Answer
$(-2,-2)$ and $(2,2)$
Work Step by Step
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Multiply the first equation with 2 and add it to the second:
$ x^{2}+2xy+y^{2}=16\quad \Rightarrow(x+y)^{2}=4^{2}\quad \Rightarrow x+y=\pm 4$
If we subtract 2(Eq.1) from Eq.2,
$x^{2}-2xy+y^{2}=0\quad \Rightarrow(x-y)^{2}=0\quad \Rightarrow x-y=0\quad \Rightarrow x=y$
Substituting into $x+y=\pm 4$,
$2x=\pm 4$
$x=\pm 2$
Solutions: $(-2,-2)$ and $(2,2)$
