# Y varies directly as x and inversely as the square of z. y=12 when x=64 and z=4. How do you find y when x=96 and z=2?

Jul 29, 2017

$y = 72$

#### Explanation:

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

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

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

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

$y = 12 \text{ when "x=64" and } z = 4$

$y = \frac{k x}{z} ^ 2 \Rightarrow k = \frac{y {z}^{2}}{x} = \frac{12 \times 16}{64} = 3$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{3 x}{z} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{when "x=96" and } z = 2$

$\Rightarrow y = \frac{3 \times 96}{4} = 72$