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Question Number 160677 by tounghoungko last updated on 04/Dec/21

{ ((((x+y)/(xyz)) = −(1/4))),((((y+z)/(xyz)) = −(1/(24)))),((((x+z)/(xyz)) = (1/(24)))) :}

{x+yxyz=−14y+zxyz=−124x+zxyz=124

Commented by bobhans last updated on 04/Dec/21

⇔ (1/(xyz))(x+y+2z)=0 ⇒x+y=−2z ⇔ ((−2z)/(xyz)) = −(1/4) ; xy=8 ⇔((−2z−x+z)/(8z)) = −(1/(24)) ⇔z+x= (1/3)z ; x=−(2/3)z ⇔ y=(8/((−(2/3)z))) = −((12)/z) ⇔ −(2/3)z−((12)/z)+2z = 0 ⇔ −((12)/z) +(4/3)z = 0 ; −(3/z) +(1/3)z=0 ⇒((z^2 −9)/z) = 0 → { ((z=3 → { ((x=−2)),((y=−4)) :})),((z=−3→ { ((x=2)),((y=4)) :})) :}

⇔1xyz(x+y+2z)=0⇒x+y=−2z⇔−2zxyz=−14;xy=8⇔−2z−x+z8z=−124⇔z+x=13z;x=−23z⇔y=8(−23z)=−12z⇔−23z−12z+2z=0⇔−12z+43z=0;−3z+13z=0⇒z2−9z=0→{z=3→{x=−2y=−4z=−3→{x=2y=4

Answered by mahdipoor last updated on 04/Dec/21

get xyz=c { ((x+y=−(c/4))),((y+z=−(c/(24)))),((x+z=(c/(24)))) :} ⇒ { ((x=−(c/(12)))),((y=−(c/6))),((z=(c/8))) :} xyz=c=(c^3 /(12×6×8))⇒c=±24 ⇒ { ((x=−2 y=−4 z=3)),((x=2 y=4 z=−3)) :}

getxyz=c{x+y=−c4y+z=−c24x+z=c24⇒{x=−c12y=−c6z=c8xyz=c=c312×6×8⇒c=±24⇒{x=−2y=−4z=3x=2y=4z=−3

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