- #1
anemone
Gold Member
MHB
POTW Director
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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
$(x,\,y)=(1,\,1),\,(-1,\,-1),\,\left(2\sqrt{\dfrac{2}{5}},\,\sqrt{\dfrac{2}{5}}\right),\,\left(-2\sqrt{\dfrac{2}{5}},\,-\sqrt{\dfrac{2}{5}}\right)$.
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
$(x,\,y)=(1,\,1),\,(-1,\,-1),\,\left(2\sqrt{\dfrac{2}{5}},\,\sqrt{\dfrac{2}{5}}\right),\,\left(-2\sqrt{\dfrac{2}{5}},\,-\sqrt{\dfrac{2}{5}}\right)$.