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! :)