# Suppose z varies directly with x and inversely with the square of y. If z=18 when x=6 and y=2, what is z when x=8 and y=9?

Aug 24, 2017

$z = \frac{32}{27}$

#### Explanation:

$\text{the initial statement here is } z \propto \frac{x}{{y}^{2}}$

$\text{to convert to an equation multiply by k the constant}$
$\text{of variation}$

$\Rightarrow z = \frac{k x}{{y}^{2}}$

$\text{to find k use the given condition}$

$z = 18 \text{ when "x=6" and } y = 2$

$z = \frac{k x}{{y}^{2}} \Rightarrow k = \frac{{y}^{2} z}{x} = \frac{4 \times 18}{6} = 12$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{z = \frac{12 x}{{y}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{when "x=8" and } y = 9$

$z = \frac{12 \times 8}{81} = \frac{32}{27}$