# system of equations - (difficult problem)

• May 4th 2008, 04:19 AM
tombrownington
system of equations - (difficult problem)
How does one solve the system of equations:

x = 3y - 4y^3
y = 3x - 4x^3

The problem statement is pretty simple but by using conventional method of replacing variables I get a 9th degree equation in y or x - depending which variable you replace - (impossible to solve normally).
The best I can do is bring that 9th degree equation down to a 6th degree equation (which is still impossible to solve).
Anybody have any pointers for me to proceed in the right direction?
Thanks a lot.
• May 4th 2008, 04:45 AM
flyingsquirrel
Hi

An idea :

$\left\{
\begin{array}{lc}
x=3y-4y^3&(1)\\
y=3x-4x^3&(2)\\
\end{array}\right. \implies \left\{
\begin{array}{lc}
x+y=3(x+y)-4(y^3+x^3)&(1)+(2)\\
x-y=3(x-y)-4(y^3-x^3)&(1)-(2)\\
\end{array}\right.$

Now, you may use that $x^3+y^3=(x+y)(\ldots)$ and that $x^3-y^3=(x-y)(\ldots)$. I let you find what the dots have to be replaced by. :D

Hope that helps
• May 4th 2008, 05:37 AM
Isomorphism
Quote:

Originally Posted by tombrownington
How does one solve the system of equations:

x = 3y - 4y^3
y = 3x - 4x^3

The problem statement is pretty simple but by using conventional method of replacing variables I get a 9th degree equation in y or x - depending which variable you replace - (impossible to solve normally).
The best I can do is bring that 9th degree equation down to a 6th degree equation (which is still impossible to solve).
Anybody have any pointers for me to proceed in the right direction?
Thanks a lot.

You can also observe that substituting $y = \sin \theta$ yields the equation $\sin \theta = \sin 9\theta$. You can then solve the trig equation.

But this may not be the complete solution set :(
For that we have to prove that $|x|,|y| \leq 1$ first...
• May 4th 2008, 11:44 PM
tombrownington
Quote:

Originally Posted by flyingsquirrel
Hi

An idea :

$\left\{
\begin{array}{lc}
x=3y-4y^3&(1)\\
y=3x-4x^3&(2)\\
\end{array}\right. \implies \left\{
\begin{array}{lc}
x+y=3(x+y)-4(y^3+x^3)&(1)+(2)\\
x-y=3(x-y)-4(y^3-x^3)&(1)-(2)\\
\end{array}\right.$

Now, you may use that $x^3+y^3=(x+y)(\ldots)$ and that $x^3-y^3=(x-y)(\ldots)$. I let you find what the dots have to be replaced by. :D

Hope that helps

Hi flying squirrel,
What you say is definitely helpful. Thanks for the insight.
By putting x = y in one case, and x = -y in another I get 5 solutions for the system of equations. But what do I do with the (x^2-xy+y^2) part of the expression in one case, and the similar expression in the other case (the one you asked me to fill up)? I can't seem to get solutions from these. Any pointers in the right direction?
Thanks!
• May 5th 2008, 12:53 AM
flyingsquirrel
Quote:

Originally Posted by tombrownington
I can't seem to get solutions from these. Any pointers in the right direction?
Thanks!

$\left\{
\begin{array}{lc}
x+y=3(x+y)-4(y^3+x^3) & (1)+(2)\\
x-y=3(x-y)-4(y^3-x^3) & (1)-(2)
\end{array}\right.
$

Simplifying and factoring gives :

$\left\{
\begin{array}{l}
-2(x+y)=-4(y^3+x^3)=-4(x+y)(y^2-xy+x^2) \\
-2(x-y)=-4(y^3-x^3)=-4(y-x)(y^2+xy+x^2)
\end{array}\right.
$

If $(x+y)(x-y)\neq 0$,

$
\left\{
\begin{array}{lc}
1=2(y^2-xy+x^2) & (3)\\
-1=2(y^2+xy+x^2) & (4)\\
\end{array}\right.$

This is where you stopped. You can take a look at $(3)+(4)$, it'll give the solutions. (we're looking for solutions in $\mathbb{R}$ aren't we ?)