Unsolvable Simultaneous Equations: A Mathematician's Dilemma

In summary, the conversation revolved around a difficult system of equations involving variables z, x, and y. The individual seeking help had attempted to eliminate the variable z and obtained another equation in terms of x and y, but realized it was not leading them in the right direction. Another individual suggested using trigonometry to solve the equations, specifically using the formulas for tan(3θ) and tan(4θ). After some calculations, it was found that the equations could be simplified to sin(13k)=cos(12k) where k represents an angle. While the equation was difficult to solve, Wolfram Alpha provided some possible solutions for the first few angles.
  • #1
anemone
Gold Member
MHB
POTW Director
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Hi MHB,

For the first time I found a system of equations where I'm at my wit's end and don't know how to solve it, no matter how hard I tried...

Problem:

Solve

\(\displaystyle z^2+2xyz=1\)

\(\displaystyle 3x^2y^2+3xy^2=1+x^3y^4\)

\(\displaystyle z+zy^4+4y^3=4y+6y^2z\)

Attempt:

I tried to eliminate the variable $z$ and obtained another equation in terms of $x$ and $y$ but I think you'll agree with me that I'm headed in the wrong direction after you saw the equation I found...

\(\displaystyle \left(\frac{4y(1-y^2)}{y^4-6y^2+1} \right)^2+2xy\left(\frac{4y(1-y^2)}{y^4-6y^2+1} \right)=1\)I'd appreciate any hints anyone could give me on this problem.

Thanks in advance.:)
 
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  • #2
anemone said:
Hi MHB,

For the first time I found a system of equations where I'm at my wit's end and don't know how to solve it, no matter how hard I tried...

Problem:

Solve

\(\displaystyle z^2+2xyz=1\)

\(\displaystyle 3x^2y^2+3xy^2=1+x^3y^4\)

\(\displaystyle z+zy^4+4y^3=4y+6y^2z\)

Attempt:

I tried to eliminate the variable $z$ and obtained another equation in terms of $x$ and $y$ but I think you'll agree with me that I'm headed in the wrong direction after you saw the equation I found...

\(\displaystyle \left(\frac{4y(1-y^2)}{y^4-6y^2+1} \right)^2+2xy\left(\frac{4y(1-y^2)}{y^4-6y^2+1} \right)=1\)I'd appreciate any hints anyone could give me on this problem.

Thanks in advance.:)

Alternatively You can eliminate z using the first equation obtaining...

$\displaystyle z = - x y \pm \sqrt{1+ x^{2} y^{2}}\ (1)$

... insert it in the third equation and then try to solve in x and y...Kind regards $\chi$ $\sigma$
 
  • #3
chisigma said:
Alternatively You can eliminate z using the first equation obtaining...

$\displaystyle z = - x y \pm \sqrt{1+ x^{2} y^{2}}\ (1)$

... insert it in the third equation and then try to solve in x and y...Kind regards $\chi$ $\sigma$

Thank you for your reply, chisigma...

I should have mentioned earlier that I used the exact same method to arrive to the equation that I showed in my first post...:eek:.

I'm sorry for not mentioning more clearly how did I end up with that new equation in terms of $x$ and $y$. Sorry!
 
  • #4
anemone said:
Solve

\(\displaystyle z^2+2xyz=1\)

\(\displaystyle 3x^2y^2+3xy^2=1+x^3y^4\)

\(\displaystyle z+zy^4+4y^3=4y+6y^2z\)
I wonder where these equations came from? To me, they smell of trigonometry, specifically the formulae \(\displaystyle \tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}\) and \(\displaystyle \tan(4\theta) = \frac{4\tan\theta(1-\tan^2\theta)}{1 - 6\tan^2\theta + \tan^4\theta}.\)

Let $u = xy$. Then the equations become

\(\displaystyle z^2+2uz=1\),

\(\displaystyle 3u^2+3uy=1+u^3y\),

\(\displaystyle z+zy^4+4y^3=4y+6y^2z\).

Now let $u = \tan\theta$. The second equation then says \(\displaystyle y = \frac{1-3u^2}{3u-u^3} = \cot(3\theta) = \tan\bigl(\tfrac\pi2 - 3\theta\bigr)\). The third equation says that \(\displaystyle z = \frac{4y(1-y^2)}{1-6y^2 + y^4} = \tan(4\arctan y) = \tan(2\pi - 12\theta).\)

The first equation is a quadratic in $z$, with solutions $z = -u \pm\sqrt{1+u^2} = -\tan\theta \pm\sec\theta.$ Comparing the two expressions for $z$, you get \(\displaystyle \boxed{\tan(2\pi - 12\theta) = -\tan\theta \pm\sec\theta}.\) That at least is an equation in just one unknown, though not one that I would want to have to solve. (A graph shows 23 solutions in the interval $[-\pi,\pi]$.) Once you know $\theta$, you can of course calculate $u,\,y,\,z$ and $x$.
 
  • #5
Opalg said:
I wonder where these equations came from? To me, they smell of trigonometry, specifically the formulae \(\displaystyle \tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}\) and \(\displaystyle \tan(4\theta) = \frac{4\tan\theta(1-\tan^2\theta)}{1 - 6\tan^2\theta + \tan^4\theta}.\)

Let $u = xy$. Then the equations become

\(\displaystyle z^2+2uz=1\),

\(\displaystyle 3u^2+3uy=1+u^3y\),

\(\displaystyle z+zy^4+4y^3=4y+6y^2z\).

Now let $u = \tan\theta$. The second equation then says \(\displaystyle y = \frac{1-3u^2}{3u-u^3} = \cot(3\theta) = \tan\bigl(\tfrac\pi2 - 3\theta\bigr)\). The third equation says that \(\displaystyle z = \frac{4y(1-y^2)}{1-6y^2 + y^4} = \tan(4\arctan y) = \tan(2\pi - 12\theta).\)

The first equation is a quadratic in $z$, with solutions $z = -u \pm\sqrt{1+u^2} = -\tan\theta \pm\sec\theta.$ Comparing the two expressions for $z$, you get \(\displaystyle \boxed{\tan(2\pi - 12\theta) = -\tan\theta \pm\sec\theta}.\) That at least is an equation in just one unknown, though not one that I would want to have to solve. (A graph shows 23 solutions in the interval $[-\pi,\pi]$.) Once you know $\theta$, you can of course calculate $u,\,y,\,z$ and $x$.

Thank you Opalg for the great reply!

I vaguely remember from where I gotten this problem but I knew it came from AoPS Forum and I did a search on my browsing history and I found it at last! I couldn't believe I didn't see the hint(trigonometric might be of help) given by the OP when I first saw the topic...shame on me!AoPS Forum - Use trigonomatric [2] ? Art of Problem Solving

And thank you for everything, your explanations, the trigonometric formulas for both \(\displaystyle \tan 3 \theta\) and \(\displaystyle \tan 4 \theta\), and everything that you said in your post...I truly appreciate it!

I have to admit it took me some time to digest your post and worked it out on my own...I see that if I let

\(\displaystyle xy=\tan k\), \(\displaystyle y=\tan 3k\), \(\displaystyle z=\tan 12k\), \(\displaystyle x=\frac{1-3\tan^2 k}{3-\tan^2 k}\)

I then managed to narrow them down to

\(\displaystyle \sin 13k=\cos 12k\)

and this equation is actually equivalent to yours(the last equation in your post) and I don't think I could solve it too. BUT, wolfram alpha does suggest some neat answers (for the first few angles for $k$).(solve sin13x=cos12x - Wolfram|Alpha)
 
  • #6
Opalg said:
\(\displaystyle \boxed{\tan(2\pi - 12\theta) = -\tan\theta \pm\sec\theta}.\) That at least is an equation in just one unknown, though not one that I would want to have to solve. (A graph shows 23 solutions in the interval $[-\pi,\pi]$.)
anemone said:
I see that if I let

\(\displaystyle xy=\tan k\), \(\displaystyle y=\tan 3k\), \(\displaystyle z=\tan 12k\), \(\displaystyle x=\frac{1-3\tan^2 k}{3-\tan^2 k}\)

I then managed to narrow them down to

\(\displaystyle \sin 13k=\cos 12k\) I seem to get 11k rather than 13k there?
When I derived that boxed equation, I thought it was too hard to solve. But your reply gives me courage to continue (on the assumption that the boxed equation is correct).

Since \(\displaystyle \tan(2\pi - 12\theta) = -\tan(12\theta)\), we can write the equation as \(\displaystyle \tan(12\theta) = \tan\theta \pm\sec\theta\), so that $$\frac{\sin(12\theta)}{\cos(12\theta)} - \frac{\sin\theta\pm1}{\cos\theta} = 0,$$$$\sin(12\theta)\cos\theta - \cos(12\theta)\sin\theta \pm\cos(12\theta) = 0,$$ $$\sin(11\theta) = \pm\cos(12\theta),$$ $$\cos(12\theta) = \pm\cos(\tfrac\pi2 - 11\theta),$$ $$12\theta = \pm11\theta \pm\tfrac\pi2 + 2k\pi.$$

If we take the plus sign, for $+11\theta$, then we get $\theta = \pm\frac\pi2+2k\pi$, and then $u = \tan\theta$ is not defined. So we need to take the minus sign, getting $23\theta = \pm\frac\pi2 + 2k\pi$, which is equivalent to $\theta = \dfrac{(2k+1)\pi}{46}\ (0\leqslant k\leqslant 22)$. That gives 23 values for $\tan\theta$, from which you can deduce the 23 solutions to the original equations for $x,\,y,\,z.$
 
  • #7
By substituting \(\displaystyle xy=\tan k\), \(\displaystyle z=\tan 12k\), \(\displaystyle x=\frac{1-3\tan^2 k}{3-\tan^2 k}\) and \(\displaystyle y=\tan 3k\) into the equation $z^2+2xyz=1$, we see that

\(\displaystyle z=\frac{-2xy\pm \sqrt{(2xy)^2-4(-1)}}{2}\)

\(\displaystyle z=\frac{-2xy\pm \sqrt{4x^2y^2+4}}{2}\)

\(\displaystyle z=\frac{-2xy\pm 2\sqrt{x^2y^2+1}}{2}\)

\(\displaystyle z=-xy\pm \sqrt{x^2y^2+1}\)

\(\displaystyle \tan 12k=-\tan k\pm \sqrt{\tan ^2k+1}\)

\(\displaystyle \tan 12k=-\tan k\pm \sqrt{\sec ^2k}\)

\(\displaystyle \frac{\sin 12k}{\cos 12k}=-\frac{\sin k}{\cos k}\pm \sec k\)

\(\displaystyle \frac{\sin 12k}{\cos 12k}=-\frac{\sin k}{\cos k}\pm \frac{1}{\cos k}\)

\(\displaystyle \frac{\sin 12k}{\cos 12k}=-\frac{\sin k\mp 1}{\cos k}\)

\(\displaystyle \sin 12k \cos k=-\sin k\cos 12k \mp \cos 12k\)

\(\displaystyle \sin 12k \cos k+\sin k\cos 12k= \mp \cos 12k\)

\(\displaystyle \sin 12k \cos k+\cos 12k \sin k= \mp \cos 12k\)

\(\displaystyle \sin (12k+k)=\mp \cos 12k\)

and hence

\(\displaystyle \sin (13k)=\mp \cos 12k\)...:eek:. I am not saying I am right though...
Opalg said:
$$ $$\sin(11\theta) = \pm\cos(12\theta),$$ $$\cos(12\theta) = \pm\cos(\tfrac\pi2 - 11\theta),$$ $$12\theta = \pm11\theta \pm\tfrac\pi2 + 2k\pi.$$

I'm stamping my foot now seeing how you magically turned the sine function into the cosine function and am disappointed in myself because really I should have thought of that.
 
  • #8
anemone said:
By substituting \(\displaystyle xy=\tan k\), \(\displaystyle z=\tan 12k\), \(\displaystyle x=\frac{1-3\tan^2 k}{3-\tan^2 k}\) and \(\displaystyle y=\tan 3k\) ...
I don't have time to double-check it right now, but I think it was $y=\cot3k$, not $\tan3k$.
 

FAQ: Unsolvable Simultaneous Equations: A Mathematician's Dilemma

What are simultaneous equations?

Simultaneous equations are a set of equations that are solved at the same time, where the solution values satisfy all of the equations.

How many variables can simultaneous equations have?

Simultaneous equations can have any number of variables, but typically there are two or three variables involved.

What is the most common method for solving simultaneous equations?

The most common method for solving simultaneous equations is the elimination method, where one variable is eliminated by adding or subtracting the equations.

Can simultaneous equations have more than one solution?

Yes, simultaneous equations can have more than one solution if there are multiple sets of values that satisfy all of the equations.

Are simultaneous equations used in real-life applications?

Yes, simultaneous equations are used in many real-life applications, such as in economics, physics, and engineering, to solve for multiple unknown variables at the same time.

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