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-

Tinku Tara June 4, 2023 Algebra 0 Comments

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

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 )

{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

no common roots means no existence for any value of x which can satify both equations ⇒ your system of equations is again wrong

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 \boldsymbol y = \frac{\boldsymbol xb (\boldsymbol x^{2} - 10 \boldsymbol a^{2})}{3 \boldsymbol a^{3}} (1) \\ {3 b}^{3} x = {ay}^{3} - {10 ab}^{2} y \Rightarrow \boldsymbol x = \frac{\boldsymbol ya (\boldsymbol y^{2} - 10 \boldsymbol b^{2})}{3 \boldsymbol b^{3}} (2) \end{cases}

\boldsymbol y^{2} = \frac{\boldsymbol x^{2} \boldsymbol b^{2} {(\boldsymbol x^{2} - 10 \boldsymbol a^{2})}^{2}}{9 \boldsymbol a^{6}}

\boldsymbol x^{2} = \frac{\boldsymbol y^{2} \boldsymbol a^{2} (\boldsymbol y^{2} - 10 \boldsymbol b^{2})}{9 \boldsymbol b^{6}}

(1) \Leftrightarrow \boldsymbol y = \frac{\boldsymbol b}{3 \boldsymbol a^{3}} [\frac{\boldsymbol ya (\boldsymbol y^{2} - 10 \boldsymbol b^{2}}{3 \boldsymbol b^{3}}] [(\frac{\boldsymbol y^{2} \boldsymbol a^{2} (\boldsymbol y^{2} - 10 \boldsymbol b^{2}) - 90 \boldsymbol a^{2} \boldsymbol b^{6}}{9 \boldsymbol b^{6}}]

81 \boldsymbol a^{3} \boldsymbol b^{9} \boldsymbol y = \boldsymbol ab (\boldsymbol y^{2} - 10 \boldsymbol b^{2}) (\boldsymbol a^{2} \boldsymbol y^{4} - 10 \boldsymbol y^{2} \boldsymbol a^{2} \boldsymbol b^{2} - 90 \boldsymbol a^{2} \boldsymbol b^{6})

= \boldsymbol a^{3} \boldsymbol b (\boldsymbol y^{2} - 10 \boldsymbol b^{2}) (\boldsymbol y^{4} - 10 \boldsymbol y^{2} \boldsymbol b^{2} - 90 \boldsymbol b^{6})

= \boldsymbol a^{3} \boldsymbol b [\boldsymbol y^{6} - 10 \boldsymbol y^{4} \boldsymbol b^{2} - 90 \boldsymbol y^{2} \boldsymbol b^{6}

- 10 \boldsymbol b^{2} \boldsymbol y^{4} + 100 \boldsymbol b^{4} \boldsymbol y^{2} + 900 \boldsymbol b^{8}

= \boldsymbol a^{3} \boldsymbol b [\boldsymbol y^{6} - 20 \boldsymbol b^{2} \boldsymbol y^{4} - (90 \boldsymbol b^{6} - 100 \boldsymbol b^{4}) \boldsymbol y^{2} + 900 \boldsymbol b^{8}]

= 81 \boldsymbol a^{3} \boldsymbol b^{9} \boldsymbol y

\Rightarrow {81 b}^{8} \boldsymbol y = \boldsymbol y^{6} - 20 \boldsymbol b^{2} \boldsymbol y^{4} - 10 \boldsymbol b^{4} (9 \boldsymbol b^{2} - 10) \boldsymbol y^{2} + 900 \boldsymbol b^{8}

\Rightarrow \boldsymbol y^{6} - 20 \boldsymbol b^{2} [\boldsymbol y^{4} - \frac{1}{2} (9 \boldsymbol b^{2} - 10) \boldsymbol b^{2} \boldsymbol y^{2} + 45 \boldsymbol b^{6}] - 81 \boldsymbol b^{8} \boldsymbol y = 0

= \boldsymbol y^{6} - 20 \boldsymbol b^{2} {(\boldsymbol y^{2} - \frac{(9 \boldsymbol b^{2} - 10) \boldsymbol b^{2}}{4})}^{2} - 81 \boldsymbol b^{8} \boldsymbol y + 45 \boldsymbol b^{6} - \frac{(9 \boldsymbol b^{2} - 10) \boldsymbol b^{4}}{16} = 0

\dots \dots \dots . \boldsymbol to \boldsymbol continious

Answered by Frix last updated on 01/Jun/23

bx^3 =10a^2 bx+3a^3 y ⇔ y=((bx(x^2 −10a^2 ))/(3a^3 )) Inserting in the 2^(nd) equation and factorizing ((b^3 x(x^2 −a^2 )(x^2 −7a^2 )(x^2 −9a^2 )(x^2 −13a^2 ))/(27a^8 ))=0 The rest is easy

{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

that would take a lot of time

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

watch this

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 \boldsymbol b = \frac{3 \boldsymbol a^{3} y}{x (x^{2} - 10 \boldsymbol a^{2})} (1) \\ {ay}^{3} - {10 ab}^{2} y = {3 b}^{3} x \Rightarrow \boldsymbol a = \frac{3 \boldsymbol b^{3} x}{y (y^{2} - {10 b}^{2})} (2) \end{cases}

\boldsymbol 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

\boldsymbol a en fonction de \boldsymbol b et \boldsymbol y en fonxtion de \boldsymbol x

9 \boldsymbol a^{2} \boldsymbol b^{2} = \boldsymbol x^{2} \boldsymbol y^{2} - 10 \boldsymbol b^{2} \boldsymbol x^{2} - 10 \boldsymbol a^{2} \boldsymbol y^{2} + 100 \boldsymbol a^{2} \boldsymbol b^{2}

91 \boldsymbol a^{2} \boldsymbol b^{2} = 10 (\boldsymbol a^{2} \boldsymbol y^{2} + \boldsymbol b^{2} \boldsymbol x^{2}) - \boldsymbol x^{2} \boldsymbol y^{2}

\boldsymbol a^{2} = \frac{\boldsymbol x^{2} ({10 b}^{2} - \boldsymbol y^{2})}{91 \boldsymbol b^{2} - 10 \boldsymbol y^{2}} (\boldsymbol si \boldsymbol 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 \dots \dots . . (i)

{a y}^{3} = 10 {a b}^{2} y + 3 b^{3} x \dots \dots . (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 (((1))/((2))) (6/3)=(4/2) true

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

but if 6=8−2 3=13−10 ⇒ 2=(6/3)=((8−2)/(13−10)) (..)!

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

what you have written here is based on if 6=4 is false 3=2 is false but if the statement is true then we can always divide two equations

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

nothing extraordinary

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 (((1))/((2))) (6/3)=(4/2) true whereas in my questions the equations are right then ((LHS_(equ (1)) )/(LHS_(equ(2)) )) =((RHS_(equ (1)) )/(RHS_(equ (2)) ))

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 −43x^2 +23x+203=0 (2) x^3 −6x^2 −13x+41=0 ================= (1) x^4 −x^3 −43x^2 −23x+210=7 (2) x^3 −6x^2 −13x+42=1 ================= (1) (x−7)(x−2)(x+3)(x+5)=7 (2) (x−7)(x−2)(x+3)=1 ================= (((1))/((2))) x+5=7 ⇒ x=2 (false) But both equations are “true”...

{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}^{″} \dots

Commented by ajfour last updated on 02/Jun/23

coz if x=2 we shudnt have divided by (x−7)(x−2)(x+3) so it doesnt deceive. we know x=2 is then false.

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

sorry but for again you are wrong because both equations canot be true simultaously (x−7)(x−2)(x+3)_(T) =1 (x−7)(x−2)(x+3)_(T) (x+5)=7 T =1 T × (x+5) =7 → x=2 That is so trivial but if you will substitute you will find that x doesnot satisfy the given system of equations then that means there are two variables to make that happen your equations has to written as (y−7)(y−2)(y+3)(x+5)=7 (y−7)(y−2)(y+3)=1 then that can happen and since you have commited such a mistake I advise you to study theory of equations specificly the concept of common roots cause those two equations that you have obtained doesnot have any common roots

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

wrong cause both equtions can not be true cause they do not have any common roots !!!! then for again the assumptions are wrong

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

who do u exactly tell?this all..

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

to frix

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 before showing that this division is legit.

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

no you can

n o y o u c a n

n o y o u c a n

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