If \(y=x+3\), then in the first equation you have
\(x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16\)
Rewriting a bit, you get
\(x^2+3x+\dfrac92=8\)
\(x^2+3x+\dfrac94+\dfrac94=8\)
\(\left(x+\dfrac32\right)^2=\dfrac{23}4\)
\(x=-\dfrac32\pm\dfrac{\sqrt{23}}2\)
Now, since \(y=x+3\), you also get
\(y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2\)
So, there are two solutions here:
\((x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)\)
\((x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)\)
