Solve for $x$ in the equation $ \frac35 \cdot \frac19 \cdot x = 6$.
So I multiply both sides by $\frac53$ and by $\frac91$, which are the reciprocals of $\frac35$ and $\frac19$, respectively. On the left side, the fractions cancel out, leaving me with just $x$. On the right side, I can multiply the numerators and the denominators of the fractions separately. So I get $x = \frac53 \cdot \frac91 \cdot 6 = \frac{5 \cdot 9 \cdot 6}{3 \cdot 1 \cdot 1} = \frac{270}{3} = 90$. 