# Solve the following system of equations - (3)

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.

• Only for $$y\ne 2$$ – Dr. Sonnhard Graubner Mar 22 at 16:24
• @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. – Infiaria Mar 22 at 16:33
• But you forgot this to say. – Dr. Sonnhard Graubner Mar 22 at 16:37
• @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.) – Infiaria Mar 22 at 16:38
• Well.... uh.... (I'm sorry.) – Lê Thành Đạt Mar 22 at 16:38

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 $$(5x+4)(x-1)^3=0$$

• Please wait. I typed the problem wrong. – Lê Thành Đạt Mar 22 at 16:33

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 $$5y^4 + 9y^3 - 11y^2 - 12y - 13=0.$$

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

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 $$y = 1$$.

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