# Real solutions (Tomato subjective 71)

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.