Question

I don't know how to solve this at all, i just need some help on how to figure it out. Thanks!

Answer 1

Answer:

$\sf \dfrac{3}{{z-3} }$

Explanation:

$\Longrightarrow \sf \dfrac{1}{z} \div \dfrac{z-3}{3z}$

apply the theorem when $\dfrac{a}{b} \div \dfrac{c}{d}$ then $\dfrac{a}{b} \times \dfrac{d}{c}$

$\Longrightarrow \sf \dfrac{1}{z} \times \dfrac{3z}{z-3}$

Join the fractions

$\Longrightarrow \sf \dfrac{3z}{z({z-3)} }$

Cancel out common factor z

$\Longrightarrow \sf \dfrac{3}{{z-3} }$