Solving System of Symmetric Equations

Question

I am to solve the system below. $$\begin{cases} (x^2+1)(y^2+1)=10 \\ (x+y)(xy-1)=3 \end{cases}$$ I have tried to factor the equations of the system but they do not factor. How can I approach the problem?

Answer 1

Let x+y=u and xy=v

Second equation becomes u(v-1)=3

First equation is ${(xy)}^2+(x+y)^2+1-2xy=10$ which is $v^2+u^2+1-2v=10 \Rightarrow (v-1)^2+u^2=10$

Eliminate v-1 from the two equations to get $\frac{9}{u^2}+u^2=10$

Solve the quadratic to get u=$\pm1$ and $\pm3$ then solve for x and y

Answer 2

We have $$x^2+y^2+x^2y^2=9$$ or $$(x+y)^2-2xy+x^2y^2=9$$ and since $$x+y=\frac{3}{xy-1},$$ we obtain: $$\frac{9}{(xy-1)^2}-2xy+x^2y^2=9$$ or $$\frac{9(2-xy)xy}{(xy-1)^2}-xy(2-xy)=0$$ or $$xy(2-xy)(4-xy)(2+xy)=0.$$ Now, $xy=0$ gives $(-3,0)$ and $(0,-3)$;

$xy=2$ gives $x+y=3$ and $(2,1)$ and $(1,2)$;

$xy=4$ does not give real solutions

and $xy=-2$ gives $x+y=-1$ and $(-2,1)$ and $(1,-2).$