Question
Solve the system of two quadratic equations with two variables
\begin{cases} xy+x+y=-13 \qquad\qquad (1)\\x^2+y^2 = 29 \qquad\qquad\qquad\: (2)\end{cases}
Answer
Observe the equations and add 2\times ( 1)+(2)
(x+y)^2+2(x+y)-3=0
Factoring the LHS
(x+y-3)(x+y+1) = 0
Then, we get
x+y-3=0 \text{ or } x+y+1=0
Substituting them to (1) gives two system of equations
\begin{cases} x+y-3=0 \\xy = -10 \end{cases}
(3)
\qquad \text{or}\qquad
\begin{cases} x+y+1=0 \\xy=14 \end{cases}
(4)
Solving (3) obtains
\begin{cases} x_1=-5 \\y_1=2 \end{cases} \qquad \text{or}\qquad\begin{cases} x_2=-2 \\y_2=-5 \end{cases}
Solving (4) obtains
\begin{cases} x_3=\dfrac{1+\sqrt{57} }{2} \\y_3=\dfrac{1-\sqrt{57} }{2} \end{cases} \qquad \text{or}\qquad \begin{cases} x_4=-\dfrac{1-\sqrt{57} }{2} \\y_4=\dfrac{1+\sqrt{57} }{2} \end{cases}