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Are there any nice/elegant ways to solve this system of equations?

$\left\{\begin{matrix} 2x^2+xy+y^2=28 & \\ x^2-xy+2y^2=32 & \end{matrix}\right.$

I can solve it by isolating one of the variables but it's too messy.


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You can add the two equations together to get rid of the $xy$ term. – crf Dec 11 '13 at 9:27
@crf - that's what i did for the isolation method. – gazok Dec 11 '13 at 9:28

There is a general way to solve such homogeneous equations.

First note that $x=0$ doesn't provide a solution. Now choose $m=\dfrac{y}x$, so that $y=mx$. Your two equations then become,

$x^2(2+m+m^2)=28 \\ x^2(1-m+2m^2)=32$

Dividing the two, we obtain:

\begin{align} &\dfrac{2+m+m^2}{1 -m+2m^2}=\dfrac78\\ &\iff 2m^2-5m-3=0\\ &\iff (2m+1)(m-3)=0. \end{align}

For $m=-\dfrac12$, $x^2=16$ while for $m=3$, we get $x^2=2$.

Thus, $(x,y) \in\{\,(4,-2),\,(-4,2),\,(\sqrt2,3\sqrt2),\,(-\sqrt2,-3\sqrt2)\,\}$.

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nice+, btw what do you mean by a general way? – gazok Dec 11 '13 at 11:00
@gazok This method of using the substitution $y=mx$, also works for other homogeneous equations. – Alraxite Dec 11 '13 at 11:17
oh by homogeneous you mean $F(tx,ty)=t^nF(x,y)$. ill check it next time! thanks. – gazok Dec 11 '13 at 11:23
@gazok Yes. Basically for a polynomial, the sum of the powers of all the variables in each term should be the same. – Alraxite Dec 11 '13 at 11:32
in a system it has to add up to the same sum of powers? like $6$ in this case for each equation? or $2$ in each term in each equation? – gazok Dec 11 '13 at 11:37

Try adding one equation to the other: eqn. 1 + eqn.2 \begin{align*} (2x^2 + xy + y^2) + (x^2 - xy + 2y^2) &= 28 + 32\\ 3x^2+3y^2 &= 60\\ x^2 + y^2 &=20\tag{*} \end{align*}

And from linear algebra we now, that points $[x,y] \in \mathbb{R}^2$ which satisfy (*) are located on a circle with center in $[0,0]$ and radius $\sqrt{20}$.

Because general equation for a circle in plane is $$ (x-x_0)^2 + (y-y_0)^2 = r^2. $$

Where $[x_0,y_0] \in \mathbb{R}^2$ is the center of the circle and $r$ is radius.

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there are infinite points that satisfy that circle. from $(*)$ i isolated one of the variable and substituted in one of the equation. that's how i got the solutions. im looking for different approach or i just didn't understand yours. – gazok Dec 11 '13 at 9:33
And what is wrong with infinitely many points that satisfy the system of equations? Consider system of equation $$y = 2,$$ but in a plane $\mathbb{R}^2$. Every pair $[t,2] \in \mathbb{R}^2$, where $t \in \mathbb{R}$ solves this system, because in fact the system is equivalent to $$ 0 \cdot x + y = 2 $$ – quapka Dec 11 '13 at 9:39
well, because there are no infinite solutions the this system.. so i don't know how you can proceed from that circle. let's say $x^2=1$ so $y^2=19$ putting that in equation 1, and it's not true. – gazok Dec 11 '13 at 10:04
My bad, sorry, I didn't finish the idea correctly. I would continue the same way you wrote (isolation and substitution). So no help at all from my side. – quapka Dec 11 '13 at 10:15
no problem but thanks for the effort. – gazok Dec 11 '13 at 10:35

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