Systems of equations in which one equation is

x + y = k or x - y = k
and the other is
xy = p

Always lead to quadratic equations, which almost always have two solutions for each variable.
Don't fall for the "obvious" positive answer for x.
And don't assume that this falls under the "two equations two variables -> sufficient" category.
The equation xy = p is NOT linear, so that idea doesn't apply here.

In this case,
xy = 12
[spoiler](y+1)y = 12[/spoiler]
[spoiler]y^2 + y - 12 = 0[/spoiler]
[spoiler](y + 4)(y - 3) = 0[/spoiler]
[spoiler]y = -4 or y = 3[/spoiler]
[spoiler]x = -3 or x = 4 Insufficient[/spoiler]

This is generally true but occasionally it isn't:

x(x-6) = -9 and any other equation that is derived from "perfect square = 0" has only one solution.