<s>[INST] Simultaneous equations How can I find the solution $(3, 3, \frac{-3}{2})$ from the following system of equations:
$x+(y-1) z = 0$
$(x-1) z+y = 0$
$x (y-1)-y+2 z = 0$
I have done eq1 - eq2 to find the other solutions. How would i get to the solution $(3, 3, \frac{-3}{2})$
 [/INST] $x+(y-1) z = 0$ : (eq1)
$(x-1)z+y = 0$ : (eq2)
$x (y-1)-y+2 z = 0$ : (eq3)
Rearranging eq1 and we have: $x=-(y-1)z$. Subbing this into eq2 and eq3 gives:
$(-(y-1)z-1)z+y=0$ : (eq4)
$-(y-1)z(y-1)-y+2z=0$ : (eq5)
Rearranging eq5 we get: $z=\frac{y}{2-(y-1)^2}$
Subbing this into eq4 gives:
$$\left(-(y-1)\frac{y}{2-(y-1)^2}-1\right)\frac{y}{2-(y-1)^2}+y=0$$
$$\left(-\frac{y(y-1)}{-y^2+2y+1}-1\right)\frac{y}{-y^2+2y+1}+y=0$$
$$(-y(y-1)+y^2-2y-1)y+y(-y^2+2y+1)^2=0$$
$$y(-y-1+y^4-4y^3+2y^2+4y+1)=0$$
$$y(y^4-4y^3+2y^2+3y)=0$$
$$y^2(y^3-4y^2+2y+3)=0$$
$$y^2(y-3)(y^2-y-1)=0$$
Hence: $y=0$, $y=3$, $y=\frac{-1\pm\sqrt{5}}{2}$
Subbing back into $x$ and $z$ gives the four solutions:
$$(x,y,z) = (0,0,0),\left(3,3,-\frac{3}{2}\right),\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},1\right), \left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2},1\right)$$
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