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 ?😃