s s t h e$ $f o l l o w i n g v a l u e s :$ $\Rightarrow z + y - x = 1 \Rightarrow z + x - y = 2 \Rightarrow y - z + x = 5 ?$ $o r c a n y o u s h a r e h o w y o u g e t t h e m ?$ $b e s i d e s, t h e o r i g i n a l e q u a t i o n s a r e 3$ $q u a d r a t i c a l r e l a t i o n s b e t w e e n x, y, z .$ $b u t y o u c o u l d r e d u c e t h e m t o 3 l i n e a r$ $r e l a t i o n s . h o w d i d y o u g e t t h i s ?$ $a n d c a n y o u f i n d t h e s o l u t i o n u s i n g$ $y o u r m e t h o d i f t h e e q u a t i o n s a r e :$ $(i) (x + y - z) (x - y + z) = 7$ $(ii) (y + z - x) (y - z + x) = 11$ $(iii) (z + x - y) (z - x + y) = 2$ $i f t h e r e i s a n e a s i e r m e t h o d, i w a n t t o$ $l e a r n .$
	Commented by jagoll last updated on 20/Jan/20

	$ha ha . {don}^{'} t setence like that$ $sir . i am not a psychic who can$ $guess the answers$
	Commented by jagoll last updated on 20/Jan/20

	Commented by mr W last updated on 20/Jan/20

	$s i n c e y o u g o t t h e c o r r e c t s o l u t i o n$ $m u c h m o r e d i r e c t l y a n d f a s t e r, i$ $w a s v e r y i n t e r e s t e d a n d w a n t e d t o$ $l e a r n h o w a n d s i n c e r e l y a s k e d y o u$ $f o r m o r e d e t a i l s .$ $b u t i f y o u {d o n}^{'} t w a n t t o t e l l m o r e a b o u t$ $y o u r s e c r e t, i r e s p e c t i t s i r . t h a n k y o u!$
	Commented by jagoll last updated on 20/Jan/20

	$look sir$
	Commented by mr W last updated on 20/Jan/20

	$t h a n k s a l o t s i r!$
	Commented by jagoll last updated on 20/Jan/20

	$yes sir . i {didn}^{'} t mean to not$ $want to explain it, but i think$ $that asking questions needs$ $a quick response so i cut some of$ $the solutions . thank you sir, for$ $this discussion . i like$
	Commented by mr W last updated on 20/Jan/20

	$t h a n k s a l o t a g a i n! n o w i u n d e r s t a n d$ $y o u r a p p r o a c h . i t h i n k i t h a s w o r k e d$ $d u e t o t h e f a c t t h a t (i) = (i i) \times (i i i), i . e .$ $10 = 5 \times 2 . i f w e h a d o t h e r v a l u e s, e . g .$ $(i) = 7, (i i) = 11, (i i i) = 2, w e {c a n}^{'} t g e t$ $t h e v a l u e o f y + z - x d i r e c t l y .$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir$
	Commented by john santu last updated on 21/Jan/20

	$(i) = \frac{7}{2} \times (i i i)$ $(x + y - z) (x - y + z) = \frac{7}{2} (z + x - y) (z - x + y)$ $2 (x + y - z) (x - y + z) = 7 (z + x - y) (z - x + y)$ $(x - y + z) {2 x + 2 y - 2 z - 7 z + 7 x - 7 y} = 0$ $(x - y + z) (9 x - 5 y - 9 z) = 0$ $x - y + z = 0 \lor 9 x - 5 y - 9 z = 0$ $(i i) = \frac{11}{2} \times (i i i)$ $(y + z - x) (y - z + x) = \frac{11}{2} (z + x - y) (z - x + y)$ $2 (y + z - x) (y - z + x) = 11 (z + x - y) (z - x + y)$ $(z + y - x) {2 x + 2 y - 2 z - 11 z - 11 x + 11 y} = 0$ $(z + y - x) (- 9 x + 13 y - 13 z) = 0$ $z + y - x = 0 \lor - 9 x + 13 y - 13 z = 0$
	Commented by john santu last updated on 21/Jan/20

	$t o M r W .$ $i a m i n t e r e s t e d i n t h e q u e s t i o n s$ $i n y o u m a d e a n d s o l v e t h e m$ $w i t h t h e c o n c e p t o f M r J a g o l l .$ $i t h i n k t h e c o n c e p t o f M r J a g o l l$ $i s n o t a m a t h e m a t i c a l c o n c e p t$ $b u t a t a c t i c a l t h i n k i n g c o n c e p t .$ $i a p p r e c i a t e M r J a g o l l, b e c a u s e$ $n o o n e i n t h e w o r l d c a n a r g u e$ $10 = 5 \times 2 . i^{'} m n o t o n {a n y o n e}^{'} s$ $s i d e, b u t i t r y t o e x p r e s s m y$ $t h o u g h t s t h a t m a t h e m a t i c s$ $m u s t b e a b l e t o m a k e s o m e o n e$ $c r e a t i v e i n s o l v i n g p r o b l e m s .$ $t h a n k s y o u$
	Commented by mr W last updated on 21/Jan/20

	$y e s s i r! b e i n g c r e a t i v e i s a l s o t h a t$ $w h a t i a p p r e c i a t e a n d w h a t i t r y t o b e .$
	Answered by jagoll last updated on 20/Jan/20

	$we get x = \frac{7}{2}, y = 3 and z = \frac{3}{2}$
	Commented by jagoll last updated on 20/Jan/20

	$check$ $x^{2} - {(y - z)}^{2} = \frac{49}{4} - {(3 - \frac{3}{2})}^{2} =$ $\frac{49}{4} - (\frac{9}{4}) = 10 . satisfy the equation$
	Answered by mr W last updated on 20/Jan/20

	$(i i) + (i i i) :$ $y^{2} + z^{2} - z^{2} + 2 x z - x^{2} - x^{2} + 2 y x - y^{2} = 7$ $2 x z - 2 x^{2} + 2 y x = 7$ $\Rightarrow y + z = \frac{7}{2 x} + x$ $(i i) - (i i i) :$ $y^{2} - z^{2} - z^{2} + 2 x z - x^{2} + x^{2} - 2 y x + y^{2} = 3$ $2 (y^{2} - z^{2}) - 2 x (y - z) = 3$ $2 (y - z) (y + z - x) = 3$ $2 (y - z) (\frac{7}{2 x}) = 3$ $\Rightarrow y - z = \frac{3 x}{7}$ $p u t t h i s i n t o (i) :$ $x^{2} - {(\frac{3 x}{7})}^{2} = 10$ $\Rightarrow x = \pm \sqrt{\frac{10 \times 7^{2}}{7^{2} - 3^{2}}} = \pm \frac{7}{2}$ $y = \frac{1}{2} (\frac{7}{2 x} + x + \frac{3 x}{7})$ $\Rightarrow y = \pm \frac{1}{2} (\frac{7 \times 2}{2 \times 7} + \frac{7}{2} + \frac{3 \times 7}{7 \times 2}) = \pm 3$ $z = \frac{1}{2} (\frac{7}{2 x} + x - \frac{3 x}{7})$ $\Rightarrow z = \pm \frac{1}{2} (\frac{7 \times 2}{2 \times 7} + \frac{7}{2} - \frac{3 \times 7}{7 \times 2}) = \pm \frac{3}{2}$
	Commented by jagoll last updated on 20/Jan/20

	$i only took 1 side of the solution . good$ $sir$
	Commented by jagoll last updated on 20/Jan/20

	$this settlement is good sir . i have$ $the view according to my$ ${father}^{'} s military upbringing$ $in solving problems to be$ $fast and on target .$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir .$
	Commented by mr W last updated on 20/Jan/20

	$m y s o l u t i o n i s v a l i d g e n e r a l l y f o r a n y$ $v a l u e s o f t h e e q u a t i o n s :$ $(i) : . . . . . = a$ $(i i) : . . . . . = b$ $(i i i) : . . . . . = c$ $\Rightarrow x = \pm \sqrt{\frac{a {(b + c)}^{2}}{4 b c}}$ $\Rightarrow y = \frac{1}{2} (\frac{b + c}{2 x} + \frac{2 b x}{b + c})$ $\Rightarrow z = \frac{1}{2} (\frac{b + c}{2 x} + \frac{2 c x}{b + c})$
	Commented by TawaTawa last updated on 20/Jan/20

	$Wow, God bless you sir$
	Commented by mr W last updated on 20/Jan/20

	$a b s o l u t e l y s i r!$
	Answered by MJS last updated on 20/Jan/20

	$let x = u - v \land y = u + v$ $(1) - (2 u - z) (2 v - z) = 10$ $(2) (2 u - z) (2 v + z) = 5$ $(3) - (2 v + z) (2 v - z) = 2$ $(1) v = \frac{2 u z - z^{2} - 10}{2 (2 u - z)}$ $(2) 2 (2 u z - z^{2} - 5) = 5 \Rightarrow u = \frac{2 z^{2} + 15}{4 z}$ $\Rightarrow v = - \frac{z}{6}$ $(3) \frac{8}{9} z^{2} = 2 \Rightarrow z = \pm \frac{3}{2}$ $\Rightarrow u = \pm \frac{13}{4} \land v = \mp \frac{1}{4}$ $\Rightarrow x = \pm \frac{7}{2} \land y = \pm 3$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir$
	Answered by behi83417@gmail.com last updated on 20/Jan/20

	$s = x + y + z$ $\Rightarrow {\begin{cases} (x + y - z) (x + z - y) = 10 \\ (y + z - x) (y + x - z) = 5 \\ (z + x - y) (z + y - x) = 2 \end{cases}$ $\Rightarrow {\begin{cases} (s - 2 z) (s - 2 y) = 10 \\ (s - 2 x) (s - 2 z) = 5 \\ (s - 2 y) (s - 2 x) = 2 \end{cases}$ $\Rightarrow (s - 2 x) (s - 2 y) (s - 2 z) = \pm 10$ $[(s - 2 x) (10) = \pm 10 \Rightarrow s - 2 x = \pm 1$ $[(s - 2 y) (5) = \pm 10 \Rightarrow s - 2 y = \pm 2$ $[(s - 2 z) (2) = \pm 10 \Rightarrow s - 2 z = \pm 5$ $\Rightarrow {\begin{cases} y + z - x = \pm 1 \\ x + z - y = \pm 2 \\ x + y - z = \pm 5 \end{cases} \Rightarrow x + y + z = s = \pm 8$ $\Rightarrow {\begin{cases} 2 x = s \mp 1 = \pm 8 \mp 1 = \pm 7 \Rightarrow x = \pm \frac{7}{2} \\ 2 y = s \mp 2 = \pm 8 \mp 2 = \pm 6 \Rightarrow y = \pm 3 \\ 2 z = s \mp 5 = \pm 8 \mp 5 = \pm 3 \Rightarrow z = \pm \frac{3}{2} \end{cases}$
	Commented by TawaTawa last updated on 20/Jan/20

	$God bless you sir$
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