Consider the following simultaneous equations in $x$ and $y$…where $a$ is a real constant: $x+y+axy=a$,$x-2y-xy^2$

Consider the following simultaneous equations in $x$ and $y$:
$$x+y+axy=a$$ $$x-2y-xy^2=0$$
where $a$ is a real constant. Show that these equations admit real solutions in $x$ and $y$.
I could not approach the problem.I have no idea how to solve these kind of problems. Please help!

[N.B.-I am not sure of the tag,please feel free to recommend]

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HINT:

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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No...tried to do that...but this does not happen easily. Not atleast with simple manipulations. – Hawk Jan 12 '14 at 16:21
@Hawk, I think we don't have to find the actual real values. This explanation should be sufficient – lab bhattacharjee Jan 12 '14 at 16:24
But, I have to reach a point where the equation is of a single variable...but I cannot reach it. – Hawk Jan 12 '14 at 16:25
@Hawk, why don't you put, $$y=\frac{a-x}{1+ay}$$ (from the first equation) in the second? – lab bhattacharjee Jan 12 '14 at 16:27
Did you mean, $x=\dfrac{a-y}{1+ay}$? – Hawk Jan 12 '14 at 16:31