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Simultaneous Equations

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anemone

MHB POTW Director
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Feb 14, 2012
3,678
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.:)
 

chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678
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!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $x$.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,678
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)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
\(\displaystyle \boxed{\tan(2\pi - 12\theta) = -\tan\theta \pm\sec\theta}.\) 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 $[-\pi,\pi]$.)
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 $x,\,y,\,z.$
 
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anemone

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Feb 14, 2012
3,678
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...


$$ $$\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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $\tan3k$.