bx^3 =10a^2 bx + 3a^3 y , ay^3 = 10ab^2 y + 3b^3 x solve for x and y in terms of (a , b) and solve for a and b in terms of (x , y )


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Question Number 192957 by York12 last updated on 31/May/23

${b x}^{3} = 10 a^{2} b x + 3 a^{3} y, {a y}^{3} = 10 {a b}^{2} y + 3 b^{3} x$ $s o l v e f o r x a n d y i n t e r m s o f (a, b)$ $a n d s o l v e f o r a a n d b i n t e r m s o f (x, y)$
Commented by York12 last updated on 02/Jun/23

Commented by York12 last updated on 02/Jun/23

$n o c o m m o n r o o t s$ $m e a n s n o e x i s t e n c e f o r a n y v a l u e$ $o f x w h i c h c a n s a t i f y b o t h e q u a t i o n s$ $\Rightarrow y o u r s y s t e m o f e q u a t i o n s i s a g a i n$ $w r o n g$
	Answered by a.lgnaoui last updated on 01/Jun/23
	${ ((bx^3 −10a^2 bx=3a^3 y⇒ y=((xb(x^2 −10a^2 ))/(3a^3 )) (1))),((3b^3 x=ay^3 −10ab^2 y⇒ x=((ya(y^2 −10b^2 ))/(3b^3 )) (2))) :} y^2 =((x^2 b^2 (x^2 −10a^2 )^2 )/(9a^6 )) x^2 =((y^2 a^2 (y^2 −10b^2 ))/(9b^6 )) (1)⇔y=(b/(3a^3 ))[((ya(y^2 −10b^2 )/(3b^3 ))][(((y^2 a^2 (y^2 −10b^2 )−90a^2 b^6 )/(9b^6 ))] 81a^3 b^9 y=ab(y^2 −10b^2 )(a^2 y^4 −10y^2 a^2 b^2 −90a^2 b^6 ) =a^3 b(y^2 −10b^2 )(y^4 −10y^2 b^2 −90b^6 ) =a^3 b[y^6 −10y^4 b^2 −90y^2 b^6 −10b^2 y^4 +100b^4 y^2 +900b^8 =a^3 b[y^6 −20b^2 y^4 −(90b^6 −100b^4 )y^2 +900b^8 ] =81a^3 b^9 y ⇒81b^8 y=y^6 −20b^2 y^4 −10b^4 (9b^2 −10)y^2 +900b^8 ⇒y^6 −20b^2 [y^4 −(1/2)(9b^2 −10)b^2 y^2 +45b^6 ]−81b^8 y=0 =y^6 −20b^2 (y^2 −(((9b^2 −10)b^2 )/4))^2 −81b^8 y+45b^6 −(((9b^2 −10)b^4 )/(16))=0 ..........to continious$
	${\begin{cases} {bx}^{3} - {10 a}^{2} bx = {3 a}^{3} y \Rightarrow y = \frac{xb (x^{2} - 10 a^{2})}{3 a^{3}} (1) \\ {3 b}^{3} x = {ay}^{3} - {10 ab}^{2} y \Rightarrow x = \frac{ya (y^{2} - 10 b^{2})}{3 b^{3}} (2) \end{cases}$ $y^{2} = \frac{x^{2} b^{2} {(x^{2} - 10 a^{2})}^{2}}{9 a^{6}}$ $x^{2} = \frac{y^{2} a^{2} (y^{2} - 10 b^{2})}{9 b^{6}}$ $(1) \Leftrightarrow y = \frac{b}{3 a^{3}} [\frac{ya (y^{2} - 10 b^{2}}{3 b^{3}}] [(\frac{y^{2} a^{2} (y^{2} - 10 b^{2}) - 90 a^{2} b^{6}}{9 b^{6}}]$ $81 a^{3} b^{9} y = ab (y^{2} - 10 b^{2}) (a^{2} y^{4} - 10 y^{2} a^{2} b^{2} - 90 a^{2} b^{6})$ $= a^{3} b (y^{2} - 10 b^{2}) (y^{4} - 10 y^{2} b^{2} - 90 b^{6})$ $= a^{3} b [y^{6} - 10 y^{4} b^{2} - 90 y^{2} b^{6}$ $- 10 b^{2} y^{4} + 100 b^{4} y^{2} + 900 b^{8}$ $= a^{3} b [y^{6} - 20 b^{2} y^{4} - (90 b^{6} - 100 b^{4}) y^{2} + 900 b^{8}]$ $= 81 a^{3} b^{9} y$ $\Rightarrow {81 b}^{8} y = y^{6} - 20 b^{2} y^{4} - 10 b^{4} (9 b^{2} - 10) y^{2} + 900 b^{8}$ $\Rightarrow y^{6} - 20 b^{2} [y^{4} - \frac{1}{2} (9 b^{2} - 10) b^{2} y^{2} + 45 b^{6}] - 81 b^{8} y = 0$ $= y^{6} - 20 b^{2} {(y^{2} - \frac{(9 b^{2} - 10) b^{2}}{4})}^{2} - 81 b^{8} y + 45 b^{6} - \frac{(9 b^{2} - 10) b^{4}}{16} = 0$ $. . . . . . . . . . to continious$
	Answered by Frix last updated on 01/Jun/23

	${b x}^{3} = 10 a^{2} b x + 3 a^{3} y \Leftrightarrow y = \frac{b x (x^{2} - 10 a^{2})}{3 a^{3}}$ $Inserting in the 2^{nd} equation and factorizing$ $\frac{b^{3} x (x^{2} - a^{2}) (x^{2} - 7 a^{2}) (x^{2} - 9 a^{2}) (x^{2} - 13 a^{2})}{27 a^{8}} = 0$ $The rest is easy$
	Commented by York12 last updated on 01/Jun/23

	$t h a t w o u l d t a k e a l o t o f t i m e$
	Commented by York12 last updated on 01/Jun/23

	$w a t c h t h i s$
	Answered by a.lgnaoui last updated on 01/Jun/23
	$(suite) { ((bx^3 −10ba^2 x=3a^3 y ⇒b=((3a^3 y)/(x(x^2 −10a^2 ))) (1))),((ay^3 −10ab^2 y=3b^3 x ⇒a=((3b^3 x)/(y(y^2 −10b^2 ))) (2))) :} Remarque: ab=((3a^3 )/((x^2 −10a^2 )))×((3b^3 )/(y^2 −10b^2 )) =((9a^3 b^3 )/((x^2 −10a^2 )(y^2 −10b^2 ))) ⇒9a^2 b^2 =(x^2 −10a^2 )(y^2 −10b^2 ) cette equation nous permet de calculer a en fonction de b et y en fonxtion de x 9a^2 b^2 =x^2 y^2 −10b^2 x^2 −10a^2 y^2 +100a^2 b^2 91a^2 b^2 =10(a^2 y^2 +b^2 x^2 )−x^2 y^2 a^2 =((x^2 (10b^2 −y^2 ))/(91b^2 −10y^2 )) (si b≠y(√((10)/(91))) etb>(y/( (√(10)))))$
	$(suite)$ ${\begin{cases} {bx}^{3} - {10 ba}^{2} x = {3 a}^{3} y \Rightarrow b = \frac{3 a^{3} y}{x (x^{2} - 10 a^{2})} (1) \\ {ay}^{3} - {10 ab}^{2} y = {3 b}^{3} x \Rightarrow a = \frac{3 b^{3} x}{y (y^{2} - {10 b}^{2})} (2) \end{cases}$ $Remarque :$ $ab = \frac{{3 a}^{3}}{(x^{2} - {10 a}^{2})} \times \frac{{3 b}^{3}}{y^{2} - {10 b}^{2}}$ $= \frac{{9 a}^{3} b^{3}}{(x^{2} - {10 a}^{2}) (y^{2} - {10 b}^{2})}$ $\Rightarrow {9 a}^{2} b^{2} = (x^{2} - {10 a}^{2}) (y^{2} - {10 b}^{2})$ $cette equation nous permet de calculer$ $a en fonction de b et y en fonxtion de x$ $9 a^{2} b^{2} = x^{2} y^{2} - 10 b^{2} x^{2} - 10 a^{2} y^{2} + 100 a^{2} b^{2}$ $91 a^{2} b^{2} = 10 (a^{2} y^{2} + b^{2} x^{2}) - x^{2} y^{2}$ $a^{2} = \frac{x^{2} ({10 b}^{2} - y^{2})}{91 b^{2} - 10 y^{2}} (si b \neq y \sqrt{\frac{10}{91}} etb > \frac{y}{\sqrt{10}})$
	Answered by York12 last updated on 01/Jun/23
	$bx^3 = 10a^2 bx + 3a^3 y ........(i) ay^3 = 10ab^2 y + 3b^3 x .......(ii) (((i))/((ii))) we obtain : ((bx^3 )/(ay^3 ))=((a^2 (10bx + 3ay))/(b^2 (10ay + 3bx))) ⇒ ((b^3 x^3 )/(a^3 y^3 ))= ((10bx + 3ay)/(10ay +3bx)) = ((((10 bx)/(ay))+3)/(10 + ((3 bx)/(ay)))) set ((bx)/(ay)) = λ ⇒ λ^3 = ((10 λ +3)/(10 + 3 λ)) ⇒ 3λ^4 + 10 λ^(3 ) = 10λ +3 3λ^4 − 3 + 10 λ^3 −10 λ = 0 (λ^2 −1)(3λ^2 + 3) + 10 λ (λ^2 −1)=0 (λ−1)(λ+1)(λ+3)(3λ+1) =0 ⇒ λ ∈ { 1 , −1 , −3 , ((−1)/3) } By substituting we obtain The required values of x , y , a and b .★$
	${b x}^{3} = 10 a^{2} b x + 3 a^{3} y . . . . . . . . (i)$ ${a y}^{3} = 10 {a b}^{2} y + 3 b^{3} x . . . . . . . (i i)$ $\frac{(i)}{(i i)} w e o b t a i n : \frac{{b x}^{3}}{{a y}^{3}} = \frac{a^{2} (10 b x + 3 a y)}{b^{2} (10 a y + 3 b x)}$ $\Rightarrow \frac{b^{3} x^{3}}{a^{3} y^{3}} = \frac{10 b x + 3 a y}{10 a y + 3 b x} = \frac{\frac{10 b x}{a y} + 3}{10 + \frac{3 b x}{a y}}$ $s e t \frac{b x}{a y} = λ \Rightarrow λ^{3} = \frac{10 λ + 3}{10 + 3 λ} \Rightarrow 3 λ^{4} + 10 λ^{3} = 10 λ + 3$ $3 λ^{4} - 3 + 10 λ^{3} - 10 λ = 0$ $(λ^{2} - 1) (3 λ^{2} + 3) + 10 λ (λ^{2} - 1) = 0$ $(λ - 1) (λ + 1) (λ + 3) (3 λ + 1) = 0$ $\Rightarrow λ \in {1, - 1, - 3, \frac{- 1}{3}}$ $B y s u b s t i t u t i n g w e o b t a i n T h e r e q u i r e d$ $v a l u e s o f x, y, a a n d b . ★$
	Commented by Frix last updated on 01/Jun/23

	$To divide 2 equations is not allowed :$ $(1) 6 = 4 false$ $(2) 3 = 2 false$ $\frac{(1)}{(2)} \frac{6}{3} = \frac{4}{2} true$
	Commented by ajfour last updated on 01/Jun/23

	$b u t i f 6 = 8 - 2$ $3 = 13 - 10$ $\Rightarrow 2 = \frac{6}{3} = \frac{8 - 2}{13 - 10} (. .)!$
	Commented by York12 last updated on 01/Jun/23

	$w h a t y o u h a v e w r i t t e n h e r e i s b a s e d o n i f$ $6 = 4 i s f a l s e$ $3 = 2 i s f a l s e$ $b u t i f t h e s t a t e m e n t i s t r u e t h e n w e c a n a l w a y s$ $d i v i d e t w o e q u a t i o n s$
	Commented by York12 last updated on 01/Jun/23

	$n o t h i n g e x t r a o r d i n a r y$
	Commented by York12 last updated on 01/Jun/23

	$To divide 2 equations is not allowed :$ $(1) 6 = 4 false$ $(2) 3 = 2 false$ $\frac{(1)}{(2)} \frac{6}{3} = \frac{4}{2} true$ $w h e r e a s i n m y q u e s t i o n s t h e e q u a t i o n s$ $a r e r i g h t t h e n$ $\frac{{L H S}_{e q u (1)}}{{L H S}_{e q u (2)}} = \frac{{R H S}_{e q u (1)}}{{R H S}_{e q u (2)}}$
	Commented by Frix last updated on 01/Jun/23

	${You}^{'} re wrong here .$ $Another example (we had some similar$ $‘ ‘ {solutions}^{″} here some time ago) :$ $(1) x^{4} - x^{3} - 43 x^{2} + 23 x + 203 = 0$ $(2) x^{3} - 6 x^{2} - 13 x + 41 = 0$ $=================$ $(1) x^{4} - x^{3} - 43 x^{2} - 23 x + 210 = 7$ $(2) x^{3} - 6 x^{2} - 13 x + 42 = 1$ $=================$ $(1) (x - 7) (x - 2) (x + 3) (x + 5) = 7$ $(2) (x - 7) (x - 2) (x + 3) = 1$ $=================$ $\frac{(1)}{(2)} x + 5 = 7 \Rightarrow x = 2 (false)$ $But both equations are ‘ ‘ {true}^{″} . . .$
	Commented by ajfour last updated on 02/Jun/23

	$c o z i f x = 2 w e s h u d n t h a v e$ $d i v i d e d b y (x - 7) (x - 2) (x + 3)$ $s o i t d o e s n t d e c e i v e . w e k n o w$ $x = 2 i s t h e n f a l s e .$
	Commented by York12 last updated on 02/Jun/23

	$s o r r y b u t f o r a g a i n y o u a r e w r o n g$ $b e c a u s e b o t h e q u a t i o n s c a n o t b e t r u e$ $s i m u l t a o u s l y$ $\underset{T}{\underset{⏟}{(x - 7) (x - 2) (x + 3)}} = 1$ $\underset{T}{\underset{⏟}{(x - 7) (x - 2) (x + 3)}} (x + 5) = 7$ $T = 1$ $T \times (x + 5) = 7 \to x = 2$ $T h a t i s s o t r i v i a l$ $b u t i f y o u w i l l s u b s t i t u t e y o u w i l l$ $f i n d t h a t x d o e s n o t s a t i s f y t h e g i v e n$ $s y s t e m o f e q u a t i o n s$ $t h e n t h a t m e a n s t h e r e a r e t w o v a r i a b l e s$ $t o m a k e t h a t h a p p e n y o u r e q u a t i o n s$ $h a s t o w r i t t e n a s$ $(y - 7) (y - 2) (y + 3) (x + 5) = 7$ $(y - 7) (y - 2) (y + 3) = 1$ $t h e n t h a t c a n h a p p e n$ $a n d s i n c e y o u h a v e c o m m i t e d$ $s u c h a m i s t a k e I a d v i s e y o u$ $t o s t u d y t h e o r y o f e q u a t i o n s$ $s p e c i f i c l y t h e c o n c e p t o f c o m m o n r o o t s$ $c a u s e t h o s e t w o e q u a t i o n s t h a t y o u h a v e$ $o b t a i n e d d o e s n o t h a v e a n y c o m m o n$ $r o o t s$
	Commented by York12 last updated on 02/Jun/23

	$w r o n g$ $c a u s e b o t h e q u t i o n s c a n n o t b e t r u e$ $c a u s e t h e y d o n o t h a v e a n y c o m m o n r o o t s!!!!$ $t h e n f o r a g a i n t h e a s s u m p t i o n s$ $a r e w r o n g$
	Commented by ajfour last updated on 02/Jun/23

	$w h o d o u e x a c t l y t e l l ? t h i s a l l . .$
	Commented by York12 last updated on 02/Jun/23

	$t o f r i x$
	Commented by ajfour last updated on 03/Jun/23
	He's long past all this, we ll never get over!
	Commented by Frix last updated on 08/Jun/23

	$I^{'} m not an idiot .$ $You cannot divide 2 equations b e f o r e$ $showing that this division is legit .$
	Commented by York12 last updated on 13/Jun/23

	$n o y o u c a n$
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