skylsn

2022-06-30

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

Govorei9b

Expert

Eliminating one of the two unknowns, say $y$ we can form a Cubic equation in $x$
Using Complex conjugate root theorem, an odd degree equation with real coefficients has an odd number of (at least one) real root(s)
From the first equation given, if $x$ is real so will be $y$ and vice versa

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