# If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

Mar 11, 2018

$y = \frac{21}{125}$

#### Explanation:

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

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

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

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

$y = \frac{1}{6} \text{ when "x=20" and } z = 6$

$y = \frac{k x}{z} ^ 2 \Rightarrow k = \frac{y {z}^{2}}{x} = \frac{\frac{1}{6} \times 36}{20} = \frac{3}{10}$

$\text{equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{3 x}{10 {z}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{when "x=14" and "z=5" then}$

$y = \frac{3 \times 14}{10 \times 25} = \frac{21}{125}$