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Lovellss

Lovellss

Answered question

2022-06-18

How to solve this system of equation?
{ 3 y 3 + 3 x 1 x = 5 1 x 2 y x 2 y 2 1 x = 2 y + 5 1 x

Answer & Explanation

victorollaa5

victorollaa5

Beginner2022-06-19Added 16 answers

Let's resurrect this old question. Two solutions to
{ 3 y 3 + 3 x z = 5 z 2 y x 2 y 2 z = 2 y + 5 z
are ( x , y ) = ( 3 , 2 ) for z = 1 x and,
x = Real root ( 7 x 3 x 2 + 7 x 3 8 x 4 + 2 x 5 x 6 + x 7 = 0 )
y = Real root ( 2 + y + 2 y 2 + 4 y 3 3 y 5 + y 7 = 0 )
for z = 1 x
Unless your system has special symmetries, I'm afraid the only way to find solutions that algebraic numbers of high degree is to use resultants.

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