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Solve the simultaneus equations

$ab$ + $c$ + $d$ = $3$

$bc$ + $d$ + $a$ = $5$

$cd$ + $a$ + $b$ = $2$

$da$ + $b$ + $c$ = $6$, where $a$,$b$,$c$,$d$ are real numbers.

I tried to add all of them and factor them as $xy$ + $x$ + $y$ = $(x+1)(y+1)$ - $1$. Thank you for your responses.

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    Did you mean $da+b+c=6?$. It would seem so by symmetry. – 2017-01-10

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Assuming the last was supposed to be $da+b+c=6$ from the symmetry of the variables, Alpha finds $a=2, b=c=0, d=3$ which is easy to verify. It can also solve the problem as written, but the solutions are not as clean.

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    Yes, sorry. I mean da+b+c=6 :) – 2017-01-10
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    Is there any Mathematical solution to it? – 2017-01-10
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    @MathFanatics: there is clearly a substitution solution for it. Solve the first for $c$, substitute that into the rest, and so on. That will get a polynomial in the last variable, which will be of no more than $8$ th degree. Then you can use the rational root theorem. It doesn't look easy. – 2017-01-10
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    @MathFanatics: Is this a problem from a course or textbook? If so, which one? – 2017-01-10
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    Its from the british math olympiad 2003 – 2017-01-10
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Using Maple's Groebner basis package, there is a unique solution in complex numbers -- namely $(a,b,c,d) = (2,0,0,3)$, as previously noted by Ross Millikan.