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 $(3, 3, \frac{-3}{2})$

$$x+(y-1) z = 0 \implies (y-1)=-\frac xz \tag{*1} $$
$$x (y-1)-y+2 z = 0  \implies (y-1)=\frac{y-2z}{x} $$
so $$(y-1)^2=- \frac yz +2 $$
$$   (x-1) z+y = 0  \implies  \frac yz = -(x-1) \tag {*2}$$
so $$(y-1)^2= x+1 \tag{*3}$$
we can get another relation between $x$ and $y$ by dividing (*1) by (*2)
$$  -\frac xy =-\frac{y-1}{x-1 } \implies y(y-1)=x(x-1) \tag{*4} $$
now solve (*3) for $x$ and plug in to (*4)
$$  y(y-1)=((y-1)^2-1)((y-1)^2-2)  $$
to simplfy the algebra let $a \equiv (y-1)$
$$a(a+1)=(a^2-1)(a^2-2)  \implies a=(a-1)(a^2-2)$$
so we end up with the cubic equation ...
$$ a^3-a^2-3a+2=0  $$
That's as far as I'm willing to take it, but clearly $a=2$ does solve the equation so $y=3$ then (*3) gives you $x=3$ and (*1) gives $z=-\frac 32$
