Simultaneous equations

Question

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

Answer 1

$$x+(y-1) z = 0 \implies (y-1)=-\frac xz \tag{*1} $$

$$x (y-1)-y+2 z = 0 \implies (y-1)=\frac{y-2z}{x} $$

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 $z=-\frac 32$

Answer 2

$x+(y-1) z = 0$ : (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:

$$(x,y,z) = (0,0,0),\left(3,3,-\frac{3}{2}\right),\left(\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},1\right), \left(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2},1\right)$$