Question
Solve the system
x^2+xy+y^2=19 \\ x^2+y^2+x+y=18
Solve the system
x^2+xy+y^2=19 \\ x^2+y^2+x+y=18
Let a = xy, b=x+y, then the system of equations is given as
Since x^2+xy+y^2=(x+y)^2-xy=19
and
x^2+y^2+x+y = (x+y)^2-2xy+x+y=18
Subtract (1) from (2)
a-b=1
Substitute a=b+1 to the first equation
Then, b=5 or b=-4
Correspondingly, a = 6, or a = -3
Case 1, if a=6, b=5
xy=6\\x+y=5
According to Vieta's formula
x,y is the roots of the quadratic equation z^2-5z+6=0
Solving the equation gives
\begin{cases} x=2 \\ y=3 \end{cases} \quad \text{or}\begin{cases} x=3 \\ y=2 \end{cases}
Case 2, if a=-3, b=-4
xy=-3\\x+y=-4
According to Vieta's formula
x,y is the roots of the quadratic equation z^2+4z-3=0
Solving the equation gives
\begin{cases} x=-2+\sqrt{7} \\ y=-2-\sqrt{7} \end{cases} \quad \text{or}\begin{cases} x=-2-\sqrt{7} \\ y=-2+\sqrt{7} \end{cases}