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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.

September 25, 2015
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