We have #G=(1,2)# and #x@y=(3xy-4x-4y+6)/(2xy-3x-3y+5)# .How you find #uin G# such that #x@u=x#?

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Answer 1 · 2017-04-28T19:12:28

#u = 3/2#

Solving #x @ u = x# we have

#(3 x u - 4 x - 4 u + 6)/(2 x u - 3 x - 3 u + 5) = x#

or

#3 x u - 4 x - 4 u + 6 - x (2 x u - 3 x - 3 u + 5)=0#

or

#6 - 4 u - 9 x + 6 u x + 3 x^2 - 2 u x^2=0#

or

#(6-4u)+(6u-9)x+(3-2u)x^2=0#

this relationship must be true for all #x# so

#{(6-4u=0),(6u-9=0),(3-2u=0):}#

and solving for #u# we have

#u = 3/2#