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2 years ago

What’s the correct why to solve

Mathematics

1 answer:

Vaselesa [24]2 years ago

6 0

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

So, there are 4 solutions.

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Step-by-step explanation:

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Hope it helps!

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