Question Number 100240 by Ar Brandon last updated on 25/Jun/20

Find  (x,y)∈R such  that;  ((x+y)/(x^2 −xy+y^2 ))=(7/2)    updated from (2/7)→(7/2). Sorry, it was a mistake.

Commented byDwaipayan Shikari last updated on 25/Jun/20

{x should be ±(1−(√2))  {y should be (√2)±1    x+y=2,  x=2−y  (x+y)^2 =7+3xy     {Assuming}  4=7+3xy   so  xy=−1⇒(2−y)y=−1                                                      ⇒y^2 −2y−1=0                                                    ⇒y=((2±(√8))/2)=1±(√2)  x(1±(√)2)=−1⇒x=((−1)/(1±(√2)))=(1±(√2))    {But it has infinite many solutions}  If we go for all solutions  x+y=2p  {p≠0  and x^2 −xy+y^2 =7p    (x+y)^2 =7p+3xy    4p^2 =7p+3xy    xy=((4p^2 −7p)/3)⇒x(2p−x)=((4p^2 −7p)/3)⇒3x^2 −6px+4p^2 −7p=0                                                          x=((6p±(√(36p^2 −48p^2 +84p)))/6)=((6p±(√(84p−12p^2 )))/6)  And   y=((8p^2 −14p)/(6p±(√(84p−12p^2 ))))

Commented bymr W last updated on 25/Jun/20

the eqn. is an ellipse without the  point(0,0). it has infinite points,  i.e. infinite solutions.

Commented byDwaipayan Shikari last updated on 25/Jun/20

Sir, you are right but i go for only one solution

Commented bymr W last updated on 25/Jun/20

Commented bymr W last updated on 25/Jun/20

you can set x=1 for example, then  you can find two values for y:  ((1+y)/(1−y+y^2 ))=(2/7)

Commented bymr W last updated on 25/Jun/20

the general solution is  x=k±(√((k(7−k))/3))  y=k∓(√((k(7−k))/3))  with 0<k≤7

Commented byDwaipayan Shikari last updated on 25/Jun/20

Yes sir i am a high school student .Thanking you for correction

Commented byAr Brandon last updated on 25/Jun/20

Thank you !

Commented byAr Brandon last updated on 25/Jun/20

Thanks for your idea !

Answered by Ar Brandon last updated on 25/Jun/20

((x+y)/(x^2 −xy+y^2 ))=(7/2)⇒((x+y)/((x−(y/2))^2 −(y^2 /4)+y^2 ))=((x+y)/((x−(y/2))^2 +((3y^2 )/4)))=(7/2)  ⇒ { ((x+y=7⇒x=7−y)),(((x−(y/2))^2 +((3y^2 )/4)=2)) :}⇒(7−y−(y/2))^2 +((3y^2 )/4)=2  ⇒(7−((3y)/2))^2 +((3y^2 )/4)=2⇒49−21y+((9y^2 )/4)+((3y^2 )/4)=2⇒3y^2 −21y+47=0  ⇒y=((21±(√(21^2 −4(3×47))))/(2(3)))=((21±(√(−123)))/6)    I arrived at this, but it′s obvious that this isn′t   correct, as y∉R. Anyone to help me out, please?

Commented bymr W last updated on 25/Jun/20

you can only say  x+y=2k  x^2 −xy+y^2 =7k≠0  there are infinite solutions for x,y∈R.  x+y=2k  (x+y)^2 −3xy=7k ⇒xy=((4k^2 −7k)/3)  x, y are roots of  z^2 −2kz+((4k^2 −7k)/3)=0  ⇒x,y=k±(√((k(7−k))/3))    (k≠0)

Commented bymr W last updated on 25/Jun/20

((x+y)/(x^2 −xy+y^2 ))=(7/2)  has also infinite solutions.

Commented byAr Brandon last updated on 25/Jun/20

I tried applying the same reasoning which you used above, but I get complex solutions √-123k²

Commented byAr Brandon last updated on 25/Jun/20

Oh my ! I'm so very sorry. You're right, it's a mistake. It's supposed to be 7/2 . Thanks for the remark

Commented byAr Brandon last updated on 25/Jun/20

Or perhabs I didn't do it in the right way.

Commented bymr W last updated on 25/Jun/20

x+y=7k  x^2 −xy+y^2 =2k≠0  (x+y)^2 −3xy=2k  ⇒xy=((49k^2 −2k)/3)  z^2 −7kz+((49k^2 −2k)/3)=0  ⇒x, y=(1/2)[7k±(√((k(8−49k))/3))]

Commented byAr Brandon last updated on 26/Jun/20

Thanks a lot for your time😃

Answered by 4635 last updated on 26/Jun/20

((x+y)/(x^2 −xy+y^2 ))=(7/2)⇒((x+y)/((x−y)^2 +xy))=(7/2)  ⇔((4(x+y))/(1+3(((x−y)/(x+y)))^(2 ) ))=(7/2)  {x+y=(7/4) et 3(((x−y)/(x+y)))^2 =1  {x+y=(7/4) et x−y=(7/(4(√3)))  x=(7/8)((((√3)+1)/(√3))) et  y=(7/8)((((√3)−1)/(√3)))

Commented bymr W last updated on 26/Jun/20

forX,Y,a,b∈R  (X/Y)=(a/b) ⇏ X=a, Y=b

Commented byAr Brandon last updated on 26/Jun/20

Hum Takou, est-ce que tu as essayé de remplacer tes réponses dans l'équation pour voir si elles sont correctes ?😃