106yeah there's more
x=y=0, z = -1
or
y=z=0, x = -1
or
x=z=0, y = -1
\textbf{Solve System of Equations: }$\\\\\left\{\begin{matrix} \mathbf{\sqrt{x^2+y^2}=z+1} \\\\ \mathbf{\sqrt{y^2+z^2}=x+1} \\\\ \mathbf{\sqrt{z^2+x^2}=y+1} \end{matrix}\right;\mathbf{x.y,z\in\mathbb{R}} .
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6 Answers
106
21continuing after man111
since the system of eq is symmetric/cyclic , we have
(x-z)(x+y+1) = 0
(y-z)(y+z+1) = 0
(x-y)(x+y+1)= 0
solutions can be broadly classified into three sets
1) x=y=z(we will get aditya's answers )
2) any two are equal, we can solve easily (we will get all the answers given by ashish)
3) none are equal. that gives us x+y+1 = 0, y+z+1 = 0 and x+z+1 = 0,which gives x=y=z whih is absurd. because we are considering none of them are equal.
1708Thanks Shubhodip.
Speciallly for nice symmetric/cyclic approach.