Solve the system 5y + 2 + 2zx = 0, 5x + 5 + 2zy = 0, x^2 + y^2 = 1.

Step-by-step solution

1. 1

   Recognise the Lagrange structure. These are exactly the stationarity conditions for optimising $f(x,y)=5xy+2x+5y$ on the unit circle: $\nabla f+z\nabla g=0$ with $g=x^{2}+y^{2}-1$, and $z$ is the multiplier. Knowing this predicts four solutions (two maxima/minima and two saddle-type points) and tells you every answer must lie on the circle.

2. 2

   Parametrise the constraint. Setting $x=\cos t$, $y=\sin t$ satisfies the third equation automatically and turns the first two into

   $5\sin t+2+2z\cos t=0,\qquad 5\cos t+5+2z\sin t=0$

3. 3

   Eliminate the multiplier $z$. Multiply the first by $\sin t$, the second by $\cos t$, and subtract; the $2z\sin t\cos t$ terms cancel:

   $5\sin^{2}t+2\sin t-5\cos^{2}t-5\cos t=0$

   This single equation in $t$ has exactly four roots in $[0,2\pi)$, which is where the count of four solutions comes from.

4. 4

   Read off the two rational solutions. Two roots give recognisable points. At $(x,y)=\left(\tfrac35,\tfrac45\right)$ the first equation gives $4+2+2z\cdot\tfrac35=0$, so $z=-5$; the second gives $3+5+2(-5)\tfrac45=8-8=0$ ✓. At $(x,y)=(-1,0)$: the first gives $0+2+2z(-1)=0$ so $z=1$, and the second gives $-5+5+0=0$ ✓.

5. 5

   Find the remaining two, which involve $\sqrt{11}$. The other two roots satisfy $20x^{2}+12x-7=0$ and $20y^{2}+24y+5=0$, giving

   $\left(x,y,z\right)=\left(\frac{-3+2\sqrt{11}}{10},\ \frac{-6-\sqrt{11}}{10},\ \frac{4+\sqrt{11}}{2}\right),\quad\left(\frac{-3-2\sqrt{11}}{10},\ \frac{-6+\sqrt{11}}{10},\ \frac{4-\sqrt{11}}{2}\right)$

6. 6

   Verify every candidate against all three equations — especially the constraint. For the $\sqrt{11}$ pair, $x^{2}+y^{2}=\dfrac{9+12\sqrt{11}+44}{100}+\dfrac{36-12\sqrt{11}+11}{100}=\dfrac{100}{100}=1$ ✓, and expanding the first two equations with $\left(\sqrt{11}\right)^{2}=11$ gives exactly $0$ in both. This check is essential: a proposed solution set that fails $x^{2}+y^{2}=1$ is wrong no matter how tidy it looks.