MHB Solving a Simultaneous Equation Problem

[anemone]

Hi MHB,

I've recently come across the following simultaneous equation problem:

$5x^2y-4xy^2+3y^3-2(x+y)=0$ and $xy( x^2+y^2)+2=(x+y)^2$.

I solved it, it isn't a hard problem, but I would think my method is way too tedious, I was wondering if there exists a more neater approach for this particular problem. If you have another method to solve it, could you please share your solution with me?

Many thanks in advance.

My solution:

Observe that we can actually factor the second equation and get:

$xy(x^2+y^2)+2=x^2+y^2+2xy$

$(x^2+y^2)(xy-1)+2(1-xy)=0$

$(x^2+y^2-2)(xy-1)=0$

So, we have either $x^2+y^2=2$ and/or $xy=1$.

We also note that if $(x,\,y)$ is a solution, then so is $(-x,\,-y)$.

It's not hard to see that $x=y=1$ and $x=y=-1$ are two possible solutions when $x^2+y^2=2$ and $xy=1$ hold.

Now, we let $x^2+y^2=2$ and $xy\ne 1$.

In this case, we have:

$x^2+y^2=2$

$\left(\dfrac{x}{\sqrt{2}}\right)^2+\left(\dfrac{y}{\sqrt{2}}\right)^2=1$

And let $x=\sqrt{2}\sin \theta$ and $y=\sqrt{2}\cos \theta$

Replacing these two into the first given equation gives:

$4\cos \theta-2\cos^3 \theta=\sin \theta+4\sin \theta \cos^2 \theta$ which is equivalent to

$\tan^3 \theta-4\tan^2 \theta+5\tan \theta-2=0$

Solving for $\tan \theta$ we see that

$\tan \theta=1\implies \sin \theta=\dfrac{1}{\sqrt{2}},\,\cos \theta=\dfrac{1}{\sqrt{2}}$, i.e. $x=\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\right)=1=y$

$\tan \theta=2\implies \sin \theta=\dfrac{2}{\sqrt{5}},\,\cos \theta=\dfrac{1}{\sqrt{5}}$, i.e. $x=\sqrt{2}\left(\dfrac{2}{\sqrt{5}}\right)=2\sqrt{\dfrac{2}{5}}$, $y=\sqrt{2}\left(\dfrac{1}{\sqrt{5}}\right)=\sqrt{\dfrac{2}{5}}$

Therefore the solutions to the given system are

$(x,\,y)=(1,\,1),\,(-1,\,-1),\,\left(2\sqrt{\dfrac{2}{5}},\,\sqrt{\dfrac{2}{5}}\right),\,\left(-2\sqrt{\dfrac{2}{5}},\,-\sqrt{\dfrac{2}{5}}\right)$.

 

[Sudharaka]

Hi MHB,

I've recently come across the following simultaneous equation problem:

$5x^2y-4xy^2+3y^3-2(x+y)=0$ and $xy( x^2+y^2)+2=(x+y)^2$.

I solved it, it isn't a hard problem, but I would think my method is way too tedious, I was wondering if there exists a more neater approach for this particular problem. If you have another method to solve it, could you please share your solution with me?

Many thanks in advance.

My solution:

Observe that we can actually factor the second equation and get:

$xy(x^2+y^2)+2=x^2+y^2+2xy$

$(x^2+y^2)(xy-1)+2(1-xy)=0$

$(x^2+y^2-2)(xy-1)=0$

So, we have either $x^2+y^2=2$ and/or $xy=1$.

We also note that if $(x,\,y)$ is a solution, then so is $(-x,\,-y)$.

It's not hard to see that $x=y=1$ and $x=y=-1$ are two possible solutions when $x^2+y^2=2$ and $xy=1$ hold.

Now, we let $x^2+y^2=2$ and $xy\ne 1$.

In this case, we have:

$x^2+y^2=2$

$\left(\dfrac{x}{\sqrt{2}}\right)^2+\left(\dfrac{y}{\sqrt{2}}\right)^2=1$

And let $x=\sqrt{2}\sin \theta$ and $y=\sqrt{2}\cos \theta$

Replacing these two into the first given equation gives:

$4\cos \theta-2\cos^3 \theta=\sin \theta+4\sin \theta \cos^2 \theta$ which is equivalent to

$\tan^3 \theta-4\tan^2 \theta+5\tan \theta-2=0$

Solving for $\tan \theta$ we see that

$\tan \theta=1\implies \sin \theta=\dfrac{1}{\sqrt{2}},\,\cos \theta=\dfrac{1}{\sqrt{2}}$, i.e. $x=\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\right)=1=y$

$\tan \theta=2\implies \sin \theta=\dfrac{2}{\sqrt{5}},\,\cos \theta=\dfrac{1}{\sqrt{5}}$, i.e. $x=\sqrt{2}\left(\dfrac{2}{\sqrt{5}}\right)=2\sqrt{\dfrac{2}{5}}$, $y=\sqrt{2}\left(\dfrac{1}{\sqrt{5}}\right)=\sqrt{\dfrac{2}{5}}$

Therefore the solutions to the given system are

$(x,\,y)=(1,\,1),\,(-1,\,-1),\,\left(2\sqrt{\dfrac{2}{5}},\,\sqrt{\dfrac{2}{5}}\right),\,\left(-2\sqrt{\dfrac{2}{5}},\,-\sqrt{\dfrac{2}{5}}\right)$.

Hi anemone, :)

I think it's not necessary to use trigonometric substitutions. After obtaining $x^2+y^2=2$ and/or $xy=1$ from the second equation we can use the first equation with,

1) $xy=1$ which will give us $(1,1)$ or $(-1,-1)$.

2) $x^2+y^2=2$ which will give us $xy=1$ or $x=2y$. Solving $x^2+y^2=2$ and $xy=1$ gives us the same two answers obtained above and solving $x^2+y^2=2$ and $x=2y$ will give us $\left(\pm 2\sqrt{\frac{2}{5}},\pm\sqrt{\frac{2}{5}}\right)$.

 

[anemone]

Hi anemone, :)

I think it's not necessary to use trigonometric substitutions. After obtaining $x^2+y^2=2$ and/or $xy=1$ from the second equation we can use the first equation with,

1) $xy=1$ which will give us $(1,1)$ or $(-1,-1)$.

2) $x^2+y^2=2$ which will give us $xy=1$ or $x=2y$. Solving $x^2+y^2=2$ and $xy=1$ gives us the same two answers obtained above and solving $x^2+y^2=2$ and $x=2y$ will give us $\left(\pm 2\sqrt{\frac{2}{5}},\pm\sqrt{\frac{2}{5}}\right)$.

Hi Sudharaka!:)

Thank you so much for your reply, I see it now, if we substitute $x^2+y^2=2$ into the first equation wisely, we will get:

$5x^2y+3y^3=2(x+y)+4xy^2$

$5x^2y+3y^3=2x+4xy^2+2y$

$5x^2y+3y^3-2y=2(x+2xy^2)$

$3y(x^2y+y^2)+2x^2y-2y=2(x+2xy^2)$

$3y(2)+2x^2y-2y=2(x+2xy^2)$

$2x^2y+4y=2(x+2xy^2)$

$x^2y+2y=x+2xy^2$

$(xy-1)(x-2y)=0$

And the rest is pretty self-explanatory...thank you Sud! I appreciate the help!

 

[Sudharaka]

Hi Sudharaka!:)

Thank you so much for your reply, I see it now, if we substitute $x^2+y^2=2$ into the first equation wisely, we will get:

$5x^2y+3y^3=2(x+y)+4xy^2$

$5x^2y+3y^3=2x+4xy^2+2y$

$5x^2y+3y^3-2y=2(x+2xy^2)$

$3y(x^2y+y^2)+2x^2y-2y=2(x+2xy^2)$

$3y(2)+2x^2y-2y=2(x+2xy^2)$

$2x^2y+4y=2(x+2xy^2)$

$x^2y+2y=x+2xy^2$

$(xy-1)(x-2y)=0$

And the rest is pretty self-explanatory...thank you Sud! I appreciate the help!

You are welcome. Yep that would do. What I did was,

\[5x^2y-4xy^2+3y^3-2(x+y)=0\]

\[y(5x^2-4xy+3y^2)-2x-2y=0\]

\[y(2x^2-4xy+6)-2x-2y=0\]

\[2x^2 y-4xy^2+4y-2x=0\]

And you group the similar terms together. :)