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