Simultaneous Diophantine equations

Write the first equation as $x^2 + y^2 + x + y = (x + y)^2 - 2xy + x + y = (x + y)^2 + (xy + x + y) - 3xy = (x + y)^2 + 3 - 3xy = 12$ Let $S = x + y$ and $P = xy$ . Then the above equation becomes $S^2 - 3P = 9$ . On the other hand, the second equation becomes $S + P = 3$ . Combining these two equations to eliminate $P$ yields $S^2 + 3S = 9 + 9 \Longrightarrow S^2 + 3S - 18 = (S + 6)(S - 3) = 0\Longrightarrow S = -6, 3$ From $S + P = 3$ , we get two cases.

Case 1: If $S = -6$ , then $P = 9$ . By Vieta's relations, this forms the equation $k^2 + 6k + 9 =0$ , giving $k = -3, -3$ as solution.

Case 2: If $S = 3$ , then $P = 0$ . This forms $k^2 - 3k = 0$ , giving $k = 0, 3$ (or its permutation).

Thus, the original system has three solutions: $(-3, -3), (0, 3)$ and $(3, 0)$ .

I wrote x in terms of y from 2nd equation and substituted it in the 1st equation and solved it by easy factorization with y having roots 0,3,-3

I solved it using x^2+y^2=(x+y)^2-2xy then solving for xy and x+y. Then it's just a matter of solving a few quadratic equations from the resulting systems of equations to find (3,0), (0,3), and (-3,-3) are the possible solutions. I'm on mobile so I'm just giving a summary and not a legit solution.