xy + x + y = 23
yz + y + z = 31
zx + z + x = 47
in real numbers.
There's bound to be something easier, but this works.
From the first equation, y = (23 - x)/(1 + x),
and from the third, z = (47 - x)/(1 + x).
Substitute into the middle eqation and tidy up.
That gets you the quadratic x^2 + 2x -35 = 0, from which x = 5 or -7.
Taking x = 5 and back substituting yields y = 3 and z = 7.
Either of the other two pairings from the original set of equations could be used instead, and a second set of solutions can be obtained from x = -7, (-7, -5, -9).