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

Substitute one function into the other and simplify each composite; both reduce to $x$.

Given $f(x) = \dfrac{3x+4}{5x-7}$ and $g(x) = \dfrac{7x+4}{5x-3}$.

Step 1 — find $f(g(x))$.

$f(g(x)) = \dfrac{3g(x)+4}{5g(x)-7}$

Numerator: $3g(x)+4 = \dfrac{3(7x+4) + 4(5x-3)}{5x-3} = \dfrac{21x+12+20x-12}{5x-3} = \dfrac{41x}{5x-3}$

Denominator: $5g(x)-7 = \dfrac{5(7x+4) - 7(5x-3)}{5x-3} = \dfrac{35x+20-35x+21}{5x-3} = \dfrac{41}{5x-3}$

So $f(g(x)) = \dfrac{41x/(5x-3)}{41/(5x-3)} = x$.

Step 2 — find $g(f(x))$.

$g(f(x)) = \dfrac{7f(x)+4}{5f(x)-3}$

Numerator: $7f(x)+4 = \dfrac{7(3x+4)+4(5x-7)}{5x-7} = \dfrac{21x+28+20x-28}{5x-7} = \dfrac{41x}{5x-7}$

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