We have $G = (1, 2)$ $G=(1,2)$ and $x \circ y = \frac{3 x y - 4 x - 4 y + 6}{2 x y - 3 x - 3 y + 5}$ $x@y=(3xy-4x-4y+6)/(2xy-3x-3y+5)$ .How you find $u \in G$ $uin G$ such that $x \circ u = x$ $x@u=x$ ?

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Answer 1 · 2017-04-28T19:12:28

$u = \frac{3}{2}$ $u = 3/2$

Solving $x \circ u = x$ $x @ u = x$ we have

$\frac{3 x u - 4 x - 4 u + 6}{2 x u - 3 x - 3 u + 5} = x$ $(3 x u - 4 x - 4 u + 6)/(2 x u - 3 x - 3 u + 5) = x$

or

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

or

$6 - 4 u - 9 x + 6 u x + 3 x^{2} - 2 u x^{2} = 0$ $6 - 4 u - 9 x + 6 u x + 3 x^2 - 2 u x^2=0$

or

$(6 - 4 u) + (6 u - 9) x + (3 - 2 u) x^{2} = 0$ $(6-4u)+(6u-9)x+(3-2u)x^2=0$

this relationship must be true for all $x$ $x$ so

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

and solving for $u$ we have

$u = 3/2$

We have G=(1,2)G=(1,2)G=(1,2) and x@y=(3xy-4x-4y+6)/(2xy-3x-3y+5)x∘y=3xy−4x−4y+62xy−3x−3y+5x@y=(3xy-4x-4y+6)/(2xy-3x-3y+5) .How you find uin Gu∈Guin G such that x@u=xx∘u=xx@u=x?