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Find the solution of the following set of equations \begin{equation} \begin{cases} yz-2z+x-1=0 \nonumber \\ zx+y-z-2=0 \nonumber \\ xy-2x-y+z+2=0 \end{cases} \end{equation}

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Start by factorizing your equations:

$$z(y-2) + (x-1) = 0$$ $$z(x-1) + (y-2) = 0$$ $$(y-2)(x-1) + z = 0$$

From this we see that we get a much simpler system if we define $\tilde{x} = x-1$, $\tilde{y} = y-2$ and $\tilde{z} = -z$ (this procedure backward is likely how the author of this problem constructed it):

$$\tilde{x} = \tilde{z}\tilde{y},~~~~~~~~~~~\tilde{y} = \tilde{z}\tilde{x},~~~~~~~~~~~\tilde{z} = \tilde{x}\tilde{y}$$

The system is now also invariant under a cyclic permutation of the variables so the solution set will also have this propery. All the combinations of any two of the three equaitions above gives us

$$\tilde{x}=\tilde{z}^2\tilde{x} = \tilde{y}^2\tilde{x}$$

$$\tilde{y}=\tilde{z}^2\tilde{y} = \tilde{x}^2\tilde{y}$$

$$\tilde{z}=\tilde{y}^2\tilde{z} = \tilde{x}^2\tilde{z}$$

From this it follows that $\tilde{x},\tilde{y},\tilde{z} \in\{-1,0,1\}$. Next we see that if one variable is $0$ then all variables must be $0$ and $(0,0,0)$ is indeed a solution so any other solution must have $\tilde{x},\tilde{y},z \in\{-1,1\}$ (which gives $8$ possible solutions). We can also rule out the possibillity that two of the variables being $1$ and the last being $-1$ or all three being $-1$ (since $\tilde{x}=\tilde{z}\tilde{y}$). This leaves us with $(\tilde{x},\tilde{y},\tilde{z})=(0,0,0)$, $(\tilde{x},\tilde{y},\tilde{z})=(1,1,1)$, $(\tilde{x},\tilde{y},\tilde{z})=(-1,-1,1)$, $(\tilde{x},\tilde{y},\tilde{z})=(1,-1,-1)$ and $(\tilde{x},\tilde{y},\tilde{z})=(-1,1,-1)$ which all satisfy the equation-set.

In terms of the original variables we get $(x,y,z)=(0,1,-1)$, $(x,y,z)=(0,3,1)$, $(x,y,z)=(1,2,0)$, $(x,y,z)=(2,1,1)$ and $(x,y,z)=(2,3,-1)$

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find sum of 2nd and 3rd equation :$$(xz+y-z-2)+(xy-2x-y+z+2)=0+0\\xz+xy-2x=0\\x(z+y-2)=0$$you can go on from there

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  • $\begingroup$ For $x=0$ we get $(x,y,z)=(0,3,1)$ and $(x,y,z)=(0,1,-1)$. The other case is difficult. $\endgroup$ – martin Feb 11 '15 at 20:29
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By summing (2) and (3) you end up with the equation: $$ x(z+y-2)=0. $$ Which has two solutions, either $x=0$ or $z+y-2=0$.

For the first case you end up with the following system: \begin{align} z(y-2)&=1\\ y-z&=2, \end{align} which is equivalent to $(y-2)^2=1$ and whose solutions are $(0,1,-1)$ and $(0,3,1)$.

For the other case you end up with the following system, given that $z=2-y$: \begin{align} y(2-y)-2(2-y) +x = -y^2 +x+4y -4 &=1\\ x(2-y)+y-(2-y)=-xy+2x +2y -2 &=2\\ xy-2x-2y+4&=0. \end{align} Observe that the second and third equation are the same.

Geomtrically the first equation is the horizontal parabola with vertex $(2,1)$: $$ (y-2)^2 = (x-1). $$ And the second is either a vertical or an horizontal line: $$ x(y-2)-2(y-2) = (x-2)(y-2)=0. $$ So the solutions are the possible intersections of them! To find them explicitly consider each possible case, $x=2$ or $y=2$, from which we obtain:

If $x=2$, then $y=1$ or $y=3$, with solutions $(2,1,1)$ and $(2,3,-1)$, respectively.

If $y=2$, then $x=1$ and $z=0$, with solution $(1,2,0)$.

These are all the possible solutions.

Plot: http://www.wolframalpha.com/input/?i=plot++%28y-2%29%5E2-%28x-1%29%3D0+and+x%3D2+and+y%3D2

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