How do I solve this system of equations $x^2-xy+2y^2=28$, $x^2-3xy-y^2=16$?

Question

How do I solve this system of equations $x^2-xy+2y^2=28$, $x^2-3xy-y^2=16$?

I started to get $x$ than to put it in equation but it didn't help me.

Answer 1

Observe that \begin{align*} x^2-xy+2y^2+2(x^2-3xy-y^2)&=28+2\cdot 16\\ 3x^2-7xy&=60\\ 3x-7y&=\frac{60}x\\ y&=\frac17\left(3x-\frac{60}x\right) \end{align*}

Then $$x^2-xy+2y^2=28\quad\iff\quad x^2-\frac17\left(3x^2-60\right)+\frac2{49}\left(9x^2-360+\frac{3600}{x^2}\right)=28$$ Last equation can be put in the form $$Ax^4+Bx^2+C=0$$ and then you can solve by making the sustitution $u=x^2$.

Answer 2

If you feed it to Alpha, you have the intersection of an ellipse with a hyperbola. There are four points. Alpha returns approximate values because the exact values (which it will provide) are a mess. It probably just uses the quadratic formula on one of the equations and substituted into the other, getting a quartic to solve. There ain't no graceful way.

Answer 3

HINT:

Put $y=mx$ as $xy\ne0$

divide the relation derived by eliminating $x$

to find the value of $m$

Answer 4

multiplying the first equation by $-1$ and adding to the first we obtain $$2xy+3y^2=12$$ and from here we get $$2xy=12-3y^2$$ thus $$x=\frac{12-3y^2}{2y}$$ from here you can eliminate $x$

Answer 5

By elimination (algorithmically by computing the Gröbner basis using e.g. wolframalpha) these two polynomials can be derived from your input:

$23 y^4 - 208 y^2 + 144, 24 x + 23 y^3 - 172 y$

Their set of zeroes is the same as the set of solutions for your original equations. From the first polynomial you get a closed form solution for all the $y$s, from the second polynomial you get a closed from solution for each corresponding $x$.