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

$$\mathrm{Find}(\mathrm{x},\mathrm{y})\in \mathbb{R}\mathrm{such}\mathrm{that};$$$$\frac{\mathrm{x}+\mathrm{y}}{{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2}=\frac{7}{2}$$$$$$$$updatedfrom\frac{2}{7}\to \frac{7}{2}.Sorry,itwasamistake.$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$\{\mathrm{x}\mathrm{should}\mathrm{be}\pm (1-\sqrt{2})$$$$\{\mathrm{y}\mathrm{should}\mathrm{be}\sqrt{2}\pm 1$$$$$$$$\mathrm{x}+\mathrm{y}=2,\mathrm{x}=2-\mathrm{y}$$$${(\mathrm{x}+\mathrm{y})}^2=7+3\mathrm{xy}\left\{\mathrm{Assuming}\right\}$$$$4=7+3\mathrm{xy}\mathrm{so}\mathrm{xy}=-1\Rightarrow (2-\mathrm{y})\mathrm{y}=-1$$$$\Rightarrow {\mathrm{y}}^2-2\mathrm{y}-1=0$$$$\Rightarrow \mathrm{y}=\frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2}$$$$\mathrm{x}(1\pm \sqrt{}2)=-1\Rightarrow \mathrm{x}=\frac{-1}{1\pm \sqrt{2}}=(1\pm \sqrt{2})$$$$$$$$\left\{\mathrm{But}\mathrm{it}\mathrm{has}\mathrm{infinite}\mathrm{many}\mathrm{solutions}\right\}$$$$\mathrm{If}\mathrm{we}\mathrm{go}\mathrm{for}\mathrm{all}\mathrm{solutions}$$$$\mathrm{x}+\mathrm{y}=2\mathrm{p}\{\mathrm{p}\ne 0$$$$\mathrm{and}{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2=7\mathrm{p}$$$${(\mathrm{x}+\mathrm{y})}^2=7\mathrm{p}+3\mathrm{xy}$$$${4\mathrm{p}}^2=7\mathrm{p}+3\mathrm{xy}$$$$\mathrm{xy}=\frac{{4\mathrm{p}}^2-7\mathrm{p}}{3}\Rightarrow \mathrm{x}(2\mathrm{p}-\mathrm{x})=\frac{{4\mathrm{p}}^2-7\mathrm{p}}{3}\Rightarrow {3\mathrm{x}}^2-6\mathrm{px}+{4\mathrm{p}}^2-7\mathrm{p}=0$$$$\mathrm{x}=\frac{6\mathrm{p}\pm \sqrt{{36\mathrm{p}}^2-{48\mathrm{p}}^2+84\mathrm{p}}}{6}=\frac{6\mathrm{p}\pm \sqrt{84\mathrm{p}-{12\mathrm{p}}^2}}{6}$$$$\mathrm{And}\mathrm{y}=\frac{{8\mathrm{p}}^2-14\mathrm{p}}{6\mathrm{p}\pm \sqrt{84\mathrm{p}-{12\mathrm{p}}^2}}$$

Commented by mr W last updated on 25/Jun/20

$$theeqn.isanellipsewithoutthe$$$$point(0,0).ithasinfinitepoints,$$$$i.e.infinitesolutions.$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$\mathrm{Sir},\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{but}\mathrm{i}\mathrm{go}\mathrm{for}\mathrm{only}\mathrm{one}\mathrm{solution}$$

Commented by mr W last updated on 25/Jun/20

Commented by mr W last updated on 25/Jun/20

$$youcansetx=1forexample,then$$$$youcanfindtwovaluesfory:$$$$\frac{1+y}{1-y+y^2}=\frac{2}{7}$$

Commented by mr W last updated on 25/Jun/20

$$thegeneralsolutionis$$$$x=k\pm \sqrt{\frac{k(7-k)}{3}}$$$$y=k\mp \sqrt{\frac{k(7-k)}{3}}$$$$with0<k⩽7$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$\mathrm{Yes}\mathrm{sir}\mathrm{i}\mathrm{am}\mathrm{a}\mathrm{high}\mathrm{school}\mathrm{student}.\mathrm{Thanking}\mathrm{you}\mathrm{for}\mathrm{correction}$$

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

Thank you !

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

Thanks for your idea !

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

$$\frac{\mathrm{x}+\mathrm{y}}{{\mathrm{x}}^2-\mathrm{xy}+{\mathrm{y}}^2}=\frac{7}{2}\Rightarrow \frac{\mathrm{x}+\mathrm{y}}{{(\mathrm{x}-\frac{\mathrm{y}}{2})}^2-\frac{{\mathrm{y}}^2}{4}+{\mathrm{y}}^2}=\frac{\mathrm{x}+\mathrm{y}}{{(\mathrm{x}-\frac{\mathrm{y}}{2})}^2+\frac{{3\mathrm{y}}^2}{4}}=\frac{7}{2}$$$$\Rightarrow \{\begin{array}{l}\mathrm{x}+\mathrm{y}=7\Rightarrow \mathrm{x}=7-\mathrm{y}\\ {(\mathrm{x}-\frac{\mathrm{y}}{2})}^2+\frac{{3\mathrm{y}}^2}{4}=2\end{array}\Rightarrow {(7-\mathrm{y}-\frac{\mathrm{y}}{2})}^2+\frac{{3\mathrm{y}}^2}{4}=2$$$$\Rightarrow {(7-\frac{3\mathrm{y}}{2})}^2+\frac{{3\mathrm{y}}^2}{4}=2\Rightarrow 49-21\mathrm{y}+\frac{{9\mathrm{y}}^2}{4}+\frac{{3\mathrm{y}}^2}{4}=2\Rightarrow {3\mathrm{y}}^2-21\mathrm{y}+47=0$$$$\Rightarrow \mathrm{y}=\frac{21\pm \sqrt{{21}^2-4\left(3\times 47\right)}}{2\left(3\right)}=\frac{21\pm \sqrt{-123}}{6}$$$$$$$$Iarrivedatthis,but{it}^{\prime }sobviousthatthis{isn}^{\prime }t$$$$correct,asy\notin \mathbb{R}.Anyonetohelpmeout,please?$$

Commented by mr W last updated on 25/Jun/20

$$youcanonlysay$$$$x+y=2k$$$$x^2-xy+y^2=7k\ne 0$$$$thereareinfinitesolutionsforx,y\in R.$$$$x+y=2k$$$${(x+y)}^2-3xy=7k\Rightarrow xy=\frac{4k^2-7k}{3}$$$$x,yarerootsof$$$$z^2-2kz+\frac{4k^2-7k}{3}=0$$$$\Rightarrow x,y=k\pm \sqrt{\frac{k(7-k)}{3}}(k\ne 0)$$

Commented by mr W last updated on 25/Jun/20

$$\frac{x+y}{x^2-xy+y^2}=\frac{7}{2}$$$$hasalsoinfinitesolutions.$$

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

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

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

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

Commented by mr W last updated on 25/Jun/20

$$x+y=7k$$$$x^2-xy+y^2=2k\ne 0$$$${(x+y)}^2-3xy=2k$$$$\Rightarrow xy=\frac{49k^2-2k}{3}$$$$z^2-7kz+\frac{49k^2-2k}{3}=0$$$$\Rightarrow x,y=\frac{1}{2}[7k\pm \sqrt{\frac{k(8-49k)}{3}}]$$

Commented by Ar Brandon last updated on 26/Jun/20

Thanks a lot for your time��

Answered by 4635 last updated on 26/Jun/20

$$\frac{x+y}{x^2-xy+y^2}=\frac{7}{2}\Rightarrow \frac{x+y}{{(x-y)}^2+xy}=\frac{7}{2}$$$$\iff \frac{4(x+y)}{1+3{\left(\frac{x-y}{x+y}\right)}^2}=\frac{7}{2}$$$$\{x+y=\frac{7}{4}et3{\left(\frac{x-y}{x+y}\right)}^2=1$$$$\{x+y=\frac{7}{4}etx-y=\frac{7}{4\sqrt{3}}$$$$x=\frac{7}{8}\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)ety=\frac{7}{8}\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)$$

Commented by mr W last updated on 26/Jun/20

$$forX,Y,a,b\in R$$$$\frac{X}{Y}=\frac{a}{b}\nRightarrow X=a,Y=b$$

Commented by Ar 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 ?��

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