An algebra problem by Paola Ramírez

$\dfrac{a}{b} = \dfrac{1}{9} \implies b=9a\\ \dfrac{b}{c}=\dfrac{1}{3} \implies c=3b=3(9a) = 27a$

$\dfrac{b-a}{c-b}\\ =\dfrac{9a-a}{27a-9a}\\ =\dfrac{8a}{18a}\\ =\boxed{\dfrac{4}{9}}$

$\dfrac{a}{b}=\dfrac{1}{9}$ $\implies$ $b=9a$

$\dfrac{b}{c}=\dfrac{1}{3}$ $\implies$ $\dfrac{9a}{c}=\dfrac{1}{3}$ $\implies$ $c=27a$

$\dfrac{b-a}{c-b}=\dfrac{9a-a}{27a-9a}=\dfrac{8a}{18a}=\boxed{\dfrac{4}{9}}$

$\dfrac{a}{b} = \dfrac{1}{9}\implies b = 9a$
$\dfrac{b}{c} = \dfrac{1}{3} \implies c = 3b$
So,
$\dfrac{b-a}{c-b}\\ =\dfrac{9a-a}{3b-b}\\ =\dfrac{8a}{2b}\\ =\boxed{\dfrac{4}{9}}$