Solve the system of quadratic equations with three variables
$$\begin{cases} xy+x+y=-13 \qquad\quad\,\, (1) \ yz+y+z = -9 \qquad\qquad (2)\ zx+z+x = 5 \qquad\qquad\,\,\,\, (3)\end{cases}$$

Question

Solve the system of quadratic equations with three variables
$$\begin{cases} xy+x+y=-13 \qquad\quad\,\, (1) \  yz+y+z = -9 \qquad\qquad (2)\ zx+z+x = 5 \qquad\qquad\,\,\,\, (3)\end{cases}$$

Uniteasy Admin · Accepted Answer

Multiply $$(1)$$ by $$(z+1)$$
$$(xy+x+y)(z+1)=-13(z+1)$$
$$xyz+xz+yz+xy+x+y = -13z-13$$4
Multiply $$(2) $$ by $$(x+1)$$
$$(yz+y+z)(x+1) = -9(x+1)$$
$$xyz+xy+xz+yz+y+z = -9x-9$$5
Subtracting (5) from (4) results in
$$3z-2x = -1$$
$$z=\dfrac{2x-1}{3} $$6
Substituting (6) into (3) obtains the quadratic equation
$$x^2+2x-8 = 0$$7
Solve the equation for $$x$$
$$x = -4 	ext{ or } x = 2$$8
Substituting $$x$$ into (6) gives the value of $$z$$
$$z = -3 	ext{ or } z = 1$$9
Substituting $$x$$ into (1) gives the value of $$y$$
$$y = 3 	ext{ or } y = -5$$10

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