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}

Collected in the board: System of quadratic equations

Steven Zheng posted 1 week ago

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 \text{ or } x = 2
(8)

Substituting x into (6) gives the value of z

z = -3 \text{ or } z = 1
(9)

Substituting x into (1) gives the value of y

y = 3 \text{ or } y = -5
(10)

Steven Zheng posted 1 week ago

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