x^2-yz=14; y^2-xz-28; z^2-xy=-14

Question

solve for x,y and z

Answer 1

x^2 - yz = 14

y^2 - xz = 28

z^2 - xy = -14

Solves first equation for y

x^2 - 14 = yz

(x^2 - 14)/z = y

Substitutes into equations 2 and 3

[(x^2 -14)/z]^2 - xz = 28

z^2 - x( x^2 - 14)/z = -14

(x^2 - 14)^2 / z^2 - xz = -28

z^2 - x^3/z +14/z = -14

multiplies first equation by z^2 and second equation by z

(x^2 -14)^2 - xz^3 = -28z^2

z^3 - x^3 + 14 = -14z

(x^2 - 14)^2 = xz^3 - 28z^2

z^3 + 14z = x^3 - 14

z^3 + 14z + 14 = x^3

(z^3 + 14z + 14)^(2/3) = x^2 <--- raises both sides to the power 2/3

(z^3 +14z +14)^(1/3) = x <---- takes square root of both sides

((z^3 + 14z + 14)^(2/3) - 14 )^2 = ((z^3 +14z +14)^(1/3)) (z^3) - 28z^2

the last equation must be solved numerically

z=-1.568

z=0894

You plug these values into the bolded equations above to find the other solutions