Recall that [tex](x+y)^2=x^2+2xy+y^2[/tex]. So we can add twice the first equation to the second one to get
[tex]x^2+y^2+2xy=10+2\cdot3\iff(x+y)^2=16\implies x+y=\pm4[/tex]
Since [tex]xy=3[/tex], we have [tex]y=\dfrac3x[/tex] [tex](x\neq0)[/tex] so
[tex]x+y=x+\dfrac3x=\pm4\implies x^2\mp4x+3=0[/tex]
If [tex]x+y=4[/tex], then
[tex]x^2-4x+3=(x-3)(x-1)=0\implies x=3,x=1\implies y=1,y=3[/tex]
If [tex]x+y=-4[/tex], then
[tex]x^2+4x+3=(x+1)(x+3)=0\implies x=-1,x=-3\implies y=-3,y=-1[/tex]
So the solution set is
[tex](x,y)\in\{(-3,1),(1,3),(-1,-3),(-3,-1)\}[/tex]
