* 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.