If `f: R-{7/5}->R-{3/5}` be defined as `f(x)=(3x+4)/(5x-7)` and `g: R-{3/5}->R-{7/5}` be defined as `g(x)=(7x+4)/(5x-3)` . Show that `gof=I_A` and `fog=I_B` , where `B=R-{3/5}` and `A=R-{7/5}` .

Explanation

Given:

`f(x)=(3x+4)/(5x-7)`,
`g(x)=(7x+4)/(5x-3)`

Note Heat f(x) and g(x) are both one one and onto in the given domain and co-domain hence both are invertible.

`fog(x)=g(g(x))=(3((7x+4)/(5x-3))+4)/(5((7x+4)/(5x-3))-7)=(21x+12+20x−12
)/(35x+20−35x+21
)=(41x)/(41)=x=I_A`

`gof(x)=g(f(x))=(7((3x+4)/(5x-7))+4)/(5((3x+4)/(5x-7))-3)=(21x+28+20x−28
...