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First, assume the numbers [tex]x,~y[/tex] such that they satisties both affirmations:
With these informations, we can set the following equations:
[tex]\begin{center}\align x^2+y^2=8\\ x\cdot y=4\\\end{center}[/tex]
Multiply both sides of the second equation by a factor of [tex]2[/tex]:
[tex]2\cdot x\cdot y = 2\cdot 4\\\\\\ 2xy=8~~~~~(2)^{\ast}[/tex]
Make [tex](1)-(2)^{\ast}[/tex]
[tex]x^2+y^2-2xy=8-8\\\\\\ x^2-2xy+y^2=0[/tex]
We can rewrite the expression on the left hand side using the binomial expansion in reverse: [tex](a-b)^2=a^2-2ab+b^2[/tex], such that:
[tex](x-y)^2=0[/tex]
The square of a number is equal to [tex]0[/tex] if and only if such number is equal to [tex]0[/tex], thus:
[tex]x-y=0\\\\\\ x=y~~~~~~(3)[/tex]
Substituting that information from [tex](3)[/tex] in [tex](2)[/tex], we get:
[tex]x\cdot x = 4\\\\\\ x^2=4[/tex]
Calculate the square root on both sides of the equation:
[tex]\sqrt{x^2}=\sqrt{4}\\\\\\ |x|=2\\\\\\ x=\pm~2[/tex]
Once again with the information in [tex](3)[/tex], we have that:
[tex]y=\pm~2[/tex]
The set of solutions of that satisfies both affirmations is:
[tex]S=\{(x,~y)\in\mathbb{R}^2~|~(x,~y)=(-2,\,-2),~(2,~2)\}[/tex]
This is the set we were looking for.