Consider the system of equations in real numbers <mtext>&#xA0;</mtext> x , y , z <mte

Joshua Foley

Joshua Foley

Answered question

2022-07-04

Consider the system of equations in real numbers   x , y , z   satisfying
x 4   x 2 + 1   =   y 5   y 2 + 1   =   z 6   z 2 + 1
and   x + y + z   =   x y z  . Find   x , y , z   ..
I substituted   x , y , z   as   tan ( θ 1 )   ,   tan ( θ 2 )   ,   tan ( θ 3 )  , where
θ 1   +   θ 2   +   θ 3   =   180 º     .

Answer & Explanation

toriannucz

toriannucz

Beginner2022-07-05Added 16 answers

to any values a , b , c instead of 1 4 , 1 5 , 1 6
If we put
(1) ϕ ( a , b , c ) = a b + a c b c ,   ψ ( a , b , c ) = a b + a c + b c
then LAcarguy’s method leads to the following explicit formulas for the solutions :
(3) x = ε ϕ ( a 2 , b 2 , c 2 ) G ( a , b , c ) ,   y = ε ϕ ( b 2 , c 2 , a 2 ) G ( a , b , c ) ,   z = ε ϕ ( c 2 , a 2 , b 2 ) G ( a , b , c )
where ε is + 1 or 1. In your initial question where a = 1 4 , b = 1 5 , c = 1 6 , one finds G = 7 921600 ,
(4) x = ε 7 3 ,   y = ε 5 7 9 ,   z = 3 ε 7
Ayaan Barr

Ayaan Barr

Beginner2022-07-06Added 6 answers

Starting with s i n t 4 = s i n u 5 = s i n w 6 , then use the law of sines to get: a 4 = b 5 = c 6 . So a = 2 c 3 , b = 5 c 6 . So apply the law of cosines to have: a 2 = b 2 + c 2 2 b c c o s A. Substituting these values of a and b into the above equation we can solve for c o s A, and then t a n B, and t a n A, and t a n C.

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