$x$ varies directly as $y$ and inversely as $z$. If $x = 6$ when $y = 4$ and $z = 9$, find $x$ when $y = 12$ and $z = 6$.

Question

Aicharya · Accepted Answer

$x = \dfrac{k y}{z}$. From the first triple: $6 = \dfrac{k \cdot 4}{9} \Rightarrow k = \dfrac{54}{4} = \dfrac{27}{2}$. Then $x = \dfrac{(27/2) 	imes 12}{6} = \dfrac{27 	imes 12}{12} = 27$.

$x$ varies directly as $y$ and inversely as $z$. If $x = 6$ when $y = 4$ and $z = 9$, find $x$ when $y = 12$…

Solution