Q.If f(x) = (3x+4)/(5x-7) (x real and x≠7/5) and g(x) = (7x+4)/(5x-3) (x real and x≠3/5), then show that f.g(x) = g.f(x).

Compute $f(g(x))$ and $g(f(x))$ directly by substitution and simplify; both collapse to $x$.

Step 1 — compute $f(g(x))$. $g(x)=\dfrac{7x+4}{5x-3}$. Then

$f(g(x))=\frac{3g(x)+4}{5g(x)-7}=\frac{\dfrac{3(7x+4)+4(5x-3)}{5x-3}}{\dfrac{5(7x+4)-7(5x-3)}{5x-3}}=\frac{21x+12+20x-12}{35x+20-35x+21}=\frac{41x}{41}=x.$

Step 2 — compute $g(f(x))$. $f(x)=\dfrac{3x+4}{5x-7}$. Then

$g(f(x))=\frac{7f(x)+4}{5f(x)-3}=\frac{\dfrac{7(3x+4)+4(5x-7)}{5x-7}}{\dfrac{5(3x+4)-3(5x-7)}{5x-7}}=\frac{21x+28+20x-28}{15x+20-15x+21}=\frac{41x}{41}=x.$

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