Solving a Simultaneous Equation Problem

  • MHB
  • Thread starter anemone
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In summary, the conversation discusses a simultaneous equation problem and a method to solve it. The problem involves two equations and the method involves factoring and substitution. The possible solutions are $(1,1), (-1,-1), (2\sqrt{\frac{2}{5}},\sqrt{\frac{2}{5}}),$ and $(-2\sqrt{\frac{2}{5}},-\sqrt{\frac{2}{5}})$. The conversation also includes a suggestion from Sudharaka to use wise substitution for the first equation.
  • #1
anemone
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MHB
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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

$(x,\,y)=(1,\,1),\,(-1,\,-1),\,\left(2\sqrt{\dfrac{2}{5}},\,\sqrt{\dfrac{2}{5}}\right),\,\left(-2\sqrt{\dfrac{2}{5}},\,-\sqrt{\dfrac{2}{5}}\right)$.
 
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  • #2
anemone said:
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 $\left(\pm 2\sqrt{\frac{2}{5}},\pm\sqrt{\frac{2}{5}}\right)$.
 
  • #3
Sudharaka said:
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!
 
  • #4
anemone said:
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. :)
 

FAQ: Solving a Simultaneous Equation Problem

What is a simultaneous equation problem?

A simultaneous equation problem is a mathematical problem that involves solving two or more equations with multiple variables at the same time.

What is the most common method for solving simultaneous equations?

The most common method for solving simultaneous equations is the substitution method. This involves substituting one equation into the other to eliminate one variable and solve for the remaining variable.

What is the difference between consistent and inconsistent simultaneous equations?

A consistent simultaneous equation has at least one solution that satisfies all of the equations. An inconsistent simultaneous equation has no solutions that satisfy all of the equations.

How do I know if a simultaneous equation problem has no solution?

If you attempt to solve a simultaneous equation problem and you reach a statement that is always false, such as 3=5, then the problem has no solution.

Can a simultaneous equation problem have more than one solution?

Yes, a simultaneous equation problem can have more than one solution. This is known as an infinite solution, where the equations intersect at multiple points or are the same equation.

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