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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.