Question 564213
Add 3xy to both sides of the 1st equation.


*[tex \LARGE x^2 + 2xy + y^2 = 7 + 3xy \Rightarrow (x+y)^2 = 7 + 3xy]


The second equation is equivalent to


*[tex \LARGE xy = x+y+1]. Substituting this into the first equation,


*[tex \LARGE (x+y)^2 = 7 + 3(x+y+1) = 10 + 3(x+y)]


*[tex \LARGE (x+y)^2 - 3(x+y) - 10 = 0]


This is a quadratic in x+y, and the solutions are x+y = 5 and x+y = -2. If x+y = 5, substituting into the second equation we obtain


*[tex \LARGE 5 - xy = -1 \Rightarrow xy = 6]


And if x+y = -2, substituting into the second equation yields


*[tex \LARGE -2 - xy = -1 \Rightarrow xy = -1]


Therefore we have two cases: x+y = 5 & xy = 6, and x+y = -2 & xy = -1. The first case yields {3,2}; the second case we can let x = -2-y, in which


*[tex \LARGE (-2-y)(y) = -1 \Rightarrow y^2 + 2y - 1 = 0]


By the quadratic formula, *[tex \LARGE y = -1 \pm \sqrt{2}], the solution obtained is *[tex \LARGE (-1+\sqrt{2}, -1-\sqrt{2})]