Troubles in solving this system of equations

Question

Well, despite using programs this "appears" simple (the solutions are numerically easy at sight), I am not able to solve this system of two equations by hand, except for the trivial solution $(0, 0)$. Can you help me perhaps?

$$\begin{cases} 3y^3 + 6xy^2 + 3x^2y - 3y + 6x^3 - 7x = 0 \\ 2y^3 - 3xy^2 + 2x^2y - y - 3x^3 + 3x = 0 \end{cases}$$

These are the other solutions:

$$\left\{x\to -\frac{1}{\sqrt{15}},y\to \sqrt{\frac{3}{5}}\right\},\left\{x\to \frac{1}{\sqrt{15}},y\to -\sqrt{\frac{3}{5}}\right\},\left\{x\to -\frac{6}{\sqrt{35}},y\to -\frac{2}{\sqrt{35}}\right\},\left\{x\to \frac{6}{\sqrt{35}},y\to \frac{2}{\sqrt{35}}\right\}$$

I tried to use some methods, like adding twice the second from the first equation, but this didn't really help. Since this exercise comes from a past exam, I cannot solve it with calculators.

Thank you for your help.

Answer 1

For brevity let \begin{eqnarray} f&=&3y^3 + 6xy^2 + 3x^2y - 3y + 6x^3 - 7x,\\ g&=&2y^3 - 3xy^2 + 2x^2y -\,\ y - 3x^3 + 3x.\\ \end{eqnarray} Then we can cancel a few similar terms in two different ways as follows: \begin{eqnarray} 2f-3g&=&21x^3+21xy^2-3y-23x,\\ f+2g&=&7y^3+7x^2y-5y-x, \end{eqnarray} and now we can cancel the higher degree terms against each other as follows: \begin{eqnarray} y(2f-3g)-3x(f+2g)&=&(21x^3y+21xy^3-3y^2-23xy)-(21xy^3+21x^3y-15xy-3x^2)\\ &=&3x^2-8xy-3y^2\\ &=&(x-3y)(3x+y) \end{eqnarray} Then $f=g=0$ implies that either $x=3y$ or $x=-\tfrac y3$. If $x=3y$ then $f=0$ simplifies to \begin{eqnarray} 0&=&f\\ &=&3y^3+6(3y)y^2+3(3y)^2y-3y+6(3y)^3-7(3y)\\ &=&210y^3-24y\\ &=&6y(35y^2-4), \end{eqnarray} and so either $y=0$ or $y=\pm2\sqrt{\frac{1}{35}}$. If $x=-\tfrac y3$ then $y=-3x$ and then $f=0$ simplifies to \begin{eqnarray} 0&=&f\\ &=&3(-3x)^3+6x(-3x)^2+3x^2(-3x)-3(-3x)+6x^3-7x\\ &=&-30x^3+2x\\ &=&-2x(15x^2-1), \end{eqnarray} and so either $x=0$ or $x=\pm\sqrt{\tfrac{1}{15}}$.

Answer 2

Put $y=kx$ so you get $$3(k+2)(k^2+1)x^3=(3k+7)x\\(2k-3)(k^2+1)x^3=(k-3)x$$ from which $k=\dfrac13$ and $k=-3$.

It follows in the first equation (or the second one if you want) taking the first value of $k$, $$\left(\frac{3}{27}+\frac69+\frac33+1+6\right)x^3-(1+7)x=0$$ so you have $$70x^2=72\Rightarrow 35x^2=36\Rightarrow x=\pm\frac{6}{\sqrt{35}}$$ and the solutions $$(x,y)=\left(\pm\frac{6}{\sqrt{35}},\pm\frac{2}{\sqrt{35}}\right)$$ Similarly, with $k=-3$, you can get the other two solutions $$(x,y)=\left(\pm\frac{1}{\sqrt{15}},\mp\frac{1}{\sqrt{15}}\right)$$

Answer 3

Gerry has given a nice hint in the comments. First, note that apart from $(0,0)$, there is no solution where one of the variables is 0. Write the first equation as $$3x^2y+3y^3+6x^3+6xy^2 -(3y+6x) = x$$ $$\iff (x^2+y^2)(3y+6x) - (3y+6x) = x$$ $$\iff (x^2+y^2-1)(3y+6x)=x \tag{1}$$ Similarly for the second equation $$2x^2y+2y^3 - 3xy^2 - 3x^3 - (2y- 3x) = -y$$ $$\iff (x^2+y^2-1)(2y-3x) = -y$$ $$\iff(x^2+y^2-1)(3x-2y) = y\tag{2}$$ Set $x^2+y^2-1 = a$. Substituting (2) in (1): $$3ax-2ay=y \iff y = \frac{3ax}{2a+1}$$ $$\implies \frac{9a^2x}{2a+1} + 6ax = x$$Since $x \neq 0$: $$\frac{9a^2}{2a+1}+6a - 1 = 0$$ $$\iff 21a^2 +4a-1 = 0 \iff (7a-1)(3a+1) = 0$$

Now, we may be tempted to plug it in the linear equations, however, that only gives the identity (why?), so we must plug this in $x^2+y^2-1 = a$, along with $y = \frac{3ax}{2a+1}$.

For $a = -1/3$: $$y= -3x$$ $$10x^2 - 1 = \frac{-1}{3} \iff 15x^2 = 1$$which gives $$\boxed{(x,y) = \left(\frac{1}{\sqrt{15}}, -\sqrt{\frac{3}{5}}\right) \text{ or } \left(-\frac{1}{\sqrt{15}}, \sqrt{\frac{3}{5}}\right)}$$ For $a = 1/7$: $$y = x/3 \iff x = 3y$$ $$10y^2 - 1 = \frac17 \iff y^2 = \frac{4}{35}$$which gives $$\boxed{(x,y) = \left(\frac{6}{\sqrt{35}}, \frac{2}{\sqrt{35}}\right) \text{ or } \left(-\frac{6}{\sqrt{35}}, -\frac{2}{\sqrt{35}}\right)}$$