<s>[INST] Triples satisfying three equations with six solutions What are all triples $(x, y, z)$ such that all three of the following equations are satisfied:$$x(y + z - 5) = 7,$$$$y(x + z - 5) = 7,$$$$x^2 + y^2 = 50.$$According to Wolfram Alpha we know that $(-7,-1,5)$, $(-5,-5,43/5)$, $(-1,-7,5)$, $(1,7,5)$, $(5,5,7/5)$, $(7,1,5)$ are the only solutions. But how do we know that there are no more solutions?
 [/INST] By subtracting the first two equations, we have $$(z-5)(x-y)=0$$
Let's consider two cases:

*

*If $z=5$, then we have $xy=7$ and $x^2+y^2=50$.

$$x^2+\frac{49}{x^2}=50$$
$$x^4-50x^2+49=0$$
$$(x^2-49)(x^2-1)=0$$
Solving this would give us $4$ solutions, corresponding to when $x=\pm 7, \pm 1$. Given $z$ and $x$, $y$ is determined.

*

*If $x=y$, then $2x^2=50$, $x=\pm 5$, this gives us another $2$ solutions. Knowing $x$ and $y$, $z$ is determined.

Hence in total, we have $6$ solutions.
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