Question

What’s the correct why to solve

Answer 1

$x^2-y^2=2\\ \underline{x^2+y^2=4}\\\\ 2x^2=6\\ x^2=3\\ x=-\sqrt3 \vee x=\sqrt3\\\\ 3+y^2=4\\ y^2=1\\ y=-1 \vee y=1\\\\ x\in\{(-\sqrt3,-1),(-\sqrt3,1),(\sqrt3,-1),(\sqrt3,1)\}$

So, there are 4 solutions.