Solve the following system of equations - (3)

Question

Solve the following system of equations: $$\large \left\{ \begin{align*} 3x^2 + xy - 4x + 2y - 2 = 0\\ x(x + 1) + y(y + 1) = 4 \end{align*} \right. $$

I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.

Answer 1

Solving the first equation for $y$ we get $$y=\frac{-3x^2+4x+2}{2+x}$$ for $$x\neq -2$$ plugging this in the second equation we get after simplifications $$(5x+4)(x-1)^3=0$$

Answer 2

The resultant of $3\,{x}^{2}-xy-4\,x+2\,y-2$ and $ x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is $$ 10\,{x}^{4}-24\,{x}^{3}-10\,{x}^{2}+42\,x-8$$ which is irreducible over the rationals. Its roots can be written in terms of radicals, but they are far from pleasant. There are two real roots, $x \approx -1.287147510$ and $x \approx 0.2049816008$, which correspond to $y \approx -2.469872787$ and $y \approx 1.500750095$ respectively.

EDIT: For the corrected question, the resultant of $3\,{x}^{2}+xy-4\,x+2\,y-2$ and $x \left( x+1 \right) +y \left( y+1 \right) -4$ with respect to $y$ is $$ 10 x^4 - 22 x^3 + 6 x^2 + 14 x - 8 = 2 (5 x + 4) (x-1)^3$$ Thus we want $x = -4/5$, corresponding to $y = -13/5$, or $x = 1$, corresponding to $y = 1$.

Answer 3

Substituting for the updated equation yields $$ x=-\frac{4}{5}, \; y=-\frac{13}{5} $$ or $(x,y)=(1,1)$. This is a very pleasant result, compared with the old one (with $-xy$ in the first equation instead of $xy$).

$$ x=\frac{5y^3 - 26y^2 - 24y + 91}{65}, $$ with $$ 5y^4 + 9y^3 - 11y^2 - 12y - 13=0. $$