Question
Solve the system of quadratic equations with three variables
\begin{cases} xy+x+y=-13 \qquad\quad\,\, (1) \\ yz+y+z = -9 \qquad\qquad (2)\\ zx+z+x = 5 \qquad\qquad\,\,\,\, (3)\end{cases}
Solve the system of quadratic equations with three variables
\begin{cases} xy+x+y=-13 \qquad\quad\,\, (1) \\ yz+y+z = -9 \qquad\qquad (2)\\ zx+z+x = 5 \qquad\qquad\,\,\,\, (3)\end{cases}
Multiply (1) by (z+1)
(xy+x+y)(z+1)=-13(z+1)
Multiply (2) by (x+1)
(yz+y+z)(x+1) = -9(x+1)
Subtracting (5) from (4) results in
3z-2x = -1
Substituting (6) into (3) obtains the quadratic equation
Solve the equation for x
Substituting x into (6) gives the value of z
Substituting x into (1) gives the value of y