Systems of 2 non linear equations (2 variables)

Question

I have a problem finding solutions to systems that contain 2 equations with 2 variables each and are second degree. Example :

\begin{cases}x^2 + x + y^2 + y - 18 = 0 &&(1)\\ x^2 + xy + y^2 - 12 = 0 & &(2)\end{cases}

What I do :

1. find the solutions of x (in terms of y) in the first equation - $ x_1,x_2 $
2. solve the second equation using $x_1,x_2$ (so i can get an equation of second degree with one variable - y)
3. substitude the solutions of the second equation in the $x_1$ and $x_2$ and find the final answer

The problem here is that to find on of the $x$ I spend $10$ minutes and to solve the problem i need $40$ minutes. I make a lot of mistakes... Is there a faster and safer way ?

Answer 1

I suggest that you should make use of symmetry the system has. In this system, if $(x,y)=(p,q)$ is the solution of the system, then so is $(x,y)=(q,p)$. This indicates that we can rewrite the system in terms of elementary symmetric polynomials, i.e. $x+y$ and $xy$. This procedure yields $$(x+y)^2+(x+y)-2xy-18=0;\quad (x+y)^2-xy-12=0.$$ This system is solved to obtain $(x+y,xy)=(3,-3),\ (-2,-8)$ and thus we obtain $(x,y)=((3+\sqrt{21})/2,(3-\sqrt{21})/2), (2,-4)$ and permutation of each.

Answer 2

HINT: eliminating $y$ we get for $x$ $$x^4-x^3-17 x^2+18 x=-24$$

Answer 3

Considering $$\begin{cases}x^2 + x + y^2 + y - 18 = 0 &&(1)\\ x^2 + xy + y^2 - 12 = 0 & &(2)\end{cases}$$ Since the term $x^2+y^2$ has the good idea to appear in both equations, subtract one from the other to get $$ -x y+x+y-6\implies x=\frac{y-6}{y-1}$$ (hoping $y\neq 1$). Plug in the first equation, reduce to same denominator and the equation reduces to $$\frac{y^4-y^3-17 y^2+18 y+24}{(y-1)^2}=0$$ By inspection, the numerator shows one root for $y=2$. So, using long division, $$y^4-y^3-17 y^2+18 y+24=(y-2)(y^3+y^2-15 y-12)=0$$ The cubic equation also shows by inspection (less trivial than the first) a root $y=-4$. Using again long division, all of that means that we are left with $$(y-2)(y+4)(y^2-3 y-3)=0$$

The quadratic equation has two real roots (with radicals). So, you have four solutions for $y$ and the corresponding $x$'s are computed using the simple formula obtained by the preliminary step.