Algebra

Let's first focus on the first equation given.

Since we know that $(x+y)^2 = x^2+2xy+y^2$, squaring both sides on the first equation, we get that

$x^2+2xy+y^2=4$

Reaaranging the second equation a bit, we find that

$x^2+xy+y^2=5$

Wait! These two equations are quite similar to each other. Subtracting the second equation from the first equation, we get that

$x^2+2xy+y^2=4\\ x^2+1xy+y^2=5 \space \space \space \space -$

______________________

$xy=-1$

Now, let's look at the second equation again. We know that

$x^2+xy+y^2=5\\ x^2+y^2=5-xy\\ x^2+y^2=5-(-1) = 6$

So the answer is 6...i think

Thanks! :)