# Stuck on system of equations

What are the solutions of this system of equations, where $$x,y \in \mathbb{R}$$?

$$\begin{cases} \frac{1}{x} + \frac{1}{2y} = (x^2+3y^2)(3x^2+y^2)\\ \frac{1}{x} - \frac{1}{2y} = 2(y^4-x^4) \end{cases}$$

First I rewrote the equations as

$$\begin{cases} \frac{1}{x} + \frac{1}{2y} = 3x^4+3y^4+10x^2y^2\\ \frac{1}{x} - \frac{1}{2y} = 2(y-x)(y+x)(y^2+x^2) \end{cases}$$

then

$$\begin{cases} \frac{1}{x} + \frac{1}{2y} = 3(x^2+y^2)^2+4x^2y^2\\ \frac{1}{x} - \frac{1}{2y} = 2(y-x)(y+x)(y^2+x^2) \end{cases}$$

I tried to add the two equations and I got:

$$\frac{2}{x}=2(y-x)(y+x)(y^2+x^2)+3(x^2+y^2)^2+4x^2y^2$$

we can rewrite this as

$$\frac{2}{x}=(y^2+x^2)[2(y-x)(y+x)+3(x^2+y^2)]+4x^2y^2$$,

$$\frac{2}{x}=(y^2+x^2)[2y^2-2x^2+3x^2+3y^2]+4x^2y^2$$

$$\frac{2}{x}=(y^2+x^2)(5y^2-x^2)+4x^2y^2$$

Im stuck from here, how should I proceed? Or is there any trick to solving this system?

• z could be anything Dec 4, 2023 at 19:36

## 3 Answers

Given,

\begin{align} \frac{1}{x} + \frac{1}{2y} &- (x^2+3y^2)(3x^2+y^2)=0\\ \frac{1}{x} - \frac{1}{2y} &- 2(y^4-x^4)=0 \end{align}

Do the substitution,

\begin{align} x &= \frac{p+1}2\\ y &= \frac{p-1}2 \end{align}

and your two equations simplify as,

\begin{align} \frac{p\,(p^5-3)}{(p^2-1)}=0\\ \frac{(p^5-3)}{(p^2-1)}=0 \end{align}

So your real solution is,

\begin{align} x &= \frac{3^{1/5}+1}2 = 1.1228654698\dots\\ y &= \frac{3^{1/5}-1}2 = 0.1228654698\dots \end{align}

• What's the motivation for this particular substitution ? It restricts $x,y$ to a specific line $y = x -1$. So shouldn't you also check that there are no other solutions outside of this line. Dec 4, 2023 at 12:24
• @Digitallis For quick results with non-linear systems, I normally use resultants. I ended up with a quintic in $x$ which wasn't obviously solvable in radicals. But checking its Galois group, it turned out to be solvable. Depressing the quintic, it simplified to binomial form $z^5-3=0$, so I then knew what substitutions to use. Dec 4, 2023 at 12:29
• I'm afraid the time when I understood Galois theory is long gone :'( Dec 4, 2023 at 12:35
• Finding the resultant, checking its Galois group and depressing the quintic seem like they should be part of the answer - that substitution will not appear on line two to most people. Dec 4, 2023 at 20:00

Numerically, by plotting the contour lines for both equations, we can guess that there is a single real solution. Observing the first plot, there is an apparent solution at $$(0,0)$$ that we must rule out, and an actual solution in the first quadrant. The second plot provides a zoom in. The approximate value of the solution is $$(x,y)=(1.1228654698077587,0.12286546980788067)$$.

• What software did you use? Your $y$ value is off by the last few digits, starting at the red numbers. \begin{align} x&\approx1.12286546980775=\frac{3^{1/5}+1}2\\ y&\approx0.12286546980775=\frac{3^{1/5}-1}2\\y&\approx0.122865469807\color{red}{88}\dots \end{align} Dec 4, 2023 at 12:23
• @TitoPiezasIII it was Wolfram Mathematica. Dec 4, 2023 at 13:01
• I guess Wolfram Mathematica may assume certain levels of accuracy. But it is strange it was more accurate with $x$ but not with $y$. Dec 4, 2023 at 13:07

Using polar coordinates

$$x = r \cos \phi$$, $$y = r \sin \phi$$

Let $$c = \cos \phi , s = \sin \phi$$ , then the two equations become

$$\dfrac{1}{c} + \dfrac{1}{2 s} = r^5 ( 3 c^4 + 3 s^4 + 10 c^2 s^2 )$$

$$\dfrac{1}{c} - \dfrac{1}{2 s} = r^5 ( 2 s^4 - 2 c^4 )$$

Dividing the two equations eliminates $$r$$:

$$\dfrac{ 2 s + c } {2 s - c } = \dfrac{ 3 c^4 + 3 s^4 + 10 c^2 s^2 }{ 2 s^4 - 2 c^4 }$$

Cross multiplication leads to

$$(2 s + c) (2 s^4 - 2 c^4) = (2 s - c) ( 3 c^4 +3 s^4 + 10 c^2 s^2 )$$

Divide by $$c^5$$ then

$$(2 t + 1 ) (2 t^4 - 2 ) = (2 t - 1) (3 + 3 t^4 + 10 t^2 )$$

where $$t = \tan \phi$$

Expanding and re-arranging gives

$$2 t^5 - 5 t^4 + 20 t^3 - 10 t^2 + 10 t - 1 = 0$$

This quintic polynomial has only one real root

$$t = 0.109421363$$

The corresponding $$\phi$$ is

$$\phi = 0.108987771$$

Now $$r$$ can be computed from the above equations, and then

$$x = r \cos \phi = 1.12286547$$

$$y = r \sin \phi = 0.12286547$$

• Your quintic is solvable in radicals and is given by, $$t=\frac{1+3^{1/5}-3^{2/5}+3^{3/5}-3^{4/5}}2 = 0.109421\dots$$ Dec 6, 2023 at 16:46