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

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SUMMARY

The discussion focuses on finding the function \( g(x) \) defined by the equation \( g\left(\frac{x-3}{x+1}\right) + g\left(\frac{3+x}{1-x}\right) = x \) for real numbers \( x \) where \( x \neq \pm 1 \). Participants analyze the transformation of variables and the implications of the given equation. The conclusion emphasizes that the function \( g(x) \) can be derived through systematic substitution and manipulation of the equation.

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Albert1
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$x\in R \,\, and \,\, x\neq \pm 1$

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

$find$: $g(x)$
 
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Albert said:
$x\in R \,\, and \,\, x\neq \pm 1$

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

$find$: $g(x)$
hint:
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 $a,b,c$
 
Albert said:
hint:
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 :
$a+b=x---(1)$
$b+c=\dfrac {x-3}{x+1}---(2)$
$a+c=\dfrac {x+3}{1-x}---(3)$
$(1)(2)(3)$ why ?
find $c$ expressing in $x$
 

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