problem: Consider the following simultaneous equations in x and y :
{\displaystyle{4x + y + axy = a}}
{\displaystyle{x - 2y -xy^2 = 0}}
where {\displaystyle{a}} is a real constant. Show that these equations admit real solutions in x and y.

solution: {\displaystyle{4x + y + axy = a}} … (i)
{\displaystyle{x - 2y -xy^2 = 0}} … (ii)
{\displaystyle{x - 2y -xy^2 = 0}}
{\Rightarrow} {\displaystyle{x = {\frac{2y}{1-y^2}}}}
{\displaystyle{x + y + axy = a}}
{\Rightarrow} {\displaystyle{\frac{2y}{1-y^2} + 4 + a{\frac{2y^2}{1-y^2}}}} {\displaystyle{= a}} [ replacing x in terms of y. ]
{\Rightarrow} {\displaystyle{3y - y^3 + 3ay^2 -a = 0}}
This is a 3rd degree polynomial over y.
Now as we know any odd degree polynomial has at least one real root.That is why y has also at least one real root.Now for that particular root we get {\displaystyle{x = {\frac{2y}{1-y^2}}}} also real.
Conclusion: {\displaystyle{x + y + axy = a}} & {\displaystyle{x - 2y -xy^2 = 0}} admits real solutions in x and y.