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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.

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  • $\begingroup$ Only for $$y\ne 2$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 22 at 16:24
  • $\begingroup$ @Dr.SonnhardGraubner For $x\neq 2$, yes. Then we can say if $x=2$, $y$ must equal $14/6$ and this does not satisfy the second equation. $\endgroup$ – Infiaria Mar 22 at 16:33
  • $\begingroup$ But you forgot this to say. $\endgroup$ – Dr. Sonnhard Graubner Mar 22 at 16:37
  • $\begingroup$ @Dr.SonnhardGraubner Good thing 4 other people already have answers. (Since OP updated the question, now $y=\frac{2+4x-3x^2}{2+x}$. For $x\neq -2$, of course.) $\endgroup$ – Infiaria Mar 22 at 16:38
  • $\begingroup$ Well.... uh.... (I'm sorry.) $\endgroup$ – Lê Thành Đạt Mar 22 at 16:38
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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 $$(5x+4)(x-1)^3=0$$

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  • $\begingroup$ Please wait. I typed the problem wrong. $\endgroup$ – Lê Thành Đạt Mar 22 at 16:33
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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 $$ 5y^4 + 9y^3 - 11y^2 - 12y - 13=0. $$

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  • $\begingroup$ I apologize. I typed the question wrong. $\endgroup$ – Lê Thành Đạt Mar 22 at 16:33
  • $\begingroup$ @LêThànhĐạt And what is the correct question? $\endgroup$ – Dietrich Burde Mar 22 at 16:34
  • $\begingroup$ I just fixed it. Thanks for asking. $\endgroup$ – Lê Thành Đạt Mar 22 at 16:35
  • $\begingroup$ @LêThànhĐạt I fixed my answer, too. $\endgroup$ – Dietrich Burde Mar 22 at 16:37
  • $\begingroup$ It is not clear how the equation of degree 4 in y was reached... $\endgroup$ – NoChance Mar 22 at 16:57
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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 $y = 1$.

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  • $\begingroup$ Could you wait for me a little bit? I typed the problem wrong. $\endgroup$ – Lê Thành Đạt Mar 22 at 16:34

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