# Assume y varies inversely as x, if y=8 when x=4, how do you find x when y=2?

Jul 6, 2017

$x = 16$

#### Explanation:

$y \text{ varies inversely as } x \implies y \propto \frac{1}{x}$

$\implies y = \frac{k}{x} \text{ where "k" is the constant of proportionality}$

$y = 8 , x = 4$

we have

$8 = \frac{k}{4} \implies k = 8 \times 4 = 32$

$\therefore y = \frac{32}{x}$

$y = 2$

$2 = \frac{32}{x} \implies x = \frac{32}{2} = 16$

$x = 16$

Jul 6, 2017

$x = 16$

#### Explanation:

$\text{the initial statement is } y \propto \frac{1}{x}$

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

$\Rightarrow y = k \times \frac{1}{x} = \frac{k}{x}$

$\text{to find k, use the condition given for x and y}$

$y = 8 \text{ when } x = 4$

$y = \frac{k}{x} \Rightarrow k = x y = 4 \times 8 = 32$

$\text{the equation is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = \frac{32}{x}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{when y = 2}$

$x y = 32 \Rightarrow x = \frac{32}{y} = \frac{32}{2} = 16$