# If y varies inversely as x and y = 5 when x = 10, how do you find y when x = 2?

Aug 7, 2016

$y = 25$

#### Explanation:

$y = \frac{k}{x}$ ;where $x$ and $y$ are variables and $k$ is constant
or
$5 = \frac{k}{10}$
or
$k = 5 \left(10\right)$
or
$k = 50$
So Equation becomes $y = \frac{50}{x}$
When $x = 2$
$y = \frac{50}{2}$
or
$y = 25$

Aug 7, 2016

The equation is $y = \frac{50}{x}$

At x=2;" "y=25

#### Explanation:

There is a relationship between $y \text{ and } x$

Mathematically this is written as $y \textcolor{w h i t e}{.} \alpha \textcolor{w h i t e}{.} \frac{1}{x}$

The $\alpha$ is stating that a relationship exists but as yet it is not totally defined

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let $k$ be the constants of variation

Write $y \textcolor{w h i t e}{.} \alpha \textcolor{w h i t e}{.} \frac{1}{x}$ as: $\text{ "y=kxx 1/x" " ->" } y = \frac{k}{x}$

You find the value $k$ by substituting the values for the known condition $\left(x , y\right) \to \left(10 , 5\right)$

Thus we have:

$y = k \times \frac{1}{x} \text{ " ->" } 5 = \frac{k}{10}$

Multiply both sides by 10

$10 \times 5 = \cancel{10} \times \frac{k}{\cancel{10}}$

$\implies k = 50$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the equation is:

$y = \frac{50}{x}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus at $x = 2$ we have:

$y = \frac{50}{2} = 25$