Let S=x+y and P=xy if rewrite both equations with S and P, problem will be solve.

Adding the two equations gives $2(x^2+y^2)^{3/2}=128\implies x^2+y^2=16$. Plug this into the equations of the original system gives $4(16+xy)=88$ and $4(16-xy)=40$, both of which give $xy=6$. Next, $(x-y)^2=x^2+y^2-2xy=16-12=4\implies x-y=\pm 2$. Case $x-y=2$ $x^2+y^2=16\implies(y+2)^2+y^2=16\implies y^2+2y-6=0\implies y=-1\pm\sqrt{7}\implies x=1\pm\sqrt{7}$. Case $x-y=-2$ $x^2+y^2=16\implies(y-2)^2+y^2=16\implies y^2-2y-6=0\implies y=1\pm\sqrt{7}\implies x=-1\pm\sqrt{7}$. None of these are extraneous. Final answer: $(1+\sqrt{7},-1+\sqrt{7})$ $(-1+\sqrt{7},1+\sqrt{7})$ $(1-\sqrt{7},-1-\sqrt{7})$ $(-1-\sqrt{7},1-\sqrt{7})$