Finding $g(x)$ for $x\in R \,\, and \,\, x\neq \pm 1$

[Albert1]

$x\in R \,\, and \,\, x\neq \pm 1$

$if$: $g(\dfrac {x-3}{x+1})+g(\dfrac {3+x}{1-x})=x$

$find$: $g(x)$

 

[Albert1]

$x\in R \,\, and \,\, x\neq \pm 1$

$if$: $g(\dfrac {x-3}{x+1})+g(\dfrac {3+x}{1-x})=x$

$find$: $g(x)$

hint:

Spoiler

let $g(\dfrac{x-3}{x+1})=a,g(\dfrac{x+3}{1-x})=b,g(x)=c$
find :$a,b,c$
so how to create 3 "Linear equations",and from them to get the solutions of $a,b,c$

 

[Albert1]

hint:

Spoiler

let $g(\dfrac{x-3}{x+1})=a,g(\dfrac{x+3}{1-x})=b,g(x)=c$
find :$a,b,c$
so how to create 3 "Linear equations",and from them to get the solutions of $a,b,c$

more hint :

Spoiler

$a+b=x---(1)$
$b+c=\dfrac {x-3}{x+1}---(2)$
$a+c=\dfrac {x+3}{1-x}---(3)$
$(1)(2)(3)$ why ?
find $c$ expressing in $x$