Stuck on a system of equations

Question

I got the following system of equations but I get stuck after a couple of steps.

$$1+x+2zy = 0$$ $$1+y+2zx = 0$$ $$y^2+x^2-1 = 0$$

From the third equation, I get:

$$x = \pm \sqrt{1-y^2}$$ and $$y = \pm \sqrt{1-x^2}$$

From the first two equations I get the following expression:

$$x =\frac{-1-y}{2z}, y =\frac{-1-x}{2z} $$

But then what? I just don't see how I can get all possible expressions for $z, y$ and $x$ where all equations hold.

I'd appreciate an hint on how to move on with this.

Paul · Accepted Answer · 2021-11-23T14:59:59.797

4

From the first and second equations, you can get $$\frac{y}{x}=\frac{1+x}{1+y}$$ or $$y^2+y=x^2+x$$ Adding 1/4 to both sides allows you to complete the square, yielding $$y+1/2 = \pm(x+1/2)$$ so $$y=x,-x-1$$ You can substitute these into the third equation (Note: there might be extraneous solutions.)

edited Nov 23 '21 at 14:59

answered Nov 23 '21 at 13:45

Paul

8,153

I understand how you get there. Didn't see that. But i'm still confused how to get to the solutions that @DietrichBurde gave. Is the third equation sufficient to find it? – Steven01123581321 Nov 23 '21 at 14:20
1

I simplified the expressions where it will be easy to substitute into the equation. – Paul Nov 23 '21 at 15:00
1

Using proportionality rules your first equality leads to: $\dfrac yx=\dfrac{1+x}{1+y}=\dfrac{y+1+x}{x+1+y}=1$ therefore $\dfrac yx=1\iff x=y$ – DINEDINE Nov 23 '21 at 16:24
@DINEDINE only if x and y are both nonzero. – Paul Nov 23 '21 at 20:55
Obviously. Mean it but di not write it. – DINEDINE Nov 23 '21 at 20:56

pieinthesky · Answer 2 · 2021-11-23T14:19:55.403

3

The last equation defines a circle of radius 1 (or a cylinder, in $xyz$ coordinates.

Substitute: $x=r\cos t$,$y=r\sin t$ to obtain: $$\begin{cases} 1+r\cos t+2zr\sin t=0\\ 1+r\sin t +2zr\cos t=0\\ r^2=1 \end{cases}$$ Subtract the 1st equation from the 2nd, $r$ is eliminated: $$\begin{cases} 2z(\cos t-\sin t)=\cos t-\sin t\\ r^2=1 \end{cases}\Rightarrow z=\frac{1}{2}$$ Now go back to the original system of equations. You'll get a circle and a line that will intersect.

Edit: the other solutions may be obtained from the last systems of equations as well, if you notice that both equations are correct when $\sin t=\cos t$. It only occurs when $\sin t=\cos t=\pm\frac{1}{\sqrt{2}}$, and from here you can also find $z$.

edited Nov 23 '21 at 14:19

answered Nov 23 '21 at 13:55

pieinthesky

115

I understand, there may be more solutions, but I'm not sure if there's a quick way to get to them from here. – pieinthesky Nov 23 '21 at 14:05
Updated the answer with a hint concerning the other solutions. – pieinthesky Nov 23 '21 at 14:20
Nice approach. Upvoated but accepted Paul's answer for the higher simplicity of the answer. Good job though! – Steven01123581321 Nov 23 '21 at 15:05

score 1 · Answer 3 · answered Nov 23 '21 at 13:58

Following Paul's approach, all solutions over $\Bbb C$ are in fact real, and given by $$ (x,y,z)=\left(-1,0,\frac{1}{2}\right), \left(0,-1,\frac{1}{2}\right), \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, \frac{- \sqrt{2} -2}{2\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, \frac{2 - \sqrt{2}}{2\sqrt{2}}\right) $$

Stuck on a system of equations

3 Answers3